Add files from ibiblio as of 2025-03-08 22:58:20HEADmain

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diff --git a/40395-t.zip b/40395-t.zip
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diff --git a/40395-t/40395-t.tex b/40395-t/40395-t.tex
index a41d5ea..2eb25af 100644
--- a/40395-t/40395-t.tex
+++ b/40395-t/40395-t.tex
@@ -13,10 +13,11 @@
 % Author: William Burnside                                                %

 %                                                                         %

 % Release Date: August 2, 2012 [EBook #40395]                             %

+% Most recently updated: June 11, 2021                         %

 %                                                                         %

 % Language: English                                                       %

 %                                                                         %

-% Character set encoding: ISO-8859-1                                      %

+% Character set encoding: UTF-8                                           %

 %                                                                         %

 % *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF GROUPS OF FINITE ORDER ***

 %                                                                         %

@@ -117,7 +118,7 @@
 \documentclass[12pt]{book}[2005/09/16]

 

 %%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

-\usepackage[latin1]{inputenc}[2006/05/05]

+\usepackage[utf8]{inputenc}[2006/05/05]

 \usepackage[T1]{fontenc}

 

 \usepackage{ifthen}[2001/05/26]

@@ -381,7 +382,7 @@
 \setcounter{tocentry}{0}

 \newcommand{\ToCAnchor}{}

 \newcommand{\SectPageLine}{%

-  \parbox{\textwidth}{\LabelBox{§§}\hfill \LabelBox{\scriptsize PAGE}}\\%

+  \parbox{\textwidth}{\LabelBox{§§}\hfill \LabelBox{\scriptsize PAGE}}\\%

 }

 

 % Chapter entries

@@ -612,8 +613,12 @@
 \newcommand{\TEntry}[1]{\multicolumn{1}{c}{#1}}

 

 % Handle degree symbols and centered dots as Latin-1 characters

-\DeclareInputText{176}{\ifmmode{{}^\circ}\else\textdegree\fi}

-\DeclareInputText{183}{\ifmmode\cdot\else\textperiodcentered\fi}

+\DeclareUnicodeCharacter{00A3}{\pounds}

+\DeclareUnicodeCharacter{00B0}{{}^\circ}

+\DeclareUnicodeCharacter{00B1}{\pm}

+\DeclareUnicodeCharacter{00B7}{\cdot}

+\DeclareUnicodeCharacter{00D7}{\times}

+\DeclareUnicodeCharacter{00F7}{\div}

 

 \renewcommand{\(}{{\upshape(}}

 \renewcommand{\)}{{\upshape)}}

@@ -696,10 +701,11 @@ Title: Theory of Groups of Finite Order
 Author: William Burnside

 

 Release Date: August 2, 2012 [EBook #40395]

+Most recently updated: June 11, 2021

 

 Language: English

 

-Character set encoding: ISO-8859-1

+Character set encoding: UTF-8     

 

 *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF GROUPS OF FINITE ORDER ***

 \end{PGtext}

@@ -794,16 +800,16 @@ developement.
 Many additions were made, mainly by French mathematicians,

 during the middle part of the century. The first

 connected exposition of the theory was given in the third

-edition of M.~Serret's ``\textit{Cours d'Algèbre Supérieure},'' which was

+edition of M.~Serret's ``\textit{Cours d'Algèbre Supérieure},'' which was

 published in 1866. This was followed in 1870 by M.~Jordan's

-``\textit{Traité des substitutions et des équations algébriques}.'' The

+``\textit{Traité des substitutions et des équations algébriques}.'' The

 greater part of M.~Jordan's treatise is devoted to a developement

 of the ideas of Galois and to their application to the

 theory of equations.

 

 No considerable progress in the theory, as apart from its

 applications, was made till the appearance in 1872 of Herr

-Sylow's memoir ``\textit{Théorèmes sur les groupes de substitutions}''

+Sylow's memoir ``\textit{Théorèmes sur les groupes de substitutions}''

 in the fifth volume of the \textit{Mathematische Annalen}. Since the

 date of this memoir, but more especially in recent years, the

 theory has advanced continuously.

@@ -885,8 +891,8 @@ published in the twentieth and twenty-second volumes of the
 \textit{Mathematische Annalen} with the title ``\textit{Gruppentheoretische

 Studien}''; three by Herr Frobenius in the \textit{Berliner Sitzungsberichte}

 for 1895 with the titles, ``\textit{Ueber endliche Gruppen},''

-``\textit{Ueber auflösbare Gruppen},'' and ``\textit{Verallgemeinerung des

-Sylow'schen Satzes}''; and one by Herr Hölder in the forty-sixth

+``\textit{Ueber auflösbare Gruppen},'' and ``\textit{Verallgemeinerung des

+Sylow'schen Satzes}''; and one by Herr Hölder in the forty-sixth

 volume of the \textit{Mathematische Annalen} with the title

 ``\textit{Bildung zusammengesetzter Gruppen}.'' Whenever a result

 is taken from an original memoir I have given a full reference;

@@ -903,8 +909,8 @@ for the care and patience with which they have read the proof-sheets.
 Without the assistance they have so generously

 given me, the errors and obscurities, which I can hardly hope

 to have entirely escaped, would have been far more numerous.

-I wish to express my grateful thanks also to Prof.\ O.~Hölder

-of Königsberg who very kindly read and criticized parts of

+I wish to express my grateful thanks also to Prof.\ O.~Hölder

+of Königsberg who very kindly read and criticized parts of

 the last chapter. Finally I must thank the Syndics of the

 University Press of Cambridge for the assistance they have

 rendered in the publication of the book, and the whole Staff

@@ -918,7 +924,7 @@ the printing has been done.
 

 \ToCChap{CHAPTER I.}{ON SUBSTITUTIONS.}

 

-%§§ PAGE

+%§§ PAGE

 

 \ToCSect{1}{Object of the chapter}{1}

 

@@ -959,7 +965,7 @@ isomorphic groups}{20-22}
 \ToCChap{CHAPTER III.}{ON THE SIMPLER PROPERTIES OF A GROUP WHICH ARE

 INDEPENDENT OF ITS MODE OF REPRESENTATION.}

 

-% §§ PAGE

+% §§ PAGE

 

 \ToCSect[22]{23}{Sub-groups; the order of a sub-group divides the order of

 the group containing it; symbol for a group}{25-27}

@@ -1008,7 +1014,7 @@ group}{55-58}
 

 \ToCChap{CHAPTER V.}{ON GROUPS WHOSE ORDERS ARE POWERS OF PRIMES.}

 

-% §§ PAGE

+% §§ PAGE

 

 \ToCSect{51}{Object of the chapter}{61}

 

@@ -1117,14 +1123,14 @@ of a group is the product of the order of
 the group and the number of the sets in which

 the symbols are interchanged transitively}{165-167}

 

-\ToCLine{Notes to §§~108,~110}{note:1}{167-170}

+\ToCLine{Notes to §§~108,~110}{note:1}{167-170}

 %% -----File: 013.png---Folio xiii-------

 

 

 \ToCChap{CHAPTER IX.}{ON SUBSTITUTION GROUPS: PRIMITIVE AND IMPRIMITIVE

 GROUPS.}

 

-% §§ PAGE

+% §§ PAGE

 

 \ToCSect{120}{Object of the chapter}{171}

 

@@ -1184,7 +1190,7 @@ every substitution of a given group}{215-217}
 

 \ToCChap{CHAPTER XI.}{ON THE ISOMORPHISM OF A GROUP WITH ITSELF.}

 

-% §§ PAGE

+% §§ PAGE

 

 \ToCSect{154}{Object of the chapter}{221}

 

@@ -1247,14 +1253,14 @@ of special groups to the general group}{256-259}
 \ToCSect[197]{198}{Limitation on the order and on the number of defining

 relations of a group of given genus}{281-284}

 

-\ToCLine{Note to §~194}{note:3}{284}

+\ToCLine{Note to §~194}{note:3}{284}

 %% -----File: 015.png---Folio xv-------

 

 

 \ToCChap{CHAPTER XIII.}{ON THE GRAPHICAL REPRESENTATION OF GROUPS: GROUPS OF

 GENUS ZERO AND UNITY: CAYLEY'S COLOUR GROUPS.}

 

-% §§ PAGE

+% §§ PAGE

 

 \ToCSect[199]{203}{Groups of genus zero; their defining relations and

 graphical representation}{285-292}

@@ -1323,7 +1329,7 @@ prime factors}{367-370}
 

 \ToCSect[261]{263}{Non-soluble composite groups}{375-378}

 

-\ToCLine{Notes to §§~257, 258,~260}{note:4}{379}

+\ToCLine{Notes to §§~257, 258,~260}{note:4}{379}

 

 %[** TN: Empty optional argument signifies unindented entry]

 \ToCLine[]{APPENDIX: On French and German technical terms}{sect:Appendix.}{380-382}

@@ -1968,7 +1974,7 @@ The operation resulting from the successive performance of
 the operations $A$, $B$,~\dots, $K$ in the sequence given is denoted by

 the symbol $AB \dots K$; and if $\Omega$~is any object on which the

 operations may be performed, the result of carrying out this

-compound operation on~$\Omega$ is denoted by $\Omega · AB \dots K$.

+compound operation on~$\Omega$ is denoted by $\Omega · AB \dots K$.

 

 If the component operations are all the same, say~$A$, and $r$~in

 number, the abbreviation~$A^{r}$ will be used for the resultant

@@ -2102,7 +2108,7 @@ or
 S^{m} = 1.

 \]

 

-Exactly as in §~8, it may be shewn that, if

+Exactly as in §~8, it may be shewn that, if

 \[

 S^{\mu} = 1,

 \]

@@ -2172,7 +2178,7 @@ are the $N$~operations of a group of order~$N$, the set of $N$~operations
 \[

 S_{r},\ S_{r}S_{1},\ S_{r}S_{2},\ \Dots,\ S_{r}S_{N-1}

 \]

-are (§~13) all distinct; and their number is equal to the order of

+are (§~13) all distinct; and their number is equal to the order of

 the group. Hence every operation of the group occurs once

 and only once in this set.

 

@@ -2500,7 +2506,7 @@ operations connected by a finite number of independent
 relations. But it is to be noted that there is no reason for

 supposing that such an origin for a group is unique; indeed,

 in general, it is not so. Thus there is no difficulty in verifying

-that the group, whose multiplication table is given in §~17, is

+that the group, whose multiplication table is given in §~17, is

 completely specified either by the system of relations

 \[

 A^{3} = 1,\quad C^{2} = 1,\quad (AC)^{2} = 1,

@@ -2512,7 +2518,7 @@ C^{2} = 1,\quad D^{2} = 1,\quad (CD)^{3} = 1.
 

 In other words, it may be generated by two operations of

 orders $2$~and~$3$, or by two operations of order~$2$. So also the

-last group of §~17 is specified either by

+last group of §~17 is specified either by

 \[

 A^{6} = 1,

 \]

@@ -2645,7 +2651,7 @@ are simply isomorphic.
 

 It may also be shewn that each of the $N$~substitutions is

 \index{Substitution group@\Topic{Substitution group}!regular, definition of}%

-regular (§~9) in the $N$~symbols. For the substitution

+regular (§~9) in the $N$~symbols. For the substitution

 \[

 \Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}

     {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}

@@ -2662,7 +2668,7 @@ of the substitution must contain the same number,~$m$, of symbols.
 

 The substitution is therefore regular in the $N$~symbols, and

 this can only be the case if $m$~is a factor of~$N$. It follows at

-once, as was stated in §~15, that the order of any operation of a

+once, as was stated in §~15, that the order of any operation of a

 group of order~$N$ must be equal to or a factor of~$N$.

 

 All the substitutions in this form of representing a group

@@ -2696,7 +2702,7 @@ independent of their mode of representation.
 If among the operations of a group~$G$ a certain set can be

 \index{Sub-group@\Topic{Sub-group}!definition of}%

 chosen which do not exhaust all the operations of the group~$G$,

-yet which at the same time satisfy all the conditions of §~12 so

+yet which at the same time satisfy all the conditions of §~12 so

 that they form another group~$H$, this group~$H$ is called a \emph{sub-group}

 of the group~$G$. Thus if $S$~be any operation, order~$m$,

 of~$G$, the operations

@@ -2796,7 +2802,7 @@ For if $S$,~$T$ are any two operations common to $G_{1}$ and~$G_{2}$,
 \index{Operations@\Topic{Operations}!common to two groups form a group}%

 \index{Symbol@\Topic{Symbol}!for a group generated by given operations}%

 $ST$~is also common to both groups; and hence the common

-operations satisfy conditions \Inum{($\alpha$)}~and~\Inum{($\beta$)} of the definition in §~12.

+operations satisfy conditions \Inum{($\alpha$)}~and~\Inum{($\beta$)} of the definition in §~12.

 But their orders are finite and they must therefore satisfy also

 condition~\Inum{($\gamma$)}, and form a group~$g$. Moreover $g$~is a sub-group

 of both $G_{1}$ and~$G_{2}$, and therefore by Theorem~I its order is a

@@ -2807,7 +2813,7 @@ also clearly a sub-group of~$G$.
 

 The set of operations, that arise by combining in every way

 the operations of the groups $G_{1}$~and~$G_{2}$, evidently satisfy the

-conditions of §~12 and form a group; but this will not necessarily

+conditions of §~12 and form a group; but this will not necessarily

 or generally be a group of finite order. If however $G_{1}$ and $G_{2}$

 are sub-groups of a group~$G$ of finite order, the group~$g'$ that

 arises from their combination will necessarily be of finite order;

@@ -2836,7 +2842,7 @@ by
 \[

 \{S\};

 \]

-and, as a further example, the sixth group of §~17 may be

+and, as a further example, the sixth group of §~17 may be

 represented by

 \[

 \{(xy), (xz)\}.

@@ -2866,7 +2872,7 @@ Two conjugate operations are always of the same order.
 For

 \begin{align*}

 (T^{-1}ST)^{n}

-  &= T^{-1}ST · T^{-1}ST \Dots T^{-1}ST \\

+  &= T^{-1}ST · T^{-1}ST \Dots T^{-1}ST \\

   &= T^{-1}S^{n}T.

 \end{align*}

 

@@ -2881,13 +2887,13 @@ and conversely, if
 \]

 then

 \[

-S^{n} = T · T^{-1} S^{n}T · T^{-1} = T(T^{-1}ST)^{n} T^{-1} = TT^{-1} = 1.

+S^{n} = T · T^{-1} S^{n}T · T^{-1} = T(T^{-1}ST)^{n} T^{-1} = TT^{-1} = 1.

 \]

 

 The operations $ST$ and $TS$ are always conjugate and

 therefore of the same order; for

 \[

-ST = T^{-1}T · ST = T^{-1} · TS · T.

+ST = T^{-1}T · ST = T^{-1} · TS · T.

 \]

 

 \begin{Ex} Shew that the operations $S_{1}S_{2}\Dots S_{n - 1}S_{n}$ and $S_{r}S_{r + 1}\Dots$

@@ -2920,7 +2926,7 @@ operation, the set of transformed operations form a group. For
 if $T_{1}$ and $T_{2}$ are any two operations of the group, so that $T_{1}T_{2}$~is

 also an operation of the group, then

 \[

-S^{-1} T_{1}S · S^{-1} T_{2}S = S^{-1} T_{1} T_{2}S;

+S^{-1} T_{1}S · S^{-1} T_{2}S = S^{-1} T_{1} T_{2}S;

 \]

 %% -----File: 045.png---Folio 29-------

 hence the product of any two operations of the transformed

@@ -3028,7 +3034,7 @@ R_{i}^{-1} TR_{i} = R_{i}^{-1} ST' S^{-1}R_{i};
 \]

 hence

 \[

-T'= S^{-1}TS = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1} R_{i}^{-1} S · T' · S^{-1}R_{i}S,

+T'= S^{-1}TS = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1} R_{i}^{-1} S · T' · S^{-1}R_{i}S,

 \]

 so that every operation of the form $S^{-1}R_{i}S$ is permutable with~$T'$.

 Hence if $H$~is the group of operations permutable with~$T$,

@@ -3257,7 +3263,7 @@ to form the groups $\{T_{1}, T_{2}, \dots, T_{r}\}$ for all conjugate sets of op
 whose orders are prime.

 

 \begin{Remark}

-With the notation of §~26 (\PageRef{p.}{32}), the order of any self-conjugate

+With the notation of §~26 (\PageRef{p.}{32}), the order of any self-conjugate

 sub-group of~$G$ must be of the form $m_{\alpha} + m_{\beta} + \Dots$;

 %% -----File: 051.png---Folio 35-------

 for if the sub-group contains any given operation, it must contain all

@@ -3273,7 +3279,7 @@ and unity, then $G$~is necessarily composite.
 \end{Remark}

 

 If $G_{1}$ and $G_{2}$ are sub-groups of~$G$, it has already been seen

-(§~23) that the operations common to $G_{1}$ and $G_{2}$ form a sub-group~$g$

+(§~23) that the operations common to $G_{1}$ and $G_{2}$ form a sub-group~$g$

 of~$G$; and it is now obvious that, when $G_{1}$ and $G_{2}$ are

 self-conjugate sub-groups, so also is~$g$. Moreover the group

 $\{G_{1}, G_{2} \}$ is a self-conjugate sub-group unless it coincides with~$G$.

@@ -3338,11 +3344,11 @@ S_{0},\ S_{1},\ S_{2},\ \dots,\ S_{n-1}
 \]

 be the set of operations of~$G$ which correspond to the identical

 operation of~$G'$. These operations must form a group, since to

-$S_{p}S_{q}$ corresponds the operation~$1 · 1$, \ie\ the identical operation

+$S_{p}S_{q}$ corresponds the operation~$1 · 1$, \ie\ the identical operation

 of~$G'$; and therefore $S_{p}S_{q}$ must belong to the set.

 

 Again, to the operation $T^{-1}S_{p}T$ of~$G$ corresponds the operation

-$T'^{-1} · 1 · T'$, that is, the identical operation of~$G'$. Hence,

+$T'^{-1} · 1 · T'$, that is, the identical operation of~$G'$. Hence,

 whatever operation of~$G$ is taken for~$T$,

 \[

 T^{-1} \{S_{0}, S_{1}, \dots, S_{n-1}\} T = \{S_{0}, S_{1}, \dots, S_{n-1}\}.

@@ -3394,7 +3400,7 @@ $\Gamma$~a self-conjugate sub-group of~$G$ consisting of the operations
 1,\ T_{1},\ T_{2},\ \dots,\ T_{n - 1}.

 \]

 

-Then, as in §~22, the operations of~$G$ can be arranged in the

+Then, as in §~22, the operations of~$G$ can be arranged in the

 scheme

 \[

 \begin{array}{*{6}{c}}

@@ -3470,10 +3476,10 @@ of~$G$, is completely defined (as an abstract group) when $G$
 and $\Gamma$ are given. This being so it is natural to use a symbol

 to denote directly the group thus defined in terms of $G$ and~$\Gamma$.

 %[** In footnote, no comma or "Vol." after journal title]

-Herr Hölder\footnote

+Herr Hölder\footnote

   {``Zur Reduction der algebraischen Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV} (1889),

   p.~31.}

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 has introduced the symbol

 \[

 \frac{G}{\Gamma}

@@ -3510,7 +3516,7 @@ the case.
 

 \Par{30.} If $G$ is multiply isomorphic with $G'$ so that the self-conjugate

 sub-group~$\Gamma$ of~$G$ corresponds to the identical

-operation of~$G'$, it was shewn, at the end of §~28, that to any

+operation of~$G'$, it was shewn, at the end of §~28, that to any

 self-conjugate sub-group of~$G'$ there corresponds a self-conjugate

 sub-group of~$G$ containing~$\Gamma$. Hence, unless $\dfrac{G}{\Gamma}$ is a simple

 group, $\Gamma$~cannot be a maximum self-conjugate sub-group of~$G$.

@@ -3665,7 +3671,7 @@ T^{-1} S' T = S',
 \]

 and if $S$~is any operation of~$G$, then

 \[

-S^{-1} T^{-1} S · S_{-1} S'S · S_{-1} TS = S^{-1} S'S,

+S^{-1} T^{-1} S · S_{-1} S'S · S_{-1} TS = S^{-1} S'S,

 \]

 so that $S_{-1} TS$ is permutable with every operation of~$G'$. Hence

 every operation of~$G$, which is conjugate to~$T$, is permutable

@@ -3762,7 +3768,7 @@ operation except identity, and every operation of~$G$ is permutable
 with every operation of~$H$.

 \end{Corollary}

 

-For, by §~33, if $G$ and~$H$ had a common sub-group, it would

+For, by §~33, if $G$ and~$H$ had a common sub-group, it would

 be a self-conjugate sub-group of both of them; and neither of

 them could then be simple, contrary to hypothesis. Consequently,

 the only sub-group common to $G$ and~$H$ is the

@@ -3858,18 +3864,18 @@ with each other, is called an \emph{Abelian}\footnote
   ``Ueber Gruppen vertauschbarer Elemente,'' \textit{Crelle's Journal}, Vol.~\VolNo{LXXXVI} (1879),

   p.~217; and a very complete discussion in the second volume of Herr Weber's

   recently published \textit{Lehrbuch der Algebra}. In the proof of the existence of a

-  set of independent generating operations (§~41) we have directly followed Herr

+  set of independent generating operations (§~41) we have directly followed Herr

   Weber.

 

-  The name ``Abelian group'' has been applied by M.~Jordan (\textit{Traité des substitutions~etc.}\

+  The name ``Abelian group'' has been applied by M.~Jordan (\textit{Traité des substitutions~etc.}\

   pp.~171 et~seq.)\ to an entirely different class of groups, whose

   operations are not permutable. Most writers, we believe, have used the phrase

   in the sense defined in the text.

 

   The connection of Abel's name with groups of permutable operations is due

   to his having been the first to investigate, with complete generality, the application

-  of such groups to the theory of equations, ``Mémoire sur une classe

-  particulière d'équations résolubles algébriquement,'' \textit{Crelle's Journal}, Vol.~\VolNo{IV}

+  of such groups to the theory of equations, ``Mémoire sur une classe

+  particulière d'équations résolubles algébriquement,'' \textit{Crelle's Journal}, Vol.~\VolNo{IV}

   (1829), p.~131; or Collected Works, 1881 edition, Vol.~\VolNo{I}, p.~478.}

 group.

 \end{Definition}

@@ -3899,7 +3905,7 @@ highest power of a prime~$p$ that divides its order. We shall
 first shew that $G$~has a single sub-group of order~$p^{m}$, consisting

 of all the operations of~$G$ whose orders are powers of~$p$.

 

-If $S_{1}$,~$S_{2}$ are any two operations of~$G$, it follows from §~33,

+If $S_{1}$,~$S_{2}$ are any two operations of~$G$, it follows from §~33,

 that, because $S_{1}$ and $S_{2}$ are permutable, the order of~$\{S_{1}, S_{2}\}$ is

 equal to or is a factor of the product of the orders of $S_{1}$~and~$S_{2}$.

 So again, if $S_{3}$~is an operation of~$G$ not contained in~$\{S_{1}, S_{2}\}$, the

@@ -3913,14 +3919,14 @@ order is divisible by~$p$, and some power of this, say~$S$, will be
 an operation of order~$p$. Now, if $m$~is greater than unity, the

 order of the factor-group $\dfrac{G}{\{S\}}$, which is also Abelian, is divisible

 by~$p$, and therefore this factor-group must have an operation of

-order~$p$. Hence $G$~will (§~28) contain a sub-group of order~$p^{2}$.

+order~$p$. Hence $G$~will (§~28) contain a sub-group of order~$p^{2}$.

 If $m$~is greater than~$2$, the same reasoning may be

 repeated to shew that $G$~has a sub-group of order~$p^{3}$, and so on.

 Hence, finally, $G$~has a sub-group of order~$p^{m}$. Let $H$~be this

 sub-group; and suppose if possible that $G$~contains an operation~$T$,

 whose order is a power of~$p$, which does not belong to~$H$.

 Then $\{H, T\}$ is a sub-group of~$G$, whose order is a power of~$p$,

-greater than~$p^{m}$; and this is impossible (§~22). The sub-group~$H$

+greater than~$p^{m}$; and this is impossible (§~22). The sub-group~$H$

 %% -----File: 064.png---Folio 48-------

 \index{Abelian group@\Topic{Abelian group}!sub-groups of}%

 must therefore contain all the operations of~$G$ whose orders

@@ -3953,13 +3959,13 @@ the operations of~$G$ whose orders are relatively prime to
 $\dfrac{N}{p_{1}^{m_{1}} p_{2}^{m_{2}}}$. For if $T$~were an operation of~$G$ of order~$p_{1}^{\alpha} p_{2}^{\beta}$, not

 contained in $\{H_{1}, H_{2}\}$, then $\{H_{1}, H_{2}, T\}$ would be a sub-group

 of~$G$, of order $p_{1}^{n_{1}} p_{2}^{n_{2}}$, where $n_{1} > m_{1}$ if $\alpha > 0$ and $n_{2} > m_{2}$ if $\beta > 0$;

-and this is impossible (§~22).

+and this is impossible (§~22).

 

 This process may clearly be continued to shew that, if

 $N = \mu\nu$, where $\mu$~and~$\nu$ are relatively prime, then $G$~contains a

 single sub-group of order~$\mu$, consisting of all the operations of~$G$

 whose orders are relatively prime to~$\nu$. Moreover $G$~itself is

-the direct product (§~31) of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

+the direct product (§~31) of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

 

 \Par{39.} The first problem of pure group-theory that presents

 itself in connection with Abelian groups is the determination

@@ -4049,7 +4055,7 @@ latter is then a cyclical group generated by the operation~$P_{1}$.
 If $m_{1} < m$, $G$~must contain other operations besides those of~$\{P_{1}\}$.

 Denoting $\{P_{1}\}$ by~$G_{1}$, let $Q$ be any operation of~$G$ not

 contained in~$G_{1}$, and let $Q^{a}$ be the lowest power of~$Q$ that is

-contained in~$G_{1}$. Then (§~33) $a$~must be a power of~$p$; and

+contained in~$G_{1}$. Then (§~33) $a$~must be a power of~$p$; and

 when for $Q$ each operation of~$G$ that is not contained in~$G_{1}$ is

 taken in turn, $a$~must have some maximum value, say~$p^{m_{2}}$.

 Since no operation of~$G$ is of greater order than~$p^{m_{1}}$ it follows

@@ -4419,7 +4425,7 @@ If
 \[

 m_{s} \geq \mu > m_{s + 1},

 \]

-then (Theorem~II, §~41)

+then (Theorem~II, §~41)

 \[

 S^{p^{\mu}}

   = P_{1}^{\alpha_{1} p^{\mu}}

@@ -4440,7 +4446,7 @@ which satisfy the equation
 S^{p} = 1,

 \]

 %% -----File: 072.png---Folio 56-------

-is~$p^{s}$ (§~42). This group must be identical with or be a sub-group

+is~$p^{s}$ (§~42). This group must be identical with or be a sub-group

 of~$G_{1}$ whose order is~$p^{r}$. Hence

 \[

 s \leq r,

@@ -4646,7 +4652,7 @@ x = \frac{p^{r + 1} - p^{r}}{p - 1}  N_{r + 1, r},
 \]

 and therefore

 \[

-N_{m, r + 1} = \frac{N_{m, r}}{N_{r + 1\DPtypo{\,}{,} r}} · \frac{p^{m - r} - 1}{p - 1}.

+N_{m, r + 1} = \frac{N_{m, r}}{N_{r + 1\DPtypo{\,}{,} r}} · \frac{p^{m - r} - 1}{p - 1}.

 \]

 \index{Abelian group@\Topic{Abelian group}!sub-groups of|)}%

 %% -----File: 076.png---Folio 60-------

@@ -4730,7 +4736,7 @@ other groups, whose order is not thus limited.
 

 \Par{52.} If $G$~is a group whose order is~$p^{m}$, where $p$~is a prime,

 the order of every sub-group of~$G$ must also be a power of~$p$; and

-therefore (§~25) the total number of operations of~$G$ which are

+therefore (§~25) the total number of operations of~$G$ which are

 conjugate with any given operation must be a power of~$p$. The

 identical operation of~$G$ is self-conjugate. Hence if the operations

 of~$G$, other than the identical operation, are distributed

@@ -4746,7 +4752,7 @@ $\alpha$, $\beta$,~\Dots\ are zero, $r_{1}$~being some integer not less than uni
 \index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!always contain self-conjugate operations}%

 \index{Self-conjugate operation@\Topic{Self-conjugate operation}!of a group whose order is the power of a prime}%

 Hence $G$~must contain $p^{r_{1}}$ self-conjugate operations, which form

-(§~27) a self-conjugate sub-group\footnotemark.

+(§~27) a self-conjugate sub-group\footnotemark.

 \footnotetext{Sylow, \textit{Math.\ Ann.}\ \VolNo{V}.\ (1872), p.~588.} %[** No comma or "Vol."]

 \index{Sylow@\Topic{Sylow}, quoted}%

 Hence:---

@@ -4793,7 +4799,7 @@ of~$r$. It by no means however follows that, when this latter
 condition is satisfied, $G$~and $G'$ are always simply isomorphic.

 This is indeed not necessarily the case; instances

 to the contrary will be found among the groups of order

-$p^{3}$ and $p^{4}$ in §§~69--72. While in general there is no limitation

+$p^{3}$ and $p^{4}$ in §§~69--72. While in general there is no limitation

 on the type of any Abelian factor-group $\dfrac{H_{r + 1}}{H_{r}}$ formed from

 two consecutive terms of the series, it is to be noted that

 the last factor-group $\dfrac{G}{H_{n}}$ cannot be cyclical\footnotemark,

@@ -4995,16 +5001,16 @@ sub-group of order~$p$, each of its operations is self-conjugate.
 

 If $p$ is~$2$, we must take

 \[

-\alpha = ±1 + k2^{s},

+\alpha = ±1 + k2^{s},

 \]

 where $k$~is odd, and we are led by the same process to the

 result

 \[

-\alpha \equiv ±1\ (\mod 2^{n - r}).

+\alpha \equiv ±1\ (\mod 2^{n - r}).

 \]

 

 \Par{57.} If $P$ is an operation of~$G$ which belongs to the sub-group~$H_{r + 1}$

-(§~53) and to no previous sub-group in the set

+(§~53) and to no previous sub-group in the set

 \[

 1,\ H_{1},\ H_{2},\ \Dots,\ H_{n},\ G,

 \]

@@ -5046,7 +5052,7 @@ S'^{-1}PS' = Ph_{s}',
 \]

 then

 \[

-S'^{-1} S^{-1} PSS' = Ph_{s}' · S'^{-1} h_{s}S,

+S'^{-1} S^{-1} PSS' = Ph_{s}' · S'^{-1} h_{s}S,

 \]

 and when both $h_{s}$ and $h_{s}'$ belong to~$H_{s}$, so also does $h_{s}' S'^{-1} h_{s}S'$.

 

@@ -5101,7 +5107,7 @@ group can therefore only exist if $s$~is even\footnotemark.
 

 The sub-group $\{P_{1}, P_{2}, Q\}$ is a group which satisfies the same

 conditions as~$G$, when $s = 2$. Its order is~$p^{r + 2}$, and it contains a

-self-conjugate operation of order~$p^{r}$. Now we shall see in §§~65,~66

+self-conjugate operation of order~$p^{r}$. Now we shall see in §§~65,~66

 that such a group is necessarily of one of two types\footnotemark,

 \footnotetext{Let

   \begin{gather*}

@@ -5109,12 +5115,12 @@ that such a group is necessarily of one of two types\footnotemark,
   f_{1}(S_{i}) = 1,\quad f_{2}(S_{i}) = 1,\ \Dots,\quad f_{k}(S_{i}) = 1,

   \end{gather*}

   where $f_{j}(S_{i})$ is an abbreviation for an expression of the form $S_{p}^{a} S_{q}^{b} \Dots S_{r}^{c}$, be a

-  set of relations, such as was considered in §~18, which completely specify a

+  set of relations, such as was considered in §~18, which completely specify a

   group. We may then, without altering the sense in which the word ``type''

-  has been used (§§~19,~44), speak of \emph{the type of group} defined by these relations.

+  has been used (§§~19,~44), speak of \emph{the type of group} defined by these relations.

   It is essential however that the relations should completely specify the group,

   as otherwise they will define more than one type. For instance, it is clear, from

-  §~56, that the equations

+  §~56, that the equations

   \[

   Q^{p} = 1,\quad P^{p^{2}} = 1,\quad Q^{-1}PQ = P^{\alpha},

   \]

@@ -5149,19 +5155,19 @@ $H_{1}$~is a cyclical group of order~$p^{r}$, and $\dfrac{G}{H_{1}}$~is an Abeli
 $(1, 1, \Dots\DPchg{\,}{,} \text{to $s$ units})$.

 \end{Remark}

 

-\Par{59.} It has been seen in §~55 that every sub-group~$G'$ of

+\Par{59.} It has been seen in §~55 that every sub-group~$G'$ of

 order~$p^{m - 1}$ of a group~$G$ of order~$p^{m}$ is self-conjugate. Suppose

 now that $G$~contains two such sub-groups $G'$~and~$G''$. Then

 since $G'$ and $G''$ are permutable with each other, while the

 order of~$\{G', G''\}$ is~$p^{m}$, the order of the greatest group~$g'$

-common to them must (§~33) be~$p^{m - 2}$; and since $g'$~is the

+common to them must (§~33) be~$p^{m - 2}$; and since $g'$~is the

 greatest common sub-group of two self-conjugate sub-groups of~$G$,

 it must itself be a self-conjugate sub-group of~$G$. The factor

 group $\dfrac{G}{g'}$ of order~$p^{2}$ contains the two distinct sub-groups $\dfrac{G'}{g'}$

 and $\dfrac{G''}{g'}$, which are of order~$p$ and permutable with each other.

 %% -----File: 086.png---Folio 70-------

 Hence $\dfrac{G}{g'}$ must be an Abelian group of type $(1, 1)$, and

-it therefore contains (§~49) $p + 1$ sub-groups of order~$p$. Hence,

+it therefore contains (§~49) $p + 1$ sub-groups of order~$p$. Hence,

 besides~$G'$, $G$~must contain $p$~other sub-groups of order~$p^{m-1}$

 which have in common with~$G'$ the sub-group~$g'$. If the $p + 1$

 sub-groups thus obtained do not exhaust the sub-groups of~$G$

@@ -5280,7 +5286,7 @@ of order~$p^{r}$. Then $\{P\}$ must be contained self-conjugately in a
 non-cyclical sub-group of order~$p^{r + 1}$.

 

 We will take first the case in which $p$~is an odd prime.

-Then (§~55) $G$~must contain an operation~$P'$ which does not

+Then (§~55) $G$~must contain an operation~$P'$ which does not

 belong to~$\{P\}$, such that

 \[

 P'^{-1} PP' = P^{\alpha},\quad P'^{p} = P^{\beta}.

@@ -5292,7 +5298,7 @@ of order~$p$ not occurring in~$\{P\}$. It has been shewn that this is
 impossible if $G_{s}$ is the only sub-group of order~$p^{s}$, and therefore

 $\alpha$~cannot be unity.

 

-We may then without loss of generality (§~56) assume that

+We may then without loss of generality (§~56) assume that

 \[

 \alpha = 1 + p^{r - 1}.

 \]

@@ -5340,7 +5346,7 @@ Q^{-1}PQ = P^{\alpha},\quad Q^{4} = P^{\beta},
 \]

 where

 \[

-\alpha = ±1 + k2^{r - 2}.

+\alpha = ±1 + k2^{r - 2}.

 \]

 Moreover $\beta$ must be a multiple of~$4$, as otherwise the order of~$Q$

 would be greater than~$2^{r}$. A simple calculation now gives

@@ -5364,13 +5370,13 @@ not contained in~$\{P\}$. But this is inconsistent with $G_{s}$ being
 the only sub-group of order~$2^{s}$; hence $\{P\}$~must contain~$Q^{2}$.

 

 If now $\{P\}$ is not a self-conjugate sub-group, there must

-(§~55) be some operation~$P'$ of order~$p^{r}$, conjugate with~$P$ and

+(§~55) be some operation~$P'$ of order~$p^{r}$, conjugate with~$P$ and

 not contained in~$\{P\}$, which transforms $\{P\}$ into itself; and

 then $\{P, P'\}$ is defined by

 \[

 P'^{-1}PP' = P^{\alpha},\quad P'^{2} = P^{2\beta},

 \]

-where $\alpha$ is $-1$ or $±1 + 2^{r - 1}$, and $\beta$~is odd.

+where $\alpha$ is $-1$ or $±1 + 2^{r - 1}$, and $\beta$~is odd.

 

 If

 \[

@@ -5512,7 +5518,7 @@ transform $P$ into one of its own powers. Hence
 \[

 Q'^{-1}PQ' = P^{\alpha},

 \]

-and since $Q'^{p}$ is permutable with~$P$ it follows, from §~56, that

+and since $Q'^{p}$ is permutable with~$P$ it follows, from §~56, that

 \[

 \alpha = 1 + kp^{m - 2}.

 \]

@@ -5646,7 +5652,7 @@ contains a cyclical self-conjugate sub-group~$\{P\}$ of order~$p^{m - 2}$,
 but no self-conjugate operation of this order.

 

 If $G$ contains an operation~$Q'$ such that $Q'^{p^{2}}$~is the lowest

-power of~$Q'$ occurring in~$\{P\}$, it follows (§~56), since $\{P\}$~is

+power of~$Q'$ occurring in~$\{P\}$, it follows (§~56), since $\{P\}$~is

 self-conjugate, that

 \[

 Q'^{-1}PQ' = P^{1 + kp^{m - 4}}.

@@ -5657,7 +5663,7 @@ Moreover, since the order of~$Q'$ cannot exceed~$p^{m - 2}$, we have
 Q'^{p^{2}} = P^{\alpha p^{2}}.

 \]

 

-A simple calculation, similar to that of §~65, will now shew

+A simple calculation, similar to that of §~65, will now shew

 that

 \[

 (P^{x}Q')^{p^{2}} = P^{(x + \alpha)p^{2}},

@@ -5728,7 +5734,7 @@ P^{p^{m - 2}} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p^{m - 3}}.
 \]

 

 If $Q$ is not a self-conjugate operation, it must be one of $p$

-conjugate operations and (§~57) these must be

+conjugate operations and (§~57) these must be

 \[

 Q, \quad QP^{p^{m - 3}}, \quad QP^{2p^{m - 3}},\ \Dots,\quad QP^{(p - 1)p^{m - 3}}.

 \]

@@ -5754,7 +5760,7 @@ PQ = QP,\quad RQ = QR.
 It is clear from these relations that the group thus arrived

 at is the direct product of the groups $\{P, R\}$ and~$\{Q\}$\footnotemark.

 \footnotetext{In each of the five distinct cases to which we have been led in the discussion

-  contained in §§~65--67, we have arrived at a set of defining relations, containing

+  contained in §§~65--67, we have arrived at a set of defining relations, containing

   no indeterminate symbols, such that in each case a set of generating operations

   can be chosen to satisfy these relations. To justify the statement, in each

   particular case, that such a set of relations gives a distinct type of group, it is

@@ -5779,7 +5785,7 @@ at is the direct product of the groups $\{P, R\}$ and~$\{Q\}$\footnotemark.
   \]

   Hence

   \[

-  Q = Q^{4 · 7 - 3 · 9} = 1,

+  Q = Q^{4 · 7 - 3 · 9} = 1,

   \]

   and the relations hold only for a group of order~$3$.

 

@@ -5791,7 +5797,7 @@ at is the direct product of the groups $\{P, R\}$ and~$\{Q\}$\footnotemark.
 

 \begin{Remark}

 It is to be expected, from the result of the corresponding case at

-the end of §~65, that the number of distinct types when $p = 2$ is

+the end of §~65, that the number of distinct types when $p = 2$ is

 much greater than when $p$~is an odd prime. There are, in fact,

 when $m > 5$, fourteen distinct types of groups of order~$2^{m}$, which

 contain a self-conjugate cyclical sub-group of order~$2^{m - 2}$ and no

@@ -5832,7 +5838,7 @@ order~$2^{m - 1}$, which has no operation except identity in common with~$\{B\}$
 Hence $G$~is a direct product of a group of order~$2$ and a group

 of order~$2^{m - 1}$. There are therefore, for this case, four types \Inum{(vii)},

 \Inum{(viii)}, \Inum{(ix)}, \Inum{(x)}, when $m > 4$, corresponding to the four groups of

-order~$2^{m - 1}$ of §~65. If $m = 4$, there are two types.

+order~$2^{m - 1}$ of §~65. If $m = 4$, there are two types.

 

 Next, let all the self-conjugate operations of~$G$ be contained in~$\{A\}$;

 and suppose that $A$~is one of two conjugate operations. Then

@@ -5850,7 +5856,7 @@ When $m = 4$, there is, for this case, no type.
 

 Lastly, suppose that $A$ is one of four conjugate operations. Then

 $G$~must contain sub-groups of order~$2^{m - 1}$, of the second and the

-third types of §~65, and a sub-group of order~$2^{m - 1}$ of either the

+third types of §~65, and a sub-group of order~$2^{m - 1}$ of either the

 first or fourth type (\lc). In this last case, there are two distinct

 types defined by

 \[

@@ -5871,7 +5877,7 @@ be distinct from that for an odd prime; for the sake of brevity
 we shall not deal in detail with this case, but shall state the results

 only and leave their verification as an exercise to the reader.

 

-It has been shewn (§~53) that all groups of order~$p^{2}$ are Abelian;

+It has been shewn (§~53) that all groups of order~$p^{2}$ are Abelian;

 and hence the only distinct types are those represented by $(2)$ and

 $(1, 1)$.

 

@@ -5880,7 +5886,7 @@ and $(1, 1, 1)$.
 %% -----File: 098.png---Folio 82-------

 

 If a non-Abelian group of order~$p^{3}$ contains an operation of

-order~$p^{2}$, the sub-group it generates is self-conjugate; hence (§~65)

+order~$p^{2}$, the sub-group it generates is self-conjugate; hence (§~65)

 in this case there is a single type of group defined by

 \[

 P^{p^{2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}.

@@ -5888,7 +5894,7 @@ P^{p^{2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}.
 

 If there is no operation of order~$p^{2}$, then since there must be a

 self-conjugate operation of order~$p$, the group comes under the head

-discussed in §~66; there is again a single type of group defined by

+discussed in §~66; there is again a single type of group defined by

 \begin{gather*}

 P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP, \\

 R^{-1}PR = P,\quad Q^{-1}PQ = P.

@@ -5901,15 +5907,15 @@ groups of order~$p^{3}$.
 are $(4)$, $(3, 1)$, $(2, 2)$, $(2, 1, 1)$ and $(1, 1, 1, 1)$.

 

 For non-Abelian groups of order~$p^{4}$ which contain operations of

-order~$p^{3}$ there is a single type, namely that given in §~65 when $m$~is

+order~$p^{3}$ there is a single type, namely that given in §~65 when $m$~is

 put equal to~$4$.

 

 For non-Abelian groups which contain a self-conjugate cyclical

 sub-group of order~$p^{2}$ and no operation of order~$p^{3}$, there are three

-distinct types, obtained by writing $4$ for~$m$ in the group of §~66 and

-in the first and the last groups of §~67. The defining relations of

+distinct types, obtained by writing $4$ for~$m$ in the group of §~66 and

+in the first and the last groups of §~67. The defining relations of

 these need not be here repeated, as they will be given in the summarizing

-table (§~73).

+table (§~73).

 

 It remains now to determine all distinct types of groups of

 order~$p^{4}$, which contain no operation of order~$p^{3}$ and no self-conjugate

@@ -5924,7 +5930,7 @@ S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S^{1 + kp}.
 \]

 

 If $R$ is any operation of~$G$, not contained in~$\{S, T\}$, then since

-$\{S\}$~is not self-conjugate, we must have (§~57)

+$\{S\}$~is not self-conjugate, we must have (§~57)

 \[

 R^{-1}SR = S^{1 + \alpha p} T^{\beta},

 \]

@@ -6055,7 +6061,7 @@ R^{p} = 1;
 with the condition that all operations of~$G$, not contained in~$\{S, T\}$,

 are of order~$p$.

 

-The formulæ for $R^{-x}SR^{x}$ and $R^{-x}TR^{x}$ enable us to calculate

+The formulæ for $R^{-x}SR^{x}$ and $R^{-x}TR^{x}$ enable us to calculate

 directly the power of any given operation of~$G$. Thus they give

 \[

 (S^{x}R)^{p} = S^{px\{1 + \efrac{1}{6} \beta\gamma(p + 1)p(p - 1) + \efrac{1}{2} \alpha p(p + 1)\}}.

@@ -6093,7 +6099,7 @@ If now $R$ is any other operation of the group, $\{P, Q, R\}$ must be
 an Abelian group of type $(1, 1, 1)$. If again $S$~is any operation not

 contained in $\{P, Q, R\}$, it cannot be permutable with~$R$; for if it

 were, $R$~would be self-conjugate. There must therefore be a relation

-of the form (§~57)

+of the form (§~57)

 \[

 S^{-1}RS = RP^{\alpha}Q^{\beta}.

 \]

@@ -6121,7 +6127,7 @@ self-conjugate in this group, it must be Abelian. Let $P$,~$Q$,~$R$ be
 generators of this group and $S$~any operation of~$G$ not contained in

 %% -----File: 102.png---Folio 86-------

 it. We may now assume that $Q$~belongs to the sub-group~$H_{2}$

-(§~53), and therefore that

+(§~53), and therefore that

 \[

 S^{-1}QS = QP^{\alpha},

 \]

@@ -6202,7 +6208,7 @@ indicating the types of $H_{1}$, $\dfrac{H_{2}}{H_{1}}$, $\dfrac{H_{3}}{H_{2}}$,
 \[

 H_{1},\ H_{2},\ H_{3},\ \Dots,\ H_{n},\ G

 \]

-is the series of self-conjugate sub-groups defined in §~53. This

+is the series of self-conjugate sub-groups defined in §~53. This

 %% -----File: 103.png---Folio 87-------

 symbol is to be read from the left so that $(a, b, \Dots)$ is the

 type of~$H_{1}$.

@@ -6213,11 +6219,11 @@ order than that denoted by~$P$.
 

 \Subsection{\textit{Table of groups of order $p^{n}$, $p$ an odd prime}\footnotemark.}

 \footnotetext{On groups of orders $p^{3}$ and~$p^{4}$, the reader may consult, in addition to

-  Young's memoir already referred to, Hölder, ``Die Gruppen der Ordnungen

+  Young's memoir already referred to, Hölder, ``Die Gruppen der Ordnungen

   $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.},\DPnote{** no "Vol."} \VolNo{XLIII}, (1893), in particular, pp.~371--410.}%

 \index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%

 \index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime}%

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 \index{Young@\Topic{Young}, quoted}%

 

 %[** TN: Layout reformatted]

@@ -6304,7 +6310,7 @@ P^{9} = 1,\ Q^{3} = 1,\ R^{3} = 1,\ Q^{-1}PQ = P,\ R^{-1}PR = PQ, \\
 R^{-1}QR = P^{-3}Q,\ (1) (1) (11).

 \end{gather*}

 

-\Par{74.} To complete the list, we add, as was promised in §~68,

+\Par{74.} To complete the list, we add, as was promised in §~68,

 \index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!table of distinct types of orders $8$ and~$16$}%

 \index{Types of group@\Topic{Types of group}, distinct, whose order is!$8$ or~$16$}%

 the types of non-Abelian groups of orders $2^{3}$ and~$2^{4}$; the

@@ -6342,7 +6348,7 @@ this group~\Inum{(vi)} being the direct product of $\{R\}$ and $\{P, Q\}$;
 \begin{Remark}

 \Par{75.} In the following examples, $G$~is a non-Abelian group whose

 order is the power of a prime~$p$; and the sub-groups $H_{1}$, $H_{2}$,~\Dots, $H_{n}$

-referred to are the series of self-conjugate sub-groups of §~53.

+referred to are the series of self-conjugate sub-groups of §~53.

 \end{Remark}

 

 \begin{Ex}[1.] If $P$ and $Q$ are any two operations of~$G$, the totality of

@@ -6402,9 +6408,9 @@ Determine in each case the number of distinct types.
 \Chapter{VI.}{On Sylow's Theorem.}

 \index{Sylow@\Topic{Sylow}, quoted}%

 

-\Par{76.} \First{It} has been proved (§~22) that the order of any sub-group

+\Par{76.} \First{It} has been proved (§~22) that the order of any sub-group

 of a group~$G$ is a factor of the order of~$G$; and it

-results at once from the investigation of §§~45--47 that, in

+results at once from the investigation of §§~45--47 that, in

 an Abelian group, there is always at least one sub-group

 whose order is any given factor of the order of the group.

 The latter result is not however generally true for groups

@@ -6420,7 +6426,7 @@ contains operations of order~$p$, is due originally to Cauchy\footnotemark.
 \footnotetext{Cauchy, \textit{Exercises d'analyse},~\VolNo{III}, (1844), p.~250.}%

 The more general result was first established by Sylow. He

 has shewn\footnote

-  {Sylow, Théorèmes sur les groupes de substitutions, \textit{Math.\ Ann.},\DPnote{** No "Vol." here, below} \VolNo{V} (1872)

+  {Sylow, Théorèmes sur les groupes de substitutions, \textit{Math.\ Ann.},\DPnote{** No "Vol." here, below} \VolNo{V} (1872)

   pp.~584 et~seq. Compare also Frobenius, Neuer Beweis des Sylow'schen Satzes,

   \textit{Crelle}, \VolNo{C.}\ (1886), p.~179.}

 \index{Frobenius@\Topic{Frobenius}, quoted}%

@@ -6460,7 +6466,7 @@ whose order is divisible by~$p^{\alpha}$.
 \end{Lemma}

 

 If the group~$G$ is Abelian, this has been already proved in

-§~37. Suppose then that $G$ of order~$N$ ($= p^{\alpha}m$, where $m$~is

+§~37. Suppose then that $G$ of order~$N$ ($= p^{\alpha}m$, where $m$~is

 prime to~$p$) is not Abelian; and let $G'$ of order~$N'$ be the sub-group

 of~$G$ formed of its self-conjugate operations. If $S$~is any

 operation of~$G$ which is not self-conjugate, and if $n$~is the order

@@ -6510,7 +6516,7 @@ order is divisible by~$p^{\alpha}$, and so on. Hence at last a sub-group
 must be arrived at whose order is~$p^{\alpha}$.

 

 \begin{Corollary}

-Since it has been seen in §~53 that a group of

+Since it has been seen in §~53 that a group of

 order~$p^{\alpha}$ contains sub-groups of every order~$p^{\beta}$ ($\beta < \alpha$), it follows

 that, if the order of a group is divisible by~$p^{\beta}$, the group must

 contain a sub-group of order~$p^{\beta}$. In particular, a group whose

@@ -6837,7 +6843,7 @@ contain $P$ self-conjugately, there must be one or more to which $g$
 belongs. Let $H$ be one of these; and suppose that $h$~and~$h'$

 are the greatest sub-groups of $H$ and $H'$ respectively that

 contain $g$ self-conjugately. The orders of both $h$ and $h'$ must

-(Theorem~II, §~55) be greater than the order of~$g$; and in consequence

+(Theorem~II, §~55) be greater than the order of~$g$; and in consequence

 of the assumption made with respect to~$H'$, every

 sub-group, having a power of~$p$ for its order and containing~$h$,

 must contain $P$ self-conjugately.

@@ -6888,7 +6894,7 @@ must, by Sylow's theorem, have a self-conjugate sub-group of order~$p_{2}$.
 If $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, the group must also

 have a self-conjugate sub-group of order~$p_{1}$. The two self-conjugate

 sub-groups of orders $p_{1}$ and $p_{2}$ can have no common operation

-except the identical operation; and therefore (Theorem~IX, §~34)

+except the identical operation; and therefore (Theorem~IX, §~34)

 every operation of one must be permutable with every operation of

 the other. Hence if $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, a group

 of order~$p_{1}p_{2}$ must be Abelian, and therefore also cyclical.

@@ -6946,7 +6952,7 @@ where $\alpha$~is a primitive root of the congruence\footnote
 \alpha^{p_{1}} \equiv 1\ (\mod p_{2}).

 \]

 

-It follows from §~33 that the order of the group defined by these

+It follows from §~33 that the order of the group defined by these

 relations cannot exceed~$p_{1} p_{2}$.  But also from the given relations it

 is clearly impossible to deduce new relations of the form

 \[

@@ -6983,18 +6989,18 @@ A group of order~$24$ must contain either $1$ or $3$~sub-groups of
 order~$8$, and either $1$ or $4$~sub-groups of order~$3$. If it has one

 sub-group of order~$8$ and one sub-group of order~$3$, the group must,

 since each of these sub-groups is self-conjugate, be their direct

-product. We have seen (§§~68, 74) that there are five distinct types

+product. We have seen (§§~68, 74) that there are five distinct types

 of group of order~$8$; there are therefore five distinct types of group

 of order~$24$, which are obtained by taking the direct product of any

 group of order~$8$ and a group of order~$3$.

 

 If there are $3$~groups of order~$8$, some two of them must

-(Theorem~II, Cor.~II, §~78) have a common sub-group of order~$4$;

-and (Theorem~III, §~80) this common sub-group must be a self-conjugate

+(Theorem~II, Cor.~II, §~78) have a common sub-group of order~$4$;

+and (Theorem~III, §~80) this common sub-group must be a self-conjugate

 sub-group of the group of order~$24$. Moreover if, in

 this case, a sub-group of order~$8$ is Abelian, each operation of the

 %% -----File: 118.png---Folio 102-------

-self-conjugate sub-group of order~$4$ must (Theorem~IV, Cor.\ §~81)

+self-conjugate sub-group of order~$4$ must (Theorem~IV, Cor.\ §~81)

 be a self-conjugate operation of the group of order~$24$.

 

 With the aid of these general considerations, it now is easy to

@@ -7208,7 +7214,7 @@ A^{4} = 1,\quad C^{3} = 1,\quad (AC)^{2} = 1.
 \]

 

 It is a good exercise to verify that these form a complete set of

-defining relations for the group. (Compare Ex.~1, §~35.)

+defining relations for the group. (Compare Ex.~1, §~35.)

 

 There are therefore, in all, fifteen distinct types of group of order~$24$.

 The last of these is the only type, which has neither a self-conjugate

@@ -7224,7 +7230,7 @@ brevity.
 %% -----File: 121.png---Folio 105-------

 

 It is to be noticed that the last type obtained gives an example,

-and indeed the simplest possible, of Theorem~V, §~82. Thus in

+and indeed the simplest possible, of Theorem~V, §~82. Thus in

 $\{A, B\}$ of order~$8$, $A^{2}$~is a self-conjugate operation and $B$~is not. In

 the group of order~$24$, the operations $A^{2}$ and $B$ are conjugate; and

 $C$~is an operation, of order prime to~$2$, such that $A^{2}$, $C^{-1}A^{2}C$, $C^{-2}A^{2}C^{2}$

@@ -7265,8 +7271,8 @@ are not self-conjugate.
 There must then be either $1$ or $3$ sub-groups of order~$4$. If there is

 only one, it must be permutable with an operation of order~$3$;

 and then $G$~contains a sub-group of order~$12$. If there are three,

-it follows, by Theorem~II, Cor.~II (§~78), that some pair of them

-must have a common sub-group of order~$2$. But (Theorem~III, §~80)

+it follows, by Theorem~II, Cor.~II (§~78), that some pair of them

+must have a common sub-group of order~$2$. But (Theorem~III, §~80)

 this sub-group of order~$2$ must be permutable with some operation

 of order prime to~$2$, which is not permutable with a sub-group of

 order~$4$. This operation must be of order~$3$; hence in this case also

@@ -7391,7 +7397,7 @@ A^{2} = 1,\quad B^{3} = 1,\quad ABA = B^{-1}.
 \]

 

 Since no operation of order $3$~or~$5$ is permutable with an operation

-of order~$2$, it follows (Theorem~III, §~80) that no two sub-groups of

+of order~$2$, it follows (Theorem~III, §~80) that no two sub-groups of

 order~$4$ can have a common operation other than identity. Hence

 there must be $5$~sub-groups of order~$4$; for if there were $3$~or~$15$,

 some of them would necessarily have common operations. Each

@@ -7414,7 +7420,7 @@ seen that the $20$~operations of order~$3$ form a conjugate set, and
 that the $24$~operations of order~$5$ form two conjugate sets of $12$

 each. Hence the $60$~operations of the group are distributed in $5$~conjugate

 sets, containing respectively $1$,~$12$, $12$,~$15$ and $20$~operations.

-It follows at once (§~27, \PageRef{p.}{35}) that the group, when it exists,

+It follows at once (§~27, \PageRef{p.}{35}) that the group, when it exists,

 is simple.

 

 A sub-group of order~$12$, the existence of which has been proved,

@@ -7479,7 +7485,7 @@ congruent to unity, $\pmod{p}$. These groups do not however, in
 general, form a single conjugate set.

 

 For a group whose order is a power of~$p$ higher than~$p^{k}$, this

-result has been already proved in §~61. Suppose now that

+result has been already proved in §~61. Suppose now that

 \[

 G_{k},\ G_{k}',\ G_{k}'',\ \Dots,

 \]

@@ -7489,7 +7495,7 @@ been seen, in the proof of Sylow's theorem, that $G$~contains at
 least one sub-group~$G_{\alpha}$ of order~$p^{\alpha}$. The above set of sub-groups

 of order~$p^{k}$ may then be divided into two classes, those

 namely which are contained in~$G_{\alpha}$, and those which are not.

-The result of §~61 shews that the number of the sub-groups

+The result of §~61 shews that the number of the sub-groups

 contained in the first of these classes is congruent to unity,

 $\pmod{p}$; and if $G_{\alpha}$ is a self-conjugate sub-group of~$G$, all the

 sub-groups must be contained in this class. If $G_{\alpha}$ is not

@@ -7602,7 +7608,7 @@ permutable and the order of~$T$ divides~$s$. Now $P$ is one of
 a set of $\dfrac{N}{p^{\lambda}r}$ conjugate operations in~$G$; and corresponding to

 each of these, there is a similar set of $kt$~operations. Moreover

 no two of these operations can be identical; for we have

-seen (§~16) that, if $m$~and~$n$ are relatively prime, an operation

+seen (§~16) that, if $m$~and~$n$ are relatively prime, an operation

 of order~$mn$ of a group~$G$, can be expressed in only one way

 as the product of two permutable operations of~$G$ of orders $m$~and~$n$.

 %% -----File: 128.png---Folio 112-------

@@ -7789,7 +7795,7 @@ If $p$~and~$q$ are distinct primes, there cannot be more than
 one type of group of order~$p^{\alpha}q^{\beta}$ which contains no operation of order~$pq$.

 \end{Ex}

 

-\Par{88.} We have seen in §~53 that a group~$G$, whose order is

+\Par{88.} We have seen in §~53 that a group~$G$, whose order is

 the power of a prime, contains a series of self-conjugate sub-groups

 \[

 H_{1},\ H_{2},\ \Dots,\ H_{n - 1},\ H_{n},\ G,

@@ -7861,7 +7867,7 @@ Hence every operation of~$G$ whose order is a power of~$p$ is
 permutable with every operation of~$G$ whose order is relatively

 prime to~$p$. The group therefore contains self-conjugate sub-groups

 of each of the orders $p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$; and it follows, from

-the definition of §~31, that $G$~is the direct product of these

+the definition of §~31, that $G$~is the direct product of these

 groups.

 

 \begin{Remark}

@@ -7887,7 +7893,7 @@ as $p$~is congruent to $1$,~$5$, $7$ or $11$, $\pmod{12}$.
 \index{Factor groups@\Topic{Factor groups}!set of, for a given group}%

 \index{Factor groups@\Topic{Factor groups}!invariance of}%

 

-\Par{89.} \First{Let} $G_{1}$~be a maximum self-conjugate sub-group (§~27)

+\Par{89.} \First{Let} $G_{1}$~be a maximum self-conjugate sub-group (§~27)

 of a given group~$G$, $G_{2}$~a maximum self-conjugate sub-group of~$G_{1}$,

 and so on. Since $G$ is a group of finite order, we must, after

 a finite number of sub-groups, arrive in this way at a sub-group

@@ -7909,7 +7915,7 @@ is called a set of \emph{factor-groups} of~$G$, and the orders of these
 groups are said to form a set of \emph{composition-factors} of~$G$.

 \end{Definitions}

 

-Each of the set of factor-groups is necessarily (§~30) a

+Each of the set of factor-groups is necessarily (§~30) a

 simple group.

 

 The set of groups forming a composition-series of $G$ is not,

@@ -7929,9 +7935,9 @@ factor-groups derived from them are identical except as regards
 the sequence in which they occur.

 

 This result, which is of great importance in the subsequent

-theory, is due to Herr Hölder\footnotemark;

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{``Zurückführung einer beliebigen algebraischen Gleichung auf eine Kette

+theory, is due to Herr Hölder\footnotemark;

+\index{Holder@\Topic{Hölder}, quoted}%

+\footnotetext{``Zurückführung einer beliebigen algebraischen Gleichung auf eine Kette

   von Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV}, (1889), p.~33.}%

 and the proof we here give does

 not differ materially from his.

@@ -7939,9 +7945,9 @@ not differ materially from his.
 The less general result, that, however the composition-series

 may be chosen, the composition-factors are always the same

 except as regards their sequence, had been proved by M.~Jordan\footnote

-  {``Traité des substitutions,'' (1870), p.~42.}

+  {``Traité des substitutions,'' (1870), p.~42.}

 \index{Jordan@\Topic{Jordan}, quoted}%

-some years before the date of Herr Hölder's memoir.

+some years before the date of Herr Hölder's memoir.

 

 \Par{90.} \begin{Theorem}[I.]

 If $H$ is any self-conjugate sub-group of a

@@ -7958,7 +7964,7 @@ Since $K$ and $K'$ are self-conjugate sub-groups of~$G$ contained
 in~$H$, $\{K, K'\}$~must also be a self-conjugate sub-group of~$G$

 contained in~$H$; and since, by supposition, there is in~$H$ no self-conjugate

 sub-group of~$G$ other than $H$ itself, which contains

-either $K$ or~$K'$, $\{K, K'\}$~must coincide with~$H$. Hence (§~33)

+either $K$ or~$K'$, $\{K, K'\}$~must coincide with~$H$. Hence (§~33)

 the product of the orders of $K$ and $K'$ is equal to the product

 of the orders of $H$ and~$L$.

 

@@ -8007,11 +8013,11 @@ K',\ S_{1}K',\ S_{2}K',\ \Dots,\ S_{m - 1}K',
 and the group $\dfrac{H}{K'}$ is defined by the laws according to which

 these sets combine. But if

 \[

-S_{p}L · S_{q}L = S_{r}L,

+S_{p}L · S_{q}L = S_{r}L,

 \]

 then necessarily

 \[

-S_{p}K' · S_{q}K' = S_{r}K'.

+S_{p}K' · S_{q}K' = S_{r}K'.

 \]

 Hence the groups $\dfrac{H}{K'}$ and $\dfrac{K}{L}$ are simply isomorphic. In precisely

 the same way it is shewn that $\dfrac{H}{K}$ and $\dfrac{K'}{L}$ are simply

@@ -8032,7 +8038,7 @@ sub-group of both $K$ and~$K'$.
 

 \Par{91.} We may now at once proceed to prove by a process of

 induction the properties of the composition-series of a group

-stated at the end of §~89. Let us suppose that, for groups

+stated at the end of §~89. Let us suppose that, for groups

 whose orders do not exceed a given number~$n$, it is already

 known that any two composition-series contain the same

 number of groups and that the factor-groups defined by them

@@ -8133,7 +8139,7 @@ of~$G$.
 It should be noticed that such a series is not necessarily

 obtained by dropping out from a composition-series those of its

 groups which are not self-conjugate in the original group.

-Thus it follows immediately, from the results of §~54, that the

+Thus it follows immediately, from the results of §~54, that the

 composition-series of a group whose order is the power of a

 prime can be chosen, either \Inum{(i)}~so that every group of the series

 is a self-conjugate sub-group, or \Inum{(ii)}~so as to contain any given

@@ -8151,7 +8157,7 @@ occur, are identical with each other.
 \end{Theorem}

 

 The formal proof of this theorem would be a mere repetition

-of the proof of §~91, Theorem~I itself being used to start from

+of the proof of §~91, Theorem~I itself being used to start from

 instead of its Corollary; it is therefore omitted.

 

 Although it is not always possible to pass from a composition-series

@@ -8226,7 +8232,7 @@ sub-group of~$I$, $S^{-1}g_{1}S$~or $g_{r}$ is a maximum self-conjugate
 sub-group of $S^{-1}IS$ or~$I$; and hence every one of the above set

 of conjugate sub-groups is a maximum self-conjugate sub-group

 of~$I$. If $g_{rs}$ represents the greatest sub-group common to $g_{r}$ and~$g_{s}$,

-then, by Theorem~I of the present chapter (§~90),

+then, by Theorem~I of the present chapter (§~90),

 \begin{gather*}

 \Dots,\ I,\ g_{1},\ g_{1r},\ \Dots \\

 \Dots,\ I,\ g_{r},\ g_{1r},\ \Dots

@@ -8240,7 +8246,7 @@ isomorphic.
 If $g_{1r}$ and $g_{1s}$ are the same group, whatever $r$~and~$s$

 may be, this group is common to the whole set of conjugate

 groups~\Inum{($\alpha$)}. If these groups have any common sub-group,

-except identity, it is (Theorem~V, §~27) a self-conjugate sub-group

+except identity, it is (Theorem~V, §~27) a self-conjugate sub-group

 of~$G$, and this self-conjugate sub-group would be contained

 in~$I$, contrary to supposition. Hence if $g_{1r}$ and $g_{1s}$ are

 the same group, for all values of $r$~and~$s$, this group consists of

@@ -8282,7 +8288,7 @@ $\nu$~groups in~$G$, no one of which has any operation except
 identity in common with any other. Since each of these $\nu$~groups

 is self-conjugate in~$I$, and since no two of them have a

 common operation except identity, it follows by Theorem~IX,

-§~34, that every operation of any one of them is permutable

+§~34, that every operation of any one of them is permutable

 with every operation of the remaining $\nu - 1$. The group generated

 by the $\nu$ groups conjugate to~$\gamma_{s-1}$, being self-conjugate in~$G$

 and contained in~$I$, must coincide with~$I$. Now, if $\gamma_{s-1}^{1}$ and

@@ -8455,7 +8461,7 @@ and
 It should be noticed that if, in the last of these series, we drop out

 the terms which are not self-conjugate in the original group, here

 the third and fifth terms, we do not arrive at a chief-series. This

-illustrates a remark made in §~92.

+illustrates a remark made in §~92.

 \end{Remark}

 \index{Composition-series@\Topic{Composition-series}!examples of}%

 

@@ -8526,7 +8532,7 @@ the composition-series of this group, with the composition-factors
 taken as proposed, there is a sub-group~$H$ of order~$p^{\alpha}$

 contained self-conjugately in a sub-group~$H_{1}$ of order~$p^{\alpha}q$. This

 sub-group~$H_{1}$ is contained self-conjugately in a group~$H_{2}$ of

-order~$p^{\alpha}q^{2}$. Hence (Theorem~VII, Cor.~IV, §~87) $H$~is

+order~$p^{\alpha}q^{2}$. Hence (Theorem~VII, Cor.~IV, §~87) $H$~is

 contained self-conjugately in~$H_{2}$. Again, $H_{2}$~is contained self-conjugately

 in a group~$H_{3}$ of order~$p^{\alpha}q^{3}$, and therefore again $H$~is

 self-conjugate in~$H_{3}$. Proceeding thus, we shew that $H$~is

@@ -8930,8 +8936,8 @@ If then we take $P_{1}$,~$P_{3}$ and~$Q^{x}$ as generating operations in the pla
 of $P_{1}$,~$P_{2}$ and~$Q$, the defining relations are reproduced with $\iota^{x}$ in the

 place of~$\iota$. The relations therefore define a single type of group\footnotemark.

 \index{Cole and Glover@\Topic{Cole} and \Topic{Glover}, quoted}%

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{On groups whose order is of the form~$p^{2}q$ the reader may consult; Hölder,

+\index{Holder@\Topic{Hölder}, quoted}%

+\footnotetext{On groups whose order is of the form~$p^{2}q$ the reader may consult; Hölder,

   ``Die Gruppen der Ordnungen $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.}\ \DPnote{** No "Vol."}\VolNo{XLIII} (1893), in

   particular pp.~335--360; and Cole and Glover, ``On groups whose orders are

   products of three prime factors,'' \textit{Amer.\ Journal}, \VolNo{XV} (1893), pp.~202--214.}%

@@ -8949,7 +8955,7 @@ the previous types it is immediately obvious.
 \Chapter{VIII.}{On Substitution-Groups: Transitive and

 Intransitive Groups.}

 

-\Par{102.} \First{It} has been proved, in the theorem in §~20, that every

+\Par{102.} \First{It} has been proved, in the theorem in §~20, that every

 group is capable of being represented as a group of substitutions

 performed on a number of symbols equal to the order of the

 group. For applications to Algebra, and in particular to the

@@ -8980,23 +8986,23 @@ representation of the group as a substitution-group.''
 The $n!$ substitutions, including the identical substitution,

 that can be performed upon $n$~distinct symbols, clearly form a

 %% -----File: 155.png---Folio 139-------

-group; for they satisfy the conditions of the definition (§~12).

+group; for they satisfy the conditions of the definition (§~12).

 Moreover they form the greatest group of substitutions that can

 be performed on the $n$~symbols, because every possible substitution

 occurs among them. When a group then is spoken of as of

 degree~$n$, it is implicitly being regarded as a sub-group of this

 most general group of order~$n!$ which can be represented by

 substitutions of the $n$~symbols; and therefore (Theorem~I,

-§~22) the order of a substitution-group of degree~$n$ must be

+§~22) the order of a substitution-group of degree~$n$ must be

 a factor of~$n!$.

 

-\Par{103.} It has been seen in §~11 that any substitution performed

+\Par{103.} It has been seen in §~11 that any substitution performed

 on $n$~symbols can be represented in various ways as

 the product of transpositions; but that the number of transpositions

 entering in any such representation of the substitution

 is either always even or always odd. In particular, the identical

 substitution can only be represented by an even number of

-transpositions. Hence if $S$~and~$S'$ are any two even (§~11)

+transpositions. Hence if $S$~and~$S'$ are any two even (§~11)

 substitutions of $n$~symbols, and $T$~any substitution at all of $n$~symbols,

 then $SS'$ and $T^{-1}ST$ are even substitutions. The

 even substitutions therefore form a self-conjugate sub-group~$H$

@@ -9026,7 +9032,7 @@ substitutions of $n$~symbols is called the \emph{alternating}\footnote
   symmetric functions.

 

   All the substitutions of the alternating group leave the square root of the

-  discriminant unaltered (§~11).}

+  discriminant unaltered (§~11).}

 group of

 degree~$n$.

 \end{Definitions}

@@ -9120,7 +9126,7 @@ differ in the substitutions that keep just one symbol unchanged.
 \begin{Remark}

 \Par{105.} We have seen that every group can be represented as a

 substitution-group whose degree is equal to its order. A reference

-to the proof of this theorem (§~20) will shew that such a substitution-group

+to the proof of this theorem (§~20) will shew that such a substitution-group

 is transitive, and that the identical substitution is the only

 one which leaves any symbol unchanged.

 

@@ -9159,12 +9165,12 @@ Suppose now that $p$ is any prime factor of~$n$, and that $p^{\alpha}$~is the
 highest power of~$p$ which divides~$n$. If $P$~is an operation whose

 order is a power of~$p$, and if $p^{\beta}\mu$~is the order of the greatest sub-group~$h$

 that contains $P$ self-conjugately, then $P$~is one of $\dfrac{nm}{p^{\beta}\mu}$

-conjugate operations. Now (§~87) the sub-group~$h$ contains~$k\mu$ ($k \nless 1$)

+conjugate operations. Now (§~87) the sub-group~$h$ contains~$k\mu$ ($k \nless 1$)

 operations whose orders are relatively prime to~$p$; and therefore

 there are $k\mu$~operations of the form~$PQ$, where $Q$~is permutable with~$P$

 and the order of~$Q$ is relatively prime to the order of~$P$. If $P'$~is

 any operation conjugate to~$P$, there are similarly $k\mu$~operations of

-the form~$P'Q'$; and (§~16) no one of these operations can be

+the form~$P'Q'$; and (§~16) no one of these operations can be

 identical with any one of those of the previous set. The group

 therefore will contain $\dfrac{nmk}{p^{\beta}}$ distinct operations, which are conjugate

 to the various operations of the set~$PQ$. Moreover since $P$~displaces

@@ -9180,7 +9186,7 @@ are divisible by~$p$, is equal to
 

 Also the number of operations, which displace all the symbols

 and the orders of which are not divisible by~$p$, is of the form $\dfrac{nk_{0}}{p^{\alpha}} - 1$

-(§~87).

+(§~87).

 

 Hence finally

 \begin{align*}

@@ -9209,10 +9215,10 @@ of degree~$n$ and order~$n(n - 1)$, whose operations other than
 identity displace all or all but one of the symbols, can exist only

 when $n$~is the power of a prime\footnotemark.

 \index{Jordan@\Topic{Jordan}, quoted}%

-\footnotetext{Jordan, ``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér.\

+\footnotetext{Jordan, ``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér.\

   Vol.~\VolNo{XVII} (1872), p.~355.}%

 Groups which satisfy these

-conditions will be discussed in §~112.

+conditions will be discussed in §~112.

 

 Similarly, a group of degree~$n$ and order~$nm$, where $m$~is not less

 than~$\sqrt{n}$, whose operations other than identity displace all or all but

@@ -9252,7 +9258,7 @@ A_{r}A_{1},\ A_{r}A_{2},\ \Dots,\ A_{r}A_{r - 1},\ A_{r}A_{r + 1},\ \Dots, A_{r}
 \]

 and since

 \[

-A_{p}A_{r} · A_{r}A_{q} = A_{p}A_{q},

+A_{p}A_{r} · A_{r}A_{q} = A_{p}A_{q},

 \]

 %% -----File: 160.png---Folio 144-------

 the product of any two of these operations is either identity or

@@ -9260,12 +9266,12 @@ another operation which displaces all the symbols. Hence the $n - 1$
 operations which displace all the symbols, with identity, form a self-conjugate

 sub-group. Now

 \[

-A_{r} · A_{p}A_{r} · A_{r} = A_{r}A_{p},

+A_{r} · A_{p}A_{r} · A_{r} = A_{r}A_{p},

 \]

 so that $A_{r}$~transforms every operation of this sub-group into its

 inverse. Hence

 \[

-A_{r}A_{p} · A_{q}A_{r} = A_{q}A_{p} = A_{q}A_{r} · A_{r}A_{q};

+A_{r}A_{p} · A_{q}A_{r} = A_{q}A_{p} = A_{q}A_{r} · A_{r}A_{q};

 \]

 \ie\ every two operations of this sub-group are permutable, and the

 sub-group is therefore Abelian.

@@ -9279,7 +9285,7 @@ and
 \[

 T = \Sub{a_{1}, a_{2}, \Dots, a_{n}}{b_{1}, b_{2}, \Dots, b_{n}},

 \]

-are any two substitutions of a group, then (§~10)

+are any two substitutions of a group, then (§~10)

 \[

 T^{-1}ST = (b_{1}b_{2} \Dots b_{i})(b_{i + 1}b_{i + 2} \Dots b_{j})\Dots.

 \]

@@ -9294,7 +9300,7 @@ for then the substitution~$T$ may be chosen so as to replace the
 $n$~symbols by any permutation of them whatever.

 

 A self-conjugate substitution of a transitive group of degree~$n$

-must be a regular substitution (§~9) changing all the $n$~symbols.

+must be a regular substitution (§~9) changing all the $n$~symbols.

 For if it did not change all the $n$~symbols, it would belong to

 one of the sub-groups that keep a symbol unchanged. Hence,

 since it is a self-conjugate substitution, it would belong to each

@@ -9348,7 +9354,7 @@ attention. If
 \[

 S_{1} (= 1),\ S_{2},\ \Dots,\ S_{n}

 \]

-are the operations of~$G$, it has been seen (§~20) that the

+are the operations of~$G$, it has been seen (§~20) that the

 substitution-group can be expressed as consisting of the $n$~substitutions

 \[

 \Sub{\PadTo{S_{1} S_{x}}{S_{1}}, \PadTo{S_{2} S_{x}}{S_{2}}, \Dots, \PadTo{S_{n} S_{x}}{S_{n}}}

@@ -9357,7 +9363,7 @@ substitution-group can be expressed as consisting of the $n$~substitutions
 performed upon the symbols of the operations of the group.

 

 Now if pre-multiplication be used in the place of post-multiplication,

-it may be verified exactly as in §~20 that the $n$~substitutions

+it may be verified exactly as in §~20 that the $n$~substitutions

 \[

 \Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \PadTo{S_{x} S_{2}}{S_{2}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}

     {S_{x} S_{1}, S_{x} S_{2}, \Dots, S_{x} S_{n}},\quad (x = 1, 2, \Dots, n)

@@ -9398,7 +9404,7 @@ permutable with every substitution of a substitution-group~$G$ of
 order~$n$, transitive in the $n$~symbols, form a group~$G'$ of order

 and degree~$n$, simply isomorphic with~$G$\footnotemark.

 \end{Theorem}

-\footnotetext{Jordan, \textit{Traité des Substitutions} (1870), p.~60.}%

+\footnotetext{Jordan, \textit{Traité des Substitutions} (1870), p.~60.}%

 \index{Jordan@\Topic{Jordan}, quoted}%

 

 It has, in fact, been seen that there is a substitution-group~$G'$

@@ -9555,7 +9561,7 @@ sub-group is contained self-conjugately in a sub-group of order~$k!m$.
 \end{Theorem}

 

 If $N$ is the order of the group, the order of the sub-group

-which keeps one symbol fixed is~$\dfrac{N}{n}$, by Theorem~I (§~104).

+which keeps one symbol fixed is~$\dfrac{N}{n}$, by Theorem~I (§~104).

 Now this sub-group is a transitive group of degree~$n - 1$; and

 therefore the order of the sub-group that keeps two symbols

 unchanged is~$\dfrac{N}{n(n - 1)}$. If $k > 2$, this sub-group again is a

@@ -9596,15 +9602,15 @@ If $m > 1$, the group will contain $m - 1$ substitutions besides
 identity, which leave unchanged any $k$ given symbols, and

 therefore displace $n - k$ symbols at most.

 

-It follows from §~105 that a $k$-ply transitive group of degree~$n$

+It follows from §~105 that a $k$-ply transitive group of degree~$n$

 and order $n(n - 1) \Dots (n - k + 1)$ can exist only if $n - k + 2$

 is the power of a prime. For such a group must contain

 sub-groups of order $(n - k + 2)(n - k + 1)$, which keep $k - 2$

 symbols unchanged and are doubly transitive in the remaining

 $n - k + 2$. When $k$~is~$n$, the group is the symmetric group; and

-when $k$~is~$n - 2$, we shall see (in §~110) that the group is the

+when $k$~is~$n - 2$, we shall see (in §~110) that the group is the

 alternating group. If $k$ is less than~$n - 2$, M.~Jordan\footnote

-  {``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér., Vol.~\VolNo{XVII}

+  {``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér., Vol.~\VolNo{XVII}

   (1872), pp.~357--363.}

 \index{Jordan@\Topic{Jordan}, quoted}%

 \PageLabel{149}%

@@ -9612,7 +9618,7 @@ has
 shewn that, with two exceptions for $n = 11$ and $n = 12$, the

 value of~$k$ cannot exceed~$3$. The actual existence of triply

 transitive groups of order $(p^{n} + 1) p^{n} (p^{n} - 1)$, for all prime

-values of~$p$, will be established in §~113.

+values of~$p$, will be established in §~113.

 %% -----File: 166.png---Folio 150-------

 

 \Par{109.} A $k$-ply transitive group, of degree~$n$ and order~$N$,

@@ -9620,7 +9626,7 @@ values of~$p$, will be established in §~113.
 is not generally contained in some $(k + 1)$-ply transitive

 group of degree~$n + 1$ and order~$N(n + 1)$. To determine

 whether this is the case for any given group, M.~Jordan\footnote

-  {\textit{Traité des Substitutions} (1870), pp.~31,~32.}

+  {\textit{Traité des Substitutions} (1870), pp.~31,~32.}

 \index{Jordan@\Topic{Jordan}, quoted}%

 has

 suggested the following tentative process, which for moderate

@@ -9671,7 +9677,7 @@ where $S_{1}$,~$S_{2}$,~$S_{3}$ are substitutions of~$G$. For instance, the
 substitution

 \begin{align*}

 AS_{p}AS_{q}AS_{r}

-  &= S_{p}'AS_{p}'' · S_{q}AS_{r}, \\

+  &= S_{p}'AS_{p}'' · S_{q}AS_{r}, \\

   &= S_{p}'AS_{t}AS_{r}, \quad\text{if}\quad S_{p}''S_{q} = S_{t}, \\

   &= S_{p}'S_{t}'AS_{t}''S_{r},

 \end{align*}

@@ -9769,7 +9775,7 @@ and
 

 Thus in the case under consideration the group contains

 one, and therefore every, circular substitution of three symbols;

-and hence (§~11) it must contain every even substitution. It

+and hence (§~11) it must contain every even substitution. It

 is therefore either the alternating or the symmetric group. If

 then a $k$-ply transitive group of degree~$n$ does not contain the

 alternating group of $n$~symbols, no one of its substitutions,

@@ -9812,7 +9818,7 @@ transitive\footnotemark.
 \index{Bochert@\Topic{Bochert}, quoted}%

 \index{Jordan@\Topic{Jordan}, quoted}%

 \footnotetext{For a further discussion of the limits of transitivity of a substitution-group,

-  compare Jordan, \textit{Traité des Substitutions}, pp.~76--87; and Bochert,

+  compare Jordan, \textit{Traité des Substitutions}, pp.~76--87; and Bochert,

   \textit{Math.\ Ann.},\DPnote{** No "Vol."} \VolNo{xxix}, (1886) pp.~27--49; \VolNo{XXXIII}, (1888) pp.~573--583.}%

 

 \Par{111.} The process used in the preceding paragraph may be

@@ -9915,7 +9921,7 @@ group except when $n = 4$.
 \end{Theorem}

 

 \begin{Remark}

-\Par{112.} It has been seen in §~108 that the order of a doubly transitive

+\Par{112.} It has been seen in §~108 that the order of a doubly transitive

 group of degree~$n$ is equal to or is a multiple of~$n(n - 1)$. If it

 is equal to this number, every substitution of the group, except

 identity, must displace either all or all but one of the symbols; for

@@ -9923,7 +9929,7 @@ a sub-group of order~$n - 1$ which keeps one symbol fixed is transitive
 in the remaining $n - 1$ symbols, and therefore all its substitutions,

 except identity, displace all the $n - 1$ symbols.

 

-Now it has been shewn in §~105 that a transitive group of degree~$n$

+Now it has been shewn in §~105 that a transitive group of degree~$n$

 and order~$n(n - 1)$, whose operations displace all or all but one of

 the symbols can exist only if $n$~is the power of a prime~$p$. The

 $n - 1$ operations displacing all the symbols are the only operations

@@ -9932,7 +9938,7 @@ of the group whose orders are powers of~$p$; and therefore with
 \index{Doubly transitive groups@\Topic{Doubly transitive groups}!of degree~$n$ and order~$n(n - 1)$|(}%

 \index{Substitution group@\Topic{Substitution group}!doubly transitive, of degree~$n$ and order~$n(n - 1)$|(}%

 identity they form a self-conjugate sub-group of order~$n$. Moreover

-it also follows from §~105 that the $n - 1$ operations of this sub-group

+it also follows from §~105 that the $n - 1$ operations of this sub-group

 other than identity form a single conjugate set. Hence this sub-group

 must be Abelian, and all its operations are of order~$p$.

 

@@ -9955,7 +9961,7 @@ P^{p} = 1,\quad S^{p-1} = 1,\quad S^{-1}PS = P^{\alpha}.
 \]

 

 It is an immediate result of a theorem, which will be proved in the

-next chapter~(§~123), that this group can be actually represented as

+next chapter~(§~123), that this group can be actually represented as

 a transitive substitution-group of degree~$p$; this may be also verified

 directly as follows.

 

@@ -9993,7 +9999,7 @@ $p^{m}(p^{m} - 1)$, in which a sub-group of order~$p^{m} - 1$ is cyclical\footno
 \index{Mathieu@\Topic{Mathieu}, quoted}%

 \footnotetext{On the subject of this and the following paragraph, the reader  should

 \PageLabel{155}%

-  consult the memoirs by Mathieu, \textit{Liouville's Journal}, \Ord{2}{me} Sér., t.~\VolNo{V} (1860),

+  consult the memoirs by Mathieu, \textit{Liouville's Journal}, \Ord{2}{me} Sér., t.~\VolNo{V} (1860),

   pp.~9--42; \textit{ib.}\ t.~\VolNo{VI} (1861), pp.~241--323; where the groups here considered were

   first shewn to exist.}

 %% -----File: 172.png---Folio 156-------

@@ -10195,11 +10201,11 @@ identity, keeps more than two symbols fixed.
 

 Finally then, since the group is triply transitive and since it

 contains no operation, except identity, that keeps more than two

-symbols fixed, its order must (§~108) be $(p^{m} + 1) p^{m} (p^{m} - 1)$.

+symbols fixed, its order must (§~108) be $(p^{m} + 1) p^{m} (p^{m} - 1)$.

 \end{Remark}

 %% -----File: 175.png---Folio 159-------

 

-\Par{114.} An intransitive substitution group, as defined in §~104,

+\Par{114.} An intransitive substitution group, as defined in §~104,

 is one which does not contain substitutions changing $a_{1}$ into

 each of the other symbols $a_{2}$, $a_{3}$,~\Dots, $a_{n}$ operated on by the

 group. Let us suppose that the substitutions of such a group

@@ -10282,11 +10288,11 @@ their number is just equal to the order of~$G_{\alpha}$. Hence $\dfrac{G_{\alpha
 defined by the laws according to which these sets of substitutions

 combine. But if

 \[

-Sg · Tg = Ug,

+Sg · Tg = Ug,

 \]

 then necessarily

 \[

-S_{\alpha} g_{\alpha} · T_{\alpha} g_{\alpha} = U_{\alpha} g_{\alpha},

+S_{\alpha} g_{\alpha} · T_{\alpha} g_{\alpha} = U_{\alpha} g_{\alpha},

 \]

 %% -----File: 177.png---Folio 161-------

 and therefore, finally, the three groups $\dfrac{G}{g}$,~$\dfrac{G_{\alpha}}{g_{\alpha}}$, and~$\dfrac{G_{\beta}}{g_{\beta}}$ are simply

@@ -10294,7 +10300,7 @@ isomorphic.
 

 \Par{115.} The relation of simple isomorphism between $\dfrac{G_{\alpha}}{g_{\alpha}}$ and $\dfrac{G_{\beta}}{g_{\beta}}$

 thus arrived at establishes between the groups $G_{\alpha}$ and $G_{\beta}$ an

-isomorphism of the most general kind (§~32).

+isomorphism of the most general kind (§~32).

 

 To every operation of~$G_{\alpha}$ correspond $n_{g_{\beta}}$~operations of~$G_{\beta}$,

 and to every operation of~$G_{\beta}$ correspond $n_{g_{\alpha}}$~operations

@@ -10310,8 +10316,8 @@ between the groups $G_{\alpha}$ and~$G_{\beta}$, each substitution of~$G_{\alpha
 multiplied by the $n_{g_{\beta}}$~substitutions that correspond to it in~$G_{\beta}$.

 The set of $n_{G_{\alpha}}n_{g_{\beta}}$ substitutions so obtained form a group, for

 \[

-S_{\alpha} S_{\beta} · S_{\alpha}' S_{\beta}'

-  = S_{\alpha} S_{\alpha}' · S_{\beta} S_{\beta}'

+S_{\alpha} S_{\beta} · S_{\alpha}' S_{\beta}'

+  = S_{\alpha} S_{\alpha}' · S_{\beta} S_{\beta}'

   = S_{\alpha}'' S_{\beta}'',

 \]

 where, if $S_{\beta}$,~$S_{\beta}'$ are substitutions corresponding to $S_{\alpha}$,~$S_{\alpha}'$, then

@@ -10350,7 +10356,7 @@ it\footnotemark.
 \footnotetext{On intransitive groups, reference may be made to Bolza, ``On the

   construction of intransitive groups,'' \textit{Amer.\ Journal}, Vol.~\VolNo{XI}, (1889), pp.~195--214.

   The general isomorphism which underlies the construction of these

-  groups is considered by Klein and Fricke, \textit{Vorlesungen über die Theorie der

+  groups is considered by Klein and Fricke, \textit{Vorlesungen über die Theorie der

   elliptischen Modulfunctionen}, Vol.~\VolNo{I}, (1890), pp.~402--406.}

 

 It is not necessary to give an independent proof of this

@@ -10567,7 +10573,7 @@ Nt_{1} = \sum_{r = 1}^{r = n} r\nu_{r}.
 \]

 

 \Par{119.} The formula just obtained is the first of a series of

-similar \DPchg{formulae}{formulæ}, due to Herr Frobenius,\footnote

+similar \DPchg{formulae}{formulæ}, due to Herr Frobenius,\footnote

   {``Ueber die Congruenz nach einem aus zwei endlichen Gruppen gebildeten

   Doppelmodul,'' \textit{Crelle}, t.~\VolNo{CI}, (1887), p.~288.}

 \index{Frobenius@\Topic{Frobenius}, quoted}%

@@ -10622,13 +10628,13 @@ t_{p} \leq t_{p + 1} \leq (n - p)t_{p}.
 unity and that the group is $(p + 1)$-ply transitive.

 \end{Ex}

 

-\begin{Ex}[2.] Apply the method of §~119 to shew that no substitution,

+\begin{Ex}[2.] Apply the method of §~119 to shew that no substitution,

 except identity, of a $k$-ply transitive group, which does not contain

 the alternating group, can displace less than $2k - 2$ letters.

 \end{Ex}

 

 

-\Note{\textit{Note to} §~108.}

+\Note{\textit{Note to} §~108.}

 

 \begin{Remark}

 The results of M.~Jordan, stated on \PageRef{p.}{149}, may be established

@@ -10683,7 +10689,7 @@ alternating group of degree~$6$.
 Next, suppose that $n_{1}$ is~$3$. The only admissible values for

 $n_{2}$, $n_{3}$,~\dots\ are $8$~and~$24$. In this case, the sub-group of~$H$ which

 keeps all the $n_{1}$ ($= 3$) symbols unchanged is a non-cyclical sub-group

-of order~$4$. But we have seen in §~105 that, if $n - 3$ is even, a

+of order~$4$. But we have seen in §~105 that, if $n - 3$ is even, a

 sub-group of order~$n - 3$ which keeps $3$~symbols unchanged can only

 have a single operation of order~$2$. Hence this case cannot occur.

 

@@ -10744,9 +10750,9 @@ $5$ are the only remaining possibilities.
 When $n$ is~$5$, $G$~is the symmetric group of degree~$5$.

 

 When $n$ is $11$, the group~$G$, if it exists, is of degree~$11$ and

-order $11 · 10 · 9 · 8$. A sub-group of order~$8$ which keeps three

+order $11 · 10 · 9 · 8$. A sub-group of order~$8$ which keeps three

 symbols fixed must contain a single operation of order~$2$; hence

-it must either be cyclical or of the type given in Theorem~V, §~63.

+it must either be cyclical or of the type given in Theorem~V, §~63.

 When expressed in $8$~symbols, this is generated by

 \[

 (1254) (3867)\quad \text{and}\quad (1758) (2643),

@@ -10763,7 +10769,7 @@ group of order~$9$, generated by
 \]

 into itself.

 

-If we now apply the method of §~109, we find that each of

+If we now apply the method of §~109, we find that each of

 these doubly transitive groups of degree~$9$ is contained in a triply

 %% -----File: 186.png---Folio 170-------

 transitive group of degree~$10$; a repetition of the same process of

@@ -10772,7 +10778,7 @@ trial shews that, while the group
 \{(123) (456) (789),\ (12835476)\}

 \]

 is not contained in a quadruply transitive group of degree~$11$ and

-order~$11·10·9·8$, the group

+order~$11·10·9·8$, the group

 \[

 \{(123) (456) (789),\ (1254) (3867),\ (1758) (2643)\}

 \]

@@ -10786,13 +10792,13 @@ A further trial will shew that this group and the substitution
 \[

 (bc) (47) (58) (69)

 \]

-generate a group of degree~$12$ and order $12·11·10·9·8$; but that

+generate a group of degree~$12$ and order $12·11·10·9·8$; but that

 there is no group of degree~$13$ which contains the last group as the

 sub-group that keeps one symbol fixed.

 

 The three substitutions, of order two, just given generate that

 sub-group of order~$24$, of the transitive group of degree~$11$ and

-order $11·10·9·8$ in the symbols

+order $11·10·9·8$ in the symbols

 \[

 a,\ b,\ c,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,

 \]

@@ -10806,7 +10812,7 @@ group of degree~$n$, it must be either the group of degree~$11$
 or the group of degree~$12$ that have been determined above.}

 

 

-\Note{\textit{Note to} §~110.}

+\Note{\textit{Note to} §~110.}

 

 It may be shewn that, with a single exception when $n = 12$, the

 inequality (\PageRef{p.}{152})

@@ -10871,7 +10877,7 @@ an imprimitive group are called \emph{imprimitive systems}.
 \end{Definition}

 

 A simple example of an imprimitive group is given by

-group~\Inum{VII} of §~17. An examination of the substitutions of this

+group~\Inum{VII} of §~17. An examination of the substitutions of this

 group will shew that they all either transform the systems of

 symbols $xyz$ and $abc$ into themselves or else interchange them,

 and that the same is true of the systems $xa$,~$yb$,~$zc$; so that, in

@@ -10893,7 +10899,7 @@ following investigation due to Herr Dyck\footnotemark.
 \index{Dyck@\Topic{Dyck}, quoted}%

 \footnotetext{Dyck, ``Gruppentheoretische Studien,~II,'' \textit{Math.\ Ann.}\DPnote{** No "Vol."} \VolNo{XXII}, (1883), pp.~86--95.}

 

-\Par{122.} In §~20 it was shewn how any group~$G$ of order~$N$

+\Par{122.} In §~20 it was shewn how any group~$G$ of order~$N$

 could be represented as a substitution group of $N$~symbols.

 This form of the group, defined as the regular form, is simply

 transitive; for all its substitutions except identity displace all

@@ -11200,7 +11206,7 @@ I_{1},\ I_{2},\ \Dots,\ I_{\nu}
 \]

 be a set of sub-groups of~$G$, such that $I_{r}$~is the greatest sub-group

 containing $H_{r}$ self-conjugately. The latter set of groups

-are not necessarily all distinct; in fact, we have seen in §~55

+are not necessarily all distinct; in fact, we have seen in §~55

 that, when the order of~$G$ is the power of a prime, they cannot

 be all distinct.

 

@@ -11249,7 +11255,7 @@ I_{1},\ I_{2},\ \Dots,\ I_{\nu},
 has no common sub-group except identity. This will be the

 case only when $I_{1}$~contains no self-conjugate sub-group of~$G$.

 

-It has been seen (§~123) that, when this condition is satisfied,

+It has been seen (§~123) that, when this condition is satisfied,

 $G$~can be represented as a transitive substitution group whose

 degree is~$\nu$, the ratio of the orders of $G$~and~$I_{1}$. That the

 form there obtained is identical with the form obtained in the

@@ -11348,7 +11354,7 @@ Hence finally, since each of the sub-groups leads to a transitive
 representation of the group, $G_{60}$~can be represented as a transitive

 substitution group in $30$,~$20$, $15$, $12$, $10$, $6$ and $5$~symbols, and in

 one distinct form in each case. The second method, as given in

-§~125, does not lead to all these modes of representation. The

+§~125, does not lead to all these modes of representation. The

 group will be found to contain $15$~conjugate operations (or sub-groups)

 of order~$2$: $10$~conjugate sub-groups and $20$~conjugate

 operations of order~$3$: $5$~conjugate sub-groups of order~$4$: $6$~conjugate

@@ -11385,7 +11391,7 @@ permute $1$,~$2$ and~$3$ among themselves, and therefore also $4$,~$5$ and~$6$
 among themselves.

 

 Let $H$~contain substitutions which leave $4$,~$5$ and~$6$ unchanged.

-These must (§~114) form a self-conjugate sub-group of~$H$, which will

+These must (§~114) form a self-conjugate sub-group of~$H$, which will

 be either

 \[

 \{(123),\ (12)\}\quad \text{or}\quad \{(123)\}.

@@ -11403,15 +11409,15 @@ In the first alternative, $H$~is the group
 \]

 for $H$~contains this group, and on the other hand, this is the most

 general group that interchanges the six symbols in two intransitive

-systems of three each. The order of this group is~$2^{2} · 3^{2}$.

+systems of three each. The order of this group is~$2^{2} · 3^{2}$.

 

 In the second alternative, $H$~contains the self-conjugate sub-group

 \[

 \{(123),\ (456)\}.

 \]

 

-Now if $H$~is of order~$2^{2} · 3^{2}$, it is necessarily of the form~\Inum{(i)}. If

-it is of order~$2 · 3^{2}$, it must contain a substitution of order~$2$ which

+Now if $H$~is of order~$2^{2} · 3^{2}$, it is necessarily of the form~\Inum{(i)}. If

+it is of order~$2 · 3^{2}$, it must contain a substitution of order~$2$ which

 transforms the sub-group just given into itself. This may be taken,

 without loss of generality, to be~$(12) (45)$; and then $H$~is the

 group

@@ -11657,7 +11663,7 @@ group is imprimitive.
 \end{Theorem}

 

 This result may also be regarded as an immediate consequence

-of Theorem~III, §~124. For if $H_{1}$~and~$H_{2}$ are two

+of Theorem~III, §~124. For if $H_{1}$~and~$H_{2}$ are two

 sub-groups of~$G$ each of which contains the sub-group~$R$, and

 if $K$~is the greatest common sub-group of $H_{1}$ and~$H_{2}$, then $K$

 contains~$R$. Now if $R$~contains no self-conjugate sub-group

@@ -11688,7 +11694,7 @@ that the symbols so grouped together form an imprimitive system.
 \end{Remark}

 

 \Par{130.} We may now discuss, more fully than was possible in

-§~106, the form of a self-conjugate sub-group of a given transitive

+§~106, the form of a self-conjugate sub-group of a given transitive

 group. Such a sub-group must clearly contain one or more

 operations displacing every symbol operated on by the group.

 For if every operation of the sub-group keeps the symbol $a_{1}$

@@ -11737,7 +11743,7 @@ of a set of imprimitive systems.
 

 If $G$~is an imprimitive group of degree~$mn$, and if there are

 $n$ imprimitive systems of $m$~symbols each, then we have seen

-in §~123 that $G$~is isomorphic with a group~$G'$ of degree~$n$. In

+in §~123 that $G$~is isomorphic with a group~$G'$ of degree~$n$. In

 particular instances, it may at once be evident, from the order of~$G$,

 that this isomorphism cannot be simple. For example, if

 the order of~$G$ has a factor which does not divide~$n!$, this is

@@ -11779,7 +11785,7 @@ except identity, which displaces less than $n - k + 3$ symbols.
 Suppose, first, that $H$~contains operations, other than identity,

 which leave one or more symbols unchanged. Then, since $H$~is

 a self-conjugate sub-group and $G$~is $k$-ply transitive, it may

-be shewn, exactly as in §~110, that $H$~must contain operations

+be shewn, exactly as in §~110, that $H$~must contain operations

 displacing not more than $2k - 2$ symbols. Hence $H_{k - 2}$ can

 consist of the identical operation alone, only if

 \[

@@ -11807,7 +11813,7 @@ $(k - 1)$-ply transitive.
 

 Suppose, next, that $H$~contains no operation, except identity,

 which leaves any symbol unchanged. Then if, with the notation

-of §~110, $j = k - 1$ for every operation of~$H$, the argument there

+of §~110, $j = k - 1$ for every operation of~$H$, the argument there

 used does not apply. For it is impossible to choose the operation~$T$

 so that $c_{k}$~is a symbol which does not occur in~$S$.

 

@@ -11842,7 +11848,7 @@ A self-conjugate sub-group of a $k$-ply
 transitive group of degree~$n$ \($2 < k < n$\), is in general at least

 $(k - 1)$-ply transitive\footnotemark.

 \index{Jordan@\Topic{Jordan}, quoted}%

-\footnotetext{Jordan, \textit{Traité des Substitutions}, p.~65; where, however, the exceptional

+\footnotetext{Jordan, \textit{Traité des Substitutions}, p.~65; where, however, the exceptional

   case is overlooked.}%

 The only exception is that a triply

 transitive group of degree~$2^{m}$ may have a self-conjugate sub-group

@@ -11860,7 +11866,7 @@ Let $G$~be the direct product of two groups $H_{1}$ and~$H_{2}$, and
 suppose that $G$~can be represented as a transitive group

 of degree~$n$. When $G$ is thus represented, we will suppose

 that $H_{1}$~is transitive in the $n$~symbols that $G$~permutes. We

-have seen in §~107 that every substitution of $n$~symbols, which

+have seen in §~107 that every substitution of $n$~symbols, which

 is permutable with each of the substitutions of a group

 transitive in the $n$~symbols, must displace all the $n$~symbols.

 It follows that every substitution of~$H_{2}$ must displace all the $n$~symbols

@@ -11869,7 +11875,7 @@ equal to or is a factor of~$n$.
 

 If the order of~$H_{2}$ is equal to~$n$, then $H_{2}$~is transitive in the

 $n$~symbols, so that the order of~$H_{1}$ cannot be greater than~$n$.

-In this case, $H_{1}$~and~$H_{2}$ must (§~107) be two simply isomorphic

+In this case, $H_{1}$~and~$H_{2}$ must (§~107) be two simply isomorphic

 groups of order~$n$, which have no self-conjugate operations

 except identity. Further, if $H_{1}$~and $H_{2}$ in this case are not

 simple groups, let $K$ be a self-conjugate sub-group of~$H_{1}$.

@@ -11896,15 +11902,15 @@ has two distinct minimum self-conjugate sub-groups $H_{1}$ and~$H_{2}$.
 Then every operation of~$H_{1}$ (or~$H_{2}$) is permutable with~$H_{2}$

 %% -----File: 207.png---Folio 191-------

 (or~$H_{1}$), and $H_{1}$,~$H_{2}$ have no common operation except

-identity. Hence (§~34) the group $\{H_{1}, H_{2}\}$, which we will call~$\Gamma$,

+identity. Hence (§~34) the group $\{H_{1}, H_{2}\}$, which we will call~$\Gamma$,

 is the direct product of $H_{1}$ and~$H_{2}$. Now $\Gamma$~is a self-conjugate

 sub-group of~$G$: it is therefore transitive in

 the $n$~symbols which $G$~permutes. Also $H_{1}$ and~$H_{2}$, being

 self-conjugate sub-groups of~$G$, are transitive. Hence, by

-Theorem~VIII (§~132), $H_{1}$~and $H_{2}$ are simply isomorphic, and $n$~is

+Theorem~VIII (§~132), $H_{1}$~and $H_{2}$ are simply isomorphic, and $n$~is

 equal to the order of~$H_{1}$. Moreover, since $H_{1}$~is a minimum

 self-conjugate sub-group of~$G$ which contains no self-conjugate

-operations except identity, it must (Theorem~IV, §~94) be either

+operations except identity, it must (Theorem~IV, §~94) be either

 a simple group of composite order, or the direct product of

 several simply isomorphic simple groups of composite order. It

 follows that $G$~cannot have two distinct minimum self-conjugate

@@ -11923,12 +11929,12 @@ group of prime order~$p$, the order of~$\Gamma$ is~$p^{\alpha}$; therefore the
 degree of~$\Gamma$, or what is the same thing, the degree of~$G$, is~$p^{\alpha}$.

 

 If $H_{1}$~is a simple group of composite order, and if $\alpha > 2$,

-then (§~132) $H_{1}$~cannot be transitive. The intransitive systems

+then (§~132) $H_{1}$~cannot be transitive. The intransitive systems

 of~$H_{1}$, since they form a set of imprimitive systems for~$\Gamma$, must

 each contain the same number $m$ of symbols. If $m$~is less than

 the order of~$H_{1}$, a sub-group of~$H_{1}$ which leaves unchanged one

 symbol of one intransitive system will leave unchanged one

-symbol of each intransitive system. Now we shall see, in §~136,

+symbol of each intransitive system. Now we shall see, in §~136,

 that the operations of an imprimitive self-conjugate sub-group

 of a doubly transitive group must displace all or all but one of

 the symbols. Hence $m$~cannot be less than the order of~$H_{1}$.

@@ -11969,7 +11975,7 @@ must be the power of a prime\footnotemark.
 \PageLabel{192}%

   others, in the letter written by Galois to his friend Chevalier on the evening

   of May~29th, 1832, the day before the duel in which he was killed. The letter

-  was first printed in the \textit{Revue Encyclopédique} (1832), p.~568; it was reprinted

+  was first printed in the \textit{Revue Encyclopédique} (1832), p.~568; it was reprinted

   in the collection of Galois's mathematical writings in \textit{Liouville's Journal}, t.~\VolNo{XI}

   (1846), pp.~381--444}

 %[** TN: See http://www.gutenberg.org/ebooks/40213]

@@ -12104,7 +12110,7 @@ among themselves the remaining symbols of each system in which
 $a_{1}$~occurs; therefore the order of~$H$ must be~$mn\mu$, where $\mu$~is a

 factor of~$m - 1$.

 

-It has been seen in §~112 that, for certain values of~$mn$, groups

+It has been seen in §~112 that, for certain values of~$mn$, groups

 satisfying these conditions actually exist. The doubly transitive

 groups, of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$ there obtained, have

 \index{Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group}%

@@ -12267,7 +12273,7 @@ doubly transitive\footnotemark.
 \index{Jordan@\Topic{Jordan}, quoted}%

 \index{Netto@\Topic{Netto}, quoted}%

 \index{Marggraff@\Topic{Marggraff}, quoted}%

-\footnotetext{The results contained in §§~139,~140 are due to Jordan (\textit{Liouville's Journal},

+\footnotetext{The results contained in §§~139,~140 are due to Jordan (\textit{Liouville's Journal},

   Vol.~\VolNo{XVI}, 1871) and Netto (\textit{Crelle's Journal}, Vol.~\VolNo{CIII}, 1889). They have been

   extended by Marggraff: ``Ueber primitiven Gruppen mit transitiven Untergruppen

   geringeren Grades,'' (\textit{Inaugural Dissertation}, Giessen, 1892). With the

@@ -12281,7 +12287,7 @@ doubly transitive\footnotemark.
 \index{Primitive groups@\Topic{Primitive groups}!with a transitive sub-group of smaller degree}%

 therefore $(n - m + 1)$-ply transitive, we at once obtain an

 inferior limit for~$m$. We have seen, in fact, in Theorem~IV,

-§~110, that a group of degree~$n$, other than the alternating or

+§~110, that a group of degree~$n$, other than the alternating or

 the symmetric group, cannot be more than $(\frac{1}{3}n + 1)$-ply transitive.

 Hence

 \[

@@ -12305,7 +12311,7 @@ be less than~$\frac{2}{3}n$, and $G$~is $(n - m + 1)$-ply transitive.
 

 \begin{Corollary}

 The order of a primitive group of degree~$n$

-cannot exceed $\dfrac{n!}{2 · 3 \Dots p}$, where $2$, $3$,~\Dots\ $p$ are the distinct

+cannot exceed $\dfrac{n!}{2 · 3 \Dots p}$, where $2$, $3$,~\Dots\ $p$ are the distinct

 primes which are less than~$\frac{2}{3}n$.

 \end{Corollary}

 

@@ -12318,7 +12324,7 @@ or the symmetric group, can contain a sub-group of order~$q^{\alpha}$;
 and therefore $q^{\alpha - 1}$ is the highest power of~$q$ that can divide

 the order of the group.

 

-\Par{142.} The ratio of $2 · 3 \Dots p$ to~$n$ increases rapidly as $n$~increases,

+\Par{142.} The ratio of $2 · 3 \Dots p$ to~$n$ increases rapidly as $n$~increases,

 and it is at once obvious that, when $n > 7$, this ratio is

 greater than unity; hence for values of~$n$ greater than~$7$, the

 symmetric group can have no primitive sub-group of order~$(n - 1)!$.

@@ -12374,7 +12380,7 @@ another and form two conjugate sets of $6$ each.
 \Par{143.} We shall now discuss certain further limitations on

 the order of a primitive group of given degree. Though it will

 be seen that these do not lead to general results, similar to that

-given by Theorem~I, §~141, yet in many special cases they are

+given by Theorem~I, §~141, yet in many special cases they are

 of considerable assistance in determining the possible existence

 of groups of given orders and degrees.

 

@@ -12382,11 +12388,11 @@ We consider first a group~$G$ of order~$N$ and of prime degree~$p$.
 If $G$~is not cyclical, it must contain substitutions which keep

 only one symbol unchanged. For let $P$ be a substitution of~$G$ of

 order~$p$. The only substitutions permutable with~$P$ are its own

-powers (§~107); and the only substitutions permutable with~$\{P\}$

+powers (§~107); and the only substitutions permutable with~$\{P\}$

 are substitutions which keep one symbol unchanged and are regular

 %% -----File: 217.png---Folio 201-------

-in the remaining $p - 1$ symbols (§~112)\footnotemark.

-\footnotetext{It is shewn in §~112 that $\{P\}$~is permutable with a circular substitution of

+in the remaining $p - 1$ symbols (§~112)\footnotemark.

+\footnotetext{It is shewn in §~112 that $\{P\}$~is permutable with a circular substitution of

 \PageLabel{201}%

   $p - 1$ symbols, which leaves one symbol~$a_{1}$ unchanged. If there are other substitutions

   which leave $a_{1}$ unchanged and are permutable with~$\{P\}$, some such

@@ -12404,7 +12410,7 @@ Hence there must be substitutions in~$G$, other than those of~$\{P\}$,
 which are permutable with~$\{P\}$; and each of these substitutions

 keeps one symbol unchanged.

 

-It follows from §~134 that $G$, if soluble, must contain a

+It follows from §~134 that $G$, if soluble, must contain a

 \index{Primitive groups@\Topic{Primitive groups}!of prime degree}%

 self-conjugate sub-group of order~$p$: therefore no group of

 prime degree~$p$, which contains more than one sub-group of

@@ -12475,13 +12481,13 @@ the symmetric group, cannot be divisible by~$p$.
 \Par{145.} As a simple example, we will shew that the order of

 a group of degree~$19$ cannot be divisible by~$7$, unless it contains

 the alternating group. It follows from Theorem~I, Corollary,

-§~141, that the order of a group of degree~$19$, which does not

+§~141, that the order of a group of degree~$19$, which does not

 contain the alternating group, cannot be divisible by a power of~$7$

 higher than the first, and that if the group contains a substitution

 of order~$7$, the substitution must consist of two cycles of $7$~symbols

 each. The sub-group of order~$7$ must therefore leave $5$~symbols

-unchanged; hence, by §~144, it must be contained self-conjugately in

-a sub-group whose order is divisible by~$5$. Now (§~83) a group of

+unchanged; hence, by §~144, it must be contained self-conjugately in

+a sub-group whose order is divisible by~$5$. Now (§~83) a group of

 order~$35$ is necessarily Abelian; so that the group of degree~$19$ must

 contain a substitution of order~$5$ which is permutable with a substitution

 of order~$7$. Such a substitution of order~$5$ must clearly

@@ -12492,8 +12498,8 @@ alternating group, its order is not divisible by~$7$.
 

 As a second example, we will determine the possible forms for

 the order of a group of degree~$13$, which does not contain the alternating

-group. It follows, from Theorem~I, Corollary, §~141, that

-the order of such a group must be of the form $2^{\alpha} · 3^{\beta} · 5^{\gamma} · 11^ {\delta} · 13$;

+group. It follows, from Theorem~I, Corollary, §~141, that

+the order of such a group must be of the form $2^{\alpha} · 3^{\beta} · 5^{\gamma} · 11^ {\delta} · 13$;

 where $\alpha$,~$\beta$, $\gamma$,~$\delta$ do not exceed $9$,~$4$, $1$,~$1$ respectively.

 

 Suppose, first, that $\gamma$~is unity, if possible. A substitution of

@@ -12506,10 +12512,10 @@ would imply that the group contains the alternating group. It
 follows that, for the group as specified, $\gamma$~must be zero.

 

 Suppose, next, that $\delta$~is unity, if possible. The group is then

-(Theorem~I, §~141) triply transitive; and the order of the sub-group,

+(Theorem~I, §~141) triply transitive; and the order of the sub-group,

 that keeps two symbols fixed and is transitive in the remaining~$11$,

-is $2^{\alpha} · 3^{\beta} · 11$; and this sub-group must contain more than one sub-group

-of order~$11$. We have seen in §~143 that no such group can

+is $2^{\alpha} · 3^{\beta} · 11$; and this sub-group must contain more than one sub-group

+of order~$11$. We have seen in §~143 that no such group can

 exist. Therefore $\delta$~must be zero.

 

 The two smallest numbers of the form $2^{m} 3^{n}$ which are congruent

@@ -12517,28 +12523,28 @@ to unity, $\pmod{13}$, are $3^{3}$ and~$2^{4} 3^{2}$; and every number of this f
 %% -----File: 220.png---Folio 204-------

 which is congruent to unity, $\pmod{13}$, can be written $(2^{4}3^{2})^{x}3^{3y}$.

 Hence the order of every group of order~$13$, which contains no odd

-substitution, must be of the form $(2^{4}3^{2})^{x} · 3^{3y} · z · 13$, where $z$~is $2$,~$3$

+substitution, must be of the form $(2^{4}3^{2})^{x} · 3^{3y} · z · 13$, where $z$~is $2$,~$3$

 or~$6$. Since $3^{4}$~is the highest power of~$3$ that can divide the order

 of the group, the only admissible values of $x$~and~$y$ are \Inum{(i)}~$x = 0$,

 $y = 1$: \Inum{(ii)}~$x = 2$, $y = 0$: \Inum{(iii)}~$x = 1$, $y = 0$.

 

 Suppose, first, that $x = 0$, $y = 1$. The order of the group is

-$2 · 3^{3} · 13$, $3^{4} · 13$, or $2 · 3^{4} · 13$. There must be $13$~sub-groups of order~$3^{4}$

+$2 · 3^{3} · 13$, $3^{4} · 13$, or $2 · 3^{4} · 13$. There must be $13$~sub-groups of order~$3^{4}$

 (or~$3^{3}$), and since $13$~is not congruent to unity, $\pmod{9}$, there must

 be sub-groups of order~$3^{3}$ (or~$3^{2}$) common to some two sub-groups of

 order~$3^{4}$ (or~$3^{3}$). Such a sub-group must be self-conjugate (Theorem~III,

-§~80). This case therefore cannot occur.

+§~80). This case therefore cannot occur.

 

 Next, suppose that $x = 2$, $y = 0$. Then $z$~must be~$2$, and the

-order of the group is $2^{9} · 3^{4} · 13$. Now it is easy to verify that $2^{9} · 3^{4}$

+order of the group is $2^{9} · 3^{4} · 13$. Now it is easy to verify that $2^{9} · 3^{4}$

 is not a possible order either for an intransitive or for an imprimitive

 group of degree~$12$. The order of a sub-group of the group of degree~$12$

-which keeps one symbol fixed is $2^{7} · 3^{3}$. This sub-group can have

+which keeps one symbol fixed is $2^{7} · 3^{3}$. This sub-group can have

 no substitution consisting of a single cycle of $3$~symbols, since no

 such substitution can occur in the original group. Hence it must

 permute the $11$ symbols in two transitive sets of $9$~and $2$~symbols

 respectively. It must therefore contain a self-conjugate sub-group

-of order $2^{6} · 3^{3}$ which keeps $3$~of the $12$~symbols unchanged; and this

+of order $2^{6} · 3^{3}$ which keeps $3$~of the $12$~symbols unchanged; and this

 sub-group must occur self-conjugately in $3$~of the $12$~sub-groups which

 keep one symbol unchanged. This however makes the group of

 degree~$12$ imprimitive, contrary to supposition. Hence this case

@@ -12546,14 +12552,14 @@ cannot occur.
 

 Finally, then, the only possible values of $x$~and~$y$ are $x = 1$,

 $y = 0$. The order of a group of degree~$13$, which has more than

-one sub-group of order~$13$ and no odd substitutions, is $2^{5} · 3^{2} · 13$,

-$2^{4} · 3^{3} · 13$, or $2^{5} · 3^{3} · 13$. The order of a group of order~$13$ with odd

+one sub-group of order~$13$ and no odd substitutions, is $2^{5} · 3^{2} · 13$,

+$2^{4} · 3^{3} · 13$, or $2^{5} · 3^{3} · 13$. The order of a group of order~$13$ with odd

 substitutions will be twice one of the preceding three numbers.

 

 A further and much more detailed examination would be necessary

 to determine whether groups of degree~$13$ correspond to any or

 all of these orders. We shall see in Chapter~XIV that there is a

-group of degree~$13$ and order~$2^{4} · 3^{3} · 13$; and M.~Jordan\footnote

+group of degree~$13$ and order~$2^{4} · 3^{3} · 13$; and M.~Jordan\footnote

   {\textit{Comptes Rendus}, t.~\VolNo{LXXV} (1872), p.~1757.}

 \index{Jordan@\Topic{Jordan}, quoted}%

 states

@@ -12581,7 +12587,7 @@ primitive.
 \Class{\Inum{(ii)}}{$n = 4$.}

 

 Since a group whose order is the power of a prime cannot

-(§~124) be represented in primitive form, a primitive group of

+(§~124) be represented in primitive form, a primitive group of

 degree~$4$ must contain $3$ as a factor of its order. Hence the

 only primitive groups of degree~$4$ are the symmetric and the

 alternating groups.

@@ -12598,7 +12604,7 @@ alternating group. Hence, besides the alternating and the
 symmetric groups, we have only sub-groups which contain a

 sub-group of order~$5$ self-conjugately. In such a group, an

 operation of order~$5$ can be permutable with its own powers

-only. Hence (§~112) the only sub-groups of the type in question

+only. Hence (§~112) the only sub-groups of the type in question

 other than cyclical sub-groups, are groups of orders $20$ and~$10$.

 These are defined by

 \begin{align*}

@@ -12611,10 +12617,10 @@ These are defined by
 \index{Primitive groups@\Topic{Primitive groups}!of degree~$6$}%

 

 If the order of a primitive group of degree~$6$ is not divisible

-by~$5$, the order must (§~141) be equal to or be a factor of~$2^{3} · 3$.

+by~$5$, the order must (§~141) be equal to or be a factor of~$2^{3} · 3$.

 The order of a sub-group that keeps one symbol fixed is equal to

 or is a factor of~$2^{2}$. Hence the sub-group must keep two symbols

-fixed, and therefore (§~128) the group cannot be primitive. Hence

+fixed, and therefore (§~128) the group cannot be primitive. Hence

 the order of every primitive group of degree~$6$ is divisible by~$5$,

 and every such group is at least doubly transitive. The

 symmetric group contains $36$~sub-groups of order~$5$; and hence,

@@ -12631,9 +12637,9 @@ self-conjugate sub-group of order~$5$. If this transitive group
 of degree~$5$ were cyclical, every operation of the doubly transitive

 group~$G$ of order~$30$ would displace all or all but one of the

 symbols. Since $6$~is not the power of a prime, this is impossible

-(§~105). Hence the sub-group of~$G$ which keeps one symbol

+(§~105). Hence the sub-group of~$G$ which keeps one symbol

 fixed must be of one of the two types given above; and the order

-of~$G$ must be $120$ or~$60$. Now we have seen, in §~142, that the

+of~$G$ must be $120$ or~$60$. Now we have seen, in §~142, that the

 symmetric group of degree~$6$ has a single conjugate set of primitive

 sub-groups of order~$120$ and a single set of order~$60$. Hence

 there is a single type of primitive group of degree~$6$, corresponding

@@ -12650,18 +12656,18 @@ that keeps one symbol unchanged.
 \index{Primitive groups@\Topic{Primitive groups}!of degree~$7$|(}%

 

 Every transitive group of degree~$7$ is primitive; and if it

-does not contain the alternating group, its order must (§~141) be

-equal to or be a factor of $7 · 6 · 5 · 4$. A cyclical sub-group of order~$7$

+does not contain the alternating group, its order must (§~141) be

+equal to or be a factor of $7 · 6 · 5 · 4$. A cyclical sub-group of order~$7$

 must (footnote, \PageRef{p.}{201}), in a group of degree~$7$ that contains

 more than one such sub-group, be self-conjugate in a group of

 order $21$ or~$42$. Now neither $20$ nor $40$ is congruent to unity,

 $\pmod{7}$; and therefore $5$~cannot be a factor of the order of such

 a group. Hence the order of a transitive group of degree~$7$,

 that does not contain the alternating group, is equal to or is a

-factor of $7 · 6 · 4$. But $8$~is the only factor of $7 · 6 · 4$ which is

+factor of $7 · 6 · 4$. But $8$~is the only factor of $7 · 6 · 4$ which is

 congruent to unity, $\pmod{7}$; and therefore, if the group contains

 more than one sub-group of order~$7$, its order must be equal to

-$7 · 6 · 4$ and it must contain $8$~sub-groups of order~$7$.

+$7 · 6 · 4$ and it must contain $8$~sub-groups of order~$7$.

 

 Such a group must be doubly transitive; for if a sub-group

 of order~$24$ interchanges the symbols in two intransitive

@@ -12672,7 +12678,7 @@ it would contain the alternating group. A sub-group of degree~$24$,
 transitive in $6$~symbols, must contain $4$~sub-groups of order~$3$.

 For if it contained only one, it would necessarily have

 circular substitutions of $6$~symbols, and the group of order

-$7 · 6 · 4$ would have a self-conjugate sub-group of order $7 · 6 · 2$;

+$7 · 6 · 4$ would have a self-conjugate sub-group of order $7 · 6 · 2$;

 which is not the case. Hence the sub-groups of order~$24$ must

 be simply isomorphic with the symmetric group of $4$~symbols.

 

@@ -12703,9 +12709,9 @@ conditions, viz.\
 It appears on trial that the group generated by the first of

 these is not permutable with the sub-group of order~$24$, while

 the groups generated by the other two are. There are therefore

-just two groups of order $7 · 6 · 4$ which contain the given group

+just two groups of order $7 · 6 · 4$ which contain the given group

 of order~$24$. Now in the symmetric group of $7$~symbols, a

-sub-group of order $7 · 6 · 4$ must, from the foregoing discussion,

+sub-group of order $7 · 6 · 4$ must, from the foregoing discussion,

 be one of a set of $30$ conjugate sub-groups. These all enter in

 the alternating group; and therefore, in that group, they must

 form two sets of $15$ conjugate sub-groups each. Each of these

@@ -12717,8 +12723,8 @@ and the alternating group contains a conjugate set of $105$ such
 such sub-groups. Hence each sub-group of this set will enter

 in two, and only in two, sub-groups of the alternating group of

 %% -----File: 224.png---Folio 208-------

-order $7 · 6 · 4$; and in the symmetric group these two sub-groups

-are conjugate. Finally then, the sub-groups of order $7 · 6 · 4$

+order $7 · 6 · 4$; and in the symmetric group these two sub-groups

+are conjugate. Finally then, the sub-groups of order $7 · 6 · 4$

 form a single conjugate set in the symmetric group. They are

 defined by

 \[

@@ -12737,7 +12743,7 @@ transitive groups of eight symbols.
 

 A group of degree~$7$, which has only one sub-group of order~$7$,

 must either be cyclical or be contained in the group of order

-$7 · 6$ given by §~112. Such groups are defined by

+$7 · 6$ given by §~112. Such groups are defined by

 \begin{align*}

 &\{(1234567),\ (243756)\}, \\

 \lintertext{or}

@@ -12754,7 +12760,7 @@ and of a distinct type from the above. It certainly cannot
 be expressed as a group of degree~$7$; and therefore it must

 have $21$~sub-groups of order~$8$. If two of these sub-groups

 have a common sub-group of order~$4$, it must be contained

-self-conjugately (§~80) in a sub-group of order $24$ or~$56$; and

+self-conjugately (§~80) in a sub-group of order $24$ or~$56$; and

 this is inconsistent with the suppositions made. If on the

 other hand, $2$~is the order of the greatest sub-group common

 to two sub-groups of order~$8$, such a common sub-group of order~$2$

@@ -12774,49 +12780,49 @@ simple group of order~$168$ distinct from the group of degree~$7$.
 \index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|(}%

 

 The order of a primitive group of degree~$8$, which does not

-contain the alternating group, cannot (§~141) be divisible by~$5$.

-Suppose, if possible, that the order of such a group is $2^{\alpha + 3} · 3$

+contain the alternating group, cannot (§~141) be divisible by~$5$.

+Suppose, if possible, that the order of such a group is $2^{\alpha + 3} · 3$

 ($\alpha = 0, 1, 2, 3$). A substitution of order~$3$ must consist of two

-cycles; and therefore the sub-group of order~$\DPchg{2^{\alpha}3}{2^{\alpha} · 3}$, which keeps

+cycles; and therefore the sub-group of order~$\DPchg{2^{\alpha}3}{2^{\alpha} · 3}$, which keeps

 one symbol fixed, must interchange the others in two intransitive

 systems of $3$ and $4$ respectively. In this sub-group, a

 sub-group of order~$3$ must be one of four conjugate sub-groups,

-and therefore $\alpha$~is either $2$ or~$3$. Now a group of order~$2^{5} · 3$

-or $2^{6} · 3$ is soluble, as is seen at once by considering the sub-groups

-of order $2^{5}$ or~$2^{6}$. Hence a primitive group of order~$2^{5} · 3$

-or $2^{6} · 3$ must contain a transitive self-conjugate sub-group of

+and therefore $\alpha$~is either $2$ or~$3$. Now a group of order~$2^{5} · 3$

+or $2^{6} · 3$ is soluble, as is seen at once by considering the sub-groups

+of order $2^{5}$ or~$2^{6}$. Hence a primitive group of order~$2^{5} · 3$

+or $2^{6} · 3$ must contain a transitive self-conjugate sub-group of

 order~$8$, whose operations are all of order~$2$.

 

 If $7$ is a factor of the order of the group, the group must be

 doubly transitive; and from the case of $n = 7$, it follows that

-the possible orders are $8 · 7$, $8 · 7 · 2$, $8 · 7 · 3$, $8 · 7 · 6$, and

-$8 · 7 · 6 · 4$. Moreover, for the orders $8 · 7 · 2$ and $8 · 7 · 6$, the

+the possible orders are $8 · 7$, $8 · 7 · 2$, $8 · 7 · 3$, $8 · 7 · 6$, and

+$8 · 7 · 6 · 4$. Moreover, for the orders $8 · 7 · 2$ and $8 · 7 · 6$, the

 group contains odd substitutions and therefore it contains self-conjugate

-sub-groups of order $8 · 7$ and $8 · 7 · 3$ respectively.

+sub-groups of order $8 · 7$ and $8 · 7 · 3$ respectively.

 

-A simple group of order~$8 · 7 · 3$ is necessarily identical in

+A simple group of order~$8 · 7 · 3$ is necessarily identical in

 type with the group of this order determined on \PageRef{p.}{208};

-and a group of order~$8 · 7 · 3$, which is not simple, is certainly

-soluble. Hence a composite group of order~$8 · 7 · 3$, and

-a group of order $8 · 7 · 6$ which does not contain a simple

-sub-group of order~$8 · 7 · 3$, must both, if expressible as primitive

+and a group of order~$8 · 7 · 3$, which is not simple, is certainly

+soluble. Hence a composite group of order~$8 · 7 · 3$, and

+a group of order $8 · 7 · 6$ which does not contain a simple

+sub-group of order~$8 · 7 · 3$, must both, if expressible as primitive

 groups of degree~$8$, contain transitive self-conjugate

 sub-groups of order~$8$ whose operations are all of order~$2$.

-With the possible exception then of groups of order $8 · 7 · 6 · 4$,

+With the possible exception then of groups of order $8 · 7 · 6 · 4$,

 the only primitive groups of degree~$8$, which do not contain a

 self-conjugate sub-group of order~$8$, are the simple group of

-order~$8 · 7 · 3$ and any group of $8 · 7 · 6$ which contains this self-conjugately.

+order~$8 · 7 · 3$ and any group of $8 · 7 · 6$ which contains this self-conjugately.

 We have seen that the simple group of order

-$8 · 7 · 3$ contains a single set of $8$ conjugate sub-groups of order~$21$,

+$8 · 7 · 3$ contains a single set of $8$ conjugate sub-groups of order~$21$,

 and therefore it can be expressed in one form only as a

-group of degree~$8$. A group of degree~$8$ and order~$8 · 7 · 6$,

+group of degree~$8$. A group of degree~$8$ and order~$8 · 7 · 6$,

 %% -----File: 226.png---Folio 210-------

 which contains this self-conjugately, can occur only in one

 form, if at all; for, if it exists, it must be triply transitive, and

 it must be given by combining the simple group with any

 operation of order~$2$ which transforms one of its operations of

 order~$7$ into its own inverse. That such a group does exist has

-been shewn in §~113. These two groups are actually given by

+been shewn in §~113. These two groups are actually given by

 \begin{align*}

 &\{(15642378),\ (1234567),\ (243756)\}, \\

 \lintertext{and}

@@ -12830,10 +12836,10 @@ primitive groups of degree~$8$, which contain a transitive self-conjugate
 sub-group of order~$8$ whose operations are all of

 order~$2$, anticipate some of the results of the next Chapter. It

 will there be shewn that, if $G$~is an Abelian group of order~$N$,

-and $L$~a group of isomorphisms (§~156) of~$G$, a group of degree~$N$

+and $L$~a group of isomorphisms (§~156) of~$G$, a group of degree~$N$

 may be formed which has a transitive self-conjugate sub-group

 simply isomorphic with $G$, while at the same time the sub-group

-that keeps one symbol fixed is isomorphic with~$L$ (§~158).

+that keeps one symbol fixed is isomorphic with~$L$ (§~158).

 The group of isomorphisms of a group of order~$8$, whose

 operations are all of order~$2$, will be shewn in Chapter~XIV to

 be identical with the simple group of order~$168$. This group

@@ -12843,55 +12849,55 @@ as a group of degree~$7$, it has a single set of conjugate sub-groups
 of order~$12$ (or~$24$) which leave no symbols unchanged.

 There are therefore primitive groups of degree~$8$ containing

 transitive self-conjugate sub-groups of order~$8$ corresponding

-to each of the orders $8 · 7$, $8 · 7 · 3$, $2^{5} · 3$, $2^{6} · 3$, and $8 · 7 · 6 · 4$;

+to each of the orders $8 · 7$, $8 · 7 · 3$, $2^{5} · 3$, $2^{6} · 3$, and $8 · 7 · 6 · 4$;

 and in each case there is a single type of such group.

 

 It remains to determine whether there can be any type

-of group, of degree~$8$ and order $8 · 7 · 6 · 4$, other than that just

+of group, of degree~$8$ and order $8 · 7 · 6 · 4$, other than that just

 obtained. Such a group must be one of $15$~conjugate sub-groups

 in the alternating group of degree~$8$, and can therefore

 itself be expressed as a group of degree~$14$. Since it

 certainly cannot be expressed as a group of degree~$7$, the

 group of degree~$14$ must be transitive. The order of the sub-group,

-in this form, that keeps one symbol fixed is~$2^{5} · 3$. If

+in this form, that keeps one symbol fixed is~$2^{5} · 3$. If

 %% -----File: 227.png---Folio 211-------

 this keeps only one symbol unchanged, it must interchange the

 remaining symbols in four intransitive systems of $3$,~$3$, $3$ and~$4$

 respectively, since a substitution of order~$3$ must clearly consist

-of $4$~cycles. A group of order~$2^{5} · 3$ cannot however be so

+of $4$~cycles. A group of order~$2^{5} · 3$ cannot however be so

 expressed; and therefore the sub-group that keeps one symbol

 fixed must keep two fixed. The group of degree~$14$ is therefore

-imprimitive, and the group must contain a sub-group of order~$2^{6} · 3$.

+imprimitive, and the group must contain a sub-group of order~$2^{6} · 3$.

 Moreover, since the group cannot be expressed as a

-transitive group of degree~$7$, this sub-group of order~$2^{6} · 3$ must

-contain (§~123) a sub-group which is self-conjugate in the group

+transitive group of degree~$7$, this sub-group of order~$2^{6} · 3$ must

+contain (§~123) a sub-group which is self-conjugate in the group

 itself. The order of this sub-group must be a power of~$2$;

 since the group is primitive, it cannot be less than~$2^{3}$. On the

 other hand, the order cannot be greater than~$2^{3}$ since the group

-contains a simple sub-group of order~$7 · 6 · 4$. Hence finally,

+contains a simple sub-group of order~$7 · 6 · 4$. Hence finally,

 there is no type of primitive group of degree~$8$ and order

-$8 · 7 · 6 · 4$ other than that already obtained.

+$8 · 7 · 6 · 4$ other than that already obtained.

 

 There is no difficulty now in actually constructing the

 primitive groups of degree~$8$ which have a self-conjugate sub-group

 of order~$8$. They are all contained in the group of order

-$8 · 7 · 6 · 4$; and this may be constructed from a sub-group

+$8 · 7 · 6 · 4$; and this may be constructed from a sub-group

 keeping one symbol fixed, which is given on \PageRef{p.}{208}, by the

-method of §~109. It will thus be found that the group in

+method of §~109. It will thus be found that the group in

 question is given by

 \[

 \{(81)(26)(37)(45),\ (1236457),\ (234)(567),\ (2763)(45)\};

 \]

-while the groups of orders $8 · 7 · 3$ and $8 · 7$ are given by omitting

+while the groups of orders $8 · 7 · 3$ and $8 · 7$ are given by omitting

 respectively the last and the two last of the four generating

 operations.

 

-The construction of the two remaining groups, of order~$2^{5} · 3$

-and $2^{6} · 3$, is left as an exercise for the reader.

+The construction of the two remaining groups, of order~$2^{5} · 3$

+and $2^{6} · 3$, is left as an exercise for the reader.

 

 It may be noticed that it has been shewn incidentally, in

 discussing above the possibility of a second type of group of

-degree~$8$ and order $8 · 7 · 6 · 4$, that the alternating group of

+degree~$8$ and order $8 · 7 · 6 · 4$, that the alternating group of

 degree~$8$ can be expressed as a doubly transitive group of

 degree~$15$.

 

@@ -12902,7 +12908,7 @@ transitive and primitive group of degree~$15$.
 \index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|)}%

 %% -----File: 228.png---Folio 212-------

 

-\Par{147.} We have seen in §~105 that a doubly transitive group,

+\Par{147.} We have seen in §~105 that a doubly transitive group,

 \index{Doubly transitive groups@\Topic{Doubly transitive groups}!the sub-groups of, which keep two symbols fixed}%

 of degree~$n$ and order~$n (n - 1)$, can exist only when $n$~is the

 power of a prime. For such a group, the identical operation is

@@ -12955,14 +12961,14 @@ no pair occurs in more than one.
 There is one class of cases in which a solution of this problem is

 given immediately by the theory of Abelian groups. Let $G$ be an

 Abelian group of order~$p^{m}$, where $p$~is a prime, and type $(1, 1, \Dots\DPchg{\,}{,}

-\text{to $m$ units})$. We have seen, in §~49, that $G$ has $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of

-order~$p$, and $\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sub-groups of order~$p^{2}$. Now any pair

+\text{to $m$ units})$. We have seen, in §~49, that $G$ has $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of

+order~$p$, and $\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sub-groups of order~$p^{2}$. Now any pair

 of sub-groups of order~$p$ generates a sub-group of order~$p^{2}$, and therefore

 every pair of sub-groups of order~$p$ occurs in one and only one

 sub-group of order~$p^{2}$. Moreover, every sub-group of order~$p^{2}$ contains

 $p + 1$ sub-groups of order~$p$. When $p$~is a prime and $m$~any integer,

 it is therefore always possible to form from $\dfrac{p^{m} - 1}{p - 1}$ symbols

-$\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sets of $p + 1$ symbols each, such that every pair of

+$\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sets of $p + 1$ symbols each, such that every pair of

 the symbols occurs in one set of $p + 1$ and no pair occurs in more

 than one set.

 

@@ -13008,9 +13014,9 @@ leaves two given symbols unchanged also leaves a third definite
 symbol unchanged.

 

 The smallest possible value of~$n$ is~$7$; and the group of Ex.~2,

-§~35, which is one of the groups obtained in §~146, satisfies all the

+§~35, which is one of the groups obtained in §~146, satisfies all the

 conditions. In fact, the group is clearly a doubly transitive group

-of degree~$7$; and since its order is $7 · 6 · 4$, the order of the sub-group

+of degree~$7$; and since its order is $7 · 6 · 4$, the order of the sub-group

 which keeps two symbols fixed is~$4$. Now in the sub-group

 \[

 \{(267) (345),\ (23) (47)\},

@@ -13060,7 +13066,7 @@ group of degree~$n$ which possesses a complete set of triplets. It is
 a good exercise to verify that there is no such group when $n$~is~$13$.

 

 The case $n = 13$, $x = 4$ is the simplest case that can occur of the

-division of $n$~symbols into sets of~$x$ in the manner of §~148 when $x$~is

+division of $n$~symbols into sets of~$x$ in the manner of §~148 when $x$~is

 greater than~$3$. We shall see in Chapter~XIV that there is a

 doubly transitive group of degree~$13$ such that from the $13$~symbols

 permuted by the group a complete set of $13$ quartets can be

@@ -13086,7 +13092,7 @@ may be made to the memoirs mentioned in the subjoined footnote\footnotemark.
 applications of substitution groups, which enable us to complete

 and extend certain earlier results.

 

-We have seen in §~107 that the substitutions of $n$~symbols,

+We have seen in §~107 that the substitutions of $n$~symbols,

 \index{Substitutions@\Topic{Substitutions} which are permutable!with a given substitution}%

 which are permutable with each of the substitutions of a

 regular substitution group~$G$ of order~$n$ of the same $n$ symbols,

@@ -13167,7 +13173,7 @@ isomorphic with the group
 \]

 \ie\ with the symmetric group of $n$~symbols, is of order~$n!$.

 

-Hence, finally, the order of~$G_{S}$ is~$m^{n} · n!$; and $G_{S}$~is generated

+Hence, finally, the order of~$G_{S}$ is~$m^{n} · n!$; and $G_{S}$~is generated

 by

 \begin{gather*}

 (a_{11} a_{12}\Dots a_{n1})

@@ -13195,11 +13201,11 @@ S_{1t} S_{2t} \Dots S_{nt},
 the set of $m$~substitutions so formed constitute an intransitive

 group~$H$ in the $mn$~symbols, simply isomorphic with~$h_{r}$.

 

-The method of §~150 can be applied directly to determine

+The method of §~150 can be applied directly to determine

 \index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%

 the group~$G_{H}$ of degree~$mn$, each of whose substitutions are\DPnote{[** [sic]]}

 permutable with every substitution of~$H$. The order of this

-group is~$m^{n} · n!$; and it can be generated by

+group is~$m^{n} · n!$; and it can be generated by

 \begin{gather*}

 (a_{11} a_{21} \Dots a_{n1})

 (a_{12} a_{22} \Dots a_{n2}) \Dots

@@ -13227,8 +13233,8 @@ $S_{1}$, $S_{2}$,~\dots\ must displace all the symbols; and therefore
 $\{H,\ S_{1},\ S_{2},\ \dots\}$ has all its substitutions regular in the $mn$~symbols.

 If its order is~$mn_{1}$ where $n = n_{1}n_{2}$, the order of the

 group formed of all the substitutions of $mn$~symbols, which are

-permutable with each of its operations, is $(mn_{1})^{n_{2}} · n_{2}!$; and this

-number is less than~$m^{n} · n!$. Thus the supposition, that $G_{H}$~has

+permutable with each of its operations, is $(mn_{1})^{n_{2}} · n_{2}!$; and this

+number is less than~$m^{n} · n!$. Thus the supposition, that $G_{H}$~has

 self-conjugate operations other than the operations of~$H$ which

 it contains, leads to an impossibility.

 

@@ -13240,7 +13246,7 @@ permutable with every operation of any given group in the $n$~symbols.
 \Par{152.} If a group, whose order is a power of a prime~$p$, be

 \index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%

 expressed as a transitive substitution group, its degree must

-also be a power of~$p$ (§~123). Moreover such a group, since it

+also be a power of~$p$ (§~123). Moreover such a group, since it

 has self-conjugate operations, must necessarily be imprimitive.

 

 The greatest value of~$m$, for which a group of order~$p^{m}$ can

@@ -13266,7 +13272,7 @@ order~$p^{\nu}$.
 

 This group may be constructed synthetically as follows.

 Since the sub-group of order~$p^{\nu - n}$, that leaves one symbol

-unchanged, is contained in a sub-group of order~$p^{\nu - n + 1}$ (§~55),

+unchanged, is contained in a sub-group of order~$p^{\nu - n + 1}$ (§~55),

 there must be imprimitive systems containing $p$~symbols each.

 If then we distribute the $p^{n}$~symbols into $p^{n - 1}$~sets of $p$~each,

 and with each set of~$p$ form a circular substitution, the $p^{n - 1}$

@@ -13304,21 +13310,21 @@ and therefore, when expressed as a substitution group in the $p^{n}$~symbols,
 it must interchange them transitively in $p^{n - r}$~sets of

 $p^{r}$ each.

 

-Now (§~151) those substitutions of the $p^{n}$~symbols, which

+Now (§~151) those substitutions of the $p^{n}$~symbols, which

 are permutable with every operation of this sub-group, form a

-group of order~$p^{rp^{n - r}} · p^{n - r}!$; this number is only divisible by~$p^{\nu}$,

+group of order~$p^{rp^{n - r}} · p^{n - r}!$; this number is only divisible by~$p^{\nu}$,

 \index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%

 as it must be, when $r = 1$.

 

 \begin{Ex} Shew that, for the group of degree~$p^{2}$ and order~$p^{p + 1}$, the

-factor groups $\dfrac{H_{r + 1}}{H_{r}}$ (of §~53) are all of type~$(1)$ except the last,

+factor groups $\dfrac{H_{r + 1}}{H_{r}}$ (of §~53) are all of type~$(1)$ except the last,

 which is of type~$(1, 1)$.

 \end{Ex}

 

 The fact that $\nu$~is a function of~$p$ when $n$~is given, explains

 why, in classifying all groups of order~$p^{n}$, some of the lower

 primes may behave in an exceptional manner. Thus we saw, in

-§~73, that for certain groups of order~$p^{4}$ it was necessary to

+§~73, that for certain groups of order~$p^{4}$ it was necessary to

 consider separately the case $p = 3$. The present article makes

 it clear that, while there may be more than one type of group

 of order~$p^{4}$ ($p > 3$), which can be expressed as a transitive group

@@ -13329,9 +13335,9 @@ which can be expressed transitively in $9$~symbols.
 \Par{153.} In the memoirs referred to in the footnote on \PageRef{p.}{155},

 M.~Mathieu has demonstrated the existence of a remarkable group,

 %% -----File: 236.png---Folio 220-------

-of degree~$12$ and order~$12 · 11 · 10 · 9 · 8$\DPtypo{.}{}, which is quintuply transitive.

+of degree~$12$ and order~$12 · 11 · 10 · 9 · 8$\DPtypo{.}{}, which is quintuply transitive.

 The generating operations of this group have been given in

-the note to §~108, at the end of Chapter~VIII\@. The verification of

+the note to §~108, at the end of Chapter~VIII\@. The verification of

 some of the more important properties of this group, as stated in

 the succeeding example, forms a good exercise on the results of this

 and the two preceding Chapters.

@@ -13345,12 +13351,12 @@ and the two preceding Chapters.
 \end{gather*}

 generate a quintuply transitive group of degree~$12$ and order

 \[

-12 · 11 · 10 · 9 · 8.

+12 · 11 · 10 · 9 · 8.

 \]

 

 Prove that this group is simple; that a sub-group of degree~$11$

-and order $11 · 10 · 9 · 8$, which leaves one symbol unchanged, is a

-simple group; and that a sub-group of degree~$10$ and order~$10 · 9 · 8$,

+and order $11 · 10 · 9 · 8$, which leaves one symbol unchanged, is a

+simple group; and that a sub-group of degree~$10$ and order~$10 · 9 · 8$,

 which leaves two symbols unchanged, contains a self-conjugate sub-group

 simply isomorphic with the alternating group of degree~$6$.

 

@@ -13369,7 +13375,7 @@ alternating group of degree~$12$.
 

 \Chapter{XI.}{On the Isomorphism of a Group with Itself.}

 

-\Par{154.} \First{It} is shewn in §~24 that, if all the operations of a

+\Par{154.} \First{It} is shewn in §~24 that, if all the operations of a

 group are transformed by one of themselves, which is not self-conjugate,

 a correspondence is thereby established among the

 operations of the group which exhibits the group as simply

@@ -13414,7 +13420,7 @@ itself is called the \emph{identical isomorphism}. In every isomorphism
 of a group with itself, the identical operation corresponds to

 itself; and the orders of two corresponding operations are

 the same. For if $1$~and~$S$ were corresponding operations, so

-also would be $1 · 1$ and~$S^{2}$; and therefore more than one operation

+also would be $1 · 1$ and~$S^{2}$; and therefore more than one operation

 would correspond to~$1$. Again, if $S$~and~$S'$, of orders $n$~and~$n'$,

 are corresponding operations, so also are $S^{n}$ and~$S'^{n}$; and therefore

 $n$~must be a multiple of~$n'$. Similarly $n'$~must be a multiple

@@ -13532,7 +13538,7 @@ S_{p} S_{q} = S_{r},
 \]

 then

 \[

-\Sigma^{-1} S_{p} \Sigma · \Sigma^{-1} S_{q}\Sigma

+\Sigma^{-1} S_{p} \Sigma · \Sigma^{-1} S_{q}\Sigma

   = \Sigma^{-1} S_{p} S_{q}\Sigma

   = \Sigma^{-1} S_{r} \Sigma;

 \]

@@ -13540,10 +13546,10 @@ and $\Sigma_{-1} S_{r}\Sigma$ is an operation of the group. An isomorphism of a
 group, which is thus formed on transforming the operations of the

 group by one of themselves, is called a \emph{cogredient} isomorphism.

 All others are called \emph{contragredient} isomorphisms\footnotemark.

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 \index{Klein@\Topic{Klein}, quoted}%

-\footnotetext{Klein, ``Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen

-  vom fünften Grade'' (1884), p.~232. Hölder, \textit{Math.\ Ann.}, Vol.~\VolNo{XLIII}

+\footnotetext{Klein, ``Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen

+  vom fünften Grade'' (1884), p.~232. Hölder, \textit{Math.\ Ann.}, Vol.~\VolNo{XLIII}

   (1893), p.~314.}%

 If $\Sub{S}{S'}$ is

 any contragredient isomorphism, the isomorphisms

@@ -13558,7 +13564,7 @@ The totality of the cogredient isomorphisms of
 a group~$G$ form a group isomorphic with~$G$; this group is

 a self-conjugate sub-group of the group of isomorphisms of~$G$\footnotemark.

 \end{Theorem}

-\footnotetext{Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

+\footnotetext{Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

   (1895), p.~326.}

 

 The product of the isomorphisms

@@ -13637,7 +13643,7 @@ within the group of isomorphisms.
 1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};

 \]

 and let $L$ be the group of isomorphisms of~$G$. We have seen

-in §~20 that $G$~may be represented as a transitive group of

+in §~20 that $G$~may be represented as a transitive group of

 substitutions performed on the $N$~symbols

 \[

 1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};

@@ -13707,7 +13713,7 @@ the set of substitutions}
 \Sub{S}{S_{x}S},\quad (x = 0, 1, 2, \Dots, N - 1).

 \]

 

-These form (§~107) a transitive group~$G''$, simply isomorphic

+These form (§~107) a transitive group~$G''$, simply isomorphic

 with~$G'$ and such that every substitution of~$G''$ is

 permutable with every substitution of~$G'$. Moreover (\lc),

 the substitutions of~$G''$ are the only substitutions of the

@@ -13838,7 +13844,7 @@ J^{2},\ JJ_{1},\ JJ_{2},\ \Dots,\ JJ_{N-1}.
 

 Now

 \[

-J^{-1} · JJ_{x} · J = J_{x}J = (JJ_{x})^{-1},

+J^{-1} · JJ_{x} · J = J_{x}J = (JJ_{x})^{-1},

 \]

 so that $J$~transforms every operation of~$G$ into its own inverse.

 But if

@@ -13978,7 +13984,7 @@ of~$K$ which is contained in~$G$ is a characteristic sub-group of~$G$.
 Suppose now a chief-series of~$K$ formed which contains~$G$.

 If $G$~has no characteristic sub-group, it must be the

 last term but one of this series, the last term being identity.

-It follows by §~94 that $G$~must be the direct product of a

+It follows by §~94 that $G$~must be the direct product of a

 number of simply isomorphic simple groups. Hence:---

 

 \begin{Theorem}[IV.]

@@ -13999,7 +14005,7 @@ be a series of such sub-groups, each containing all that follow it
 and chosen so that, for each consecutive pair $G_{r}$ and $G_{r+1}$, there

 is no characteristic sub-group of~$G$ contained in~$G_{r}$ and containing~$G_{r+1}$,

 except $G_{r+1}$~itself. Such a series is called\footnote

-  {Frobenius, ``Ueber auflösbare Gruppen, II,'' \textit{Berliner Sitzungsberichte}, 1895,

+  {Frobenius, ``Ueber auflösbare Gruppen, II,'' \textit{Berliner Sitzungsberichte}, 1895,

   p.~1027.}

 a \emph{characteristic series} of~$G$.

 %% -----File: 249.png---Folio 233-------

@@ -14045,7 +14051,7 @@ the order~$p^{m}$, ($s = 1, 2, \dots, r$), while}
 m_{1} > m_{2} > \Dots > m_{r}.

 \]

 

-The sub-group~$G_{\mu}$ (§~42), formed of the operations of~$G$

+The sub-group~$G_{\mu}$ (§~42), formed of the operations of~$G$

 which satisfy the relation

 \[

 S^{p^{\mu}} = 1,

@@ -14060,7 +14066,7 @@ for this is a set of characteristic sub-groups such that each
 contains the one that follows it.

 %% -----File: 250.png---Folio 234-------

 

-Now the sub-group~$H_{\nu}$ (§~45), formed of the distinct

+Now the sub-group~$H_{\nu}$ (§~45), formed of the distinct

 operations that remain when every operation of~$G$ is raised

 to the power~$p^{\nu}$, is also a characteristic sub-group; and since

 the operations common to two characteristic sub-groups also

@@ -14128,7 +14134,7 @@ consider the series of sub-groups
 \[

 G,\ H_{n},\ H_{n - 1},\ \Dots,\ H_{1},\ 1

 \]

-of §~53. Each of these is clearly a characteristic sub-group, and

+of §~53. Each of these is clearly a characteristic sub-group, and

 each contains the succeeding. Moreover, $\dfrac{H_{r + 1}}{H_{r}}$ is an Abelian

 group; and, by the process that we have just investigated, a

 characteristic series may be formed for it. To each group in

@@ -14185,9 +14191,9 @@ A group, which contains no self-conjugate
 \index{Group@\Topic{Group}!complete}%

 operation except identity and which admits of no contragredient

 isomorphism, is called\footnote

-  {Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

+  {Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

   (1895), p.~325.}

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 a \emph{complete} group.

 \end{Definition}

 

@@ -14207,11 +14213,11 @@ operations except identity; and if the group of cogredient isomorphisms
 of~$G$ is a characteristic sub-group of~$L$, the group of

 isomorphisms of~$G$; then $L$~is a complete group\footnotemark.

 \end{Theorem}

-\footnotetext{Hölder (\textit{loc.~cit.}\ p.~331) gives a theorem which is similar but not quite

+\footnotetext{Hölder (\textit{loc.~cit.}\ p.~331) gives a theorem which is similar but not quite

   equivalent to Theorem~VI.}%

 %% -----File: 253.png---Folio 237-------

 

-With the notation of §~158, the operations of~$L$ may be

+With the notation of §~158, the operations of~$L$ may be

 represented by the substitutions

 \[

 \Sub{S}{S'}.

@@ -14276,13 +14282,13 @@ For suppose, if possible, in this case that $G'$~is not a
 characteristic sub-group of~$L$; and that, by a contragredient

 isomorphism of~$L$, $G$~is transformed into~$G''$. Then $G''$~is a

 self-conjugate sub-group of~$L$, and each of the groups $G'$ and~$G''$

-transforms the other into itself. Hence (§~34) either every

+transforms the other into itself. Hence (§~34) either every

 operation of~$G'$ is permutable with every operation of~$G''$, or $G'$

 and~$G''$ must have a common sub-group. The former supposition

 is impossible since no operation of~$L$ is permutable with

 every operation of~$G$. On the other hand, if $G'$~and~$G''$ have a

 common sub-group, it is a self-conjugate sub-group of~$L$ and it

-therefore is a characteristic sub-group of $G'$. Now (§~162) $G$~has

+therefore is a characteristic sub-group of $G'$. Now (§~162) $G$~has

 no characteristic sub-groups, and therefore the second supposition

 is also impossible. It follows that, in this case, $G'$~is a characteristic

 sub-group of~$L$, and that $L$~is a complete group.

@@ -14293,11 +14299,11 @@ and if $K$~is the holomorph of~$G$; then when $G$~is a characteristic
 sub-group of~$K$, the latter group is a complete group.

 \end{Theorem}

 

-If $N$~is the order of~$G$, then $K$~can be expressed (§~158) as a

+If $N$~is the order of~$G$, then $K$~can be expressed (§~158) as a

 transitive group of degree~$N$. When $K$~is so expressed, those

 operations of~$K$ which leave one symbol unchanged form a

 sub-group~$H$, which is simply isomorphic with the group of

-isomorphisms of~$G$. Now (§~160) an Abelian group of odd order

+isomorphisms of~$G$. Now (§~160) an Abelian group of odd order

 admits of a single isomorphism of order two, which changes every

 operation into its own inverse. The corresponding substitution

 of~$H$ is a self-conjugate substitution in~$H$, and is one of $N$~conjugate

@@ -14309,7 +14315,7 @@ substitutions of order two, into itself. Also, no substitution of~$K$
 can be permutable with each one of these $N$~substitutions,

 since each of them keeps just one symbol unchanged. Hence

 %% -----File: 255.png---Folio 239-------

-(Theorem~III, Cor.~II, §~161) the group of isomorphisms of~$K$

+(Theorem~III, Cor.~II, §~161) the group of isomorphisms of~$K$

 can be expressed as a transitive group of degree~$N$, which

 contains $G$ as a transitive self-conjugate sub-group. But

 when the group of isomorphisms of~$K$ is so expressed, $K$~itself

@@ -14336,7 +14342,7 @@ of isomorphisms of~$K$, is conjugate to~$G$. Then $G$~and~$G'$ being
 both self-conjugate in~$K$ must have a common sub-group, since

 $K$~cannot contain their direct product. But the common sub-group

 of $G$ and~$G'$, being self-conjugate in~$K$, is a characteristic

-sub-group of~$G$. This is impossible (§~162); hence $G$~is a

+sub-group of~$G$. This is impossible (§~162); hence $G$~is a

 characteristic sub-group of~$K$.

 

 \begin{Ex} Shew that the holomorph of an Abelian group of degree~$2^{m}$

@@ -14401,7 +14407,7 @@ S^{p - 1} = 1,\quad S^{-1}PS = P^{\alpha},
 where $\alpha$~is a primitive root of~$p$, $\{S, P\}$~is the holomorph of~$G$.

 

 The reader will at once observe that this group of order~$p(p - 1)$

-is identical with the doubly transitive group of §~112. It is a

+is identical with the doubly transitive group of §~112. It is a

 complete group.

 

 \Par{169.} We shall consider next the case of any cyclical group.

@@ -14498,7 +14504,7 @@ $p_{1}^{m_{1}}$,~etc.
 

 When the order of~$G$ is odd, the holomorph~$K$ is easily shewn

 to be a complete group. Suppose it to be expressed transitively,

-as in §~158, in $N$~symbols, where $N$~is the order of~$G$; if $G$~is

+as in §~158, in $N$~symbols, where $N$~is the order of~$G$; if $G$~is

 not a characteristic sub-group of~$K$, let $G'$ be a group into which

 $G$~is transformed by a contragredient isomorphism of~$K$. Then $G'$~is

 %% -----File: 258.png---Folio 242-------

@@ -14510,7 +14516,7 @@ operation of~$G'$ into its own inverse therefore each keep one symbol
 fixed; hence each of them must transform every operation of~$G$

 into its inverse. But there is only one such set of operations

 of order~$2$, and therefore $G'$~cannot differ from~$G$. It follows by

-Theorem~VII, §~167 that, as $G$~is a characteristic sub-group of~$K$, the

+Theorem~VII, §~167 that, as $G$~is a characteristic sub-group of~$K$, the

 group~$K$ itself is complete.

 

 If the order of~$G$ is even, $K$~must contain a self-conjugate

@@ -14713,7 +14719,7 @@ N_{n} = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n-1}).
 

 The reader will notice that an independent proof of this result

 \index{Holomorph@\Topic{Holomorph}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$}%

-has already been obtained in §~48. The discussion there given of

+has already been obtained in §~48. The discussion there given of

 the number of distinct ways, in which independent generating

 operations of an Abelian group of type $(1, 1, \Dots, 1)$ may be

 \index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!holomorph of}%

@@ -14761,22 +14767,22 @@ of the Abelian group, it follows that the group of transformations
 must be simply isomorphic with the holomorph of the Abelian

 group. In the simplest instance, where $p^{n}$~is~$2^{2}$, the holomorph is

 simply isomorphic with the alternating group of four symbols. In

-any case the holomorph, when expressed as in §~158, is a doubly

+any case the holomorph, when expressed as in §~158, is a doubly

 transitive group of degree~$p^{n}$.

 \end{Remark}

 

-\Par{173.} It has been seen in §~142 that, except when $n = 6$, the

+\Par{173.} It has been seen in §~142 that, except when $n = 6$, the

 symmetric group of degree~$n$ has $n$~and only~$n$ sub-groups

 of order~$n - 1!$,\DPnote{** No () in original} which form a conjugate set. Hence by

-Theorem~III, §~161, the group of isomorphisms of the symmetric

+Theorem~III, §~161, the group of isomorphisms of the symmetric

 group of degree~$n$ can be expressed, except when $n = 6$,

 %% -----File: 262.png---Folio 246-------

 as a transitive group of degree~$n$. The symmetric group of $n$~symbols

 however consists of all possible substitutions that can

 be performed on the $n$~symbols, and therefore it must coincide

 with its group of isomorphisms. Hence\footnotemark:---

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}, (1895), p.~345.}%

+\index{Holder@\Topic{Hölder}, quoted}%

+\footnotetext{Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}, (1895), p.~345.}%

 

 \begin{Theorem}[VIII.]

 The symmetric group of $n$~symbols is a

@@ -14800,7 +14806,7 @@ has just $n$ sub-groups of order~$\frac{1}{2} (n - 1)!$.
 \Par{174.} The alternating group of degree~$6$ occurs as a special

 case of another class of groups of which we will determine the

 groups of isomorphisms. These are the doubly and the triply transitive

-groups that have been defined in §§~112,~113.

+groups that have been defined in §§~112,~113.

 

 The doubly transitive group of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$

 \index{Group of isomorphisms@\Topic{Group of isomorphisms}!of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$|(}%

@@ -14810,7 +14816,7 @@ as a transitive group of degree~$p^{m}$. Let $p^{m}(p^{m} - 1) \mu$~be the
 order of the group of isomorphisms. The order of a sub-group that

 keeps two symbols fixed is~$\mu$; and every operation of this sub-group

 must transform a cyclical sub-group of order~$p^{m} - 1$ into itself. With

-the notation of §~112, we will consider the sub-group which keeps $0$

+the notation of §~112, we will consider the sub-group which keeps $0$

 and $i$ fixed. Every operation of this sub-group must transform the

 cyclical sub-group generated by the congruence\footnote

   {This and all subsequent congruences in the present section are to be taken,

@@ -14914,7 +14920,7 @@ defined by the congruences
 \[

 x' \equiv \frac{\alpha x + \beta}{\gamma x + \delta},

 \]

-where $\alpha$,~$\beta$, $\gamma$,~$\delta$ satisfy the conditions of §~113, is the group of order

+where $\alpha$,~$\beta$, $\gamma$,~$\delta$ satisfy the conditions of §~113, is the group of order

 $(p^{m} + 1) p^{m} (p^{m} - 1) m$ obtained by combining the previous congruences

 with

 \[

@@ -14924,7 +14930,7 @@ x' \equiv x^{p}.
 In fact, it may be immediately verified that the operation given

 by this congruence is permutable with the group, and does not give

 a cogredient isomorphism of the group. Moreover, by Theorem~III,

-§~161, the group of isomorphisms of the given group can be expressed

+§~161, the group of isomorphisms of the given group can be expressed

 as a transitive group of degree~$p^{m} + 1$; and therefore, among a class of

 contragredient isomorphisms, there must be some transforming into

 itself a sub-group which keeps one symbol fixed. Hence the order

@@ -14953,7 +14959,7 @@ as a transitive group of degree~$p^{m} + 1$, and must clearly contain the
 triply transitive group of order $(p^{m} + 1)p^{m}(p^{m} - 1)$ self-conjugately.

 It therefore coincides with the group of order $(p^{m} + 1)p^{m}(p^{m} - 1)m$,

 which has just been determined. The latter group is therefore

-(Theorem~VI, Cor.\ §~166) a complete group. Further, if the simple

+(Theorem~VI, Cor.\ §~166) a complete group. Further, if the simple

 group be denoted by~$G$ and the group of isomorphisms by~$L$, the

 factor group~$\dfrac{L}{G}$ will be determined by

 \[

@@ -14971,8 +14977,8 @@ The group of isomorphisms of the alternating group of degree~$6$
 is therefore a group of order~$1440$; and the symmetric group of

 degree~$6$ admits a single class of contragredient isomorphisms\footnotemark.

 \end{Remark}

-\footnotetext{Hölder, \textit{loc.~cit.}\ p.~343.}%

-\index{Holder@\Topic{Hölder}, quoted}%

+\footnotetext{Hölder, \textit{loc.~cit.}\ p.~343.}%

+\index{Holder@\Topic{Hölder}, quoted}%

 

 \Par{175.} Let $P$~be any group whose order is the power of a

 prime, and let

@@ -14983,7 +14989,7 @@ of orders
 \[

 p^{\alpha},\ p^{\alpha_{1}},\ p^{\alpha_{2}},\ \Dots,\ P^{\alpha_{n}},\ 1

 \]

-be a characteristic series (§~163) of~$P$. Every isomorphism

+be a characteristic series (§~163) of~$P$. Every isomorphism

 of~$P$ must transform $P_{r}$ and $P_{r + 1}$ into themselves, and therefore

 also $\dfrac{P_{r}}{P_{r + 1}}$ into itself. Suppose now that an isomorphism~$I$

 of~$P$ transforms every operation of $P_{r + 1}$ into itself and

@@ -15027,7 +15033,7 @@ prime to~$p$, is a factor of
 \]

 

 Herr Frobenius\footnote

-  {``Ueber auflösbare Gruppen,~\VolNo{II},'' \textit{Berliner Sitzungsberichte}, 1895, p.~1028.}

+  {``Ueber auflösbare Gruppen,~\VolNo{II},'' \textit{Berliner Sitzungsberichte}, 1895, p.~1028.}

 \index{Frobenius@\Topic{Frobenius}, quoted}%

 \index{Symbol@\Topic{Symbol}!theta@$\vartheta(P)$ and $\theta(P)$, definition of}%

 has introduced the symbol~$\vartheta(P)$ to denote

@@ -15086,7 +15092,7 @@ a new symbol~$\theta(P)$ by the equation
 \theta(P) = (p - 1) (p^{2} - 1) \Dots (p^{\rho} - 1).

 \]

 

-In forming the characteristic series of~$P$ in §~164, we

+In forming the characteristic series of~$P$ in §~164, we

 commenced with the series of groups~$G_{r}$ ($r = 1, 2, \Dots$), such

 that $G_{r}$~consists of all the operations of~$P$ satisfying the

 relation

@@ -15114,7 +15120,7 @@ the order of~$\dfrac{G_{r + 1}}{G_{r}}$ is~$p^{\rho}$. Hence, in this case, $\va
 factor of~$\theta(P)$.

 

 If now $P'$~is any sub-group of~$P$, then it has been seen

-(§~46) that the number of independent generating operations of~$P'$

+(§~46) that the number of independent generating operations of~$P'$

 is equal to or less than~$\rho$. Hence $\theta(P')$~is equal to or is a

 factor of~$\theta(P)$.

 

@@ -15168,7 +15174,7 @@ relatively prime to~$N$, is the order of an isomorphism of~$G$, all
 the numbers $q^{r}$, $q'^{r'}$,~\Dots\ are factors of~$\vartheta(G)$, and so also is

 their product.

 

-\Par{178.} The method, by which it has been shewn in §~175 that

+\Par{178.} The method, by which it has been shewn in §~175 that

 the order of any isomorphism of~$P_{r}$ which transforms every

 operation of each of the groups $\dfrac{P_{r}}{P_{r + 1}}$ and $P_{r + 1}$ into itself is a

 %% -----File: 269.png---Folio 253-------

@@ -15238,12 +15244,12 @@ order of the group of isomorphisms.
 \end{Ex}

 

 \begin{Ex}[3.] Shew that the group of isomorphisms of the group of

-order~$2^{n}$, defined by (§~63)

+order~$2^{n}$, defined by (§~63)

 \[

 P^{2^{n - 1}} = 1,\quad Q^{2} = P^{2^{n - 2}}, \quad Q^{-1}PQ = P^{-1},

 \]

 is of order~$2^{2n - 3}$, when $n > 3$. If $n = 3$, its order is~$24$ and it is

-simply isomorphic with the last type but one of §~84.

+simply isomorphic with the last type but one of §~84.

 \end{Ex}

 

 \begin{Ex}[4.] If $G$~is a complete group of order~$N$, shew that the order

@@ -15482,7 +15488,7 @@ S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}H,
 since $H$~is a self-conjugate sub-group; and, because

 \[

 S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}

-  = S_{1}^{p + p'} · S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q} · S_{2}^{q + q'},

+  = S_{1}^{p + p'} · S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q} · S_{2}^{q + q'},

 \]

 while $S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q}$ belongs to~$H$, the class $S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'} H$ is the

 same as $S_{1}^{p + p'} S_{2}^{q + q'} H$. Hence the operations of any two given classes,

@@ -15527,7 +15533,7 @@ S^{x} (ST^{2})^{y} (TST)^{z} H,\quad (x, y, z = 0, 1, 2).
 The group~$\Bar{G}$ will be defined by the laws according to which

 these $27$~classes combine among themselves; and the reader will

 have no difficulty in verifying that it is isomorphic with the non-Abelian

-group of order~$27$, whose operations are all of order~$3$ (§~73).

+group of order~$27$, whose operations are all of order~$3$ (§~73).

 \end{Remark}

 

 \begin{Ex} If $S_{1} (= 1)$, $S_{2}$,~\Dots, $S_{N}$ are the operations of a group~$G$

@@ -15664,7 +15670,7 @@ derived from~$P$ by an even or an odd number of inversions. The
 %% -----File: 280.png---Folio 264-------

 latter we shall suppose coloured black, and the former (including~$P$)

 left white. Every white polygon will be surrounded

-by black polygons and vice~versâ. Since there is only one

+by black polygons and vice~versâ. Since there is only one

 definite set of inversions that will transform~$P$ into any other

 white polygon~$P'$, the $n + 1$ corners of~$P'$ will correspond one

 by one to the $n + 1$ corners of~$P$; and when the perimeters of

@@ -15718,7 +15724,7 @@ For an inversion at~$A_{r}A_{r + 1}$, followed by an inversion at~$A_{s}A_{s + 1
 is given by~$T_{r}T_{s}$; and

 \begin{align*}

 T_{r}T_{s}

-  &= T_{r}T_{r + 1} · T_{r + 1}T_{r + 2} · \Dots · T_{s - 1}T_{s} \\

+  &= T_{r}T_{r + 1} · T_{r + 1}T_{r + 2} · \Dots · T_{s - 1}T_{s} \\

   &= S_{r + 1}S_{r + 2}\Dots S_{s}.

 \end{align*}

 

@@ -15837,7 +15843,7 @@ relations
 F_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).

 \]

 

-It has been seen (§~181) that to the identical operation of the

+It has been seen (§~181) that to the identical operation of the

 special group there corresponds a self-conjugate sub-group~$H$ of

 the general group; or in other words, that the set of operations

 $\Sigma H$ of the general group give one and only one operation in the

@@ -16117,7 +16123,7 @@ S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1.
 \]

 

 Again, if $P'$~is any white polygon of the figure, which can be

-derived from~$P$ by the operation~$\Sigma$, a positive rotation (§~185) of~$P'$

+derived from~$P$ by the operation~$\Sigma$, a positive rotation (§~185) of~$P'$

 round its corner~$A_{r}'$ is effected by the operation $\Sigma^{-1}S_{r}\Sigma$;

 and, if $P''$~is the polygon so obtained, $P''$~is derived from~$P$ by

 the operation~$S_{r}\Sigma$. It is to be observed that a positive rotation

@@ -16151,7 +16157,7 @@ established between the complete set of double-polygons in the
 figure and the operations of the group, the relations which the

 generating operations satisfy being directly indicated by the

 configuration of the figure. Moreover, as with the general

-group (§~185), a simple rule may be stated for determining,

+group (§~185), a simple rule may be stated for determining,

 from an inspection of the figure, the polygon that corresponds

 to any given operation of the group.

 

@@ -16172,7 +16178,7 @@ the further $m$~relations
 f_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).

 \]

 

-As has been seen in §~182, if $H$~is the self-conjugate sub-group

+As has been seen in §~182, if $H$~is the self-conjugate sub-group

 of~$G$ which is formed by combining all possible operations

 of the form

 \[

@@ -16307,7 +16313,7 @@ S_{1}^{4} = 1,\quad S_{2}^{4} = 1,\quad S_{3}^{4} = 1, \\
 S_{1}S_{2}S_{3} = 1,\quad S_{2}^{-1}S_{1}S_{2}S_{1} = 1.

 \end{gather*}

 

-This group (§~74) is a non-Abelian group of order~$8$, containing

+This group (§~74) is a non-Abelian group of order~$8$, containing

 a single operation of order~$2$. The reader will have no difficulty in

 verifying that the plane figure for this group is given by fig.~7; and

 that opposite sides of the octagonal boundary correspond. The

@@ -16409,7 +16415,7 @@ that the number of sides is
 Nn.

 \]

 

-Using these values for $A$~and~$E$ in the formula of §~196,

+Using these values for $A$~and~$E$ in the formula of §~196,

 we obtain the equation

 \[

 2(p - 1) = N\left(n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}\right).

@@ -16531,14 +16537,14 @@ division into $n$-sided polygons the group is represented.
 

 The $2p$~relations given by $2p$~independent paths on the

 surface are not, however, necessarily independent. In fact we

-have already had an example to the contrary in §~195. On the

+have already had an example to the contrary in §~195. On the

 closed surface, by the regular division of which the group there

 considered is represented, four independent closed paths can be

 drawn. Any three of the corresponding relations can be derived

 from the fourth by transformation.

 

 The only known cases in which the $2p$~relations are independent

-are those of a class of groups of genus one (§~205).

+are those of a class of groups of genus one (§~205).

 %% -----File: 300.png---Folio 284-------

 

 \begin{Ex} Draw the figure of the group generated by $S_{1}$,~$S_{2}$,~$S_{3}$, where

@@ -16556,10 +16562,10 @@ regular division of a surface of connectivity~$5$.
 \end{Ex}

 

 

-\Note{\textit{Note to} §~194.}

+\Note{\textit{Note to} §~194.}

 

 \begin{Remark}

-If in the process of bending and stretching, described in §~194,

+If in the process of bending and stretching, described in §~194,

 by means of which the plane figure~$C$ is changed into an unbounded

 surface, the angles of the polygons all remain unaltered, the circles

 of the plane figure will become continuous curves on the surface.

@@ -16572,7 +16578,7 @@ circle of the figure leaves the figure unchanged geometrically but
 interchanges the black and white polygons. Each circle is, in fact, a

 line of symmetry for the figure such that, in respect of it, there

 is corresponding to every white polygon a symmetric black polygon

-and vice~versâ.

+and vice~versâ.

 

 Similarly on the surface a circle which does not cut itself may

 be a line of symmetry, such that a reflection at it is an operation of

@@ -16611,7 +16617,7 @@ there is a single type of group corresponding to each solution
 of the equation; and that, when $p = 1$, there is an infinite

 number of types of group, all characterized by a common

 property, corresponding to each solution of the equation.

-When $p = 0$, the groups are (§~196) of genus zero; and all

+When $p = 0$, the groups are (§~196) of genus zero; and all

 possible groups of genus zero are found by putting $p = 0$

 %% -----File: 302.png---Folio 286-------

 in the equation. The groups thus obtained are of special

@@ -16712,7 +16718,7 @@ stereographic projection of the latter. The five distinct types
 are represented graphically by the following figures.

 

 The first is a cyclical group, and the figure (fig.~9) does not

-differ essentially from fig.~2 in §~183.

+differ essentially from fig.~2 in §~183.

 

 The group given by the second solution of the equation

 \index{Group@\Topic{Group}!dihedral}%

@@ -16745,7 +16751,7 @@ which will bring respectively a double pyramid on an $n$-sided
 base, a tetrahedron, an octohedron, and an icosahedron to coincidence

 with itself in each case\footnotemark.

 \index{Klein@\Topic{Klein}, quoted}%

-\footnotetext{Klein, ``Vorlesungen über das Ikosaeder,'' Chap.~\DPchg{\textsc{i}}{I}.}%

+\footnotetext{Klein, ``Vorlesungen über das Ikosaeder,'' Chap.~\DPchg{\textsc{i}}{I}.}%

 

 When the figures are drawn on a sphere, and the three circles

 of the original triangle and therefore also all the circles of the

@@ -16865,7 +16871,7 @@ the symmetric group of four symbols.
 

 The icosahedral group is simple. It is, in fact, simply

 isomorphic with the alternating group of five symbols which

-has been shewn (§~111) to be a simple group. Thus if

+has been shewn (§~111) to be a simple group. Thus if

 \[

 S_{1} = (12)(34),\quad S_{2} = (135),

 \]

@@ -16979,7 +16985,7 @@ S_{1}S_{2}\quad \text{and}\quad S_{2}S_{3};
 \]

 for

 \begin{gather*}

-S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3}, \\

+S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3}, \\

 S_{2}S_{1} = (S_{1}S_{2})^{-1},\quad

 S_{3}S_{2} = (S_{2}S_{3})^{-1},\quad

 S_{3}S_{1} = (S_{2}S_{3})^{-1} (S_{1}S_{2})^{-1};

@@ -16988,7 +16994,7 @@ and therefore every operation containing an even number of
 factors can be represented in terms of $S_{1}S_{2}$ and~$S_{2}S_{3}$. Lastly,

 these two operations are permutable with each other; for

 \[

-S_{2}S_{3} · S_{1}S_{2} = S_{2}S_{3} · S_{4}S_{3} = S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3};

+S_{2}S_{3} · S_{1}S_{2} = S_{2}S_{3} · S_{4}S_{3} = S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3};

 \]

 and therefore every operation of the group is contained, once

 and only once, in the form

@@ -17128,16 +17134,16 @@ the two latter are $S_{1} S_{2}^{2}$ and $S_{2}S_{1} S_{2}$ respectively.
 Hence the three operations generate a self-conjugate sub-group;

 and since

 \[

-S_{1} S_{2}^{2} · S_{2} S_{1} S_{2} · S_{2}^{2} S_{1} = 1,

+S_{1} S_{2}^{2} · S_{2} S_{1} S_{2} · S_{2}^{2} S_{1} = 1,

 \]

 this sub-group is generated by $S_{1} S_{2}^{2}$ and $S_{2} S_{1} S_{2}$.

 

 These two operations are permutable; for

 \[

-S_{1} S_{2}^{2} · S_{2} S_{1} S_{2}

+S_{1} S_{2}^{2} · S_{2} S_{1} S_{2}

   = S_{1}^{2} S_{2}

-  = S_{2} S_{1} S_{2} S_{1} S_{2} · S_{2}

-  = S_{2} S_{1} S_{2} · S_{1} S_{2}^{2}.

+  = S_{2} S_{1} S_{2} S_{1} S_{2} · S_{2}

+  = S_{2} S_{1} S_{2} · S_{1} S_{2}^{2}.

 \]

 

 Hence finally, every operation of the group is represented,

@@ -17382,8 +17388,8 @@ S_{3}^{\alpha} (S_{2}^{2} S_{3}^{2})^{\beta} (S_{3} S_{2}^{2} S_{3})^{\gamma}, \
 and the order of the group is $6(b^{2} + bc + c^{2})$.

 

 \Par{209.} There are thus four distinct classes of groups\footnote

-  {Dyck, ``Ueber Aufstellung und Untersuchung von Gruppe und Irrationalität

-  regulärer Riemann'scher Flächen,'' \textit{Math.\ Ann.}\ \VolNo{XVII}, (1880), pp.~501--509.}

+  {Dyck, ``Ueber Aufstellung und Untersuchung von Gruppe und Irrationalität

+  regulärer Riemann'scher Flächen,'' \textit{Math.\ Ann.}\ \VolNo{XVII}, (1880), pp.~501--509.}

 \index{Dyck@\Topic{Dyck}, quoted}%

 \index{Defining relations@\Topic{Defining relations} of a group!for groups of genus unity}%

 of

@@ -17437,7 +17443,7 @@ three; viz.\ the groups defined by
 \Par{210.} As a final illustration of the present method of

 \PageLabel{302}%

 graphical representation, we will consider the simple group of

-order~$168$ (§~146), given by

+order~$168$ (§~146), given by

 \[

 \{(1236457),\ (234) (567),\ (2763) (45)\}.

 \]

@@ -17467,7 +17473,7 @@ This gives
 \[

 p = 3;

 \]

-it follows from Theorem~II, §~197, that the genus of the

+it follows from Theorem~II, §~197, that the genus of the

 group is~$3$.

 

 The figure for the general group, generated by $S_{2}$,~$S_{3}$, and~$S_{7}$,

@@ -17568,7 +17574,7 @@ the form of the condition may be simplified. Thus it may be
 written

 \begin{align*}

 S_{7}^{4} S_{2} S_{7}^{2} S_{2}

-  &= S_{2} S_{7} S_{2} S_{7} · S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5} \\

+  &= S_{2} S_{7} S_{2} S_{7} · S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5} \\

   &= S_{7}^{6} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5},

 \end{align*}

 or

@@ -17578,8 +17584,8 @@ S_{2} S_{7}^{2} S_{2} = S_{7}^{2} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5
 

 Now

 \begin{align*}

-S_{2} S_{7} · S_{7} S_{2}

-  &= S_{7}^{6} S_{2} S_{7}^{6} S_{2} · S_{2} S_{7}^{6} S_{2} S_{7}^{6} \\

+S_{2} S_{7} · S_{7} S_{2}

+  &= S_{7}^{6} S_{2} S_{7}^{6} S_{2} · S_{2} S_{7}^{6} S_{2} S_{7}^{6} \\

   &= S_{7}^{6} S_{2} S_{7}^{5} S_{2} S_{7}^{6}.

 \end{align*}

 

@@ -17589,7 +17595,7 @@ S_{7}^{4} S_{2} S_{7}^{5} S_{2} S_{7} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},
 \]

 or

 \[

-S_{7}^{4} S_{2} S_{7}^{4} · S_{2} S_{7}^{6} S_{2} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},

+S_{7}^{4} S_{2} S_{7}^{4} · S_{2} S_{7}^{6} S_{2} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},

 \]

 or

 \[

@@ -17771,7 +17777,7 @@ simplified since, so far as a graphical definition of the group
 is concerned, a large number of the lines are always redundant.

 

 If in the diagram consisting of $N$~points and $\frac{1}{2}N(N - 1)$

-coloured lines, which satisfies the conditions of §~212, all the

+coloured lines, which satisfies the conditions of §~212, all the

 lines of one or more colours are omitted, two cases may occur.

 We may still have a figure in which it is possible to pass

 along lines from one point to any other; or the points may

@@ -17832,7 +17838,7 @@ diagram required.
 As an illustration of this form of graphical representation,

 \index{Colour-groups@\Topic{Colour-groups}!examples of}%

 \index{Graphical representation@\Topic{Graphical representation}!examples of}%

-we may consider the octohedral group (§~201), defined by

+we may consider the octohedral group (§~201), defined by

 \begin{gather*}

 S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{4} = 1, \\

 S_{1} S_{2} S_{3} = 1.

@@ -17871,7 +17877,7 @@ the group\footnotemark.
 \Chapter{XIV.}{On the Linear Group\protect\footnotemark.}

 

 \footnotetext{The homogeneous linear group and its sub-groups forms the subject of the

-  greater part of Jordan's \textit{Traité des Substitutions}. The investigation of its

+  greater part of Jordan's \textit{Traité des Substitutions}. The investigation of its

   composition-series, given in the text, is due to Jordan.

 

   The complete analysis of the fractional linear group, defined by

@@ -17883,7 +17889,7 @@ the group\footnotemark.
   \alpha\delta - \beta\gamma \equiv 1,

   \]

   is due originally to Gierster, ``Die Untergruppen der Galois'schen Gruppe der

-  Modulargleichungen für den Fall eines primzahligen Transformationsgrades,''

+  Modulargleichungen für den Fall eines primzahligen Transformationsgrades,''

   \textit{Math.\ Ann.}\ Vol.~\VolNo{XVIII}, (1881), pp.~319--365. With a few unimportant

   modifications, the investigation in the text follows the lines of Gierster's

   memoir.

@@ -17899,7 +17905,7 @@ discuss and give an analysis of certain special groups. The first
 that we choose for this purpose is the group of isomorphisms

 of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.

 This group has been defined and its order determined in

-§§~171,~172. It is there shewn that the group is simply

+§§~171,~172. It is there shewn that the group is simply

 isomorphic with the homogeneous linear group defined by all

 sets of congruences

 \[

@@ -18235,7 +18241,7 @@ For a discussion of the properties of~$H$, some concrete
 representation of the group itself is necessary; this may be

 obtained in the following way. Instead of the pair of homogeneous

 congruences that define each operation of~$\Gamma$, let us,

-as in §~113, consider the single non-homogeneous congruence

+as in §~113, consider the single non-homogeneous congruence

 \[

 y \equiv \frac{\alpha x + \beta} {\gamma x + \delta},\ (\mod p),

 \]

@@ -18324,12 +18330,12 @@ $\left(\dfrac{\alpha x + \beta}{\delta}\right)$ for which $\alpha\delta \equiv 1
 residue, while $\beta$~may have any value whatever. Hence the

 order of the sub-group is~$\frac{1}{2}(p - 1)p$; and $H$~therefore contains

 $p + 1$ sub-groups of order~$p$. Since $H$~is a simple group, it

-follows (§~125) that it can be represented as a transitive

+follows (§~125) that it can be represented as a transitive

 substitution group of degree~$p + 1$.

 %% -----File: 338.png---Folio 320-------

 

 This representation of the group can be directly derived, as

-in §~113, from the congruences already used to define it. Thus

+in §~113, from the congruences already used to define it. Thus

 if, in

 \[

 y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},

@@ -18443,7 +18449,7 @@ S^{-1} Q S = \left(\frac{z^{-1} x}{z}\right) = Q^{-1}.
 

 The group of order~$p - 1$, which contains self-conjugately a

 cyclical sub-group of order~$\frac{1}{2}(p - 1)$ that keeps two symbols fixed,

-is therefore a group of dihedral (§~202) type. Moreover, if $t$~is

+is therefore a group of dihedral (§~202) type. Moreover, if $t$~is

 any factor of~$p - 1$, this investigation shews that $\{S, Q\}$~is the

 greatest sub-group that contains $\{Q^{t}\}$ self-conjugately.

 

@@ -18618,27 +18624,27 @@ sub-group of order~$4$, whose operations, except identity, are all
 of order~$2$. Now every group of order~$2^{m + 1}$ contains one self-conjugate

 operation of order~$2$, and $2^{m}$~operations of order~$2$

 falling into $2$~conjugate sets of $2^{m - 1}$ each. Moreover, the group

-of order~$p ± 1$, which has a cyclical sub-group of order~$2^{m}$ and

+of order~$p ± 1$, which has a cyclical sub-group of order~$2^{m}$ and

 contains the operation~$A$ of order~$2$ of this cyclical group

-self-conjugately, has $\frac{1}{2}(p ± 1)$ other operations of order~$2$; and

-therefore it contains $\dfrac{p ± 1}{2^{m + 1}}$ sub-groups of order~$2^{m + 1}$, each of

+self-conjugately, has $\frac{1}{2}(p ± 1)$ other operations of order~$2$; and

+therefore it contains $\dfrac{p ± 1}{2^{m + 1}}$ sub-groups of order~$2^{m + 1}$, each of

 which has $A$ for its self-conjugate operation. If now $B$~is

-any operation of order~$2$ of this sub-group of order~$p ± 1$, and

+any operation of order~$2$ of this sub-group of order~$p ± 1$, and

 if it is distinct from~$A$, then $B$~enters into a sub-group of

 order~$2^{m + 1}$ that contains $A$ self-conjugately. But since $A$~is

 permutable with~$B$, $A$~must belong to the sub-group of order

-$p ± 1$, which contains $B$ self-conjugately; hence $A$~enters into

+$p ± 1$, which contains $B$ self-conjugately; hence $A$~enters into

 a sub-group of order~$2^{m + 1}$ which contains $B$ self-conjugately.

 The sub-group $\{A, B\}$ is therefore common to two distinct

 sub-groups of order~$2^{m + 1}$. Now no group of order~$2^{r}$ ($r > 2$)

 can be common to two sub-groups of order~$2^{m + 1}$; and therefore

-$\{A, B\}$~must (§~80) be permutable with some operation~$S$ whose

+$\{A, B\}$~must (§~80) be permutable with some operation~$S$ whose

 order~$s$ is prime to~$2$. If $s$~is not~$3$, $S$~must be permutable with

 $A$ and~$B$: and then $\{A, S\}$ and $\{B, S\}$ would be two distinct

 sub-groups of orders~$2s$, whose operations are permutable with

 each other. It has been seen that $H$~does not contain such

 sub-groups. Hence $s = 3$; and $S$~transforms $A$,~$B$ and~$AB$ cyclically,

-or $\{S, A, B\}$ is a sub-group of tetrahedral type (§~202).

+or $\{S, A, B\}$ is a sub-group of tetrahedral type (§~202).

 

 The number of quadratic\footnote

   {A non-cyclical group of order~$4$ is called a \emph{quadratic} group.}

@@ -18647,16 +18653,16 @@ sub-groups contained in~$H$ may
 be directly enumerated. A group of order~$2^{m + 1}$ contains $2^{m - 1}$

 such sub-groups, which fall into $2$ conjugate sets of $2^{m - 2}$ each;

 a single group of order~$8$ containing each quadratic group self-conjugately.

-The quadratic groups, contained in the $\dfrac{p ± 1}{2^{m + 1}}$

-sub-groups of order~$2^{m + 1}$ of a sub-group of order~$p ± 1$, are

+The quadratic groups, contained in the $\dfrac{p ± 1}{2^{m + 1}}$

+sub-groups of order~$2^{m + 1}$ of a sub-group of order~$p ± 1$, are

 %% -----File: 345.png---Folio 327-------

 clearly all distinct, and each quadratic group belongs to just $3$

-groups of order~$p ± 1$; thus $\{A, B\}$ belongs to the $3$~groups

+groups of order~$p ± 1$; thus $\{A, B\}$ belongs to the $3$~groups

 which contain $A$,~$B$ and~$AB$ respectively as self-conjugate

 operations. Hence the total number of quadratic groups contained

 in~$H$ is

 \[

-\tfrac{1}{2} p (p \mp 1) \frac{p ± 1}{2^{m + 1}} 2^{m - 1} \tfrac{1}{3}

+\tfrac{1}{2} p (p \mp 1) \frac{p ± 1}{2^{m + 1}} 2^{m - 1} \tfrac{1}{3}

   = \frac{\frac{1}{2}p(p^{2} - 1)}{12}.

 \]

 

@@ -18670,7 +18676,7 @@ group self-conjugately is~$24$, and the $\dfrac{\frac{1}{2}p(p^{2} - 1)}{12}$ qu
 groups fall into two conjugate sets of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$ each. The

 group of order~$24$, that contains a quadratic group self-conjugately,

 contains also a self-conjugate tetrahedral sub-group,

-while the sub-groups of order~$8$ are dihedral. Hence (§~84) this

+while the sub-groups of order~$8$ are dihedral. Hence (§~84) this

 group must be of octohedral type.

 

 Since every tetrahedral sub-group of~$H$ contains a quadratic

@@ -18681,13 +18687,13 @@ conjugate sets of octohedral sub-groups, the number in each set
 being $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$.

 

 \Par{230.} We have hitherto supposed $m > 1$, or what is the

-same thing, $p \equiv ±1\ (\mod 8)$. If now $m = 1$, so that $p \equiv ±3\ (\mod 8)$,

+same thing, $p \equiv ±1\ (\mod 8)$. If now $m = 1$, so that $p \equiv ±3\ (\mod 8)$,

 the highest power of~$2$ that divides the order of~$H$ is~$2^{2}$;

-and, since $2^{2}$~is not a factor of~$\frac{1}{2}(p ± 1)$, the sub-groups of

+and, since $2^{2}$~is not a factor of~$\frac{1}{2}(p ± 1)$, the sub-groups of

 order~$2^{2}$ are quadratic. Moreover, since $2^{2}$~is the highest power

 of~$2$ dividing the order of~$H$, the quadratic sub-groups form

-a single conjugate set. Each sub-group of order~$p ± 1$, which

-has a self-conjugate operation of order~$2$, contains $\frac{1}{4}(p ± 1)$

+a single conjugate set. Each sub-group of order~$p ± 1$, which

+has a self-conjugate operation of order~$2$, contains $\frac{1}{4}(p ± 1)$

 %% -----File: 346.png---Folio 328-------

 sub-groups of order~$4$, and each of the latter belongs to $3$ of the

 former. The total number is as before

@@ -18707,7 +18713,7 @@ therefore the question of octohedral sub-groups does not arise.
 

 \Par{231.} The group~$H$ always contains tetrahedral sub-groups;

 when its order is divisible by~$24$, it contains also octohedral

-sub-groups. Now if $p \equiv ±1\ (\mod 5)$, the order of~$H$ is

+sub-groups. Now if $p \equiv ±1\ (\mod 5)$, the order of~$H$ is

 divisible by~$60$: it may be shewn as follows that, in these cases,

 $H$~contains sub-groups of icosahedral type.

 

@@ -18722,7 +18728,7 @@ $B$~denotes $\left(\dfrac{\alpha x + \beta}{\gamma x - \alpha}\right)$, since ea
 \[

 AB = \left(\frac{\alpha jx + \beta j^{-1}}{\gamma jx - \alpha j^{-1}}\right);

 \]

-and (§~203) if $A$~and~$B$ generate an icosahedral group,

+and (§~203) if $A$~and~$B$ generate an icosahedral group,

 \[

 (AB)^{3} = 1.

 \]

@@ -18779,14 +18785,14 @@ number of icosahedral sub-groups in~$H$ is therefore
   = \frac{\frac{1}{2}p(p^{2} - 1)}{30}.

 \]

 The group of isomorphisms of the icosahedral group is the

-symmetric group of degree~$5$ (§~173). Now $H$~can contain no

+symmetric group of degree~$5$ (§~173). Now $H$~can contain no

 sub-group simply isomorphic with the symmetric group of

 degree~$5$. For if it contained such a sub-group, an operation of

 %% -----File: 348.png---Folio 330-------

 order~$5$ would be conjugate to its own square; and this is not

 the case.

 

-Hence (§~165), if an icosahedral sub-group~$K$ of~$H$ is contained

+Hence (§~165), if an icosahedral sub-group~$K$ of~$H$ is contained

 self-conjugately in a greater sub-group~$L$, then $L$~must be

 the direct product of~$K$ and some other sub-group. This also is

 impossible; for the greatest sub-group of~$H$ in which any cyclical

@@ -18830,7 +18836,7 @@ Suppose next that $g$~is a sub-group of~$H$, whose order~$n$
 is not divisible by~$p$, and let $S_{1}$ be an operation of~$g$ whose

 order~$q_{1}$ is not less than the order of any other operation of~$g$.

 In~$H$ the sub-group~$\{S_{1}\}$ is self-conjugate in a dihedral group of

-order~$p ± 1$; and the greatest sub-group of this group, which

+order~$p ± 1$; and the greatest sub-group of this group, which

 contains no operation of order greater than~$q_{1}$, is a dihedral

 group of order~$2q_{1}$. Hence in~$g$ the sub-group~$\{S_{1}\}$ is self-conjugate

 in a group of order $q_{1}$ or~$2q_{1}$, and therefore it forms

@@ -18943,7 +18949,7 @@ conjugate set. No operation of this group is permutable with
 each of the $4$ sub-groups of order~$3$; and therefore, if the group

 exists, it can be represented as a transitive group of $4$~symbols.

 On the other hand, the order of the symmetric group of~$4$~symbols,

-which (§~203) is simply isomorphic with the octohedral

+which (§~203) is simply isomorphic with the octohedral

 group, is~$24$; and its cyclical sub-groups are distributed as

 above. Hence to this solution there correspond the octohedral

 sub-groups of~$H$.

@@ -18951,7 +18957,7 @@ sub-groups of~$H$.
 Solution~\Inum{($\delta$)} gives a group of order~$60$, with $6$~conjugate

 sub-groups of order~$5$, $10$~conjugate sub-groups of order~$3$, and

 a conjugate set of~$15$ operations of order~$2$. It has been shewn,

-in §~85, that there is only one type of group of order~$60$ that

+in §~85, that there is only one type of group of order~$60$ that

 has $6$~sub-groups of order~$5$; viz.\ the alternating group of

 degree~$5$: and that, in this group, the distribution of sub-groups

 in conjugate sets agrees with that just given. Moreover, the

@@ -18992,7 +18998,7 @@ of the triply transitive groups of degree~$p + 1$, defined by
 \[

 y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p),

 \]

-the existence of which was demonstrated in §~113. In fact,

+the existence of which was demonstrated in §~113. In fact,

 since $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$

 and $\left(\dfrac{k\alpha x + k\beta}{k\gamma x + k\delta}\right)$ represent the same transformation,

 the determinant, $\alpha\delta - \beta\gamma$, of any transformation may

@@ -19000,7 +19006,7 @@ always be taken as either unity or a given non-residue; and it
 follows at once that the transformations of determinant unity

 form a self-conjugate sub-group of the whole group of transformations.

 

-If, as in §~113, $\alpha$,~$\beta$, $\gamma$,~$\delta$, are powers of~$i$, where $i$~is a primitive

+If, as in §~113, $\alpha$,~$\beta$, $\gamma$,~$\delta$, are powers of~$i$, where $i$~is a primitive

 root of the congruence

 \[

 i^{p^{n} - 1} \equiv 1,\ (\mod p),

@@ -19010,7 +19016,7 @@ the triply transitive group~$G$ of degree~$p^{n} + 1$, which is defined
 by the transformations, has again, when $p$~is an odd prime, a

 self-conjugate sub-group~$H$ of order~$\frac{1}{2}p^{n}(p^{2n} - 1)$, which is

 given by the transformations of determinant unity. It follows

-from Theorem~IX, §~134, that $G$, being a triply transitive group

+from Theorem~IX, §~134, that $G$, being a triply transitive group

 of degree~$p^{n} + 1$, must have, as a self-conjugate sub-group, a

 doubly transitive simple group; and it is easy to shew that $H$

 is this sub-group.

@@ -19054,7 +19060,7 @@ below\footnotemark.
 of transformations of $n$~symbols, taken to a prime modulus~$p$;

 and consider it more directly as the group of isomorphisms of

 an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.

-As in §~156, it may be expressed in the form of a substitution

+As in §~156, it may be expressed in the form of a substitution

 group performed on the $p^{n} - 1$ symbols of the operations, other

 than identity, of the Abelian group. In this form it is clearly

 transitive, since there are isomorphisms changing any operation

@@ -19103,7 +19109,7 @@ those given by the powers of
 \]

 where $\alpha$ is a primitive root of~$p$. This operation is the same as

 %% -----File: 355.png---Folio 337-------

-that denoted by $A$ in §~217. It follows immediately that the

+that denoted by $A$ in §~217. It follows immediately that the

 \index{Homogeneous linear group@\Topic{Homogeneous linear group}!represented as a transitive substitution group}%

 factor-group $\dfrac{G}{\{A\}}$ can be represented as a transitive group in

 $\dfrac{p^{n} - 1}{p - 1}$ symbols. In fact, the operations of~$\{A\}$ are the only

@@ -19139,10 +19145,10 @@ of
 shall be unity; and this isomorphism still changes the sub-groups

 $\{P\}$ and $\{P'\}$ into $\{Q\}$ and $\{Q'\}$ respectively.

 

-The lowest power of~$A$ contained in~$\Gamma$ is (§~217) $A^{\efrac{p - 1}{d}}$.

+The lowest power of~$A$ contained in~$\Gamma$ is (§~217) $A^{\efrac{p - 1}{d}}$.

 Hence the group $\dfrac{\Gamma}{\{A^{\efrac{p - 1}{d}}\}}$ can be represented as a doubly

 %% -----File: 356.png---Folio 338-------

-transitive group of degree $\dfrac{p^{n} - 1}{p - 1}$. This group is (§~220) simply

+transitive group of degree $\dfrac{p^{n} - 1}{p - 1}$. This group is (§~220) simply

 isomorphic with the simple group of order $\dfrac{N}{(p - 1)d}$, which is

 defined by the composition-series of~$G$.

 

@@ -19167,12 +19173,12 @@ $n$ units})$. Now every pair of sub-groups of such an Abelian
 group enters in one, and only in one, sub-group of order~$p^{2}$;

 and every sub-group of order~$p^{2}$ contains $p + 1$ sub-groups of

 order~$p$. Hence from the $\dfrac{p^{n} - 1}{p - 1}$ symbols permuted by the

-doubly transitive group, $\dfrac{p^{n} - 1 · p^{n-1} - 1}{p - 1 · p^2 - 1}$\DPnote{** TN: No () in original} sets of $p + 1$ symbols

+doubly transitive group, $\dfrac{p^{n} - 1 · p^{n-1} - 1}{p - 1 · p^2 - 1}$\DPnote{** TN: No () in original} sets of $p + 1$ symbols

 each may be formed, such that every pair of symbols occurs in

 one set and no pair in more than one set, while the sets are

 permuted transitively by the operations of the group. These

 groups therefore belong to the class of groups referred to in

-§~148. The sub-group, that leaves two symbols unchanged,

+§~148. The sub-group, that leaves two symbols unchanged,

 permutes the remaining symbols in two transitive systems of

 $p - 1$ and $p^{n - 1} + p^{n - 2} + \Dots + p^{2}$; and the sub-group, that leaves

 unchanged each of the symbols of one of the sets of $p + 1$, is

@@ -19196,9 +19202,9 @@ itself a simple group.
 

 If $n = 3$, then $N = 168$; so that the group of isomorphisms of a

 group of order~$8$, whose operations are all of order~$2$, is the simple

-group of order~$168$ (§~146).

+group of order~$168$ (§~146).

 

-If $n = 4$, then $N = 2^{6} · 3^{2} · 5 · 7$. This is the order of the

+If $n = 4$, then $N = 2^{6} · 3^{2} · 5 · 7$. This is the order of the

 alternating group of $8$~symbols; and it may be shewn that this

 group is simply isomorphic with the group of isomorphisms.

 

@@ -19245,9 +19251,9 @@ with the alternating group of degree~$7$. Since this must be one of
 $8$~conjugate sub-groups, the group of substitutions itself is simply

 isomorphic with the alternating group of degree~$8$.

 

-If $p^{n} = 3^{3}$, then $p^{n - 1} + \Dots + p + 1 = 13$, $d = 1$, and $N = 2^{4} · 3^{3} · 13$.

+If $p^{n} = 3^{3}$, then $p^{n - 1} + \Dots + p + 1 = 13$, $d = 1$, and $N = 2^{4} · 3^{3} · 13$.

 There is therefore a doubly transitive simple group of degree~$13$ and

-order~$2^{4} · 3^{3} · 13$ (§§~145,~149).

+order~$2^{4} · 3^{3} · 13$ (§§~145,~149).

 \end{Remark}

 

 \Par{239.} The homogeneous linear group may be generalized by

@@ -19271,7 +19277,7 @@ $G_{p, n, \nu}$, defined by all sets of transformations
 \end{aligned}

 \quad (\mod p),

 \]

-whose determinant differs from zero, may be shewn, as in §~172,

+whose determinant differs from zero, may be shewn, as in §~172,

 to be

 \[

 N = (p^{n\nu} - 1) (p^{n\nu} - p^{\nu}) (p^{n\nu} - p^{2 \nu}) \Dots (p^{n\nu} - p^{(n - 1)\nu});

@@ -19282,7 +19288,7 @@ self-conjugate operations of~$\Gamma$ are the operations of the sub-group
 generated by $(ix_{1}, ix_{2}, \Dots, ix_{n})$, which are contained in~$\Gamma$.

 If $\delta$~is the greatest common factor of $p^{\nu} - 1$ and~$n$, these

 self-conjugate operations of~$\Gamma$ form a cyclical sub-group~$\gamma$ of

-order~$\delta$. Finally, the argument of §~219 maybe repeated to

+order~$\delta$. Finally, the argument of §~219 maybe repeated to

 shew that $\dfrac{\Gamma}{\gamma}$~is a simple group.

 

 The homogeneous linear group $G_{p, n, \nu}$, when values of~$\nu$

@@ -19321,7 +19327,7 @@ where
   i^{s_{n}} &\equiv a_{n1}i^{r_{1}} + a_{n2}i^{r_{2}} + \Dots + a_{nn}i^{r_{n}}.

 \end{align*}

 

-Every operation of~$G_{p, n, \nu}$ as defined in §~239, is therefore

+Every operation of~$G_{p, n, \nu}$ as defined in §~239, is therefore

 permutable with the Abelian group, and gives a distinct isomorphism

 of it; or in other words, as stated above, $G_{p, n, \nu}$~is simply isomorphic

 with a sub-group of~$G_{p, n\nu, 1}$.

@@ -19372,7 +19378,7 @@ with a group of isomorphisms of the Abelian group which permutes
 among themselves such a set of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ sub-groups of order~$p^{\nu}$.

 \end{Remark}

 

-\begin{Ex}[1.] Shew that the $\dfrac{p^{n\nu} - 1 · p^{n\nu} - p\Dots p^{n\nu} - p^{\nu}}{p - 1 · p^{2} - 1\Dots p^{\nu} - 1}$\DPnote{** TN: No () in original} sub-groups of

+\begin{Ex}[1.] Shew that the $\dfrac{p^{n\nu} - 1 · p^{n\nu} - p\Dots p^{n\nu} - p^{\nu}}{p - 1 · p^{2} - 1\Dots p^{\nu} - 1}$\DPnote{** TN: No () in original} sub-groups of

 order~$p^{\nu}$ of an Abelian group of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to

 $n\nu$ units})$ can be divided into sets of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ each, such that each set

 contains every operation of the group, other than identity, once and

@@ -19450,12 +19456,12 @@ simple; $G$~is therefore not simple.
 

 \begin{CorollaryFN}

 If a prime~$p$ divides~$N$, and if $m$~is the

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 greatest factor of~$N$ which is congruent to unity~$(\mod p)$, $G$~will

 be composite unless $N$~is a factor of~$m!$.

 \end{CorollaryFN}

 \footnotetext

-  {Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), p.~57.}

+  {Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), p.~57.}

 

 There cannot, in fact, be more than $m$ conjugate sub-groups

 whose order is the highest power of~$p$ that divides~$N$; so that

@@ -19466,7 +19472,7 @@ In dealing with a given integer as order, it may happen that,
 though no single application of this theorem will prove the corresponding

 group to be composite, a repeated application of Sylow's

 theorem, on which the theorem depends, will lead to that result.

-As an example, let us consider groups whose order is $3^{2} · 5 · 11$. Any

+As an example, let us consider groups whose order is $3^{2} · 5 · 11$. Any

 such group must contain either $1$ or $45$ sub-groups of order~$11$, and

 either $1$ or $11$ sub-groups of order~$5$. Hence, if the group is simple,

 it must contain $45$~sub-groups of order~$11$, and $11$~sub-groups of

@@ -19487,7 +19493,7 @@ Every group whose order is the power
 of a prime is soluble.

 \end{Theorem}

 

-This has already been proved in §~54.

+This has already been proved in §~54.

 

 \begin{Theorem}[III\protect\footnotemark.]

 A group~$G$ whose order is~$p^{\alpha} q^{\beta}$, where $\alpha$~is

@@ -19504,7 +19510,7 @@ If $\alpha < m$, $G$~contains a self-conjugate sub-group~$H$ of order~$q^{\beta}
 If $\alpha = m$, $G$~contains either $1$ or $p^{m}$ sub-groups of order~$q^{\beta}$.

 In the latter case, if $q^{r}$~is the order of the greatest sub-group

 common to two groups of order~$q^{\beta}$, and if $h$~is such a sub-group,

-$h$~must (§~80) be contained self-conjugately in a

+$h$~must (§~80) be contained self-conjugately in a

 sub-group~$k$ of order~$p^{x}q^{r + s}$, $x > 0$, $s > 0$; and this sub-group

 must contain more than one sub-group of order~$q^{r + s}$. Hence $x$~must

 be~$m$; and therefore $h$~must be common to all the $p^{m}$

@@ -19556,7 +19562,7 @@ of this order cannot enter in a group of order~$p^{\alpha - m} q^{\beta}$. There
 $G$~contains $q^{\beta}$~sub-groups of order~$p^{\alpha}$.

 

 Since the sub-groups of~$G$, of order~$p^{\alpha}$, are transitive, their

-self-conjugate operations must (§~106) displace all the symbols,

+self-conjugate operations must (§~106) displace all the symbols,

 and they cannot therefore be permutable with any operation

 whose order is a power of~$q$. Suppose now that every

 operation of a sub-group~$Q$, of order~$q^{x}$, is permutable with

@@ -19577,7 +19583,7 @@ is therefore either a power of~$p$ or a power of~$q$.
 

 Suppose now that $p^{r}$~is the order of the greatest sub-group

 that is common to any two sub-groups of order~$p^{\alpha}$, and that $h$~is

-such a sub-group. Then (§~80) $h$~must be permutable with

+such a sub-group. Then (§~80) $h$~must be permutable with

 an operation of order~$q$. Since no operation of~$h$ is permutable

 with any operation of order~$q$, it follows that $p^{r} - 1$ is divisible

 by~$q$, and therefore that $r = m$.

@@ -19649,12 +19655,12 @@ immediately from the theorem, except for the cases $2^{4}3^{\alpha}$ and
 $2^{5}3^{\alpha}$.

 

 If in these cases there are $16$ sub-groups of order~$3^{\alpha}$ ($\alpha > 1$),

-there must (§~78) be sub-groups of order~$3^{\alpha - 1}$ common to two

+there must (§~78) be sub-groups of order~$3^{\alpha - 1}$ common to two

 sub-groups of order~$3^{\alpha}$; and, in the groups of order $2^{4}3^{\alpha}$~or

 $2^{5}3^{\alpha}$, such a sub-group, if not self-conjugate, must be one of

 either $4$ or~$8$ conjugate sub-groups. The groups must therefore

 be isomorphic with groups of degree $4$ or~$8$; from this it follows

-immediately that they must be soluble (§~146).

+immediately that they must be soluble (§~146).

 

 \Par{244.} \begin{Theorem}[IV.]

 Groups of order~$p_{1}^{\alpha} p_{2}^{2}$ are soluble.

@@ -19672,7 +19678,7 @@ therefore a group of order~$p_{2}^{2}$ is self-conjugate. If the operations
 of the $p_{2}^{2}$~sub-groups of order~$p_{1}^{\alpha}$ are not all distinct, let $p_{1}^{r}$ be

 the order of the greatest sub-group common to any two of them;

 %% -----File: 367.png---Folio 349-------

-and let $h$ be such a sub-group. Then (§~80) $h$~must be self-conjugate

+and let $h$ be such a sub-group. Then (§~80) $h$~must be self-conjugate

 in a group~$k$ of order~$p_{1}^{r + s}p_{2}^{\beta}$ ($s > 0$, $\beta = 1$ or~$2$). If

 $\beta = 2$, $h$~is self-conjugate in~$G$; and if $\beta = 1$, $k$~must contain $p_{2}$~sub-groups

 of order~$p_{1}^{r + s}$, and therefore

@@ -19680,7 +19686,7 @@ of order~$p_{1}^{r + s}$, and therefore
 p_{2} \equiv 1\ (\mod p_{1}).

 \]

 

-Now (§~78)

+Now (§~78)

 \[

 p_{2}^{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),

 \]

@@ -19690,7 +19696,7 @@ p_{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),
 \]

 unless $p_{1} = 2$; but, if $p_{1} = 2$, then

 \[

-p_{2} \equiv ± 1\ (\mod p_{1}^{\alpha - r - 1}).

+p_{2} \equiv ± 1\ (\mod p_{1}^{\alpha - r - 1}).

 \]

 

 Again, if  $s = \alpha - r$, $h$~is one of $p_{2}$~conjugate sub-groups; as

@@ -19709,7 +19715,7 @@ Let
 \[

 H,\ H_{1},\ \Dots,\ H_{p_{2} - 1}

 \]

-be the $p_{2}$~sub-groups (§~80) of order~$p_{1}^{\alpha}$ that contain~$h$; and

+be the $p_{2}$~sub-groups (§~80) of order~$p_{1}^{\alpha}$ that contain~$h$; and

 suppose that $\Gh$~is not contained in~$h$. Every operation of each

 of the $p_{2}$~sub-groups

 \[

@@ -19729,7 +19735,7 @@ A group of order~$p_{2}^{2}$ cannot be expressed in less than $2p_{2}$
 symbols; a cyclical group of order~$p_{2}^{2}$ cannot be expressed in

 %% -----File: 368.png---Folio 350-------

 $2(p_{2} + 1)$~symbols; and a group of degree~$2(p_{2} + 1)$, which

-contains a non-cyclical sub-group of order~$p_{2}^{2}$, must (§~141)

+contains a non-cyclical sub-group of order~$p_{2}^{2}$, must (§~141)

 contain the alternating group. Hence again, in this case, $G$~must

 have a self-conjugate sub-group.

 

@@ -19777,7 +19783,7 @@ is a set of $p_{2}$~sub-groups of order~$p_{2}^{\alpha}$, which have the common
 sub-group $Q'^{-1} hQ'$.

 

 Now, since by supposition $h$~is not a self-conjugate sub-group

-of~$H$, there must (§~55) be in~$H$ a sub-group~$h'$ conjugate

+of~$H$, there must (§~55) be in~$H$ a sub-group~$h'$ conjugate

 to and permutable with~$h$. Let this be the sub-group of order~$p_{1}^{r}$,

 common to

 \[

@@ -19825,14 +19831,14 @@ transform the $p_{2}^{\gamma}$~sub-groups of order~$p_{1}^{\alpha}$ transitively
 themselves; and therefore, in the isomorphism between the

 given group and the substitution group of degree~$p_{2}^{\gamma}$, there

 corresponds to an Abelian sub-group of order~$p_{2}^{\beta}$ a transitive

-substitution group of degree~$p_{2}^{\gamma}$. But (§~124) an Abelian group

+substitution group of degree~$p_{2}^{\gamma}$. But (§~124) an Abelian group

 can only be represented as a transitive substitution group of a

 number of symbols equal to its order. Hence the isomorphism

 between the given group and the substitution group is multiple;

 and the given group is not simple.

 

 Suppose, next, that there are $p_{2}^{\beta}$~sub-groups of order~$p_{1}^{\alpha}$, and

-therefore (§~81) $p_{1}^{\alpha} - 1$ distinct sets of conjugate operations

+therefore (§~81) $p_{1}^{\alpha} - 1$ distinct sets of conjugate operations

 %% -----File: 370.png---Folio 352-------

 whose orders are powers of~$p_{1}$. Let $P$~be an operation whose

 order is a power of~$p_{1}$; and let $H$, of order~$p_{1}^{\alpha} p_{2}^{\delta}$, be the greatest

@@ -19869,7 +19875,7 @@ soluble\footnotemark.
 

 If the number of operations of the group, whose orders divide

 $p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$, is exactly equal to this number, it follows

-from Theorem~VII, Cor.~I, §~87, that the same is true for the

+from Theorem~VII, Cor.~I, §~87, that the same is true for the

 number of operations of the group whose orders divide

 $p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$. Now, when $r = 1$, this relation obviously holds;

 and therefore it is true for all values of~$r$. The group therefore

@@ -19917,10 +19923,10 @@ order of the group.
 \begin{Remark}

 \Par{247.} A special case of the class of groups under consideration

 is that in which the order contains no repeated prime factor. These

-groups have formed the subject of a memoir by Herr O.~Hölder\footnotemark.

-\index{Holder@\Topic{Hölder}, quoted}%

+groups have formed the subject of a memoir by Herr O.~Hölder\footnotemark.

+\index{Holder@\Topic{Hölder}, quoted}%

 \index{Groups whose orders contain@\Topic{Groups} whose orders contain!no squared factor}%

-\footnotetext{``Die Gruppen mit quadratfreier Ordnungszahl'' \textit{Göttingen Nachrichten},

+\footnotetext{``Die Gruppen mit quadratfreier Ordnungszahl'' \textit{Göttingen Nachrichten},

   1895: pp.~211--229. Compare also Frobenius, \textit{Berliner Sitzungsberichte}, 1895,

   pp.~1043, 1044.}%

 He shews that they are capable of a specially simple form of representation,

@@ -19939,7 +19945,7 @@ the orders of those cyclical sub-groups of prime order which are
 contained self-conjugately in~$G$. They generate a self-conjugate

 sub-group~$H$ of order $p_{l} p_{m} \Dots p_{n}$, say~$\mu$. Each of the cyclical

 sub-groups $\{P_{l}\}$, $\{P_{m}\}$,~\Dots, $\{P_{n}\}$ of~$H$, generated by an operation of

-prime order, is permutable with all the rest; and therefore (§~34)

+prime order, is permutable with all the rest; and therefore (§~34)

 the operations $P_{l}$, $P_{m}$,~\Dots, $P_{n}$ are all permutable. Hence, since

 their orders are distinct primes, the sub-group~$H$ is cyclical. Let

 now $K$~be a sub-group of~$G$ of order~$\nu$, where $N = \mu\nu$. No self-conjugate

@@ -19948,14 +19954,14 @@ such a sub-group of order $p_{\alpha} p_{\beta} \Dots p_{\gamma}$ ($\alpha < \be
 would contain a single cyclical sub-group of order~$p_{\gamma}$; and $G$~would

 contain a self-conjugate sub-group of order~$p_{\gamma}$, contrary to

 supposition. Since then $K$~contains no self-conjugate sub-group of~$G$,

-it follows (§~123) that $G$~can be expressed transitively in $\mu$~symbols;

+it follows (§~123) that $G$~can be expressed transitively in $\mu$~symbols;

 when it is so expressed, an operation~$M$ of order~$\mu$, that

 generates~$H$, will be a circular substitution of $\mu$~symbols. The

 only substitutions performed on the $\mu$~symbols, which are permutable

 with~$M$, are its own powers; and therefore no operation of~$G$

 is permutable with~$M$ except its own powers. Every operation

 of~$K$ must therefore give a distinct isomorphism of~$\{M\}$. Now

-(§~169) the group of isomorphisms of a cyclical group is Abelian.

+(§~169) the group of isomorphisms of a cyclical group is Abelian.

 Hence $K$~must be Abelian; and since its order contains no repeated

 prime factor, it must be cyclical. The group~$G$ is therefore completely

 defined by the relations

@@ -19965,10 +19971,10 @@ M^{\mu} = 1,\quad N^{\nu} = 1,\quad N^{-1}MN = M^{\alpha},
 where $\alpha$~belongs to index $\nu, \pmod{\mu}$.

 \end{Remark}

 

-\Par{248.} The theorem of §~246 is a particular case of the

+\Par{248.} The theorem of §~246 is a particular case of the

 following more general one, due to Herr Frobenius\footnotemark,

 \index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{``Über auflösbare Gruppen,~II'': \textit{Berliner Sitzungsberichte}, 1895, p.~1035.}%

+\footnotetext{``Über auflösbare Gruppen,~II'': \textit{Berliner Sitzungsberichte}, 1895, p.~1035.}%

 from which

 a number of interesting and important results may be deduced.

 

@@ -19993,7 +19999,7 @@ factor of~$m$, it is also true for~$G$.
 Suppose that the greatest group, which contains $R$ self-conjugately,

 is a group~$H'$ of order~$p^{\alpha'} q^{\beta'} \Dots r^{\gamma}n'$, where $n'$~is a

 factor of~$n$. Let $N'$~be any operation of this sub-group, whose

-order~$\nu'$ is a factor of~$n'$. Then (§~177) since $\theta(R)$ and~$\nu'$ are

+order~$\nu'$ is a factor of~$n'$. Then (§~177) since $\theta(R)$ and~$\nu'$ are

 relatively prime, $N'$~is permutable with every operation of~$R$.

 Hence every operation of~$H'$, whose order divides~$n'$, is permutable

 with every operation of~$R$.

@@ -20002,9 +20008,9 @@ Let now $S$~be any operation of~$R$; and suppose that the
 order of~$K$, the greatest sub-group which contains $S$ self-conjugately,

 is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}n_{1}$. If the order of~$S$ is~$r^{\delta}$, the

 order of~$\dfrac{K}{\{S\}}$ is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma - \delta} n_{1}$, say~$m_{1}n_{1}$. We have seen in

-§~176 that, if $P_{1}$~is a sub-group of~$P$, then $\theta(P_{1})$~is equal to or

+§~176 that, if $P_{1}$~is a sub-group of~$P$, then $\theta(P_{1})$~is equal to or

 is a factor of~$\theta(P)$. Hence if $P_{1}$~is a sub-group of order~$p^{\alpha_{1}}$ of~$\dfrac{K}{\{S\}}$,

-$\theta(P_{1})$~and $\dfrac{N}{p^{\alpha}}$ are relatively prime; therefore, \DPtypo{à~fortiori}{a~fortiori},

+$\theta(P_{1})$~and $\dfrac{N}{p^{\alpha}}$ are relatively prime; therefore, \DPtypo{à~fortiori}{a~fortiori},

 $\theta(P_{1})$ and $\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}}$ are relatively prime. Similarly $\theta(Q_{1})$ and

 $\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}q^{\beta_{1}}}$ are relatively prime, and so on. Moreover $m_{1}$~is a factor

 of~$m$. We may therefore assume that the theorem holds for~$\dfrac{K}{\{S\}}$.

@@ -20018,7 +20024,7 @@ of~$N$ divides~$n$.
 

 Now $S$~is one of~$p^{\alpha - \alpha_{1}}q^{\beta - \beta_{1}} \Dots \dfrac{n}{n_{1}}$ conjugate operations in~$G$;

 corresponding to each of these, there is such a set of $n_{1}$

-operations of the form~$SN$, while (§~16) no two of these

+operations of the form~$SN$, while (§~16) no two of these

 operations can be identical. Hence $G$~contains $p^{\alpha - \alpha_{1}} q^{\beta - \beta_{1}} \Dots n$

 operations of the form~$SN$, arising from the conjugate set to

 which $S$~belongs. If then we sum for the distinct conjugate

@@ -20038,7 +20044,7 @@ operation of~$R$; and therefore $\nu = n'$. Hence the number of
 operations of~$H'$, excluding identity, whose orders are powers of~$r$,

 is $\sum' p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots$, where the summation is extended to all the

 distinct conjugate sets of operations of~$H'$ whose orders are

-powers of~$r$. It has been shewn (Theorem~IV, Corollary, §~81)

+powers of~$r$. It has been shewn (Theorem~IV, Corollary, §~81)

 that the number of distinct conjugate sets of operations in~$G$,

 whose orders are powers of~$r$, is the same as the number in~$H'$.

 Hence the symbols $\sum$ and $\sum'$ represent summations of the same

@@ -20151,7 +20157,7 @@ an exception\footnotemark.
 \footnotetext{Frobenius, \textit{loc.~cit.}, p.~1041.}%

 

 If an Abelian sub-group~$P_{r}$ of order~$p_{r}^{\alpha_{r}}$ is generated by a

-single operation, it is cyclical and $\theta(P_{r})$~is (§~176) equal to $p_{r} - 1$.

+single operation, it is cyclical and $\theta(P_{r})$~is (§~176) equal to $p_{r} - 1$.

 If $P_{r}$~is generated by two independent and permutable operations,

 $\theta(P_{r})$~is equal to~$(p_{r} - 1)(p_{r}^{2} - 1)$. Now no prime greater

 than~$p_{r}$ can divide $(p_{r} - 1)(p_{r}^{2} - 1)$ unless $p_{r} + 1$ be a prime;

@@ -20190,7 +20196,7 @@ A,\ A_{1},\ \Dots,\ A_{r},\ A_{r + 1},\ \dots
 \]

 be a characteristic series of the group~$A$ of order~$2^{\alpha_{1}}$; and

 suppose that $A_{r + 1}$~is the greatest of these groups with every

-one of whose operations~$B$ is permutable. Then (§~175) $B$~is

+one of whose operations~$B$ is permutable. Then (§~175) $B$~is

 not permutable with every operation of~$\dfrac{A_{r}}{A_{r + 1}}$. Hence $\dfrac{A_{r}}{A_{r + 1}}$~is

 a quadratic group; and its three operations of order~$2$ must

 be permuted cyclically when transformed by~$B$. It follows

@@ -20228,27 +20234,27 @@ of $2^{\alpha}$~symbols each. This is an odd substitution; and therefore
 $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. In this sub-group

 there are cyclical operations of order~$2^{\alpha - 1}$. Hence the

 same reasoning will apply to it, and it contains a self-conjugate

-sub-group of order~$2^{\alpha - 2}n$. Now (Theorem~VII, Cor.~I. §~87) $G$~contains

+sub-group of order~$2^{\alpha - 2}n$. Now (Theorem~VII, Cor.~I. §~87) $G$~contains

 exactly $2^{\alpha - 2}n$ operations whose orders are not divisible

 by~$2^{\alpha - 1}$. Hence the sub-group of order~$2^{\alpha - 2}n$ must be self-conjugate

 in~$G$. Proceeding thus, it may be shewn that $G$~contains

 self-conjugate (and characteristic) sub-groups of every

 order~$2^{\alpha - r}n$ ($r = 1, 2, \Dots, \alpha$)\footnotemark.

 \index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, ``Über auflösbare \DPtypo{Grüppen}{Gruppen},~II'' \textit{Berliner Sitzungsberichte},

+\footnotetext{Frobenius, ``Über auflösbare \DPtypo{Grüppen}{Gruppen},~II'' \textit{Berliner Sitzungsberichte},

   1895, p.~1039.}%

 

 \Par{251.} Suppose, secondly, that, under the same conditions, $G$~contains

 a self-conjugate sub-group~$\scrQ$ of order~$2^{\alpha}$. Since $\theta(\scrQ)$

 and $n$ are relatively prime, every operation of~$\scrQ$ is permutable

-with every operation of~$G$ (§~177). If $\scrQ'$~is a sub-group of~$\scrQ$

+with every operation of~$G$ (§~177). If $\scrQ'$~is a sub-group of~$\scrQ$

 of order~$2^{\alpha - 1}$, $\scrQ'$~must therefore be self-conjugate in~$G$. Now

 the order of the factor group $\dfrac{G}{\scrQ'}$ is~$2n$; hence, by the preceding

 result, it has a self-conjugate sub-group of order~$n$. It follows

 that $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. This

 group, again, has a self-conjugate sub-group of order~$2^{\alpha - 2}n$,

 and so on. Hence $G$~has a sub-group of order~$n$. Now by

-Theorem~VII (§~248), $G$~has just $n$~operations whose

+Theorem~VII (§~248), $G$~has just $n$~operations whose

 %% -----File: 379.png---Folio 361-------

 divide~$n$. Hence the sub-group~$H$, of order~$n$, must be self-conjugate;

 and $G$~is the direct product of the two groups $\scrQ$

@@ -20257,7 +20263,7 @@ and~$H$.
 \Par{252.} Suppose next that $\scrQ$~is not self-conjugate in~$G$; and

 let~$I$, of order~$2^{\alpha}n'$, be the greatest group that contains $\scrQ$ self-conjugately.

 Every operation of~$I$ is permutable with every

-operation of~$\scrQ$; and therefore (§~81) $G$~contains $2^{\alpha - 1}$ distinct

+operation of~$\scrQ$; and therefore (§~81) $G$~contains $2^{\alpha - 1}$ distinct

 sets of conjugate operations whose orders are powers of~$2$.

 The case, in which $\scrQ$~is cyclical, has been already dealt with

 and will now be excluded; we may thus assume that $\scrQ$~contains

@@ -20313,7 +20319,7 @@ and therefore $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$.
 In the same way it may be shewn that this sub-group has a

 self-conjugate sub-group of order~$2^{\alpha - 2}n$; and so on. Hence $G$~has

 a sub-group $H$ of order~$n$. But it follows from Theorem~VII,

-§~248, that $G$~has exactly $n$ operations whose orders divide~$n$;

+§~248, that $G$~has exactly $n$ operations whose orders divide~$n$;

 and therefore $H$~is a self-conjugate sub-group. Moreover,

 since $\dfrac{G}{H}$~is an Abelian group of order~$2^{\alpha}$, it must contain self-conjugate

 sub-groups of every order~$2^{r}$ ($r = 1, 2, \Dots, \alpha - 1$);

@@ -20356,7 +20362,7 @@ sub-group~$G'$ of~$G$ whose order is divisible by~$2$. In
 this case, we may deal with~$\dfrac{G}{G'}$ exactly as we have been

 dealing with~$G$.

 

-The results of §§~250--254 may be summed up in the

+The results of §§~250--254 may be summed up in the

 following form:---

 

 \begin{Theorem}[IX.]

@@ -20395,7 +20401,7 @@ namely $\{A^{2^{\alpha - 2}}\}$.
 We will consider first the case where $\scrQ$~is of type~\Inum{(i)}. In

 this case, $\{A^{2{\alpha - 3}}\}$~and $\{B\}$ are sub-groups of~$G$ of order~$4$, the

 former being self-conjugate in~$\scrQ$ while the latter is not. If

-they are conjugate sub-groups in~$G$, we have seen (§~82)

+they are conjugate sub-groups in~$G$, we have seen (§~82)

 that $G$~must contain an operation of odd order~$S$, such that

 the sub-groups $S^{-n}\{B\}S^{n}$ ($n = 0, 1, 2, \dots$) are permutable with

 each other. This is impossible, since every operation that

@@ -20431,7 +20437,7 @@ $G$~has therefore a self-conjugate sub-group of order~$2^{\alpha - 1}n$. The
 groups of order~$2^{\alpha - 1}$ contained in this self-conjugate sub-group

 are either of type~\Inum{(i)}, with $\alpha - 1$ written for~$\alpha$: or they are

 cyclical; for, like~$\scrQ$, they can only contain a single operation of

-order~$2$. If they are cyclical, the reasoning of §~250 may

+order~$2$. If they are cyclical, the reasoning of §~250 may

 be applied; and if they are of type~\Inum{(i)} the same reasoning

 will apply to the self-conjugate sub-group of order~$2^{\alpha - 1}n$

 that has been used for~$G$. Finally, then, $G$~must contain

@@ -20453,9 +20459,9 @@ order~$2^{\alpha - 1}$ are cyclical. Hence, again, there is a self-conjugate
 sub-group of order~$n$.

 

 \Par{256.} The only non-Abelian groups of order~$2^{3}$ are those of

-types \Inum{(i)}~and~\Inum{(ii)} of §~255, when $\alpha = 3$. The Abelian groups

+types \Inum{(i)}~and~\Inum{(ii)} of §~255, when $\alpha = 3$. The Abelian groups

 of order $2^{2}$~and~$2^{3}$ are all included in the types considered in

-Theorem~IX, §~254. For an Abelian group~$\scrQ$, of order~$2^{3}$ and

+Theorem~IX, §~254. For an Abelian group~$\scrQ$, of order~$2^{3}$ and

 type~$(2, 1)$, $\theta(\scrQ)$~is~$3$; and for one of order~$2^{3}$ and type $(1, 1, 1)$,

 $\theta(\scrQ)$~is~$21$. Hence Theorems IX~and~X shew that, if $2$,~$2^{2}$

 or~$2^{3}$ divide the order of a group, but not~$2^{4}$, then the

@@ -20488,7 +20494,7 @@ Let $h$ be a sub-group of~$G$ whose order
 is a power of~$p$; and let $S$ be an operation of~$G$ whose order is not

 divisible by~$p$. Then since $m$~is relatively prime to $p(p - 1) \Dots

 (p^{\alpha} - 1)$, if $S$~is permutable with~$h$, it is permutable with every

-operation of~$h$. Hence it follows from §~82 that no operation of~$H$,

+operation of~$h$. Hence it follows from §~82 that no operation of~$H$,

 which is not self-conjugate in~$H$, can be conjugate in~$G$ to

 a self-conjugate operation of~$H$. Suppose now that $S_{1}$~and $S_{2}$ are

 two operations of~$H$, each of which in~$H$ is one of $p$ conjugate

@@ -20510,7 +20516,7 @@ transforms $S_{1}$ into $S_{2}$ and $H_{1}$~into $H_{2}$, must transform $\Gh$ i
 Hence $S_{1}$ and $S_{2}$ are conjugate operations in that sub-group,~$G'$, of~$G$

 which contains $\Gh$ self-conjugately. Now in $\dfrac{G'}{\Gh}$ the operations, that

 correspond to $S_{1}$~and~$S_{2}$, are self-conjugate operations of~$\dfrac{H}{\Gh}$. Since

-then these operations are conjugate in~$\dfrac{G'}{\Gh}$, they must (§~82) be

+then these operations are conjugate in~$\dfrac{G'}{\Gh}$, they must (§~82) be

 conjugate in~$\dfrac{I}{\Gh}$. This however is impossible, since every operation

 of~$\dfrac{I}{\Gh}$ whose order is a power of~$p$ is self-conjugate. Finally, then, no

 two operations of~$H$, which are not conjugate in~$H$, can be conjugate

@@ -20538,7 +20544,7 @@ $p_{1} p_{2} p_{3}^{2}$ or~$p_{1} p_{2} p_{3} p_{4}$. If $p_{1}$~is an odd prime
 these orders have already been shewn to be soluble; while if $p_{1}$~is~$2$,

 the only case which can give a simple group is~$2^{2} p_{2} p_{3}$. If

 a group of this order is simple, it follows from Theorem~VIII,

-Cor.\ (§~249), that $p_{2}$~must be~$3$; and the order of the group~$12p_{3}$.

+Cor.\ (§~249), that $p_{2}$~must be~$3$; and the order of the group~$12p_{3}$.

 A cyclical sub-group of order~$p_{3}$ must be one of $1 + kp_{3}$ conjugate

 sub-groups. Hence $1 + kp_{3}$ must be a factor of~$12$, so that $p_{3}$~is

 either $5$~or~$11$. If $p_{3}$~were~$11$, the $12$~conjugate sub-groups of

@@ -20547,7 +20553,7 @@ the tetrahedral sub-group, which the group (if simple) must
 contain, would be self-conjugate. Hence $p_{3}$~must be~$5$ and the

 order of the group is~$60$. We have already seen that a simple

 group of order~$60$ actually exists, namely, the icosahedral group;

-and that there is only one type for such a group (§~85).

+and that there is only one type for such a group (§~85).

 

 If the order of a group contains five prime factors, it must

 be of one of the forms:---

@@ -20569,7 +20575,7 @@ must be all distinct, or else they must all have a common
 \index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|(}%

 sub-group, for the order of the group contains no factor

 congruent to unity $(\mod p_{2})$, except $p_{3}$~or $p_{1} p_{3}$. Now the group

-contains (§~248) just $p_{2}^{m} p_{3}$ operations whose orders are not

+contains (§~248) just $p_{2}^{m} p_{3}$ operations whose orders are not

 divisible by~$p_{1}$; and if there are $p_{3}$~sub-groups of order~$p_{2}^{m}$

 whose operations are all distinct, there are $(p_{2}^{m} - 1)p_{3}$ operations

 whose orders are powers of~$p_{2}$. Hence, in this case, a sub-group

@@ -20585,7 +20591,7 @@ same is true, when $p_{1} = 2$, if $p_{2}$~is not equal to~$3$. We shall
 therefore first deal with a group of order~$p_{1}^{3} p_{2} p_{3}$ on the supposition

 that $p_{2}$~is not~$3$.

 

-If neither $p_{2}$~nor $p_{3}$ divides~$p_{1}^{3} - 1$, it follows from §§~248,~257

+If neither $p_{2}$~nor $p_{3}$ divides~$p_{1}^{3} - 1$, it follows from §§~248,~257

 that there are just $p_{2} p_{3}$ operations in the group whose orders

 divide~$p_{2} p_{3}$. When this is the case, the group clearly cannot

 be simple.

@@ -20634,11 +20640,11 @@ not be simple.
 

 Hence, finally, the only possibility is a group of order~$168$.

 That a simple group of this order actually exists is shewn in

-§~146; also, there is only one type of such group.

+§~146; also, there is only one type of such group.

 

 When $p_{1}$~is equal to~$2$, the only case, that requires discussion

 in addition to those we have dealt with, is $p_{1}^{2} p_{2} p_{3} p_{4}$.

-A group of this order can only be simple (§~249) when it

+A group of this order can only be simple (§~249) when it

 contains a tetrahedral sub-group. In this case, the operations

 of order~$2$ form a single conjugate set, and the group contains

 just $3p_{3} p_{4}$ operations whose orders are divisible by~$2$. If a sub-group

@@ -20673,7 +20679,7 @@ x &= 6,\quad &y &= 14,\quad &p_{3} &= 7,\quad &p_{4} &= 13.
 \end{alignat*}

 

 That simple groups actually exist corresponding to the

-two orders $660$ and $1092$ thus arrived at, is shewn in §~221.

+two orders $660$ and $1092$ thus arrived at, is shewn in §~221.

 There is also in each case a single type; the verification

 of this statement is left to the reader.

 

@@ -20685,8 +20691,8 @@ determination of all possible simple groups. A complete solution
 of this latter problem is not to be expected; but for orders, which

 do not exceed some given limit, the problem may be attacked

 directly. The first determination of this kind was due to

-Herr O.~Hölder\footnotemark,

-\index{Holder@\Topic{Hölder}, quoted}%

+Herr O.~Hölder\footnotemark,

+\index{Holder@\Topic{Hölder}, quoted}%

 \footnotetext{\textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), pp.~55--88.}%

 who examined all possible orders up to~$200$:

 he proved that none of them, except $60$~and $168$, correspond to

@@ -20716,7 +20722,7 @@ continued without substantial difficulty up to~$2000$.
 %% -----File: 389.png---Folio 371-------

 

 We shall here be content with verifying the result of Herr

-Hölder's and Mr~Cole's investigations for orders up to~$660$. The

+Hölder's and Mr~Cole's investigations for orders up to~$660$. The

 method used in dealing with particular numbers may suggest

 to the reader how the determination might be continued.

 

@@ -20724,7 +20730,7 @@ to the reader how the determination might be continued.
 can be the order of a simple group, it must be the product of at

 least $6$~prime factors. Moreover, by previous theorems, we have

 seen that $3^{6}$,~$3^{5}p$ and~$3^{4}p^{2}$ cannot be orders of simple groups.

-Hence certainly no odd number less than $3^{4} · 5 · 7$ or~$2835$\footnote

+Hence certainly no odd number less than $3^{4} · 5 · 7$ or~$2835$\footnote

   {It is not, of course, here suggested that $2835$~is the order of a simple

   group. It may, in fact, be shewn with little difficulty that no number of the

   form $3^{4}pq$,~$3^{3}p^{2}q$, or $3^{2}p^{3}q$, where $p$~and~$q$ are odd primes, can be the order of a

@@ -20742,7 +20748,7 @@ $240$, $336$, $360$, $480$, $504$, $528$, $540$, $560$,~$660$. We will first
 deal with such of these numbers as do not actually correspond

 to simple groups.

 

-\Subsection{$240 = 2^{4} · 3 · 5$.}

+\Subsection{$240 = 2^{4} · 3 · 5$.}

 

 %[** TN: No indentation in the original]

 A simple group of this order would contain $6$~or $16$ sub-groups

@@ -20752,30 +20758,30 @@ and there is no group of degree~$6$ and order~$240$. If there
 were $16$~sub-groups of order~$5$, each would be self-conjugate in

 a group of order~$15$, which must be cyclical. The group then

 would be a doubly transitive group of degree~$16$ and order

-$16 · 15$. Such a group (§~105) contains a self-conjugate sub-group

+$16 · 15$. Such a group (§~105) contains a self-conjugate sub-group

 of order~$16$.

 

-\Subsection{$336 = 2^{4} · 3 · 7$.}

+\Subsection{$336 = 2^{4} · 3 · 7$.}

 

 %[** TN: No indentation in the original]

 A simple group of this order would contain $8$~sub-groups of

 order~$7$, each self-conjugate in a group of order~$42$. We have

-seen (§~146) that there is no simple group of degree~$8$ and order

-$8 · 7 · 6$.

+seen (§~146) that there is no simple group of degree~$8$ and order

+$8 · 7 · 6$.

 \index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|)}%

 %% -----File: 390.png---Folio 372-------

 

-\Subsection{$480 = 2^{5} · 3 · 5$.}

+\Subsection{$480 = 2^{5} · 3 · 5$.}

 

 Since $2^{5}$~is not a factor of~$5!$, a group of order~$480$ must, if

 simple, contain $15$~sub-groups of order~$2^{5}$. Now $2^{5}$~is not congruent

-to unity, $\pmod{4}$, and therefore (§~78) some two sub-groups

+to unity, $\pmod{4}$, and therefore (§~78) some two sub-groups

 of order~$2^{5}$ must have a common sub-group of order~$2^{4}$. Such

-a sub-group, of order~$2^{4}$, must (§~80) either be self-conjugate, or

+a sub-group, of order~$2^{4}$, must (§~80) either be self-conjugate, or

 it must be contained self-conjugately in a sub-group of order

-$2^{5} · 3$~or $2^{5} · 5$. In either case the group is composite.

+$2^{5} · 3$~or $2^{5} · 5$. In either case the group is composite.

 

-\Subsection{$528 = 2^{4} · 3 · 11$.}

+\Subsection{$528 = 2^{4} · 3 · 11$.}

 

 There must be $12$~sub-groups of order~$11$, each being

 self-conjugate in a group of order~$44$. A group of order~$44$

@@ -20783,44 +20789,44 @@ necessarily contains operations of order~$22$, and such an operation
 cannot be represented as a substitution of $12$~symbols.

 The group is therefore not simple.

 

-\Subsection{$540 = 2^{2} · 3^{3} · 5$.}

+\Subsection{$540 = 2^{2} · 3^{3} · 5$.}

 

 There must be $10$~sub-groups of order~$3^{3}$, each self-conjugate

-in a group of order~$2 · 3^{3}$. There must also be $36$~sub-groups of

+in a group of order~$2 · 3^{3}$. There must also be $36$~sub-groups of

 order~$5$, each self-conjugate in a group of order~$15$. Now when

 the group is expressed as transitive in $10$~symbols, the substitutions

 of order~$5$ consist of $2$~cycles of $5$~symbols each, and

 no such substitution can be permutable with a substitution of

 order~$3$. Hence the group is not simple.

 

-\Subsection{$560 = 2^{4} · 5 · 7$.}

+\Subsection{$560 = 2^{4} · 5 · 7$.}

 

 There must be $8$~sub-groups of order~$7$. A transitive group

-of degree~$8$ and order~$8 · 7 · 10$ does not exist (§~146).

+of degree~$8$ and order~$8 · 7 · 10$ does not exist (§~146).

 

 That there are actually simple groups of orders $60$,~$168$, $360$,

 and~$660$, has already been seen; the verification that there is

 only a single type of simple group of order~$360$ may be left

 to the reader. It remains then to consider the order~$504$.

 

-\Subsection{$504 = 2^{3} · 3^{2} · 7$.}

+\Subsection{$504 = 2^{3} · 3^{2} · 7$.}

 

 A simple group of this order must contain $8$~or $36$ sub-groups

-of order~$7$. There is no group of degree~$8$ and order~$504$ (§~146).

+of order~$7$. There is no group of degree~$8$ and order~$504$ (§~146).

 Hence there must be $36$~sub-groups of order~$7$.

 %% -----File: 391.png---Folio 373-------

 

 Again, there must be $7$~or $28$ sub-groups of order~$3^{2}$; and

 since there is no group of degree~$7$ and order~$504$ (\lc), there

 are $28$~sub-groups of order~$3^{2}$. If two of these have a common

-operation~$P$ of order~$3$, it must (§~80) be self-conjugate in a group

+operation~$P$ of order~$3$, it must (§~80) be self-conjugate in a group

 containing more than one sub-group of order~$3^{2}$. If the order

-of this sub-group is~$2^{2} · 3^{2}$, it must contain a sub-group of order

+of this sub-group is~$2^{2} · 3^{2}$, it must contain a sub-group of order

 $2^{2}$ self-conjugately; and therefore this sub-group of order~$2^{2}$ must

-be self-conjugate in a group of order~$2^{3} · 3^{2}$ at least. Such a

+be self-conjugate in a group of order~$2^{3} · 3^{2}$ at least. Such a

 group would be one of $7$~conjugate sub-groups, and this is

 impossible. Similarly, $P$~cannot be self-conjugate in a group

-whose order is greater than~$2^{2} · 3^{2}$. Hence no two sub-groups

+whose order is greater than~$2^{2} · 3^{2}$. Hence no two sub-groups

 of order~$3^{2}$ have a common operation, except identity.

 

 There are therefore in the group $216$~operations of order~$7$,

@@ -20829,7 +20835,7 @@ $64$ other operations.
 

 Suppose now that $A$~is an operation of order~$2$, self-conjugate

 in a sub-group of order~$2^{3}$. If $A$~is self-conjugate in a

-group of order~$2^{3}x$ ($x$~being a factor of~$3^{2} · 7$), this group must

+group of order~$2^{3}x$ ($x$~being a factor of~$3^{2} · 7$), this group must

 contain at least $x$~operations of odd order and therefore at least $x$~operations

 of the form~$AS$, where $S$~is an operation of odd order

 permutable with~$A$. There are also $x$~operations of this form,

@@ -20844,10 +20850,10 @@ no operations whose orders are powers of~$2$ except those of
 the conjugate set to which $A$~belongs. The sub-groups of order~$2^{3}$

 are therefore Abelian groups whose operations are all of order~$2$;

 and since the $7$~operations of order~$2$, in a group of order~$2^{3}$,

-are all conjugate in the group of order~$504$, they must (§~81) be

+are all conjugate in the group of order~$504$, they must (§~81) be

 conjugate in the sub-group within which the group of order~$2^{3}$

 is self-conjugate. Hence, finally, there must be $9$~sub-groups of

-order~$2^{3}$, each self-conjugate in a group of order~$2^{3} · 7$, in which

+order~$2^{3}$, each self-conjugate in a group of order~$2^{3} · 7$, in which

 the $7$~operations of order~$2$ form a single conjugate set.

 %% -----File: 392.png---Folio 374-------

 

@@ -20896,13 +20902,13 @@ brief consideration will shew that all these conditions cannot be
 satisfied, and therefore that no self-conjugate sub-group exists.

 

 That there is a triply transitive group of degree~$9$ and order

-$9 · 8 · 7$, has already been seen in §~113; and the actual formation

+$9 · 8 · 7$, has already been seen in §~113; and the actual formation

 %% -----File: 393.png---Folio 375-------

 of the substitutions generating this group will verify that it

 satisfies the conditions just obtained.

 

 We may however very simply construct the group by the

-method of §~109; and this process has the advantage of shewing

+method of §~109; and this process has the advantage of shewing

 at the same time that there is only one type.

 

 There is no difficulty in constructing the sub-group of order~$56$

@@ -20917,7 +20923,7 @@ contains the transposition~$(89)$, and satisfies the conditions
 \[

 AsA = S_{1},\quad AtA = S_{2}AS_{3},

 \]

-where $S_{1}$,~$S_{2}$,~$S_{3}$ belong to~$\{s, t\}$: then (§~109) it follows that

+where $S_{1}$,~$S_{2}$,~$S_{3}$ belong to~$\{s, t\}$: then (§~109) it follows that

 $\{A, s, t\}$ is a triply transitive group of degree~$9$ and order~$504$.

 

 Now every operation of order~$2$ in the group, which interchanges

@@ -20952,10 +20958,10 @@ number, there is always at least one non-soluble group of order~$Nn$;
 for, $G$~being a simple group of order~$N$ and $H$~any group

 of order~$n$, the direct product of $G$~and~$H$ is a non-soluble group

 %% -----File: 394.png---Folio 376-------

-of order~$Nn$. Herr Hölder\footnote

+of order~$Nn$. Herr Hölder\footnote

   {``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}., (1895),

   pp. 321--422; in particular, p.~420.}

-\index{Holder@\Topic{Hölder}, quoted}%

+\index{Holder@\Topic{Hölder}, quoted}%

 \index{Composite groups@\Topic{Composite groups}!non-soluble|(}%

 has determined all distinct types of

 non-soluble groups whose orders are less than~$480$. He shews

@@ -20976,7 +20982,7 @@ the following table.
 

 For the proof of all these results except the first, and for

 the very interesting and suggestive methods that lead to them,

-the reader is referred to Herr Hölder's memoir. The case

+the reader is referred to Herr Hölder's memoir. The case

 of a non-soluble group of order~$120$ is susceptible of simple

 treatment; and we will consider it here as exemplifying how

 composite groups may be constructed when their factor-groups

@@ -20988,7 +20994,7 @@ are $2$~and~$60$. If these may be taken in either order, then $G$~contains
 a self-conjugate sub-group of order~$2$ and a self-conjugate

 sub-group of order~$60$. The latter, being simple, must be an

 icosahedral group and cannot contain the former. Hence, in this

-case, $G$~must (§~34) be the direct product of an icosahedral group

+case, $G$~must (§~34) be the direct product of an icosahedral group

 and a group of order~$2$.

 

 Next, suppose that the composition-factors can only be taken in

@@ -20996,16 +21002,16 @@ the order $2$,~$60$. Then $G$~contains a self-conjugate sub-group~$H$ of
 icosahedral type and no self-conjugate sub-group of order~$2$. Hence

 if $S$~is any operation of~$G$ which is not contained in~$H$, the

 isomorphism of~$H$ which arises on transforming its operations by~$S$

-must (§~165) be contragredient. It follows that $G$~is simply isomorphic

+must (§~165) be contragredient. It follows that $G$~is simply isomorphic

 with a group of isomorphisms of~$H$. Now $H$~is isomorphic

 with the alternating group of degree five, and its group of isomorphisms

-is therefore (§~173) simply isomorphic with the symmetric

+is therefore (§~173) simply isomorphic with the symmetric

 group of degree five. Hence, in this case, $G$~is simply isomorphic

 with the symmetric group of degree five.

 

 Lastly, suppose that the composition-factors can only be taken

 in the order $60$,~$2$. Then $G$~has a self-conjugate sub-group of order~$2$

-and no self-conjugate sub-group of order~$60$. Hence (§~35) $G$~has

+and no self-conjugate sub-group of order~$60$. Hence (§~35) $G$~has

 %% -----File: 395.png---Folio 377-------

 no sub-group of order~$60$. Suppose that $A$~is the self-conjugate

 operation of order~$2$, and arrange the operations of~$G$ in the sets

@@ -21013,7 +21019,7 @@ operation of order~$2$, and arrange the operations of~$G$ in the sets
 1,\ A;\quad S_{1},\ S_{1}A;\quad S_{2},\ S_{2}A;\ \Dots;\quad S_{59},\ S_{59}A.

 \]

 Since $\dfrac{G}{\{A\}}$ is simply isomorphic with an icosahedral group, it must

-(§~203) be possible to choose three sets

+(§~203) be possible to choose three sets

 \[

 S',\ S'A;\quad S'',\ S''A;\quad S''',\ S'''A;

 \]

@@ -21059,7 +21065,7 @@ products can be formed. It follows that the relations just given are
 the defining relations of a non-soluble group of order~$120$, which has

 a self-conjugate sub-group of order~$2$ and no sub-group of order~$60$;

 and that there is only one type of such group. We have already

-seen (§~221) that there must be at least one such type of group; viz.\

+seen (§~221) that there must be at least one such type of group; viz.\

 the group defined by

 \[

 \begin{aligned}

@@ -21073,14 +21079,14 @@ x_{2}' &\equiv \gamma x_{1} + \delta x_{2}, \\
 

 \begin{Ex} If $p$~is an odd prime, shew that a group, whose composition-factors

 are $60$~and~$p$, must be the direct product of an icosahedral

-group and a group of order~$p$. \Attrib{Hölder.}

+group and a group of order~$p$. \Attrib{Hölder.}

 \end{Ex}

 

 \Par{263.} Let $G$~be a composite group; and suppose that the

 factor-groups of the composition-series of~$G$ are the two non-cyclical

 simple groups $H'$~and~$H$ in the order given. Then

 %% -----File: 396.png---Folio 378-------

-(§~165) if $H_{1}$~is the group formed of all the operations of~$G$

+(§~165) if $H_{1}$~is the group formed of all the operations of~$G$

 which are permutable with every operation of~$H$, the direct

 product $\{H, H_{1}\}$ of $H$~and $H_{1}$ is a self-conjugate sub-group of~$G$.

 Now the composition-factors of $\{H, H_{1}\}$ are composition-factors

@@ -21110,7 +21116,7 @@ groups which are isomorphic with $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.
 If, however, several of these groups are of the same type, the

 inference is no longer necessarily true. Thus if $G$~is the direct

 product of five icosahedral groups, and if $L$~is the group of

-isomorphisms of~$G$, the order of~$\dfrac{L}{G}$ is~$2^{5} · 5!$, and this group

+isomorphisms of~$G$, the order of~$\dfrac{L}{G}$ is~$2^{5} · 5!$, and this group

 contains icosahedral sub-groups. A group of order~$(60)^{6}$, the

 factor-groups of whose composition-series are all of order~$60$,

 is not therefore necessarily the direct product of $6$~icosahedral

@@ -21119,7 +21125,7 @@ groups.
 \index{Composite groups@\Topic{Composite groups}!non-soluble|)}%

 %% -----File: 397.png---Folio 379-------

 

-\Note{\textit{Note to} §~257.}

+\Note{\textit{Note to} §~257.}

 

 \begin{Remark}

 The property stated in the footnote on \PageRef{p.}{366} may be proved as

@@ -21129,7 +21135,7 @@ not permutable with each other; and suppose that
 S_{2}^{-1} S_{1}S_{2} = S_{1}\Sigma.

 \]

 The sub-group of order~$p^{n - 1}$, which contains $S_{1}$ self-conjugately, also

-contains $S_{1}\Sigma$ self-conjugately (§~55); therefore it contains $\Sigma$ self-conjugately.

+contains $S_{1}\Sigma$ self-conjugately (§~55); therefore it contains $\Sigma$ self-conjugately.

 Similarly the sub-group, which contains $S_{2}$ self-conjugately,

 contains $S_{2}\Sigma^{-1}$ and therefore $\Sigma$ self-conjugately. Hence $\Sigma$,

 being contained self-conjugately in two distinct sub-groups of order~$p^{n - 1}$,

@@ -21138,7 +21144,7 @@ $\dfrac{H}{\Gh}$~is Abelian.
 \end{Remark}

 

 

-\Note{\textit{Note to} §~258.}

+\Note{\textit{Note to} §~258.}

 

 \begin{Remark}

 The statement, on \PageRef{p.}{368}, that the congruences $p_{1}^{3} \equiv 1\ (\mod p_{2})$

@@ -21157,7 +21163,7 @@ cannot however be satisfied; the group is therefore composite.
 \end{Remark}

 

 

-\Note{\textit{Note to} §~260.}

+\Note{\textit{Note to} §~260.}

 

 \begin{Remark}

 No simple group of odd order is at present known to exist. An

@@ -21166,7 +21172,7 @@ would undoubtedly lead, whatever the conclusion might be, to
 results of importance; it may be recommended to the reader as

 well worth his attention. Also, there is no known simple group

 whose order contains fewer than three different primes. This

-suggests that Theorems III~and~IV, §§~243,~244, may be capable of

+suggests that Theorems III~and~IV, §§~243,~244, may be capable of

 generalisation. Investigation in this direction is also likely to lead

 to results of interest and importance.

 \end{Remark}

@@ -21195,12 +21201,12 @@ term, in the subjoined table.
 \begin{longtable}{*{2}{>{\hangindent1em}p{0.45\textwidth}}}

 Group & Groupe \\

       & Gruppe \\

-Abelian group & Groupe des opérations échangeables \\

+Abelian group & Groupe des opérations échangeables \\

       & Abel'sche Gruppe \\

-Alternating group & Groupe alterné \\

+Alternating group & Groupe alterné \\

       & Alternierende Gruppe \\

 Complete group & Vollkommene Gruppe \\

-Composite group & Groupe composé \\

+Composite group & Groupe composé \\

       & Zusammengesetzte Gruppe \\

 Factor-group & Groupe facteur \\

       & Factorgruppe \\

@@ -21210,22 +21216,22 @@ Transitive or intransitive group & Groupe transitif ou intransitif \\
       & Transitive oder intransitive Gruppe \\

 Simple group & Groupe simple \\

       & Einfache Gruppe \\

-Soluble group & Groupe résoluble \\

-      & Auflösbare Gruppe \\

+Soluble group & Groupe résoluble \\

+      & Auflösbare Gruppe \\

       & Metacyclische Gruppe\footnotemark\ (\textit{Weber})

 \footnotetext{This term is used by other German writers to denote the holomorph of a

 group of prime order.} \\

-Substitution group & Groupe des substitutions \textit{or} système

-des substitutions conjuguées \\

+Substitution group & Groupe des substitutions \textit{or} système

+des substitutions conjuguées \\

       & Substitutionengruppe \\

-Symmetric group & Groupe symétrique \\

+Symmetric group & Groupe symétrique \\

       & Symmetrische Gruppe \\

 Group of isomorphisms & Gruppe der Isomorphismen \\

 %% -----File: 399.png---Folio 381-------

 Sub-group & Sousgroupe \\

       & Untergruppe \\

       & Theiler (\textit{Weber}) \\

-Conjugate sub-groups & Sousgroupes conjugués \\

+Conjugate sub-groups & Sousgroupes conjugués \\

       & Gleichberechtigte Untergruppen \\

       & Conjugirte Theiler (\textit{Weber}) \\

 Self-conjugate sub-group & Sousgroupe invariant \\

@@ -21233,7 +21239,7 @@ Self-conjugate sub-group & Sousgroupe invariant \\
       & Normaltheiler (\textit{Weber}) \\

 Maximum self-conjugate sub-group & Sousgroupe invariant maximum \\

       & Ausgezeichnete \textit{or} invariante Maximaluntergruppe \\

-      & Grösster Normaltheiler (\textit{Weber}) \\

+      & Grösster Normaltheiler (\textit{Weber}) \\

 Characteristic sub-group & Charakteristische Untergruppe \\

 \hline

 \Strut

@@ -21243,21 +21249,21 @@ Composition-series & Suite de composition \\
 Composition factors & Facteurs de composition \\

       & Factoren der Zusammensetzung \\

 Chief series & Hauptreihe \\

-Characteristic series & Lückenlose Reihe charakteristischer

+Characteristic series & Lückenlose Reihe charakteristischer

   Untergruppen \\

 \hline

 \Strut

-Simple isomorphism & Isomorphisme holoédrique \\

-      & Holoëdrischer \textit{or} einstufiger Isomorphismus \\

-Multiple isomorphism\footnotemark & Isomorphisme meriédrique

+Simple isomorphism & Isomorphisme holoédrique \\

+      & Holoëdrischer \textit{or} einstufiger Isomorphismus \\

+Multiple isomorphism\footnotemark & Isomorphisme meriédrique

 \footnotetext{It is to be noticed that, if $G$~is multiply isomorphic with~$H$, then $H$~is

-  ``meriédriquement isomorphe'' with~$G$.} \\

-      & Meriëdrischer \textit{or} mehrstufiger Isomorphismus \\

+  ``meriédriquement isomorphe'' with~$G$.} \\

+      & Meriëdrischer \textit{or} mehrstufiger Isomorphismus \\

 Isomorphism of a group with itself & Isomorphismus der Gruppe in sich \\

 Cogredient or contragredient isomorphism & Cogredient oder contragredient Isomorphismus \\

 \hline

 \Strut

-Degree (of a group) & Degré \\

+Degree (of a group) & Degré \\

       & Grad \\

 Order \Ditto{(of a} \Ditto{group)} & Ordre \\

       & Ordnung \\

@@ -21270,12 +21276,12 @@ Substitution & Substitution \\
       & Substitution \textit{or} Buchstabenvertauschung \\

 Circular substitution & Substitution circulaire \\

       & Cirkularsubstitution \\

-Even and odd substitutions & Substitutions positives et négatives \\

+Even and odd substitutions & Substitutions positives et négatives \\

       & Gerade und ungerade Substitutionen \\

-Regular substitution & Substitution régulière \\

-      & Reguläre Substitution \\

+Regular substitution & Substitution régulière \\

+      & Reguläre Substitution \\

 Similar substitutions & Substitutions semblables \\

-      & Ähnliche Substitutionen \\

+      & Ähnliche Substitutionen \\

 Transposition & Transposition \\

       & Transposition \\

 \hline

@@ -21560,7 +21566,7 @@ Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ ar
 Heffter@\Topic{Heffter}, quoted#Heffter 215

 

 

-Holder@\Topic{Hölder}, quoted#Hölder 38, 87, 119, 137, 224, 236, 246, 249, 344, 353, 370, 376

+Holder@\Topic{Hölder}, quoted#Hölder 38, 87, 119, 137, 224, 236, 246, 249, 344, 353, 370, 376

 

 

 Holomorph@\Topic{Holomorph}

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-%                                                                         %

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-%                                                                         %

-%                                                                         %

-% Title: Theory of Groups of Finite Order                                 %

-%                                                                         %

-% Author: William Burnside                                                %

-%                                                                         %

-% Release Date: August 2, 2012 [EBook #40395]                             %

-%                                                                         %

-% Language: English                                                       %

-%                                                                         %

-% Character set encoding: ISO-8859-1                                      %

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-%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%

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-%%%% PG BOILERPLATE %%%%

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-\begin{center}

-\begin{minipage}{\textwidth}

-\small

-\begin{PGtext}

-Project Gutenberg's Theory of Groups of Finite Order, by William Burnside

-

-This eBook is for the use of anyone anywhere at no cost and with

-almost no restrictions whatsoever.  You may copy it, give it away or

-re-use it under the terms of the Project Gutenberg License included

-with this eBook or online at www.gutenberg.org

-

-

-Title: Theory of Groups of Finite Order

-

-Author: William Burnside

-

-Release Date: August 2, 2012 [EBook #40395]

-

-Language: English

-

-Character set encoding: ISO-8859-1

-

-*** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF GROUPS OF FINITE ORDER ***

-\end{PGtext}

-\end{minipage}

-\end{center}

-\clearpage

-

-%%%% Credits and transcriber's note %%%%

-\begin{center}

-\begin{minipage}{\textwidth}

-\begin{PGtext}

-Produced by Andrew D. Hwang, Brenda Lewis, and the Online

-Distributed Proofreading Team at http://www.pgdp.net (This

-file was produced from images generously made available

-by The Internet Archive/American Libraries.)

-\end{PGtext}

-\end{minipage}

-\vfill

-\TranscribersNote{\TransNoteText}

-\end{center}

-%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%

-%% -----File: 001.png---Folio i-------

-\FrontMatter

-\cleardoublepage

-\null\vfill

-\begin{center}

-\Huge THEORY OF GROUPS \\[18pt]

-\large OF \\[18pt]

-\Huge FINITE ORDER.

-\end{center}

-\vfill\vfill

-%% -----File: 002.png---Folio ii-------

-\newpage

-\null\vfill

-\begin{center}

-\textgoth{London}: \textsc{C. J. CLAY and SONS}, \\[6pt]

-\small CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, \\[6pt]

-AVE MARIA LANE. \\[6pt]

-\footnotesize \textgoth{Glasgow}: 263, ARGYLE STREET. \\[6pt]

-%[** CUP device]

-\textgoth{Leipzig}: F. A. BROCKHAUS. \\[6pt]

-\textgoth{New York}: THE MACMILLAN COMPANY.

-\end{center}

-\vfill\vfill

-%% -----File: 003.png---Folio iii-------

-\newpage

-\begin{center}

-\Huge THEORY OF GROUPS \\[18pt]

-\large OF \\[18pt]

-\Huge FINITE ORDER

-\vfill

-

-\large BY \\[18pt]

-\Large W. BURNSIDE, M.A., F.R.S., \\[12pt]

-\scriptsize LATE FELLOW OF PEMBROKE COLLEGE, CAMBRIDGE; \\[6pt]

-PROFESSOR OF MATHEMATICS AT THE ROYAL NAVAL COLLEGE, GREENWICH.

-\vfill\vfill

-

-\large

-CAMBRIDGE: \\[6pt]

-AT THE UNIVERSITY PRESS. \\[6pt]

-1897 \\[6pt]

-\footnotesize [\textit{All Rights reserved}.]

-\end{center}

-%% -----File: 004.png---Folio iv-------

-\newpage

-\null\vfill

-\begin{center}

-\textgoth{Cambridge}: \\[6pt]

-\scriptsize PRINTED BY J. AND C. F. CLAY, \\[6pt]

-AT THE UNIVERSITY PRESS.

-\end{center}

-\vfill\vfill

-%% -----File: 005.png---Folio v-------

-

-

-\Section{Preface.}

-

-\First{The} theory of groups of finite order may be said to date

-from the time of Cauchy. To him are due the first

-attempts at classification with a view to forming a theory from

-a number of isolated facts. Galois introduced into the theory

-the exceedingly important idea of a self-conjugate sub-group,

-and the corresponding division of groups into simple and composite.

-Moreover, by shewing that to every equation of finite

-degree there corresponds a group of finite order on which

-all the properties of the equation depend, Galois indicated

-how far reaching the applications of the theory might be, and

-thereby contributed greatly, if indirectly, to its subsequent

-developement.

-

-Many additions were made, mainly by French mathematicians,

-during the middle part of the century. The first

-connected exposition of the theory was given in the third

-edition of M.~Serret's ``\textit{Cours d'Algèbre Supérieure},'' which was

-published in 1866. This was followed in 1870 by M.~Jordan's

-``\textit{Traité des substitutions et des équations algébriques}.'' The

-greater part of M.~Jordan's treatise is devoted to a developement

-of the ideas of Galois and to their application to the

-theory of equations.

-

-No considerable progress in the theory, as apart from its

-applications, was made till the appearance in 1872 of Herr

-Sylow's memoir ``\textit{Théorèmes sur les groupes de substitutions}''

-in the fifth volume of the \textit{Mathematische Annalen}. Since the

-date of this memoir, but more especially in recent years, the

-theory has advanced continuously.

-

-In 1882 appeared Herr Netto's ``\textit{Substitutionentheorie und}

-%% -----File: 006.png---Folio vi-------

-\textit{ihre Anwendungen auf die Algebra},'' in which, as in M.~Serret's

-and M.~Jordan's works, the subject is treated entirely from the

-point of view of groups of substitutions. Last but not least

-among the works which give a detailed account of the subject

-must be mentioned Herr Weber's ``\textit{Lehrbuch der Algebra},'' of

-which the first volume appeared in 1895 and the second in

-1896. In the last section of the first volume some of the more

-important properties of substitution groups are given. In the

-first section of the second volume, however, the subject is

-approached from a more general point of view, and a theory of

-finite groups is developed which is quite independent of any

-special mode of representing them.

-

-The present treatise is intended to introduce to the reader

-the main outlines of the theory of groups of finite order apart

-from any applications. The subject is one which has hitherto

-attracted but little attention in this country; it will afford

-me much satisfaction if, by means of this book, I shall succeed

-in arousing interest among English mathematicians in a branch

-of pure mathematics which becomes the more fascinating the

-more it is studied.

-

-Cayley's dictum that ``a group is defined by means of the

-laws of combination of its symbols'' would imply that, in dealing

-purely with the theory of groups, no more concrete mode of

-representation should be used than is absolutely necessary.

-It may then be asked why, in a book which professes to leave

-all applications on one side, a considerable space is devoted to

-substitution groups; while other particular modes of representation,

-such as groups of linear transformations, are not

-even referred to. My answer to this question is that while, in

-the present state of our knowledge, many results in the pure

-theory are arrived at most readily by dealing with properties

-of substitution groups, it would be difficult to find a result that

-could be most directly obtained by the consideration of groups

-of linear transformations.

-

-The plan of the book is as follows. The first Chapter has

-been devoted to explaining the notation of substitutions. As

-this notation may not improbably be unfamiliar to many

-English readers, some such introduction is necessary to make

-%% -----File: 007.png---Folio vii-------

-the illustrations used in the following chapters intelligible.

-Chapters II to VII deal with the more important properties of

-groups which are independent of any special form of representation.

-The notation and methods of substitution groups

-have been rigorously excluded in the proofs and investigations

-contained in these chapters; for the purposes of illustration,

-however, the notation has been used whenever convenient.

-Chapters VIII to X deal with those properties of groups which

-depend on their representation as substitution groups. Chapter~XI

-treats of the isomorphism of a group with itself. Here,

-though the properties involved are independent of the form of

-representation of the group, the methods of substitution groups

-are partially employed. Graphical modes of representing a

-group are considered in Chapters XII and XIII\@. In Chapter~XIV

-the properties of a class of groups, of great importance in

-analysis, are investigated as a general illustration of the

-foregoing theory. The last Chapter contains a series of results

-in connection with the classification of groups as simple,

-composite, or soluble.

-

-A few illustrative examples have been given throughout

-the book. As far as possible I have selected such examples

-as would serve to complete or continue the discussion in the

-text where they occur.

-

-In addition to the works by Serret, Jordan, Netto and

-Weber already referred to, I have while writing this book

-consulted many original memoirs. Of these I may specially

-mention, as having been of great use to me, two by Herr Dyck

-published in the twentieth and twenty-second volumes of the

-\textit{Mathematische Annalen} with the title ``\textit{Gruppentheoretische

-Studien}''; three by Herr Frobenius in the \textit{Berliner Sitzungsberichte}

-for 1895 with the titles, ``\textit{Ueber endliche Gruppen},''

-``\textit{Ueber auflösbare Gruppen},'' and ``\textit{Verallgemeinerung des

-Sylow'schen Satzes}''; and one by Herr Hölder in the forty-sixth

-volume of the \textit{Mathematische Annalen} with the title

-``\textit{Bildung zusammengesetzter Gruppen}.'' Whenever a result

-is taken from an original memoir I have given a full reference;

-any omission to do so that may possibly occur is due to an

-oversight on my part.

-%% -----File: 008.png---Folio viii-------

-

-To Mr A.~R. Forsyth, Sc.D., F.R.S., Fellow of Trinity

-College, Cambridge, and Sadlerian Professor of Mathematics,

-and to Mr G.~B. Mathews, M.A., F.R.S., late Fellow of St~John's

-College, Cambridge, and formerly Professor of Mathematics in

-the University of North Wales, I am under a debt of gratitude

-for the care and patience with which they have read the proof-sheets.

-Without the assistance they have so generously

-given me, the errors and obscurities, which I can hardly hope

-to have entirely escaped, would have been far more numerous.

-I wish to express my grateful thanks also to Prof.\ O.~Hölder

-of Königsberg who very kindly read and criticized parts of

-the last chapter. Finally I must thank the Syndics of the

-University Press of Cambridge for the assistance they have

-rendered in the publication of the book, and the whole Staff

-of the Press for the painstaking and careful way in which

-the printing has been done.

-

-\Signature{W. BURNSIDE.}{\textit{July}, 1897.}

-%% -----File: 009.png---Folio ix-------

-

-\Section{Contents.}

-

-\ToCChap{CHAPTER I.}{ON SUBSTITUTIONS.}

-

-%§§ PAGE

-

-\ToCSect{1}{Object of the chapter}{1}

-

-\ToCSect{2}{Definition of a substitution}{1}

-

-\ToCSect[3]{6}{Notation for substitutions; cycles; products of substitutions}{1-4}

-

-\ToCSect[7]{8}{Identical substitution; inverse substitutions; order of a substitution}{4-6}

-

-\ToCSect[9]{10}{Circular, regular, similar, and permutable substitutions}{7, 8}

-

-\ToCSect{11}{Transpositions; representation of a substitution as a product

-of transpositions; odd and even substitutions}{9, 10}

-

-

-\ToCChap{CHAPTER II.}{THE DEFINITION OF A GROUP.}

-

-\ToCSect{12}{Definition of a group}{11, 12}

-

-\ToCSect{13}{Identical operation}{12, 13}

-

-\ToCSect{14}{Continuous, discontinuous, and mixed groups}{13, 14}

-

-\ToCSect[15]{16}{Order of an operation; products of operations}{14-16}

-

-\ToCSect{17}{Examples of groups of operations; multiplication table of a

-group}{17-20}

-

-\ToCSect[18]{19}{Generating operations of a group; defining relations; simply

-isomorphic groups}{20-22}

-

-\ToCSect{20}{Dyck's theorem}{22-24}

-

-\ToCSect{21}{Various modes of representing groups}{24}

-%% -----File: 010.png---Folio x-------

-

-

-\ToCChap{CHAPTER III.}{ON THE SIMPLER PROPERTIES OF A GROUP WHICH ARE

-INDEPENDENT OF ITS MODE OF REPRESENTATION.}

-

-% §§ PAGE

-

-\ToCSect[22]{23}{Sub-groups; the order of a sub-group divides the order of

-the group containing it; symbol for a group}{25-27}

-

-\ToCSect{24}{Transforming one operation by another; conjugate operations

-and sub-groups; self-conjugate operations and

-sub-groups; simple and composite groups}{27-29}

-

-\ToCSect[25]{26}{The operations of a group which are permutable with a

-given operation or sub-group form a group; complete

-sets of conjugate operations or sub-groups}{29-33}

-

-\ToCSect{27}{Theorems on self-conjugate sub-groups; maximum sub-groups;

-maximum self-conjugate sub-groups}{33-35}

-

-\ToCSect[28]{31}{Multiply isomorphic groups; factor groups; direct product

-of two groups}{35-40}

-

-\ToCSect{32}{General isomorphism between two groups}{40, 41}

-

-\ToCSect[33]{35}{Permutable groups; Examples}{41-45}

-

-

-\ToCChap{CHAPTER IV.}{ON ABELIAN GROUPS.}

-

-\ToCSect[36]{38}{Sub-groups of Abelian groups; every Abelian group is the

-direct product of Abelian groups whose orders are

-powers of different primes}{46-48}

-

-\ToCSect{39}{Limitation of the discussion to Abelian groups whose orders

-are powers of primes}{48}

-

-\ToCSect[40]{44}{Existence of a set of independent generating operations of

-such a group; invariance of the orders of the generating

-operations; symbol for Abelian group of given

-type}{49-55}

-

-\ToCSect[45]{47}{Determination of all types of sub-groups of a given Abelian

-group}{55-58}

-

-\ToCSect[48]{49}{Properties of an Abelian group of type $(1, 1, \dots, 1)$}{58-60}

-

-\ToCSect{50}{Examples}{60}

-%% -----File: 011.png---Folio xi-------

-

-

-\ToCChap{CHAPTER V.}{ON GROUPS WHOSE ORDERS ARE POWERS OF PRIMES.}

-

-% §§ PAGE

-

-\ToCSect{51}{Object of the chapter}{61}

-

-\ToCSect{52}{Every group whose order is the power of a prime contains

-self-conjugate operations}{61, 62}

-

-\ToCSect[53]{58}{General properties of groups whose orders are powers of

-primes}{62-69}

-

-\ToCSect[59]{61}{The number of sub-groups of order~$p^{s}$ of a group of order~$p^{m}$,

-where $p$~is a prime, is congruent to unity,

-$\pmod{p}$}{69-71}

-

-\ToCSect[62]{63}{Groups of order~$p^{m}$ with a single sub-group of order~$p^{s}$}{71-75}

-

-\ToCSect[64]{67}{Groups of order~$p^{m}$ with a self-conjugate cyclical sub-group

-of order~$p^{m-2}$}{75-81}

-

-\ToCSect{68}{Distinct types of groups of orders $p^{2}$ and~$p^{3}$}{81, 82}

-

-\ToCSect[69]{72}{Distinct types of groups of order~$p^{4}$}{82-86}

-

-\ToCSect[73]{74}{Tables of groups of orders $p^{2}$, $p^{3}$, and~$p^{4}$}{86-89}

-

-\ToCSect{75}{Examples}{89}

-

-

-\ToCChap{CHAPTER VI.}{ON SYLOW'S THEOREM.}

-

-\ToCSect{76}{Object of the chapter}{90, 91}

-

-\ToCSect[77]{78}{Proof of Sylow's theorem}{91-94}

-

-\ToCSect[79]{82}{Direct consequences of Sylow's theorem}{95-100}

-

-\ToCSect[83]{85}{Distinct types of groups of orders~$pq$ ($p$~and $q$ being

-different primes), $24$, and~$60$}{100-108}

-

-\ToCSect{86}{Generalization of Sylow's theorem}{108-110}

-

-\ToCSect{87}{Frobenius's theorem}{110-115}

-

-\ToCSect{88}{Groups with properties analogous to those of groups

-whose orders are powers of primes}{115-117}

-

-

-\ToCChap{CHAPTER VII.}{ON THE COMPOSITION-SERIES OF A GROUP.}

-

-\ToCSect{89}{The composition-series, composition-factors, and factor-groups

-of a given group}{118, 119}

-

-\ToCSect[90]{91}{Invariance of the composition-series of a group}{119-122}

-%% -----File: 012.png---Folio xii-------

-

-\ToCSect{92}{The chief composition-series, or chief series of a group;

-its invariance; construction of a composition-series

-from a chief series}{122, 123}

-

-\ToCSect[93]{95}{Types of the factor-groups of a chief series; minimum

-self-conjugate sub-groups}{124-127}

-

-\ToCSect[96]{97}{Examples of composition-series}{128, 129}

-

-\ToCSect[98]{100}{Soluble groups}{129-132}

-

-\ToCSect{101}{Distinct types of groups of order~$p^{2}q$, where $p$~and $q$

-are different primes}{132-137}

-

-\ToCChap{CHAPTER VIII.}{ON SUBSTITUTION GROUPS: TRANSITIVE AND INTRANSITIVE

-GROUPS.}

-

-\ToCSect{102}{Substitution groups; degree of a group}{138, 139}

-

-\ToCSect{103}{The symmetric and the alternating groups}{139, 140}

-

-\ToCSect{104}{Transitive and intransitive groups; the degree of a

-transitive group is a factor of its order}{140, 141}

-

-\ToCSect{105}{Transitive groups whose substitutions displace all or

-all but one of the symbols}{141-144}

-

-\ToCSect[106]{107}{Self-conjugate operations and sub-groups of transitive

-groups; transitive groups of which the order is

-equal to the degree}{144-148}

-

-\ToCSect[108]{109}{Multiply transitive groups; the order of a $k$-ply transitive

-group of degree~$n$ is divisible by $n(n - 1) \dots

-(n - k + 1)$; construction of multiply transitive

-groups}{148-151}

-

-\ToCSect{110}{Groups of degree~$n$, which do not contain the alternating

-group, cannot be more than $(\frac{1}{3}n + 1)$-ply

-transitive}{151-153}

-

-\ToCSect{111}{The alternating group of degree~$n$ is simple, except

-when $n$ is~$4$}{153, 154}

-

-\ToCSect[112]{113}{Examples of doubly and triply transitive groups}{154-158}

-

-\ToCSect[114]{116}{Properties of intransitive groups}{159-162}

-

-\ToCSect{117}{Intransitive groups of degree~$7$}{163, 164}

-

-\ToCSect[118]{119}{Number of symbols left unchanged by all the substitutions

-of a group is the product of the order of

-the group and the number of the sets in which

-the symbols are interchanged transitively}{165-167}

-

-\ToCLine{Notes to §§~108,~110}{note:1}{167-170}

-%% -----File: 013.png---Folio xiii-------

-

-

-\ToCChap{CHAPTER IX.}{ON SUBSTITUTION GROUPS: PRIMITIVE AND IMPRIMITIVE

-GROUPS.}

-

-% §§ PAGE

-

-\ToCSect{120}{Object of the chapter}{171}

-

-\ToCSect{121}{Imprimitive and primitive groups; imprimitive systems}{171, 172}

-

-\ToCSect[122]{125}{Representation of any group in transitive form; primitivity

-or imprimitivity of the group so represented}{172-179}

-

-\ToCSect{126}{Number of distinct modes of representing the alternating

-group of degree~$5$ in transitive form}{179, 180}

-

-\ToCSect{127}{Imprimitive groups of degree~$6$}{180-183}

-

-\ToCSect[128]{129}{Tests of primitivity: properties of imprimitive systems}{183-186}

-

-\ToCSect{130}{Self-conjugate sub-groups of transitive groups; a self-conjugate

-sub-group of a primitive group must be

-transitive}{186-188}

-

-\ToCSect{131}{Self-conjugate sub-groups of $k$-ply transitive groups are

-in general $(k - 1)$-ply transitive}{188, 189}

-

-\ToCSect[132]{136}{Further properties of self-conjugate sub-groups of primitive

-groups}{190-194}

-

-\ToCSect{137}{Examples}{194, 195}

-

-

-\ToCChap{CHAPTER X.}{ON SUBSTITUTION GROUPS: TRANSITIVITY AND PRIMITIVITY:

-(CONCLUDING PROPERTIES).}

-

-\ToCSect{138}{References to tables of primitive groups}{196}

-

-\ToCSect[139]{141}{Primitive groups with transitive sub-groups of smaller

-degree: limit to the order of a primitive group of

-given degree}{196-199}

-

-\ToCSect{142}{Properties of the symmetric group}{199, 200}

-

-\ToCSect[143]{145}{Further limitations on the orders of primitive groups

-of given degree}{200-204}

-

-\ToCSect{146}{Primitive groups whose degrees do not exceed~$8$}{205-211}

-

-\ToCSect[147]{149}{Sub-groups of doubly transitive groups which leave two

-symbols unchanged; complete sets of triplets}{212-215}

-

-\ToCSect[150]{151}{The most general groups each of whose substitutions

-is permutable with a given substitution, or with

-every substitution of a given group}{215-217}

-

-\ToCSect{152}{Transitive groups whose orders are powers of primes}{218, 219}

-

-\ToCSect{153}{Example}{220}

-%% -----File: 014.png---Folio xiv-------

-

-

-\ToCChap{CHAPTER XI.}{ON THE ISOMORPHISM OF A GROUP WITH ITSELF.}

-

-% §§ PAGE

-

-\ToCSect{154}{Object of the chapter}{221}

-

-\ToCSect[155]{156}{Isomorphism of a group with itself; the group of isomorphisms}{222-224}

-

-\ToCSect{157}{Cogredient and contragredient isomorphisms; the group

-of cogredient isomorphisms is contained self-conjugately

-in the group of isomorphisms}{224-226}

-

-\ToCSect{158}{The holomorph of a group}{226-228}

-

-\ToCSect[159]{161}{Properties of isomorphisms; representation of the group

-of isomorphisms in transitive form}{228-232}

-

-\ToCSect{162}{Characteristic sub-groups; groups with no characteristic

-sub-groups}{232}

-

-\ToCSect[163]{164}{Characteristic series of a group; its invariance; characteristic

-series of a group whose order is the power

-of a prime}{232-235}

-

-\ToCSect[165]{167}{Complete groups}{235-239}

-

-\ToCSect[168]{170}{The group of isomorphisms and the holomorph of a

-cyclical group}{239-242}

-

-\ToCSect[171]{172}{The group of isomorphisms and the holomorph of an

-Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$;

-the homogeneous linear group}{242-245}

-

-\ToCSect{173}{The group of isomorphisms of the alternating group}{245, 246}

-

-\ToCSect{174}{The group of isomorphisms of doubly transitive groups

-of degree~$p^{n} + 1$ and order~$\frac{1}{2} p^{n} (p^{2n} - 1)$}{246-249}

-

-\ToCSect[175]{178}{Further properties of isomorphisms; the symbols $\vartheta (P)$

-and~$\theta (P)$}{249-253}

-

-\ToCSect{179}{Examples}{253, 254}

-

-

-\ToCChap{CHAPTER XII.}{ON THE GRAPHICAL REPRESENTATION OF A GROUP.}

-

-\ToCSect{180}{Groups with an infinite number of operations}{255}

-

-\ToCSect[181]{182}{The most general discontinuous group that can be

-generated by a finite number of operations; relation

-of special groups to the general group}{256-259}

-

-\ToCSect{183}{Graphical representation of a cyclical group}{260-262}

-

-\ToCSect[184]{187}{Graphical representation of a general group}{262-267}

-

-\ToCSect[188]{190}{Graphical representation of a special group}{267-272}

-

-\ToCSect[191]{195}{Graphical representation of groups of finite order}{272-280}

-

-\ToCSect{196}{The genus of a group}{280}

-

-\ToCSect[197]{198}{Limitation on the order and on the number of defining

-relations of a group of given genus}{281-284}

-

-\ToCLine{Note to §~194}{note:3}{284}

-%% -----File: 015.png---Folio xv-------

-

-

-\ToCChap{CHAPTER XIII.}{ON THE GRAPHICAL REPRESENTATION OF GROUPS: GROUPS OF

-GENUS ZERO AND UNITY: CAYLEY'S COLOUR GROUPS.}

-

-% §§ PAGE

-

-\ToCSect[199]{203}{Groups of genus zero; their defining relations and

-graphical representation}{285-292}

-

-\ToCSect[204]{209}{Groups of genus unity; their defining relations and

-graphical representation}{293-302}

-

-\ToCSect{210}{The graphical representation and the defining relations

-of the simple group of order~$168$}{302-305}

-

-\ToCSect[211]{214}{Cayley's colour groups}{306-310}

-

-

-\ToCChap{CHAPTER XIV.}{ON THE LINEAR GROUP.}

-

-\ToCSect{215}{The homogeneous linear group}{311, 312}

-

-\ToCSect[216]{220}{Its composition-series}{312-317}

-

-\ToCSect{221}{The simple group which it defines}{317-319}

-

-\ToCSect[222]{234}{The fractional linear group; determination of the orders

-of its operations and of their distribution in conjugate

-sets; determination of all of its sub-groups

-and of their distribution in conjugate sets; its

-representation as a doubly transitive group}{319-334}

-

-\ToCSect{235}{Generalization of the fractional linear group}{334, 335}

-

-\ToCSect[236]{238}{Representation of the simple group, defined by the

-homogeneous linear group, as a doubly transitive

-group; special cases}{336-340}

-

-\ToCSect[239]{240}{Generalization of the homogeneous linear group}{340-342}

-

-

-\ToCChap{CHAPTER XV.}{ON SOLUBLE AND COMPOSITE GROUPS.}

-

-\ToCSect{241}{Object of the chapter}{343, 344}

-

-\ToCSect{242}{Direct applications of Sylow's theorem often shew that

-a group of given order must be composite}{344, 345}

-

-\ToCSect[243]{245}{Soluble groups whose orders are $p^{\alpha} q^{\beta}$, where $p$ and~$q$

-are primes}{345-352}

-%% -----File: 016.png---Folio xvi-------

-

-\ToCSect{246}{Groups whose sub-groups of order~$p^{\alpha}$ are all cyclical,

-$p^{\alpha}$~being any power of a prime which divides the

-order}{352, 353}

-

-\ToCSect{247}{Groups whose orders contain no square factor}{353, 354}

-

-\ToCSect[248]{249}{Further tests of solubility; groups whose orders contain

-no cube factor}{354-360}

-

-\ToCSect[250]{257}{Groups of even order in which the operations of odd

-order form a self-conjugate sub-group; either $12$,

-$16$, or $56$ must divide the order of a simple group

-if it is even}{360-366}

-

-\ToCSect{258}{The simple groups whose orders contain less than $6$

-prime factors}{367-370}

-

-\ToCSect[259]{260}{The simple groups whose orders do not exceed~$660$}{370-375}

-

-\ToCSect[261]{263}{Non-soluble composite groups}{375-378}

-

-\ToCLine{Notes to §§~257, 258,~260}{note:4}{379}

-

-%[** TN: Empty optional argument signifies unindented entry]

-\ToCLine[]{APPENDIX: On French and German technical terms}{sect:Appendix.}{380-382}

-

-\ToCLine[]{INDEX}{sect:Index.}{383-388}

-

-\iffalse

-%[** TN: Applied in the text; q.v. \Erratum macro]

-\Section{Errata.}

-

-p.~60, line~18, \emph{for} $7$ \emph{read} $56$.

-

-p.~68, line~10, \emph{after} $p^{r+s}$ \emph{add} where $p$ is an odd prime.

-

-p.~150, line~18, \emph{after} $G$ \emph{add} which displaces $a_{1}$.

-

-p.~150, line~26, \emph{after} $S$, \emph{add} either $ASA$ belongs to $G$ or.

-

-p.~151, line~6, \emph{after} form \emph{add} The reduction is here carried out

-on the supposition that $S_{p}$ and $S_{t}$ displace $a_{1}$.

-The modification, when this is not the case, is

-obvious.

-\fi

-%% -----File: 017.png---Folio 1-------

-\MainMatter

-\index{Abelian group@\Topic{Abelian group}|seealso{\Topic{Homogeneous linear group}}}%

-\index{Group@\Topic{Group}|seealso{\Topic{Substitution group}}}%

-\index{Representation@\Topic{Representation}, graphical|see{\Topic{Graphical representation}}}%

-\index{Sub-group@\Topic{Sub-group}|seealso{\Topic{conjugate}, \Topic{self-conjugate} and \Topic{characteristic sub-group}}}%

-\index{Type of a group@\Topic{Type of a group}|see{\Topic{Simply isomorphic groups}}}%

-

-

-\Chapter{I.}{On Substitutions.}

-\index{Substitution@\Topic{Substitution}!definition of}%

-

-\Par{1.} \First{Among} the various notations used in the following

-pages, there is one of such frequent recurrence that a certain

-readiness in its use is very desirable in dealing with the

-subject of this treatise. We therefore propose to devote a

-preliminary chapter to explaining it in some detail.

-

-\Par{2.} Let $a_{1}$, $a_{2}$,~\dots, $a_{n}$ be a set of $n$~distinct letters. The

-operation of replacing each letter of the set by another, which

-may be the same letter or a different one, when carried out

-under the condition that no two letters are replaced by one and

-the same letter, is called a \emph{substitution} performed on the $n$~letters.

-Such a substitution will change any given arrangement

-\[

-a_{1},\ a_{2},\ \dots,\ a_{n}

-\]

-of the $n$~letters into a definite new arrangement

-\[

-b_{1},\ b_{2},\ \dots,\ b_{n}

-\]

-of the same $n$~letters.

-

-\Par{3.} One obvious form in which to write the substitution is

-\[

-\Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}}

-\]

-thereby indicating that each letter in the upper line is to be

-replaced by the letter standing under it in the lower. The

-disadvantage of this form is its unnecessary complexity, each

-%% -----File: 018.png---Folio 2-------

-of the $n$~letters occurring twice in the expression for the

-substitution; by the following process, the expression of the

-substitution may be materially simplified.

-

-Let $p$ be any one of the $n$~letters, and $q$ the letter in the

-lower line standing under $p$ in the upper. Suppose now that $r$

-is the letter in the lower line that stands under $q$ in the upper,

-and so on. Since the number of letters is finite, we must arrive

-at last at a letter~$s$ in the upper line under which $p$ stands.

-If the set of $n$~letters is not thus exhausted, take any letter~$p'$

-in the upper line, which has not yet occurred, and let $q'$, $r'$,~\dots\

-follow it as $q$, $r$,~\dots\ followed~$p$, till we arrive at~$s'$ in the upper

-line with $p'$ standing under it. If the set of $n$~letters is still not

-exhausted, repeat the process, starting with a letter~$p''$ which

-has not yet occurred. Since the number of letters is finite, we

-must in this way at last exhaust them; and the $n$~letters are

-thus distributed into a number of sets

-\[

-\begin{array}{*{5}{l}}

-p, & q, & r, & \dots, & s; \\

-p', & q', & r', & \dots, & s'; \\

-p'', & q'', & r'', & \dots, & s''; \\

-\hdotsfor{5}; \\

-\end{array}

-\]

-such that the substitution replaces each letter of a set by the

-one following it in that set, the last letter of each set being replaced

-by the first of the same set.

-

-If now we represent by the symbol

-\[

-(pqr \dots s)

-\]

-the operation of replacing $p$ by~$q$, $q$ by $r$,~\dots, and $s$ by~$p$, the

-substitution will be completely represented by the symbol

-\[

-(pqr \dots s)(p'q'r' \dots s')(p''q''r'' \dots s'')\Dots\Add{.}

-\]

-The advantage of this mode of expressing the substitution is

-that each of the letters occurs only once in the symbol.

-

-\Par{4.} The separate components of the above symbol, such as

-\index{Substitution@\Topic{Substitution}!cycles of}%

-$(pqr \dots s)$ are called the \emph{cycles} of the substitution. In particular

-cases, one or more of the cycles may contain a single letter;

-when this happens, the letters so occurring singly are unaltered

-by the substitution. The brackets enclosing single letters may

-%% -----File: 019.png---Folio 3-------

-clearly be omitted without risk of ambiguity, as also may the

-unaltered letters themselves. Thus the substitution

-\[

-\Sub{a, b, c, d, e}{c, b, d, a, e}

-\]

-may be written $(acd)(b)(e)$, or $(acd)be$, or simply~$(acd)$. If for

-any reason it were desirable to indicate that substitutions of the

-five letters $a$, $b$, $c$, $d$, $e$ were under consideration, the second of

-these three forms would be used.

-

-\Par{5.} The form thus obtained for a substitution is not unique.

-The symbol $(qr \dots sp)$ clearly represents the same substitution

-as $(pqr \dots s)$, if the letters that occur between $r$ and $s$ in the two

-symbols are the same and occur in the same order; so that, as

-regards the letters inside the bracket, any one may be chosen

-to stand first so long as the cyclical order is preserved unchanged.

-

-Moreover the order in which the brackets are arranged is

-clearly immaterial, since the operation denoted by any one

-bracket has no effect on the letters contained in the other

-brackets. This latter property is characteristic of the particular

-expression that has been obtained for a substitution;

-it depends upon the fact that the expression contains each

-of the letters once only.

-

-\Par{6.} When we proceed to consider the effect of performing

-two or more substitutions successively, it is seen at once that

-the order in which the substitutions are carried out in general

-affects the result. Thus to give a very simple instance, the

-substitution~$(ab)$ followed by~$(ac)$ changes $a$ into~$b$, since $b$~is

-unaltered by the second substitution. Again, $(ab)$~changes $b$

-into~$a$ and $(ac)$~changes $a$ into~$c$, so that the two substitutions

-performed successively change $b$ into~$c$. Lastly, $(ab)$ does not

-affect~$c$ and $(ac)$~changes $c$ into~$a$. Hence the two substitutions

-performed successively change $a$ into~$b$, $b$ into~$c$, $c$ into~$a$, and

-affect no other symbols. The result of the two substitutions

-performed successively is therefore equivalent to the substitution~$(abc)$;

-and it may be similarly shewn that $(ac)$~followed

-by $(ab)$ gives~$(acb)$ as the resulting substitution. To avoid

-ambiguity it is therefore necessary to assign, once for all, the

-%% -----File: 020.png---Folio 4-------

-meaning to be attached to such a symbol as~$s_{1}s_{2}$, where $s_{1}$ and

-$s_{2}$ are the symbols of two given substitutions. We shall always

-understand by the symbol~$s_{1}s_{2}$ \emph{the result of carrying out first

-the substitution~$s_{1}$ and then the substitution~$s_{2}$}. Thus the two

-simple examples given above may be expressed in the form

-\begin{align*}

-(ab) (ac) &= (abc), \\

-(ac) (ab) &= (acb),

-\end{align*}

-the sign of equality being used to represent that the substitutions

-are equivalent to each other.

-

-If now

-\[

-s_{1}s_{2} = s_{4} \quad\text{and}\quad s_{2}s_{3} = s_{5},

-\]

-the symbol $s_{1}s_{2}s_{3}$ may be regarded as the substitution $s_{4}$ followed

-by~$s_{3}$ or as $s_{1}$~followed by~$s_{5}$. But if $s_{1}$ changes \emph{any} letter $a$

-into~$b$, while $s_{2}$ changes $b$ into~$c$ and $s_{3}$ changes $c$ into~$d$, then $s_{4}$

-changes $a$ into~$c$ and $s_{5}$ changes $b$ into~$d$. Hence $s_{4}s_{3}$ and $s_{1}s_{5}$

-both change $a$ into~$d$; and therefore, $a$~being any letter operated

-upon by the substitutions,

-\[

-s_{4}s_{3} = s_{1}s_{5}.

-\]

-

-Hence the meaning of the symbol $s_{1}s_{2}s_{3}$ is definite; it

-\index{Substitution@\Topic{Substitution}!identical}%

-\index{Substitution@\Topic{Substitution}!inverse}%

-\index{Substitution@\Topic{Substitution}!symbol for the product of two or more}%

-\index{Symbol@\Topic{Symbol}!for the product of two or more substitutions}%

-depends only on the component substitutions $s_{1}$, $s_{2}$, $s_{3}$ and their

-sequence, and it is independent of the way in which they are

-associated when their sequence is assigned. And the same

-clearly holds for the symbol representing the successive performance

-of any number of substitutions. To avoid circumlocution,

-it is convenient to speak of the substitution $s_{1}s_{2} \dots s_{n}$

-as the \emph{product} of the substitutions $s_{1}$, $s_{2}$,~\dots, $s_{n}$ in the sequence

-given. The product of a number of substitutions, thus defined,

-always obeys the associative law but does not in general obey

-the commutative law of algebraical multiplication.

-

-\Par{7.} The substitution which replaces every symbol by itself

-is called the \emph{identical substitution}. The \emph{inverse} of a given

-substitution is that substitution which, when performed after

-the given substitution, gives as result the identical substitution.

-Let $s_{-1}$ be the substitution inverse to~$s$, so that, if

-\[

-s = \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}},

-\]

-then

-\[

-s_{-1} = \Sub{b_{1}, b_{2}, \dots, b_{n}}{a_{1}, a_{2}, \dots, a_{n}}.

-\]

-%% -----File: 021.png---Folio 5-------

-Let $s_{0}$ denote the identical substitution which can be represented

-by

-\[

-\Sub{a_{1}, a_{2}, \dots, a_{n}}{a_{1}, a_{2}, \dots, a_{n}}.

-\]

-

-Then

-\[

-ss_{-1} = s_{0}\quad \text{and}\quad s_{-1}s = s_{0},

-\]

-so that $s$ is the substitution inverse to~$s_{-1}$.

-

-Now if

-\[

-ts = t' s,

-\]

-then

-\[

-tss_{-1} = t' ss_{-1},

-\]

-or

-\[

-ts_{0} = t' s_{0}.

-\]

-

-But $ts_{0}$ is the same substitution as~$t$, since $s_{0}$~produces no

-change; and therefore

-\[

-t = t'.

-\]

-

-In exactly the same way, it may be shewn that the relation

-\[

-st = st'

-\]

-involves

-\[

-t = t'.

-\]

-

-\Par{8.} The result of performing $r$~times in succession the same

-substitution~$s$ is represented symbolically by~$s^{r}$. Since, as

-has been seen, products of substitutions obey the associative

-law of multiplication, it follows that

-\[

-s^{\mu} s^{\nu} = s^{\mu + \nu} = s^{\nu} s^{\mu}.

-\]

-

-Now since there are only a finite number of distinct

-substitutions that can be performed on a given finite set of

-symbols, the series of substitutions $s$, $s^{2}$, $s^{3}$,~\dots\ cannot be all

-distinct. Suppose that $s^{m + 1}$~is the first of the series which is

-the same as~$s$, so that

-\[

-s^{m + 1} = s.

-\]

-

-Then

-\[

-s^{m} ss_{-1} = ss_{-1},

-\]

-or

-\[

-s^{m} = s_{0}.

-\]

-

-There is no index~$\mu$ smaller than~$m$ for which this relation

-holds. For if

-\[

-s^{\mu} = s_{0},

-\]

-then

-\[

-s^{\mu + 1} = ss_{0} = s,

-\]

-contrary to the supposition that $s^{m+1}$ is the first of the series

-which is the same as~$s$.

-%% -----File: 022.png---Folio 6-------

-

-Moreover the $m - 1$ substitutions $s$, $s^{2}$,~\dots, $s^{m - 1}$ must be all

-distinct. For if

-\[

-s^{\mu} = s^{\nu},\quad \nu < \mu < m,

-\]

-then

-\[

-s^{\mu - \nu} s^{\nu} (s^{\nu})_{-1} = s^{\nu} (s^{\nu})_{-1},

-\]

-or

-\[

-s^{\mu - \nu} = s_{0},

-\]

-which has just been shewn to be impossible.

-

-The number~$m$ is called the \emph{order} of the substitution~$s$.

-\index{Substitution@\Topic{Substitution}!order of}%

-In connection with the order of a substitution, two properties

-are to be noted. First, if

-\[

-s^{n} = s_{0},

-\]

-it may be shewn at once that $n$~is a multiple of $m$ the order of~$s$;

-and secondly, if

-\[

-s^{\alpha} = s^{\beta},

-\]

-then

-\[

-\alpha - \beta \equiv 0\ (\mod m).

-\]

-

-If now the equation

-\[

-s^{\mu + \nu} = s^{\mu} s^{\nu}

-\]

-be assumed to hold, when either or both of the integers $\mu$ and $\nu$

-is a negative integer, a definite meaning is obtained for the

-symbol~$s^{-r}$, implying the negative power of a substitution; and

-a definite meaning is also obtained for~$s^{0}$. For

-\[

-s^{\mu} s^{-\nu} = s^{\mu - \nu} = s^{\mu - \nu} s^{\nu} (s^{\nu})_{-1} = s^{\mu} (s^{\nu})_{-1},

-\]

-so that

-\[

-s^{-\nu} = (s^{\nu})_{-1}.

-\]

-Similarly it can be shewn that

-\[

-s^{0} = s_{0}.

-\]

-

-Since every power of~$s_{0}$ is the same as~$s_{0}$, and since

-wherever $s_{0}$ occurs in the symbol $s_{1} s_{2} \dots s_{n}$ of a compound

-substitution it may be omitted without affecting the result, it

-is clear that no ambiguity will result from replacing $s_{0}$ everywhere

-by~$1$; in other words, we may use~$1$ to represent the

-identical substitution which leaves every letter unchanged.

-But when this is done, it must of course be remembered that

-the equation

-\[

-s^{m} = 1

-\]

-is not a reducible algebraical equation, which is capable of being

-written in the form

-\[

-(s - 1) (s^{m - 1} + s^{m - 2} + \dots + 1) = 0.

-\]

-%% -----File: 023.png---Folio 7-------

-

-Indeed the symbol $s + s'$, where $s$ and $s'$ are any two substitutions,

-has no meaning.

-

-\Par{9.} If the cycles of a substitution

-\[

-s = (pqr\dots s)(p'q'\dots s')(p''q''\dots s'')\dots

-\]

-contain $m$, $m'$, $m''$,~\dots\ letters respectively, and if

-\[

-s^{\mu} = 1,

-\]

-$\mu$ must be a common multiple of $m$, $m'$, $m''$,~\dots. For $s^{\mu}$ changes~$p$

-into a letter $\mu$~places from it in the cyclical set $p$,~$q$, $r$,~\dots,~$s$;

-and therefore, if it changes $p$ into itself, $\mu$~must be a multiple

-of~$m$. In the same way, it must be a multiple of $m'$, $m''$,~\dots.

-Hence the order of~$s$ is the least common multiple of $m$, $m'$,

-$m''$,~\dots.

-

-In particular, when a substitution consists of a single

-\index{Substitution@\Topic{Substitution}!circular}%

-\index{Substitution@\Topic{Substitution}!regular}%

-\index{Substitution@\Topic{Substitution}!similar}%

-cycle, its order is equal to the number of letters which it interchanges.

-Such a substitution is called a \emph{circular substitution}.

-

-A substitution, all of whose cycles contain the same

-number of letters, is said to be \emph{regular} in the letters which it

-interchanges; the order of such a substitution is clearly equal

-to the number of letters in one of its cycles.

-

-\Par{10.} Two substitutions, which contain the same number of

-cycles and the same number of letters in corresponding cycles,

-are called \emph{similar}. If $s$,~$s'$ are similar substitutions, so also

-clearly are $s^{r}$,~$s'^{r}$; and the orders of $s$~and $s'$ are the same.

-

-Let now

-\[

-s = (a_{p}a_{q}\dots a_{s})(a_{p'}a_{q'}\dots a_{s'})\dots

-\]

-and

-\[

-t = \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}}

-\]

-be any two substitutions. Then

-\begin{align*}

-t^{-1}st &= \Sub{b_{1}, b_{2}, \dots, b_{n}}{a_{1}, a_{2}, \dots, a_{n}}

-  (a_{p}a_{q}\dots a_{s})(a_{p'}a_{q'}\dots a_{s'}) \dots

-  \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}} \\

-  &= (b_{p}b_{q}\dots b_{s})(b_{p'}b_{q'}\dots b_{s'})\dots,

-\end{align*}

-the latter form of the substitution being obtained by actually

-carrying out the component substitutions of the earlier form.

-Hence $s$ and $t^{-1}st$ are similar substitutions.

-%% -----File: 024.png---Folio 8-------

-

-Since

-\[

-s_{2}s_{1} = s_{1}^{-1}s_{1}s_{2}s_{1},

-\]

-it follows that $s_{1}s_{2}$ and $s_{2}s_{1}$ are similar substitutions and therefore

-that they are of the same order. Similarly it may be shewn

-that $s_{1}s_{2}s_{3} \dots s_{n}$, $s_{2}s_{3} \dots s_{n}s_{1}$,~\dots, $s_{n}s_{1} \dots s_{2}s_{3}$ are all similar substitutions.

-

-It may happen in particular cases that $s$ and $t^{-1}st$ are the

-\index{Substitution@\Topic{Substitution}!permutable}%

-same substitution. When this is so, $t$~and $s$ are \emph{permutable},

-that is, $st$ and $ts$ are equivalent to one another; for if

-\[

-s = t^{-1}st,

-\]

-then

-\[

-ts = st.

-\]

-

-This will certainly be the case when none of the symbols

-that are interchanged by~$t$ are altered by~$s$; but it may happen

-when $s$ and $t$ operate on the same symbols. Thus if

-\[

-s = (ab)(cd), \quad t = (ac)(bd),

-\]

-then

-\[

-st = (ad)(bc) = ts.

-\]

-

-\begin{Ex}[1.] Shew that every regular substitution is some power of a

-circular substitution.

-\end{Ex}

-

-\begin{Ex}[2.] If $s$,~$s'$ are permutable regular substitutions of the same

-$mn$~letters of orders $m$ and~$n$, these numbers being relatively prime,

-shew that $ss'$~is a circular substitution in the $mn$~letters.

-\end{Ex}

-

-\begin{Ex}[3\protect\footnotemark.]

-\footnotetext{It is often convenient to use digits rather than letters for the purpose of

-  illustration.}

-If

-\begin{alignat*}{3}

-&s     &&= (123) &&(456)(789), \\

-&s_{1} &&= (147) &&(258)(369), \\

-&s_{2} &&=       &&(456)(798),

-\end{alignat*}

-shew that $s$~is permutable with both $s_{1}$ and~$s_{2}$, and that it can be

-formed by a combination of $s_{1}$ and~$s_{2}$.

-\end{Ex}

-

-\begin{Ex}[4.] Shew that the only substitutions of $n$ given letters

-which are permutable with a circular substitution of the $n$~letters

-are the powers of the circular substitution.

-\end{Ex}

-

-\begin{Ex}[5.] Determine all the substitutions of the ten symbols

-involved in

-\[

-s = (abcde)(\alpha\beta\gamma\delta\epsilon)

-\]

-which are permutable with~$s$.

-\end{Ex}

-

-The determination of all the substitutions which are

-permutable with a given substitution will form the subject

-of investigation in Chapter~X\@.

-%% -----File: 025.png---Folio 9-------

-

-\Par{11.} A circular substitution of order two is called a \emph{transposition}.

-\index{Transpositions@\Topic{Transpositions}!definition of}%

-\index{Transpositions@\Topic{Transpositions}!representation of a substitution by means of}%

-It may be easily verified that

-\[

-(pqr\dots s) = (pq)(pr)\dots(ps),

-\]

-so that every circular substitution can be represented as a

-product of transpositions; and thence, since every substitution

-is the product of a number of circular substitutions, every

-substitution can be represented as a product of transpositions.

-It must be remembered, however, that, in general, when a

-substitution is represented in this way, some of the letters will

-occur more than once in the symbol, so that the order in which

-the constituent transpositions occur is essential. There is thus

-a fundamental difference from the case when the symbol of a

-substitution is the product of circular substitutions, no two of

-which contain a common letter.

-

-Since

-\[

-(p'q') = (pp')(pq')(pp'),

-\]

-every transposition, and therefore every substitution of $n$~letters,

-can be expressed in terms of the $n - 1$ transpositions

-\[

-(a_{1}a_{2}),\ (a_{1}a_{3}),\ \dots,\ (a_{1}a_{n}).

-\]

-

-The number of different ways in which a given substitution

-may be represented as a product of transpositions is evidently

-unlimited; but it may be shewn that, however the representation

-is effected, the number of transpositions is either always even

-or always odd. To prove this, it is sufficient to consider the

-effect of a transposition on the square root of the discriminant

-of the $n$~letters, which may be written

-\[

-D = \prod_{r = 1}^{r = n - 1} \left\{\prod_{s = r + 1}^{s = n} (a_{r} - a_{s})\right\}.

-\]

-

-The transposition $(a_{r}a_{s})$ changes the sign of the factor

-$a_{r} - a_{s}$. When $q$~is less than either $r$~or~$s$, the transposition

-interchanges the factors $a_{q} - a_{r}$ and $a_{q} - a_{s}$; and when $q$~is

-greater than either $r$~or~$s$, it interchanges the factors $a_{r} - a_{q}$ and

-$a_{s} - a_{q}$. When $q$~lies between $r$ and~$s$, the pair of factors

-$a_{r} - a_{q}$ and $a_{q} - a_{s}$ are interchanged and are both changed in

-sign. Hence the effect of the single transposition on~$D$ is

-to change its sign. Since any substitution can be expressed as

-the product of a number of transpositions, the effect of any

-substitution on~$D$ must be either to leave it unaltered or to

-%% -----File: 026.png---Folio 10-------

-change its sign. If a substitution leaves $D$ unaltered it must,

-when expressed as a product of transpositions in any way,

-contain an even number of transpositions; and if it changes

-the sign of~$D$, every representation of it, as a product of

-transpositions, must contain an odd number of transpositions.

-Hence no substitution is capable of being expressed both by an

-even and by an odd number of transpositions.

-

-A substitution is spoken of as \emph{odd} or \emph{even}, according as the

-\index{Substitution@\Topic{Substitution}!even and odd}%

-\index{Transpositions@\Topic{Transpositions}!number of, which enter in the representation of a substitution is either always even or always odd}%

-transpositions which enter into its representation are odd or

-even in number.

-

-Further, an even substitution can always be represented as

-a product of circular substitutions of order three. For any

-even substitution of $n$~letters can be represented as the product

-of an even number of the $n - 1$ transpositions

-\[

-(a_{1}a_{2}),\ (a_{1}a_{3}),\ \dots,\ (a_{1}a_{n}),

-\]

-in appropriate sequence and with the proper number of

-occurrences; and the product of any consecutive pair of these

-$(a_{1}a_{r})(a_{1}a_{s})$ is the circular substitution $(a_{1}a_{r}a_{s})$.

-

-Now

-\begin{align*} %[** TN: Set on three lines in the original]

-(a_{1}a_{2}a_{s})(a_{1}a_{2}a_{r})(a_{1}a_{2}a_{s})^{2}

-  &= (a_{1}a_{2}a_{s})(a_{1}a_{2}a_{r})(a_{1}a_{s}a_{2}) \\

-  &= (a_{1}a_{r}a_{s}),

-\end{align*}

-so that every circular substitution of order three displacing~$a_{1}$,

-and therefore every even substitution of $n$~letters, can be

-expressed in terms of the $n - 2$ substitutions

-\[

-(a_{1}a_{2}a_{3}),\ (a_{1}a_{2}a_{4}),\ \dots,\ (a_{1}a_{2}a_{n})

-\]

-and their powers.

-

-\begin{Ex}[1.] Shew that every even substitution of $n$~letters can be

-expressed in terms of

-\[

-(a_{1}a_{2}a_{3}),\ (a_{1}a_{4}a_{5}),\ \Dots,\ (a_{1}a_{n-1}a_{n}),

-\]

-when $n$~is odd; and in terms of

-\[

-(a_{1}a_{2}a_{3}),\ (a_{1}a_{4}a_{5}),\ \Dots,\ (a_{1}a_{n - 2}a_{n - 1}),\ (a_{1}a_{2}a_{n}),

-\]

-when $n$~is even.

-\end{Ex}

-

-\begin{Ex}[2.] If $n + 1$~is odd, shew that every even substitution of

-$mn + 1$ letters can be expressed in terms of

-\[

-(a_{1}a_{2}\Dots a_{n + 1}),\ (a_{1}a_{n + 2}\Dots a_{2n + 1}),\ \Dots,\ (a_{1}a_{(m - 1)n + 2}\Dots a_{mn + 1});

-\]

-and if $n + 1$~is even, that every substitution of $mn + 1$ letters can

-be expressed in terms of this set of $m$~circular substitutions.

-\end{Ex}

-%% -----File: 027.png---Folio 11-------

-

-

-\Chapter{II.}{The Definition of a Group.}

-\index{Group@\Topic{Group}!definition of}%

-

-\Par{12.} \First{In} the present chapter we shall enter on our main

-subject and we shall begin with definitions, explanations and

-examples of what is meant by a group.

-

-\begin{Definition}

-Let

-\[

-A,\ B,\ C,\ \dots

-\]

-represent a set of operations, which can be performed on the

-same object or set of objects. Suppose this set of operations

-has the following characteristics.

-

-\Inum{($\alpha$)} The operations of the set are all distinct, so that no

-two of them produce the same change in every possible application.

-

-\Inum{($\beta$)} The result of performing successively any number of

-operations of the set, say $A$, $B$,~\dots, $K$, is another definite

-operation of the set, which depends only on the component

-operations and the sequence in which they are carried out, and

-not on the way in which they may be regarded as associated.

-Thus $A$~followed by~$B$ and $B$~followed by~$C$ are operations of

-the set, say $D$~and~$E$; and $D$~followed by~$C$ is the same operation

-as $A$~followed by~$E$.

-

-\Inum{($\gamma$)} $A$ being any operation of the set, there is always

-another operation~$A_{-1}$ belonging to the set, such that $A$

-followed by~$A_{-1}$ produces no change in any object.

-%% -----File: 028.png---Folio 12-------

-

-The operation~$A_{-1}$ is called the inverse of~$A$.

-

-\index{Group@\Topic{Group}!definition of}%

-The set of operations is then said to form a \emph{Group}.

-\end{Definition}

-

-From the definition of the inverse of~$A$ given in~\Inum{($\gamma$)}, it

-follows directly that $A$~is the inverse of~$A_{-1}$. For if $A$~changes

-any object~$\Omega$ into~$\Omega'$, $A_{-1}$~must change $\Omega'$ into~$\Omega$. Hence $A_{-1}$

-followed by~$A$ leaves $\Omega'$, and therefore every object, unchanged.

-

-The operation resulting from the successive performance of

-the operations $A$, $B$,~\dots, $K$ in the sequence given is denoted by

-the symbol $AB \dots K$; and if $\Omega$~is any object on which the

-operations may be performed, the result of carrying out this

-compound operation on~$\Omega$ is denoted by $\Omega · AB \dots K$.

-

-If the component operations are all the same, say~$A$, and $r$~in

-number, the abbreviation~$A^{r}$ will be used for the resultant

-operation, and it will be called the $r$th~\emph{power} of~$A$.

-

-\begin{Definition}

-Two operations, $A$~and~$B$, are said to be

-\index{Permutable operations@\Topic{Permutable operations}!definition of}%

-\emph{permutable} when $AB$ and $BA$ are the same operation.

-\end{Definition}

-

-\Par{13.} If $AB$ and $AC$ are the same operation, so also are

-\index{Symbol@\Topic{Symbol}!for the product of two or more operations}%

-$A_{-1}AB$ and~$A_{-1}AC$. But the operation~$A_{-1}A$ produces no

-change in any object and therefore $A_{-1}AB$ and~$B$, producing

-the same change in every object, are the same operation. Hence

-$B$~and~$C$ are the same operation.

-

-This is expressed symbolically by saying that, if

-\[

-AB = AC,

-\]

-then

-\[

-B = C;

-\]

-the sign of equality being used to imply that the symbols

-represent the same operation.

-

-In a similar way, if

-\[

-BA = CA,

-\]

-it follows that

-\[

-B = C.

-\]

-

-From conditions \Inum{($\beta$)} and \Inum{($\gamma$)}, $AA_{-1}$~must be a definite

-operation of the group. This operation, by definition, produces

-no change in any possible object, and it must, by

-\index{Identical operation@\Topic{Identical operation}!definition of}%

-condition~\Inum{($\alpha$)}, be unique. It is called the \emph{identical} operation.

-%% -----File: 029.png---Folio 13-------

-If it is represented by~$A_{0}$ and if $A$~be any other operation,

-then

-\[

-A_{0} A = A = AA_{0},

-\]

-and for every integer~$r$,

-\[

-A_{0}^{r} = A_{0}.

-\]

-

-Hence $A_{0}$ may, without ambiguity, be replaced by~$1$,

-wherever it occurs.

-

-\Par{14.} The number of distinct operations contained in a

-\index{Group@\Topic{Group}!continuous, discontinuous, or mixed}%

-group may be either finite or infinite. When the number is

-infinite, the group may contain operations which produce an

-infinitesimal change in every possible object or operand.

-

-Thus the totality of distinct displacements of a rigid body

-evidently forms a group, for they satisfy conditions \Inum{($\alpha$)},~\Inum{($\beta$)} and~\Inum{($\gamma$)}

-of the definition. Moreover this group contains operations

-of the kind in question, namely infinitesimal twists; and

-each operation of the group can be constructed by the continual

-repetition of a suitably chosen infinitesimal twist.

-

-Next, the set of translations, that arise by shifting a cube

-parallel to its edges through distances which are any multiples

-of an edge, forms a group containing an infinite number

-of operations; but this group contains no operation which

-effects an infinitesimal change in the position of the cube.

-

-As a third example, consider the set of displacements by

-which a complete right circular cone is brought to coincidence

-with itself. It consists of rotations through any angle about

-the axis of the cone, and rotations through two right angles

-about any line through the vertex at right angles to the

-axis. Once again this set of displacements satisfies the conditions

-\Inum{($\alpha$)},~\Inum{($\beta$)} and~\Inum{($\gamma$)} of the definition and forms a group.

-

-This last group contains infinitesimal operations, namely

-rotations round the axis through an infinitesimal angle; and

-every finite rotation round the axis can be formed by the

-continued repetition of an infinitesimal rotation. There is

-however in this case no infinitesimal displacement of the group

-by whose continued repetition a rotation through two right

-angles about a line through the vertex at right angles to the

-axis can be constructed. Of these three groups with an infinite

-%% -----File: 030.png---Folio 14-------

-number of operations, the first is said to be a \emph{continuous} group,

-\index{Group@\Topic{Group}!continuous, discontinuous, or mixed}%

-the second a \emph{discontinuous} group, and the third a \emph{mixed} group.

-

-Continuous groups and mixed groups lie entirely outside the

-plan of the present treatise; and though, later on, some of the

-properties of discontinuous groups with an infinite number of

-operations will be considered, such groups will be approached

-from a point of view suggested by the treatment of groups

-containing a finite number of operations. It is not therefore

-necessary here to deal in detail with the classification of infinite

-groups which is indicated by the three examples given above;

-and we pass on at once to the case of groups which contain a

-finite number only of distinct operations.

-

-\Par{15.} \begin{Definition}

-If the number of distinct operations contained

-\index{Group@\Topic{Group}!order of}%

-\index{Order@\Topic{Order}!of a group, definition of}%

-in a group be finite, the number is called the \emph{order} of the

-group.

-\end{Definition}

-

-Let $S$ be an operation of a group of finite order~$N$. Then

-the infinite series of operations

-\[

-S,\ S^{2},\ S^{3},\ \Dots

-\]

-must all be contained in the group, and therefore a finite

-number of them only can be distinct. If $S^{m + 1}$~is the first of

-the series which is the same as~$S$, and if $S_{-1}$~is the operation

-inverse to~$S$, then

-\[

-S^{m + 1} S_{-1} = SS_{-1} = 1,

-\]

-or

-\[

-S^{m} = 1.

-\]

-

-Exactly as in §~8, it may be shewn that, if

-\[

-S^{\mu} = 1,

-\]

-$\mu$ must be a multiple of~$m$, and that the operations $S$, $S^{2}$,~\Dots,

-$S^{m - 1}$ are all distinct.

-

-Since the group contains only $N$~distinct operations, $m$~must

-be equal to or less than~$N$. It will be seen later that, if

-$m$~is less than~$N$, it must be a factor of~$N$.

-

-The integer $m$ is called the \emph{order} of the operation~$S$. The

-\index{Order@\Topic{Order}!of an operation, definition of}%

-order~$m'$ of the operation~$S^{x}$ is the least integer for which

-\[

-S^{xm'} = 1,

-\]

-that is, for which

-\[

-xm' \equiv 0\ (\mod m).

-\]

-%% -----File: 031.png---Folio 15-------

-

-Hence, if $g$ is the greatest common factor of $x$ and~$m$,

-\[

-m' = \frac{m}{g};

-\]

-and, if $m$~is prime, all the powers of~$S$, whose indices are less

-than~$m$, are of order~$m$.

-

-Since

-\[

-S^{x} S^{m-x} = S^{m} = 1,\quad (x < m),

-\]

-and

-\[

-S^{x} (S^{x})_{-1} = 1,

-\]

-it follows that

-\[

-(S^{x})_{-1} = S^{m-x}.

-\]

-

-If now a meaning be attached to~$S^{-x}$, by assuming that

-the equation

-\[

-S^{x+y} = S^{x} S^{y}

-\]

-holds when either $x$ or $y$ is a negative integer, then

-\[

-S^{m-x} = S^{m} S^{-x} = S^{-x},

-\]

-and

-\[

-(S^{x})_{-1} = S^{-x},

-\]

-so that $S^{-x}$ denotes the inverse of the operation~$S^{x}$.

-

-\begin{Ex} If $S_{a}$, $S_{b}$,~\Dots, $S_{c}$, $S_{d}$ are operations of a group, shew that

-the operation inverse to $S_{a}^{\alpha} S_{b}^{\beta} \Dots S_{c}^{\gamma} S_{d}^{\delta}$ is $S_{d}^{-\delta} S_{c}^{-\gamma} \Dots S_{b}^{-\beta} S_{a}^{-\alpha}$.

-\end{Ex}

-

-\Par{16.} If

-\[

-1,\ S_{1},\ S_{2},\ \Dots,\ S_{N-1}

-\]

-are the $N$~operations of a group of order~$N$, the set of $N$~operations

-\[

-S_{r},\ S_{r}S_{1},\ S_{r}S_{2},\ \Dots,\ S_{r}S_{N-1}

-\]

-are (§~13) all distinct; and their number is equal to the order of

-the group. Hence every operation of the group occurs once

-and only once in this set.

-

-Similarly every operation of the group occurs once and only

-once in the set

-\[

-S_{r},\ S_{1}S_{r},\ S_{2}S_{r},\ \Dots,\ S_{N-1}S_{r}.

-\]

-

-Every operation of the group can therefore be represented

-as the product of two operations of the group, and either the

-first factor or the second factor can be chosen at will.

-

-A relation of the form

-\[

-S_{p} = S_{q} S_{r}

-\]

-between three operations of the group will not in general

-involve any necessary relation between the order of~$S_{p}$ and the

-%% -----File: 032.png---Folio 16-------

-orders of $S_{q}$ and~$S_{r}$. If however the two latter are permutable,

-the relation requires that, for all values of~$x$,

-\[

-S_{p}^{x} = S_{q}^{x} S_{r}^{x};

-\]

-and in that case the order of~$S_{p}$ is the least common multiple

-of the orders of $S_{q}$ and~$S_{r}$.

-

-Suppose now that $S$, an operation of the group, is of order~$mn$,

-where $m$~and~$n$ are relatively prime. Then we may shew

-that, of the various ways in which $S$~may be represented as the

-product of two operations of the group, there is just one in

-which the operations are permutable and of orders $m$ and $n$

-respectively.

-

-Thus let

-\[

-S^{n} = M,

-\]

-and

-\[

-S^{m} = N,

-\]

-so that $M$,~$N$ are operations of orders $m$ and~$n$. Since $S^{m}$ and

-$S^{n}$ are permutable, so also are $M$ and~$N$, and powers of $M$ and~$N$.

-

-If $x_{0}$,~$y_{0}$ are integers satisfying the equation

-\[

-xn + ym = 1,

-\]

-every other integral solution is given by

-\[

-x = x_{0} + tm,\quad y = y_{0} - tn,

-\]

-where $t$ is an integer.

-

-Now

-\[

-M^{x} N^{y} = S^{xn + ym} = S;

-\]

-and since $x$~and~$m$ are relatively prime, as also are $y$~and~$n$,

-$M^{x}$~and $N^{y}$ are permutable operations of orders $m$~and~$n$, so

-that $S$~is expressed in the desired form.

-

-Moreover, it is the only expression of this form; for let

-\[

-S = M_{1} N_{1},

-\]

-where $M_{1}$ and $N_{1}$ are permutable and of orders $m$ and~$n$.

-

-Then $S^{n} = M_{1}^{n}$, since $N_{1}^{n} = 1$.

-

-Hence

-\[

-M_{1}^{n} = M,

-\]

-or

-\[

-M_{1}^{xn} = M^{x},

-\]

-or

-\[

-M_{1}^{1-ym} = M^{x}.

-\]

-But $M_{1}^{m} = 1$, and therefore $M_{1}^{-ym} = 1$; hence

-\[

-M_{1} = M^{x}.

-\]

-In the same way it is shewn that $N_{1}$ is the same as~$N^{y}$. The

-representation of~$S$ in the desired form is therefore unique.

-%% -----File: 033.png---Folio 17-------

-

-\Par{17.} Two given operations of a group successively performed

-give rise to a third operation of the group which, when the

-operations are of known concrete form, may be determined by

-actually carrying out the two given operations. Thus the set

-of finite rotations, which bring a regular solid to coincidence

-with itself, evidently form a group; and it is a purely geometrical

-problem to determine that particular rotation of the

-group which arises from the successive performance of two

-given rotations of the group.

-

-When the operations are represented by symbols, the relation

-in question is represented by an equation of the form

-\[

-AB = C;

-\]

-but the equation indicates nothing of the nature of the actual

-operations. Now it may happen, when the operations of two

-groups of equal order are represented by symbols,

-\begin{alignat*}{2}

-&\Inum{(i)}       && 1,\ A,\ B,\ C, \Dots \\

-&\Inum{(ii)}\quad && 1,\ A',\ B',\ C', \Dots

-\end{alignat*}

-that, to every relation of the form

-\[

-AB = C

-\]

-between operations of the first group, there corresponds the

-relation

-\[

-A'B' = C'

-\]

-between operations of the second group. In such a case,

-although the nature of the actual operations in the first group

-may be entirely different from the nature of those in the

-second, the laws according to which the operations of each

-group combine among themselves are identical. The following

-series of groups of operations, of order six, will at once illustrate

-the possibility just mentioned, and will serve as concrete examples

-to familiarize the reader with the conception of a group of

-operations.

-

-\Example{\Inum{I.} Group of inversions.} Let $P$, $Q$, $R$ be three circles with

-a common radical axis and let each pair of them intersect at an

-angle~$\frac{1}{3} \pi$. Denote the operations of inversion with respect to

-$P$,~$Q$,~$R$ by $C$,~$D$,~$E$; and denote successive inversions at $P$,~$R$

-and at $P$,~$Q$ by $A$~and~$B$. The object of operation may be any

-point in the plane of the circles, except the two common points

-%% -----File: 034.png---Folio 18-------

-in which they intersect. Then it is easy to verify, from the

-geometrical properties of inversion, that the operations

-\[

-1,\ A,\ B,\ C,\ D,\ E

-\]

-are all distinct, and that they form a group. For instance, $DE$~represents

-successive inversions at $Q$~and~$R$. But successive

-inversions at $Q$ and $R$ produce the same displacement of points

-as successive inversions at $P$ and~$Q$, and therefore

-\[

-DE = B.

-\]

-

-\Example{\Inum{II.} Group of rotations.} Let $POP'$, $QOQ'$, $ROR'$ be three

-concurrent lines in a plane such that each of the angles $POQ$

-and $QOR$ is~$\frac{1}{3} \pi$, and let $IOI'$ be a perpendicular to their plane.

-Denote by~$A$ a rotation round~$II'$ through~$\frac{2}{3} \pi$ bringing $PP'$ to~$RR'$;

-and by $B$ a rotation round~$II'$ through~$\frac{4}{3} \pi$ bringing $PP'$

-to~$Q'Q$. Denote also by $C$,~$D$,~$E$ rotations through two right

-angles round $PP'$, $QQ'$,~$RR'$. The object of the rotations may

-be any point or set of points in space. Then it may again be

-verified, by simple geometrical considerations, that the operations

-\[

-1,\ A,\ B,\ C,\ D,\ E

-\]

-are distinct and that they form a group.

-

-\Example{\Inum{III.} Group of linear transformations of a single variable.}

-The operation of replacing $x$ by a given function~$f(x)$ of itself

-is sometimes represented by the symbol $\bigl(x, f(x)\bigr)$. With this

-notation, if

-\begin{alignat*}{4}

- A &= \left(x, \frac{1}{1 - x} \right),\quad

-&B &= \left(x, \frac{x - 1}{x} \right),\quad

-&C &= \left(x, \frac{1}{x} \right),\quad

-&D &= (x, 1 - x), \\

-%

-&&

- E &= \left(x, \frac{x}{x - 1} \right),\quad

-&1 &= (x, x),

-\end{alignat*}

-it may again be verified without difficulty that these six

-operations form a group.

-

-\Example{\Inum{IV.} Group of linear transformations of two variables.}

-With a similar notation, the six operations

-\begin{alignat*}{3}

- A &= \left(x, \frac{y}{x};\ y, \frac{1}{x} \right),\quad

-&B &= \left(x, \frac{1}{y};\ y, \frac{x}{y} \right),\quad

-&C &= (x, y;\ y, x), \\

-%

- D &= \left(x, \frac{1}{x};\ y, \frac{y}{x} \right),\quad

-&E &= \left(x, \frac{x}{y};\ y, \frac{1}{y} \right),\quad

-&1 &= (x, y;\ x, y)

-\end{alignat*}

-form a group.

-%% -----File: 035.png---Folio 19-------

-

-\Example{\Inum{V.} Group of linear transformations to a prime modulus.}

-

-The six operations defined by

-\begin{alignat*}{3}

-A &= (x, x + 1),       & B &= (x, x + 2),       & C &= (x, 2x), \\

-D &= (x, 2x + 2),\quad & E &= (x, 2x + 1),\quad & 1 &= (x, x),

-\end{alignat*}

-where each transformation is taken to modulus~$3$, form a group.

-

-\Example{\Inum{VI.} Group of substitutions of $3$~symbols.} The six substitutions

-\[

-1,\quad A = (xyz),\quad B = (xzy),\quad C = x(yz),\quad D = y(zx),\quad E = z(xy)

-\]

-are the only substitutions that can be formed with three symbols;

-they must therefore form a group.

-

-\Example{\Inum{VII.} Group of substitutions of $6$~symbols.} The substitutions

-\begin{gather*}

-1,\quad A = (xyz)(abc),\quad B = (xzy)(acb),\quad C = (xa)(yc)(zb), \\

-D = \DPtypo{xb(ya)(zc)}{(xb)(ya)(zc)},\quad E = (xc)(yb)(za)

-\end{gather*}

-may be verified to form a group.

-

-\Example{\Inum{VIII.} Group of substitutions of $6$~symbols.} The substitutions

-\begin{gather*}

-1,\quad A = (xaybzc),\quad B = (xyz)(abc),\quad C = (xb)(yc)(za), \\

-D = (xzy)(acb),\quad E = (xczbya)

-\end{gather*}

-form a group.

-

-The operations in the first seven of these groups, as well as

-the objects of operation, are quite different from one group to

-another; but it may be shewn that the laws according to which

-the operations, denoted by the same letters in the different

-groups, combine together are identical for all seven. There is

-no difficulty in verifying that in each instance

-\[

-A^{3} = 1,\quad C^{2} = 1,\quad B = A^{2},\quad D = AC = CA^{2},\quad E = A^{2}C = CA;

-\]

-and from these relations the complete system, according to

-which the six operations in each of the seven groups combine

-together, may be at once constructed. This is given by the

-following multiplication table, where the left-hand vertical

-column gives the first factor and the top horizontal line the

-%% -----File: 036.png---Folio 20-------

-\index{Group@\Topic{Group}!multiplication table of}%

-\index{Multiplication table@\Topic{Multiplication table} of a group}%

-second factor in each product; thus the table is to be read

-$A1 = A$, $AB = 1$, $AC = D$, and so on.

-\[

-\begin{array}{c|*{6}{c}}

-\multicolumn{1}{c}{}

-  & 1 & A & B & C & D & E \\

-\cline{2-7}\Strut

-1 & 1 & A & B & C & D & E \\

-A & A & B & 1 & D & E & C \\

-B & B & 1 & A & E & C & D \\

-C & C & E & D & 1 & B & A \\

-D & D & C & E & A & 1 & B \\

-E & E & D & C & B & A & 1

-\end{array}

-\]

-

-But, though the operations of the seventh and eighth

-groups are of the same nature and though the operands are

-identical, the laws according to which the six operations

-combine together are quite distinct for the two groups. Thus,

-for the last group, it may be shewn that

-\[

-B = A^{2},\quad C = A^{3},\quad D = A^{4},\quad E = A^{5},\quad A^{6} = 1,

-\]

-so that the operations of this group may, in fact, be represented

-by

-\[

-1,\ A,\ A^{2},\ A^{3},\ A^{4},\ A^{5}.

-\]

-

-\Par{18.} If we pay no attention to the nature of the actual

-operations and operands, and consider only the number of the

-former and the laws according to which they combine, the first

-seven groups of the preceding paragraph are identical with

-each other. From this point of view a group, abstractly

-considered, is completely defined by its multiplication table;

-and, conversely, the multiplication table must implicitly

-contain all properties of the group which are independent of

-any special mode of representation.

-

-It is of course obvious that this table cannot be arbitrarily

-constructed. Thus, if

-\[

-AB = P \quad\text{and}\quad BC = Q,

-\]

-the entry in the table for~$PC$ must be the same as that for~$AQ$.

-Except in the very simplest cases, the attempt to form a

-consistent multiplication table, merely by trial, would be most

-laborious.

-%% -----File: 037.png---Folio 21-------

-

-The very existence of the table shews that the symbols

-\index{Defining relations@\Topic{Defining relations} of a group!definition of}%

-\index{Group@\Topic{Group}!defining relations of}%

-\index{Group@\Topic{Group}!fundamental or generating operations of}%

-denoting the different operations of the group are not all

-independent of each other; and since the number of symbols is

-finite, it follows that there must exist a set of symbols

-$S_{1}$, $S_{2}$,~\dots, $S_{n}$ no one of which can be expressed in terms of the

-remainder, while every operation of the group is expressible in

-terms of the set. Such a set is called a set of \emph{fundamental} or

-\emph{generating} operations of the group. Moreover though no one

-of the generating operations can be expressed in terms of the

-remainder, there must be relations of the general form

-\[

-S_{m}^{a} S_{n}^{b} \dots S_{p}^{c} = 1

-\]

-among them, as otherwise the group would be of infinite order;

-and the number of these relations, which are independent

-of one another, must be finite. Among them there necessarily

-occur the relations

-\[

-S_{1}^{a_{1}} = 1,\quad S_{2}^{a_{2}} = 1,\ \Dots,\quad S_{n}^{a_{n}} = 1

-\]

-giving the orders of the fundamental operations.

-

-We thus arrive at a virtually new conception of a group;

-it can be regarded as arising from a finite number of fundamental

-operations connected by a finite number of independent

-relations. But it is to be noted that there is no reason for

-supposing that such an origin for a group is unique; indeed,

-in general, it is not so. Thus there is no difficulty in verifying

-that the group, whose multiplication table is given in §~17, is

-completely specified either by the system of relations

-\[

-A^{3} = 1,\quad C^{2} = 1,\quad (AC)^{2} = 1,

-\]

-or by the system

-\[

-C^{2} = 1,\quad D^{2} = 1,\quad (CD)^{3} = 1.

-\]

-

-In other words, it may be generated by two operations of

-orders $2$~and~$3$, or by two operations of order~$2$. So also the

-last group of §~17 is specified either by

-\[

-A^{6} = 1,

-\]

-or by

-\[

-B^{3} = 1,\quad C^{2} = 1,\quad BC = CB.

-\]

-

-\Par{19.} \begin{Definition}

-Let $G$ and $G'$ be two groups of equal

-order. If a correspondence can be established between the

-operations of $G$~and~$G'$, so that to every operation of~$G$ there

-%% -----File: 038.png---Folio 22-------

-\index{Simply isomorphic groups@\Topic{Simply isomorphic groups}!definition of}%

-\index{Simply isomorphic groups@\Topic{Simply isomorphic groups}!said to be of the same type}%

-corresponds a single operation of~$G'$ and to every operation of~$G'$

-there corresponds a single operation of~$G$, while to the

-product~$AB$ of any two operations of~$G$ there corresponds the

-product~$A'B'$ of the two corresponding operations of~$G$', the

-groups $G$ and $G'$ are said to be \emph{simply isomorphic}\footnotemark.

-\end{Definition}

-\footnotetext{We shall sometimes use the phrase that two groups are of the same \emph{type}

-  to denote that they are simply isomorphic.}

-

-Two simply isomorphic groups are, abstractly considered,

-identical. In discussing the properties of groups, some definite

-mode of representation is, in general, indispensable; and as

-long as we are dealing with the properties of a group \textit{per~se},

-and not with properties which depend on the form of

-representation, the group may, if convenient, be replaced by

-any group which is simply isomorphic with it. For the discussion

-of such properties, it would be most natural to suppose

-the group given either by its multiplication table or by its

-fundamental operations and the relations connecting them;

-and as far as possible we shall follow this course. Unfortunately,

-however, these purely abstract modes of representing a

-group are by no means the easiest to deal with. It thus

-becomes an important question to determine as far as possible

-what different concrete forms of representation any particular

-group may be capable of; and we shall accordingly end the

-present chapter with a demonstration of the following general

-theorem bearing on this question.

-

-\Par{20.} \begin{Theorem}

-\index{Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$|(}%

-\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group}%

-\index{Substitution group@\Topic{Substitution group}!representation of any group as a regular}%

-Every group of finite order~$N$ is capable of

-representation as a group of substitutions of $N$~symbols\footnotemark.

-\end{Theorem}

-\footnotetext{Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XX} (1882), p.~30.}

-\index{Dyck@\Topic{Dyck}, quoted}%

-

-Let

-\[

-1,\ S_{1},\ S_{2},\ \dots,\ S_{i},\ \dots,\ S_{N - 1}

-\]

-be the $N$~operations of the group; and form the complete

-multiplication table

-\[

-\begin{array}{*{7}{c}}

-1,     & S_{1},      & S_{2},      & \dots, & S_{i},      & \dots, & S_{N - 1}, \\

-S_{1}, & S_{1}^{2},  & S_{2}S_{1}, & \dots, & S_{i}S_{1}, & \dots, & S_{N - 1}S_{1}, \\

-\hdotsfor{7} \\

-S_{i}, & S_{1}S_{i}, & S_{2}S_{i}, & \dots, & S_{i}^{2},  & \dots, & S_{N - 1}S_{i}, \\

-\hdotsfor{7} \\

-S_{N - 1}, & S_{1}S_{N - 1}, & S_{2}S_{N - 1}, & \dots, & S_{i}S_{N - 1}, & \dots, & S_{N - 1}^{2},

-\end{array}

-\]

-%% -----File: 039.png---Folio 23-------

-that results from multiplying the symbols in the first horizontal

-line by $1$,~$S_{1}$, $S_{2}$,~\dots, $S_{N-1}$ in order. Each horizontal line in the

-table so obtained contains the same $N$ symbols as the original

-line, but any given symbol occupies a different place in each

-line; for the supposition that the two symbols $S_{i} S_{p}$ and $S_{i} S_{q}$

-were identical would involve

-\[

-S_{p} = S_{q},

-\]

-which is not true.

-

-It thus appears that the first line in the table, taken

-with the $(i + 1)$th line, defines a substitution

-\[

-\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}

-    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}

-\]

-performed on the $N$~symbols $1$,~$S_{1}$, $S_{2}$,~\dots, $S_{N-1}$; and by taking

-the first line with each of the others, including itself, a set of $N$~substitutions

-performed on these $N$~symbols is obtained. Now

-this set of substitutions forms a group simply isomorphic

-to the given group. For if

-\[

-S_{p} S_{i} = S_{q} \quad\text{and}\quad S_{q} S_{j} = S_{r},

-\]

-the substitutions

-\[

-\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}

-    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}

-\]

-and

-\[

-\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}

-    {S_{j}, S_{1} S_{j}, S_{2} S_{j}, \dots, S_{N-1} S_{j}},

-\]

-successively performed, change $S_{p}$ into~$S_{r}$.

-

-But

-\[

-S_{p} S_{i} S_{j} = S_{r},

-\]

-and therefore the product of the two substitutions, in the order

-given, is the substitution

-\[

-\Sub{\PadTo{S_{i} S_{j}}{1},

-     \PadTo{S_{1} S_{i} S_{j}}{S_{1}},

-     \PadTo{S_{2} S_{i} S_{j}}{S_{2}}, \dots,

-     \PadTo{S_{N-1} S_{i} S_{j}}{S_{N-1}}}

-    {S_{i} S_{j}, S_{1} S_{i} S_{j}, S_{2} S_{i} S_{j}, \dots, S_{N-1} S_{i} S_{j}},

-\]

-or the product of any two of the $N$~substitutions is again one

-of the substitutions of the set. Moreover the above reasoning

-shews that the result of performing successively the substitutions

-corresponding to the operations $S_{i}$~and~$S_{j}$ gives the

-substitution corresponding to the operation~$S_{i} S_{j}$. Hence,

-since the number of substitutions is equal to the number of

-%% -----File: 040.png---Folio 24-------

-operations, the given group and the group of substitutions

-are simply isomorphic.

-

-It may also be shewn that each of the $N$~substitutions is

-\index{Substitution group@\Topic{Substitution group}!regular, definition of}%

-regular (§~9) in the $N$~symbols. For the substitution

-\[

-\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}

-    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}

-\]

-changes $S_{r}$ into $S_{r} S_{i}$, $S_{r} S_{i}$ into $S_{r} S_{i}^{2}$, and so on. If then $m$~is

-the order of~$S_{i}$, the cycle of the substitution which contains~$S_{r}$

-will be

-\[

-(S_{r},\ S_{r} S_{i},\ S_{r} S_{i}^{2},\ \dots,\ S_{r} S_{i}^{m-1}).

-\]

-

-Now $S_{r}$ may be any symbol of the set; hence all the cycles

-of the substitution must contain the same number,~$m$, of symbols.

-

-The substitution is therefore regular in the $N$~symbols, and

-this can only be the case if $m$~is a factor of~$N$. It follows at

-once, as was stated in §~15, that the order of any operation of a

-group of order~$N$ must be equal to or a factor of~$N$.

-

-All the substitutions in this form of representing a group

-being regular, the group itself is said to be expressed as a

-regular substitution group.

-\index{Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$|)}%

-

-\Par{21.} The form, in which it has just been shewn that every

-group can be represented, is by no means the only form of

-representation possessing this property. Thus Hurwitz\footnote

-  {``Algebraische Gebilde mit eindeutigen Transformationen in sich,'' \textit{Math.\

-  Ann.}, Vol.~\VolNo{XLI} (1893), p.~421.}

-\index{Hurwitz@\Topic{Hurwitz}, quoted}%

-has

-shewn that every group can be expressed as a group of rational

-and reversible transformations which change a suitably chosen

-algebraic curve (or Riemann's surface) into itself. We shall see

-in Chapter~XIV that every group can also be expressed as a

-group of linear transformations of a finite set of variables to a

-prime modulus.

-%% -----File: 041.png---Folio 25-------

-

-

-\Chapter{III.}{On the Simpler Properties of a Group which

-are Independent of its Mode of Representation.}

-

-\Par{22.} \First{In} this chapter we proceed to discuss some of the

-simplest of the properties of groups of finite order which are

-independent of their mode of representation.

-

-If among the operations of a group~$G$ a certain set can be

-\index{Sub-group@\Topic{Sub-group}!definition of}%

-chosen which do not exhaust all the operations of the group~$G$,

-yet which at the same time satisfy all the conditions of §~12 so

-that they form another group~$H$, this group~$H$ is called a \emph{sub-group}

-of the group~$G$. Thus if $S$~be any operation, order~$m$,

-of~$G$, the operations

-\[

-1,\ S,\ S^{2},\ \Dots,\ S^{m-1}

-\]

-evidently form a group; and when the order of~$G$ is greater

-than~$m$, this group is a sub-group of~$G$. A sub-group of this

-nature, which consists of the different powers of a single

-\index{Group@\Topic{Group}!cyclical}%

-operation, is called a \emph{cyclical sub-group}; and a group, which

-consists of the different powers of a single operation, is called

-a \emph{cyclical group}.

-

-\begin{Theorem}[I.]

-If $H$ is a sub-group of~$G$, the order~$n$ of~$H$ is

-a factor of the order~$N$ of~$G$.

-\end{Theorem}

-

-Let

-\[

-1,\ T_{1},\ T_{2},\ \Dots,\ T_{n-1}

-\]

-be the $n$~operations of~$H$; and let $S_{1}$ be any operation of~$G$

-which is not contained in~$H$.

-%% -----File: 042.png---Folio 26-------

-\index{Sub-group@\Topic{Sub-group}!order of a, divides order of group}%

-

-Then the operations

-\[

-S_{1},\ T_{1} S_{1},\ T_{2} S_{1},\ \Dots,\ T_{n-1} S_{1}

-\]

-are all distinct from each other and from the operations of~$H$.

-

-For if

-\[

-T_{p} S_{1} = T_{q} S_{1},

-\]

-then

-\[

-T_{p} = T_{q},

-\]

-contrary to supposition; and if

-\[

-T_{p} = T_{q} S_{1},

-\]

-then

-\[

-S_{1} = T_{q}^{-1} T_{p},

-\]

-and $S_{1}$~would be contained among the operations of~$H$.

-

-If the $2n$~operations thus obtained do not exhaust all the

-operations of~$G$, let $S_{2}$ be any operation of~$G$ not contained

-among them.

-

-Then it may be shewn, by repeating the previous reasoning,

-that the $n$~operations

-\[

-S_{2},\ T_{1} S_{2},\ T_{2} S_{2},\ \Dots,\ T_{n-1} S_{2}

-\]

-are all different from each other and from the previous $2n$

-operations. If the group~$G$ is still not exhausted, this process

-may be repeated; so that finally the $N$~operations of~$G$ can be

-exhibited in the form

-\[

-\begin{array}{*{5}{c}}

-1,     & T_{1},       & T_{2},       & \Dots, & T_{n-1}\DPchg{:}{,} \\

-S_{1}, & T_{1} S_{1}, & T_{2} S_{1}, & \Dots, & T_{n-1} S_{1}\DPchg{:}{,} \\

-S_{2}, & T_{1} S_{2}, & T_{2} S_{2}, & \Dots, & T_{n-1} S_{2}\DPchg{:}{,} \\

-\hdotsfor{5} \\

-S_{m-1}, & T_{1} S_{m-1}, & T_{2} S_{m-1}, & \Dots, & T_{n-1} S_{m-1}.

-\end{array}

-\]

-Hence $N = mn$, and $n$~is therefore a factor of~$N$.

-

-\begin{Remark}

-When $N$~is a prime~$p$, the group~$G$ can have no sub-group other

-than one of order unity consisting of the identical operation alone.

-Every operation~$S$ of the group, other than the identical operation,

-is of order~$p$, and the group consists of the operations

-\[

-1,\ S,\ S^{2},\ \Dots,\ S^{p-1}.

-\]

-A group whose order is prime is therefore necessarily cyclical.

-\end{Remark}

-

-\Par{23.} \begin{Theorem}[II.]

-The operations common to two groups

-$G_{1}$,~$G_{2}$ themselves form a group~$g$, whose order is a factor of the

-orders of $G_{1}$ and~$G_{2}$.

-\end{Theorem}

-%% -----File: 043.png---Folio 27-------

-

-For if $S$,~$T$ are any two operations common to $G_{1}$ and~$G_{2}$,

-\index{Group@\Topic{Group}!symbol for a}%

-\index{Operations@\Topic{Operations}!common to two groups form a group}%

-\index{Symbol@\Topic{Symbol}!for a group generated by given operations}%

-$ST$~is also common to both groups; and hence the common

-operations satisfy conditions \Inum{($\alpha$)}~and~\Inum{($\beta$)} of the definition in §~12.

-But their orders are finite and they must therefore satisfy also

-condition~\Inum{($\gamma$)}, and form a group~$g$. Moreover $g$~is a sub-group

-of both $G_{1}$ and~$G_{2}$, and therefore by Theorem~I its order is a

-factor of the orders of both these groups.

-

-If $G_{1}$ and $G_{2}$ are sub-groups of a third group~$G$, then $g$~is

-also clearly a sub-group of~$G$.

-

-The set of operations, that arise by combining in every way

-the operations of the groups $G_{1}$~and~$G_{2}$, evidently satisfy the

-conditions of §~12 and form a group; but this will not necessarily

-or generally be a group of finite order. If however $G_{1}$ and $G_{2}$

-are sub-groups of a group~$G$ of finite order, the group~$g'$ that

-arises from their combination will necessarily be of finite order;

-it may either coincide with~$G$ or be a sub-group of~$G$. In

-either case, the order of~$g'$ will be a multiple of the orders of

-$G_{1}$~and~$G_{2}$.

-

-It is convenient here to explain a notation that enables

-us to avoid an otherwise rather cumbrous phraseology. Let

-$S_{1}$,~$S_{2}$, $S_{3}$,~\Dots\ be a given set of operations, and $G_{1}$, $G_{2}$,~\Dots\ a

-set of groups. Then the symbol

-\[

-\{S_{1}, S_{2}, S_{3}, \Dots, G_{1}, G_{2}, \Dots\}

-\]

-will be used to denote the group that arises by combining in

-every possible way the given operations and the operations of

-the given groups.

-

-Thus, for instance, the group~$g'$ above would be represented

-by

-\[

-\{G_{1}, G_{2}\};

-\]

-the cyclical group that arises from the powers of an operation~$S$

-by

-\[

-\{S\};

-\]

-and, as a further example, the sixth group of §~17 may be

-represented by

-\[

-\{(xy), (xz)\}.

-\]

-

-\Par{24.} \begin{Definition}

-If $S$ and $T$ are any two operations of a

-\index{Conjugate operations@\Topic{Conjugate operations}!definition of}%

-\index{Transforming an operation@\Topic{Transforming an operation}!definition of}%

-group, the operations $S$ and $T^{-1}ST$ are called \emph{conjugate} operations;

-while $T^{-1}ST$~is spoken of as the result of \emph{transforming}

-the operation~$S$ by~$T$.

-\end{Definition}

-%% -----File: 044.png---Folio 28-------

-

-The two operations $S$ and $T^{-1}ST$ are identical only when $S$

-and $T$ are permutable. For if

-\[

-S = T^{-1}ST,

-\]

-then

-\[

-TS = ST.

-\]

-

-Two conjugate operations are always of the same order.

-For

-\begin{align*}

-(T^{-1}ST)^{n}

-  &= T^{-1}ST · T^{-1}ST \Dots T^{-1}ST \\

-  &= T^{-1}S^{n}T.

-\end{align*}

-

-Therefore, if

-\begin{align*}

-S^{n} &= 1, \\

-(T^{-1}ST)^{n} &= T^{-1}T = 1;

-\end{align*}

-and conversely, if

-\[

-(T^{-1}ST)^{n} = 1,

-\]

-then

-\[

-S^{n} = T · T^{-1} S^{n}T · T^{-1} = T(T^{-1}ST)^{n} T^{-1} = TT^{-1} = 1.

-\]

-

-The operations $ST$ and $TS$ are always conjugate and

-therefore of the same order; for

-\[

-ST = T^{-1}T · ST = T^{-1} · TS · T.

-\]

-

-\begin{Ex} Shew that the operations $S_{1}S_{2}\Dots S_{n - 1}S_{n}$ and $S_{r}S_{r + 1}\Dots$

-$S_{n}S_{1}\Dots S_{r - 1}$ are conjugate within the group $\{S_{1}, S_{2}, \Dots, S_{n}\}$.

-\end{Ex}

-

-\begin{Definition}

-An operation~$S$ of a group~$G$, which is

-\index{Self-conjugate operation@\Topic{Self-conjugate operation}!definition of}%

-identical with all its conjugate operations, is called a \emph{self-conjugate}

-operation. Such an operation must evidently be

-permutable with each of the operations of~$G$.

-\end{Definition}

-

-\begin{Remark}

-In every group the identical operation is self-conjugate; and in

-a group, whose operations are all permutable, every operation is

-self-conjugate. A simple example of a group, which contains self-conjugate

-operations other than the identical operation, while at the

-same time its operations are not all self-conjugate, is given by

-\[

-\{(1234), (13)\}.

-\]

-It is easy to shew that the order of this group is~$8$, and that $(13)(24)$

-is a self-conjugate operation.

-\end{Remark}

-

-If all the operations of a group be transformed by a given

-operation, the set of transformed operations form a group. For

-if $T_{1}$ and $T_{2}$ are any two operations of the group, so that $T_{1}T_{2}$~is

-also an operation of the group, then

-\[

-S^{-1} T_{1}S · S^{-1} T_{2}S = S^{-1} T_{1} T_{2}S;

-\]

-%% -----File: 045.png---Folio 29-------

-hence the product of any two operations of the transformed

-set is another operation belonging to the transformed set, and

-the set therefore forms a group. Moreover the preceding

-equation shews that the new group is simply isomorphic

-to the original group. If $G$ is the given group, the symbol

-$S^{-1}GS$ will be used for the new group. When $S$~belongs to the

-group~$G$, the groups $G$ and $S^{-1}GS$ are evidently the same.

-

-Now unless $S$ is a self-conjugate operation of~$G$, the pairs of

-operations $T$ and $S^{-1}TS$ will not all be identical when for $T$ the

-different operations of~$G$ are put in succession. Hence the

-process of transforming all the operations of a group by one of

-themselves is equivalent to establishing a correspondence

-between the operations of the group, which exhibits it as

-simply isomorphic with itself.

-

-\begin{Definitions}

-When $H$~is a sub-group of~$G$ and $S$~is any

-\index{Composite groups@\Topic{Composite groups}!definition of}%

-\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!definition of}%

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!definition of}%

-operation of~$G$, the groups $H$ and $S^{-1}HS$ are called \emph{conjugate}

-sub-groups of~$G$.

-

-If $H$ and $S^{-1}HS$ are identical, $S$~is said to be permutable

-with the sub-group~$H$. This does not necessarily involve that

-$S$~is permutable with each of the operations of~$H$.

-

-If $H$ and $S^{-1}HS$ are identical, whatever operation $S$ is of~$G$,

-$H$~is said to be a \emph{self-conjugate} sub-group of~$G$.

-

-A group is called \emph{composite} or \emph{simple}, according as it does

-\index{Group@\Topic{Group}!simple and composite}%

-\index{Simple groups@\Topic{Simple groups}!definition of}%

-or does not possess at least one self-conjugate sub-group other

-than that formed of the identical operation alone.

-\end{Definitions}

-

-\Par{25.} \begin{Theorem}[III.]

-The operations of a group~$G$, which

-are permutable with a given operation~$T$, form a sub-group~$H$;

-and the order of~$G$ divided by the order of~$H$ is the number of

-operations conjugate to~$T$\footnotemark.

-\end{Theorem}

-\footnotetext{Among these $T$ of course occurs.}

-

-If $R_{1}$ and $R_{2}$ are any two operations permutable with~$T$, so

-\index{Operations@\Topic{Operations}!of a group, which are permutable with a given operation or sub-group, form a group}%

-\index{Permutability@\Topic{Permutability} of an operation with a group!definition of}%

-that

-\[

-R_{1}T = TR_{1} \quad\text{and}\quad R_{2}T = TR_{2};

-\]

-then

-\[

-R_{1} R_{2}T = R_{1} TR_{2} = TR_{1} R_{2},

-\]

-and therefore $R_{1}R_{2}$ is permutable with~$T$. The operations

-permutable with~$T$ therefore form a group~$H$. Let $n$ be its

-order and

-\[

-1,\ R_{1},\ R_{2},\ \Dots,\ R_{n - 1}

-\]

-%% -----File: 046.png---Folio 30-------

-its operations. Then if $S$~is any operation of~$G$ not contained

-in~$H$, the operations

-\[

-S,\ R_{1}S,\ R_{2}S,\ \Dots,\ R_{n - 1}S

-\]

-all transform $T$ into the same operation~$T'$.

-

-For

-\[

-(R_{i}S)^{-1} TR_{i}S = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1}TS.

-\]

-

-Also the $n$~operations thus obtained are the only operations

-which transform $T$ into~$T'$; for if

-\[

-S'^{-1}TS' = T',

-\]

-then

-\[

-SS'^{-1} TS'S^{-1} = ST'S^{-1} = T;

-\]

-and therefore $S'S^{-1}$ belongs to~$H$. The number of operations

-which transform $T$ into any operation conjugate to it is

-therefore equal to the number that transform $T$ into itself, that

-is, to the order of~$H$. If then $N$~is the order of~$G$, the

-operations of~$G$ may be divided into $\dfrac{N}{n}$~sets of $n$~each, such

-that the operations of each set transform $T$ into a distinct

-operation, those of the first set, namely the operations of~$H$,

-transforming $T$ into itself. The number of operations conjugate

-to~$T$, including itself, is therefore~$\dfrac{N}{n}$.

-

-Since

-\[

-T = ST'S^{-1},

-\]

-therefore

-\[

-R_{i}^{-1} TR_{i} = R_{i}^{-1} ST' S^{-1}R_{i};

-\]

-hence

-\[

-T'= S^{-1}TS = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1} R_{i}^{-1} S · T' · S^{-1}R_{i}S,

-\]

-so that every operation of the form $S^{-1}R_{i}S$ is permutable with~$T'$.

-Hence if $H$~is the group of operations permutable with~$T$,

-and if

-\[

-S^{-1}TS = T',

-\]

-then $S^{-1}HS$ is the group of operations permutable with~$T'$.

-

-It is convenient to have a symbol to represent the set of

-operations

-\[

-S,\ R_{1}S,\ R_{2}S,\ \dots,\ R_{n - 1}S,

-\]

-where

-\[

-1,\ R_{1},\ R_{2},\ \dots,\ R_{n - 1}

-\]

-form a group~$H$. We shall in future represent this set of

-operations by~$HS$; and we shall use $SH$ to represent the set

-\[

-S,\ SR_{1},\ SR_{2},\ \dots,\ SR_{n - 1}.

-\]

-%% -----File: 047.png---Folio 31-------

-

-\begin{Theorem}[IV.]

-The operations of a group~$G$ which are

-permutable with a sub-group~$H$ form a sub-group~$I$, which is

-either identical with~$H$ or contains~$H$ as a self-conjugate sub-group.

-The order of~$G$ divided by the order of~$I$ is the number

-of sub-groups conjugate to~$H$\footnotemark.

-\end{Theorem}

-\footnotetext{Among these sub-groups $H$~itself occurs.}

-

-If $S_{1}$,~$S_{2}$ are any two operations of~$G$ which are permutable

-\index{Operations@\Topic{Operations}!of a group, which are permutable with a given operation or sub-group, form a group}%

-with~$H$, then

-\[

-S_{1}^{-1} HS_{1} = H,\quad S_{2}^{-1} HS_{2} = H,

-\]

-and therefore

-\[

-S_{2}^{-1} S_{1}^{-1} HS_{1} S_{2} = H,

-\]

-so that $S_{1}S_{2}$ is permutable with~$H$. The operations of~$G$

-which are permutable with~$H$ therefore form a group~$I$, which

-may be identical with~$H$ and, if not identical with~$H$, must

-contain it. Also, if $S$~is any operation of~$I$,

-\[

-S^{-1}HS = H,

-\]

-and therefore $H$ is a self-conjugate sub-group of~$I$.

-

-If now $\Sigma$ is any operation of~$G$ not contained in~$I$, it may

-be shewn, exactly as in the proof of Theorem~III, that the

-operations $I\Sigma$ and no others transform $H$ into a conjugate

-sub-group~$H'$ which is not identical with~$H$; and therefore

-that the number of sub-groups in the conjugate set to which

-$H$~belongs is the order of~$G$ divided by the order of~$I$.

-

-The operations of~$G$ which are permutable with~$H'$ may

-also be shewn to form the group~$\Sigma^{-1}I\Sigma$.

-

-It is perhaps not superfluous to point out that two

-distinct conjugate sub-groups may have some operations in

-common with one another.

-

-\Par{26.} Let $S_{1}$ be any operation of~$G$, and

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{m}

-\]

-the distinct operations obtained on transforming $S_{1}$ by every

-operation of~$G$. The number,~$m$, of these operations is, by

-Theorem~III, a factor of~$N$, the order of~$G$. Moreover if,

-instead of transforming~$S_{1}$, we transform any other operation of

-the set,~$S_{r}$, by every operation of~$G$, the same set of $m$~distinct

-operations of~$G$ will result. Such a set of operations we call a

-\index{Conjugate operations@\Topic{Conjugate operations}!complete set of}%

-\emph{complete set} of conjugate operations. If $T$~is any operation of~$G$

-%% -----File: 048.png---Folio 32-------

-\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!complete set of}%

-\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!simple, definition of}%

-which does not belong to this complete set of conjugate

-operations, no operation that is conjugate to~$T$ can belong to

-the set. Hence the operations of $G$ may be distributed into a

-number of distinct sets such that every operation belongs to

-one set and no operation belongs to more than one set; while

-any set forms by itself a complete set of conjugate operations.

-If $m_{1}$, $m_{2}$,~\Dots, $m_{s}$ are the numbers of operations in the different

-sets, then

-\[

-N = m_{1} + m_{2} + \Dots + m_{s};

-\]

-and, since the identical operation is self-conjugate, one at least

-\PageLabel{32}%

-of the $m$'s must be unity.

-

-Similarly, if $H_{1}$ is any sub-group of~$G$, and

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{p}

-\]

-the distinct sub-groups obtained on transforming $H_{1}$ by every

-operation of~$G$, we call the set a \emph{complete set} of conjugate

-sub-groups. If $K$~is a sub-group of~$G$ not contained in the set,

-no sub-group conjugate to~$K$ can belong to the set. If the

-operation~$S_{1}$ belongs to one or more of a complete set of

-conjugate sub-groups, $\Sigma^{-1} S_{1}\Sigma$~must also belong to one or more

-sub-groups of the set, $\Sigma$~being any operation of~$G$. Hence

-among the operations contained in the complete set of conjugate

-sub-groups, the complete set of conjugate operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{m}

-\]

-occurs.

-

-No sub-group of~$G$ can contain operations belonging to

-every one of the complete sets of conjugate operations of~$G$.

-For if such a sub-group~$H$ existed, the complete set of

-conjugate sub-groups, to which $H$ belongs, would contain all

-the operations of~$G$. Let $m$ be the order of~$H$ and $n$~($\geq m$) the

-order of the sub-group~$I$ formed of those operations of~$G$ which

-are permutable with~$H$. Then $H$~is one of $\dfrac{N}{n}$~conjugate sub-groups,

-each of which contains $m$~operations. The identical

-operation is common to all these sub-groups, and they therefore

-cannot contain more than

-\[

-1 + \frac{N(m - 1)}{n}

-\]

-%% -----File: 049.png---Folio 33-------

-distinct operations in all. This number is less than~$N$, and

-therefore the complete set of conjugate sub-groups cannot

-contain all the operations of~$G$.

-

-\Par{27.} If a group contains self-conjugate operations, it must

-\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!operations common to or permutable with a complete set of, form a self-conjugate sub-group}%

-\index{Operations@\Topic{Operations}!common to or permutable with each of a complete set of conjugate sub-groups form a self-conjugate sub-group}%

-contain self-conjugate sub-groups. For the cyclical sub-group

-generated by any self-conjugate operation must be self-conjugate.

-The only exception to this statement is the case of

-the cyclical groups of prime order. Every operation of such a

-group is clearly self-conjugate; but since the cyclical sub-group

-generated by any operation coincides with the group itself,

-there can be no self-conjugate sub-group\footnotemark.

-\footnotetext{Strictly speaking, this statement should be qualified by the addition

-  ``except that formed by the identical operation alone.'' No real ambiguity

-  however will be introduced by always leaving this exception unexpressed.}

-

-If every operation of a group is not self-conjugate, or, in

-other words, if the operations of a group are not all permutable

-with each other, the totality of the self-conjugate operations

-forms a self-conjugate sub-group. For, if $S_{1}$~and~$S_{2}$ are

-permutable with every operation of the group, so also is~$S_{1}S_{2}$.

-

-\begin{Theorem}[V.]

-The operations common to a complete set of

-conjugate sub-groups form a self-conjugate sub-group.

-\end{Theorem}

-

-It is an immediate consequence of Theorem~II that the

-operations common to a complete set of conjugate sub-groups

-form a sub-group. Also the set of conjugate sub-groups, when

-transformed by any operation of the group, is changed into

-itself. Hence their common sub-group must be self-conjugate.

-

-It may of course happen that the identical operation is the

-only one which is common to every sub-group of the set.

-

-\begin{Corollary}

-The operations permutable with each of a

-complete set of conjugate sub-groups form a self-conjugate

-sub-group.

-\end{Corollary}

-

-For, if the operations permutable with the sub-group~$H$

-form a sub-group~$I$, the operations permutable with every

-sub-group of the conjugate set to which $H$ belongs are the

-operations common to every sub-group of the conjugate set to

-which $I$ belongs.

-%% -----File: 050.png---Folio 34-------

-

-Further, the operations which are permutable with every

-operation of a complete set of conjugate sub-groups form a

-self-conjugate sub-group.

-

-\begin{Theorem}[VI.]

-If $T_{1}$, $T_{2}$,~\dots, $T_{r}$ are a complete set of conjugate

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!generated by a complete set of conjugate operations}%

-operations of~$G$, the group $\{T_{1}, T_{2}, \dots, T_{r}\}$, if it does not coincide

-with~$G$, is a self-conjugate sub-group of~$G$; and it is the self-conjugate

-sub-group of smallest order that contains~$T_{1}$.

-\end{Theorem}

-

-Since the operations $T_{1}$, $T_{2}$,~\dots, $T_{r}$ are merely rearranged

-in a new sequence when the set is transformed by any operation

-of~$G$, it follows that

-\[

-S^{-1} \{T_{1}, T_{2}, \dots, T_{r}\}S = \{T_{1}, T_{2}, \dots, T_{r}\},

-\]

-whatever operation of~$G$ may be represented by~$S$. Hence

-$\{T_{1}, T_{2}, \dots, T_{r}\}$ is a self-conjugate sub-group. Also any self-conjugate

-sub-group of~$G$ that contains~$T_{1}$ must contain

-$T_{2}$, $T_{3}$,~\dots, $T_{r}$; and therefore any self-conjugate sub-group of~$G$

-which contains~$T_{1}$ must contain $\{T_{1}, T_{2}, \dots, T_{r}\}$.

-

-In exactly the same way it may be shewn that, if

-\[

-H_{1},\ H_{2},\ \dots,\ H_{s}

-\]

-{\Loosen are a complete set of conjugate sub-groups of~$G$, the group

-$\{H_{1}, H_{2}, \dots, H_{s}\}$,} if it does not coincide with~$G$, is the smallest

-self-conjugate sub-group of~$G$ which contains the sub-group~$H_{1}$.

-

-The theorem just proved suggests a process for determining

-whether any given group is simple or composite. To this end,

-the groups $\{T_{1}, T_{2}, \dots, T_{r}\}$ corresponding to each set of conjugate

-operations in the group are formed. If any one of them

-differs from the group itself, it is a self-conjugate sub-group

-and the group is composite; but if each group so formed

-coincides with the original group, the latter is simple. If

-the order of~$T_{1}$ contains more than one prime factor and

-if $T_{1}^{m}$~is of prime order, it is easy to see that the distinct

-operations of the set $T_{1}^{m}$, $T_{2}^{m}$,~\dots, $T_{r}^{m}$ form a complete set of

-conjugate operations, and that the group $\{T_{1}, T_{2}, \dots, T_{r}\}$ contains

-the group $\{T_{1}^{m}, T_{2}^{m}, \dots, T_{r}^{m}\}$. Hence practically it is sufficient

-to form the groups $\{T_{1}, T_{2}, \dots, T_{r}\}$ for all conjugate sets of operations

-whose orders are prime.

-

-\begin{Remark}

-With the notation of §~26 (\PageRef{p.}{32}), the order of any self-conjugate

-sub-group of~$G$ must be of the form $m_{\alpha} + m_{\beta} + \Dots$;

-%% -----File: 051.png---Folio 35-------

-for if the sub-group contains any given operation, it must contain all

-the operations conjugate with it. Moreover one at least of the

-numbers $m_{\alpha}$, $m_{\beta}$,~\Dots\ must be unity, since the sub-group must

-contain the identical operation. It may happen that the numbers~$m_{r}$

-are such that the only factors of~$N$ of the form $m_{\alpha} + m_{\beta} + \Dots$,

-one of the~$m$'s being unity, are $N$~itself and unity. When this is

-\PageLabel{35}%

-the case, $G$~is necessarily a simple group. It must not however

-be inferred that, if $N$~has factors of this form, other than $N$~itself

-and unity, then $G$~is necessarily composite.

-\end{Remark}

-

-If $G_{1}$ and $G_{2}$ are sub-groups of~$G$, it has already been seen

-(§~23) that the operations common to $G_{1}$ and $G_{2}$ form a sub-group~$g$

-of~$G$; and it is now obvious that, when $G_{1}$ and $G_{2}$ are

-self-conjugate sub-groups, so also is~$g$. Moreover the group

-$\{G_{1}, G_{2} \}$ is a self-conjugate sub-group unless it coincides with~$G$.

-For

-\[

-S^{-1} \{G_{1}, G_{2}\} S = \{S^{-1}G_{1}S, S^{-1}G_{2}S\} = \{G_{1}, G_{2}\}.

-\]

-

-Again, with the same notation, if $T_{1}$ is an operation of~$G$

-not contained in the self-conjugate sub-group~$G_{1}$, and if

-$T_{1}$, $T_{2}$,~\dots, $T_{r}$ is a complete set of conjugate operations, the

-group $\{G_{1}, T_{1}, T_{2}, \dots, T_{r}\}$ is a self-conjugate sub-group, unless

-it coincides with~$G$.

-

-\begin{Definitions}

-If $G_{1}$, a self-conjugate sub-group of~$G$, is

-\index{Maximum sub-group@\Topic{Maximum sub-group}, definition of}%

-\index{Maximum self-conjugate sub-group@\Topic{Maximum self-conjugate sub-group}!definition of}%

-such that the group

-\[

-\{G_{1}, T_{1}, T_{2}, \dots, T_{r} \}

-\]

-coincides with~$G$, when $T_{1}$, $T{2}$,~\dots, $T_{r}$ is any complete set of

-conjugate operations not contained in~$G_{1}$, then $G_{1}$~is said to be

-a \emph{maximum} self-conjugate sub-group of~$G$. This does not

-imply that $G_{1}$~is the self-conjugate sub-group of~$G$ of absolutely

-greatest order; but that there is no self-conjugate sub-group

-of~$G$, distinct from $G$~itself, which contains $G_{1}$ and is of

-greater order than~$G_{1}$.

-

-If $H$ is a sub-group of~$G$, and if, for every operation~$S$ of~$G$

-which does not belong to~$H$, the group $\{H, S\}$ coincides with~$G$,

-$H$~is said to be a \emph{maximum} sub-group of~$G$.

-\end{Definitions}

-

-\Par{28.}

-\begin{Definition}

-When a correspondence can be established

-between the operations of a group~$G$ and the operations

-of a group~$G'$, whose order is smaller than the order of~$G$, such

-that to each operation~$S$ of~$G$ there corresponds a single

-%% -----File: 052.png---Folio 36-------

-\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!multiple, definition of}%

-\index{Multiply isomorphic groups@\Topic{Multiply isomorphic groups}!definition of}%

-operation~$S'$ of~$G'$, while to the operation $S_{p}S_{q}$ there corresponds

-the operation $S_{p}'S_{q}'$, the group~$G$ is said to be \emph{multiply isomorphic}

-with the group~$G'$.

-\end{Definition}

-

-\begin{Theorem}[VII.]

-If a group $G$ is multiply isomorphic with

-a group~$G'$, then \Inum{(i)}~the operations of~$G$, which correspond to

-the identical operation of~$G'$, form a self-conjugate sub-group of~$G$;

-\Inum{(ii)}~to each operation of~$G'$ there correspond the same number

-of operations of~$G$; and \Inum{(iii)}~the order of~$G$ is a multiple of the

-order of~$G'$.

-\end{Theorem}

-

-Let

-\[

-S_{0},\ S_{1},\ S_{2},\ \dots,\ S_{n-1}

-\]

-be the set of operations of~$G$ which correspond to the identical

-operation of~$G'$. These operations must form a group, since to

-$S_{p}S_{q}$ corresponds the operation~$1 · 1$, \ie\ the identical operation

-of~$G'$; and therefore $S_{p}S_{q}$ must belong to the set.

-

-Again, to the operation $T^{-1}S_{p}T$ of~$G$ corresponds the operation

-$T'^{-1} · 1 · T'$, that is, the identical operation of~$G'$. Hence,

-whatever operation of~$G$ is taken for~$T$,

-\[

-T^{-1} \{S_{0}, S_{1}, \dots, S_{n-1}\} T = \{S_{0}, S_{1}, \dots, S_{n-1}\}.

-\]

-

-The sub-group~$\Gamma$ of~$G$ formed of the operations

-\[

-S_{0},\ S_{1},\ S_{2},\ \dots,\ S_{n-1}

-\]

-is therefore self-conjugate.

-

-Again, if $T$ and $T_{1}$ are two operations of~$G$ which correspond

-to the operation~$T'$ of~$G'$, the operation $T^{-1}T_{1}$ corresponds to

-the identical operation of~$G'$, and therefore belongs to~$\Gamma$.

-Hence the operations that correspond to~$T'$ are all contained

-in the set~$T\Gamma$. The operations of this set are all distinct and

-equal in number to the order of~$\Gamma$. Hence if $n$~is the order of~$\Gamma$,

-to each operation of~$G'$ there correspond $n$~operations of~$G$.

-

-Finally, since to each operation of~$G$ there corresponds only

-one of~$G'$, while to each operation of~$G'$ there correspond $n$ of~$G$,

-the order of~$G$ is $n$~times the order of~$G'$.

-

-To any sub-group of~$G'$ of order~$\mu$, there corresponds a

-sub-group of~$G$ of order~$\mu n$. For if $T_{p}'T_{q}'$ forms one of the

-set

-\[

-1,\ T_{1}',\ \dots,\ T_{p}',\ \dots,\ T_{q}',\ \dots,\ T_{\mu - 1}',

-\]

-%% -----File: 053.png---Folio 37-------

-at least one, and therefore all, of the set $T_{p} T_{q} \Gamma$ must occur

-among

-\[

-\Gamma,\ T_{1} \Gamma,\ \dots,\ T_{p}\Gamma,\ \dots,\ T_{q}\Gamma,\ \dots,\ T_{\mu - 1}\Gamma,

-\]

-and hence these operations form a group. Moreover, if the

-sub-group of~$G'$ is self-conjugate, so also is the corresponding

-sub-group of~$G$.

-

-It should be noticed that no correspondence is thus

-established between a sub-group of~$G$ which does not contain~$\Gamma$

-and any sub-group of~$G'$.

-

-\Par{29.} The relation of multiple isomorphism between two

-groups can be presented in a manner rather different from that

-of the last paragraph. Let $G$~be any composite group, and

-$\Gamma$~a self-conjugate sub-group of~$G$ consisting of the operations

-\[

-1,\ T_{1},\ T_{2},\ \dots,\ T_{n - 1}.

-\]

-

-Then, as in §~22, the operations of~$G$ can be arranged in the

-scheme

-\[

-\begin{array}{*{6}{c}}

-1,         & T_{1},          & T_{2},          & \dots, & T_{n - 1}\DPchg{;}{,} \\

-S_{1},     & T_{1}S_{1},     & T_{2}S_{1},     & \dots, & T_{n - 1}S_{1}\DPchg{;}{,} \\

-\hdotsfor{5} \\

-S_{i},     & T_{1}S_{i},     & T_{2}S_{i},     & \dots, & T_{n - 1}S_{i}\DPchg{;}{,} \\

-\hdotsfor{5} \\

-S_{m - 1}, & T_{1}S_{m - 1}, & T_{2}S_{m - 1}, & \dots, & T_{n - 1}S_{m - 1}.

-\end{array}

-\]

-

-Now $\Gamma$ being a self-conjugate sub-group of~$G$, it follows that

-\[

-S_{i}^{-1} \Gamma S_{i} = \Gamma,

-\]

-and therefore the two sets of operations $\Gamma S_{i}$ and $S_{i} \Gamma$ coincide

-except as regards arrangement. Hence

-\[

-T_{\alpha} S_{i} T_{\beta} S_{j} = T_{\alpha} T_{\beta'} S_{i} S_{j} = T_{\gamma} S_{i} S_{j},

-\]

-where

-\[

-S_{i} T_{\beta} S_{i}^{-1} = T_{\beta'},

-\]

-and

-\[

-T_{\alpha}T_{\beta'} = T_{\gamma};

-\]

-so that both $T_{\beta'}$ and $T_{\gamma}$ belong to~$\Gamma$. Now all the operations of

-the set~$\Gamma S_{i}S_{j}$ occur in a single line, say the $(k + 1)$th, of the

-above scheme. Hence if any operation of the $(i + 1)$th~line be

-followed by any operation of the $(j + 1)$th~line, the result is

-some operation of the $(k + 1)$th~line. If then we regard the

-set of operations contained in each line of the scheme as a

-%% -----File: 054.png---Folio 38-------

-single entity, they will by their laws of combination define a

-new group of order~$m$. In fact, if we denote these entities by

-\[

-S_{0}',\ S_{1}',\ \dots,\ S_{m-1}',

-\]

-a relation of the form

-\[

-S_{i}' S_{j}' = S_{k}'

-\]

-has been proved to hold for every pair. Moreover, these relations

-necessarily obey the associative law; for if

-\[

-S_{j}' S_{p}' = S_{q}',

-\]

-then the relation

-\[

-S_{k}' S_{p}' = S_{i}' S_{q}'

-\]

-follows in consequence of the symbols of $G$ itself obeying the

-associative law. The symbol~$S_{0}'$, corresponding to the first line

-of the scheme, is clearly the symbol of the identical operation in

-the new group thus defined.

-

-If now $G$ and $\Gamma$ coincide with the groups of the preceding

-paragraph which are represented by the same symbols, then $G'$

-of the preceding paragraph must be simply isomorphic with the

-group whose operations are

-\[

-1,\ S_{1}',\ S_{2}',\ \dots,\ S_{m-1}'.

-\]

-

-It follows that a group~$G'$ with which a group~$G$ is

-\index{Factor groups@\Topic{Factor groups}!definition of}%

-multiply isomorphic, in such a way that to the identical operation

-of~$G'$ there corresponds a given self-conjugate sub-group~$\Gamma$

-of~$G$, is completely defined (as an abstract group) when $G$

-and $\Gamma$ are given. This being so it is natural to use a symbol

-to denote directly the group thus defined in terms of $G$ and~$\Gamma$.

-%[** In footnote, no comma or "Vol." after journal title]

-Herr Hölder\footnote

-  {``Zur Reduction der algebraischen Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV} (1889),

-  p.~31.}

-\index{Holder@\Topic{Hölder}, quoted}%

-has introduced the symbol

-\[

-\frac{G}{\Gamma}

-\]

-to represent this group; he calls it the \emph{quotient} of $G$ by~$\Gamma$,

-and a \emph{factor-group} of~$G$. We shall in the sequel make constant

-use both of the symbol and of the phrase thus defined.

-

-It may not be superfluous to notice that the symbol~$\dfrac{G}{\Gamma}$ has

-%% -----File: 055.png---Folio 39-------

-no meaning\footnotemark,

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Herr Frobenius has extended the use of the symbol to the case in which $\Gamma$~is

-  any group, whether contained in~$G$ or not, with which every operation of~$G$ is

-  permutable: ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte}, 1895, p.~169.

-  We shall always use the symbol in the sense defined in the text.}%

-unless $\Gamma$~is a self-conjugate sub-group of~$G$.

-Moreover, it may happen that $G$ has two simply isomorphic

-self-conjugate sub-groups $\Gamma$~and~$\Gamma'$. When this is

-the case, there is no necessary relation between the factor-groups

-$\dfrac{G}{\Gamma}$ and $\dfrac{G}{\Gamma'}$ (except of course that their orders are equal);

-in other words, the type of the factor-group $\dfrac{G}{\Gamma}$ depends on the

-actual self-conjugate sub-group of~$G$ which is chosen for~$\Gamma$ and

-not merely on the type of~$\Gamma$.

-

-Further, though in relation to its definition by means of $G$

-and $\Gamma$ we call $\dfrac{G}{\Gamma}$ a factor-group of~$G$, we may without ambiguity,

-since the symbol represents a group of definite type, omit the

-word factor and speak of the group~$\dfrac{G}{\Gamma}$.

-

-It is also to be observed that $G$ has not necessarily a

-sub-group simply isomorphic with~$\dfrac{G}{\Gamma}$. This may or may not be

-the case.

-

-\Par{30.} If $G$ is multiply isomorphic with $G'$ so that the self-conjugate

-sub-group~$\Gamma$ of~$G$ corresponds to the identical

-operation of~$G'$, it was shewn, at the end of §~28, that to any

-self-conjugate sub-group of~$G'$ there corresponds a self-conjugate

-sub-group of~$G$ containing~$\Gamma$. Hence, unless $\dfrac{G}{\Gamma}$ is a simple

-group, $\Gamma$~cannot be a maximum self-conjugate sub-group of~$G$.

-If $g_{1}$~is any self-conjugate sub-group of~$\dfrac{G}{\Gamma}$, and $G_{1}$~the

-corresponding (necessarily self-conjugate) sub-group of~$G$, containing~$\Gamma$,

-we may form the factor-group~$\dfrac{G}{G_{1}}$, and determine

-again whether this group is simple or composite. By continuing

-this process a maximum self-conjugate sub-group of~$G$, containing~$\Gamma$,

-must at last be reached.

-%% -----File: 056.png---Folio 40-------

-

-\Par{31.} Though $\dfrac{G}{\Gamma}$ is completely defined by $G$~and~$\Gamma$, where $\Gamma$~is

-any given self-conjugate sub-group of~$G$, the reader will

-easily verify that $G$~is not in general determined when $\Gamma$ and

-$\dfrac{G}{\Gamma}$ are given.

-

-We shall have in the sequel to consider the solution of

-this problem in various particular cases. There is, however, in

-every case one solution of it which is immediately obvious.

-We may take any two groups $G_{1}$ and $G_{2}$, simply isomorphic

-with the given groups $\Gamma$ and $\dfrac{G}{\Gamma}$, such that $G_{1}$ and $G_{2}$ have

-no common operation except identity, while each operation

-of one is permutable with each operation of the other. The

-group $\{G_{1}, G_{2}\}$, formed by combining these two, is clearly such

-that $\dfrac{\{G_{1}, G_{2}\}}{G_{1}}$ is simply isomorphic with~$\dfrac{G}{\Gamma}$; it therefore gives a

-solution of the problem.

-

-\begin{Definition}

-If two groups $G_{1}$,~$G_{2}$ have no common operation

-\index{Direct product@\Topic{Direct product} of two groups!definition of}%

-except identity, and if each operation of~$G_{1}$ is permutable

-with each operation of~$G_{2}$, the group $\{G_{1}, G_{2}\}$ is called the

-\emph{direct product} of $G_{1}$~and~$G_{2}$.

-\end{Definition}

-

-\begin{Ex} If $H$,~$h$ are self-conjugate sub-groups of~$G$, and if $h$~is

-contained in~$H$, so that $\dfrac{H}{h}$~is a self-conjugate sub-group of~$\dfrac{G}{h}$, shew

-that the quotient of~$\dfrac{G}{h}$ by~$\dfrac{H}{h}$ is simply isomorphic with~$\dfrac{G}{H}$.

-\end{Ex}

-

-\begin{Remark}

-\Par{32.} If $H$ is a self-conjugate sub-group of~$G$, of order~$n$, and if

-$H'$~is a self-conjugate sub-group of~$G'$, of order~$n'$, and if $\dfrac{G}{H}$ and $\dfrac{G'}{H'}$

-are simply isomorphic, a correspondence of the most general kind

-may be established between the operations of $G$~and~$G'$. To every

-operation of~$G$ (or~$G'$) there will correspond $n'$ (or~$n$) operations of~$G'$

-(or~$G$), in such a way that to the product of any two operations

-of~$G$ (or~$G'$) there corresponds a definite set of $n'$ (or~$n$) operations

-of~$G'$ (or~$G$). Let

-\[

-G = H,\ S_{1}H,\ S_{2}H,\ \Dots,\ S_{m - 1}H,

-\]

-and

-\[

-G' = H',\ S_{1}'H',\ S_{2}'H',\ \Dots,\ S_{m - 1}'H';

-\]

-{\Loosen and in the simple isomorphism between $\dfrac{G}{H}$ and~$\dfrac{G'}{H'}$, let $S_{r}$~and~$S_{r}'$

-%% -----File: 057.png---Folio 41-------

-($r = 0, 1, \dots, m - 1$)} be corresponding operations. Then if we take

-the set~$S_{r}' H'$ as the $n'$~operations of~$G'$ that correspond to any

-operation of the set~$S_{r}H$ of~$G$, and the set~$S_{r}H$ as the $n$~operations

-of~$G$ that correspond to any operation of the set~$S_{r}' H'$ of~$G'$, the

-correspondence is, in fact, established.

-

-For, if $h_{1}'$ and $h_{2}'$ are any two operations of~$H'$, the set of operations

-\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!general, definition of}%

-$S_{r}' h_{1}' S_{s}' h_{2}'$ includes $n'$~distinct operations only, namely those of

-the set~$S_{r}' S_{s}' H'$. Hence to the product of any given operation of

-the set~$S_{r}H$ by any given operation of the set~$S_{s}H$, there corresponds

-the set of $n'$~operations $S_{r}' S_{s}' H'$; at the same time the product of the

-two given operations belongs $\Bigl($in consequence of the isomorphism

-between $\dfrac{G}{H}$ and~$\dfrac{G'}{H'}${}$\Bigr)$ to the set~$S_{r} S_{s} H$. The same statements clearly

-hold when we interchange accented and unaccented symbols.

-

-We still speak of $G$ and $G'$ as isomorphic groups, and the

-correspondence between their operations is said to give an $n$-to-$n'$

-isomorphism of the two groups. We shall return to this general

-form of isomorphism in dealing with intransitive substitution groups.

-\end{Remark}

-

-\Par{33.} \begin{Definition}

-Two groups $G$ and $G'$ are said to be

-\index{Permutable groups@\Topic{Permutable groups}!definition of}%

-\DPchg{permutable}{\emph{permutable}} with each other when the distinct operations of

-the set~$S_{i} S_{j}'$, where for $S_{i}$ every operation of the group~$G$ is

-put in turn and for $S_{j}'$ every operation of the group~$G'$, coincide

-with the distinct operations of the set~$S_{j}' S_{i}$ except possibly

-as regards arrangement.

-\end{Definition}

-

-If the two groups $G$ and $G'$ are permutable, the group

-$\{G, G'\}$ must be of finite order. For, by the definition, every

-operation

-\[

-\dots S_{p} S_{q}' S_{r} S_{t}' \dots

-\]

-can be reduced to the form~$S_{i} S_{j}'$; and therefore the number of

-distinct operations of the group $\{G, G'\}$ cannot exceed the

-product of the orders of $G$~and~$G'$. Let $g$~be the group formed

-of the common operations of $G$~and~$G'$. Divide the operations

-of these groups into the sets

-\[

-g,\ \Sigma_{1}g,\ \Sigma_{2}g,\ \dots,\ \Sigma_{m - 1}g

-\]

-and

-\[

-g,\ g \Sigma_{1}',\ g \Sigma_{2}',\ \DPchg{\Dots\Dots}{\Dots},\ g \Sigma_{m'-1}'.

-\]

-

-Then every operation of the set~$S_{i} S_{j}'$ can clearly be expressed

-in the form

-\[

-\Sigma_{p} \gamma \Sigma_{q}',

-\]

-%% -----File: 058.png---Folio 42-------

-where $\gamma$~is some operation of~$g$. And no two operations of this

-form can be identical, for if

-\[

-\Sigma_{p} \gamma_{1} \Sigma_{q}' = \Sigma_{r} \gamma_{2} \Sigma_{t}',

-\]

-then

-\[

-\gamma_{2}^{-1} \Sigma_{r}^{-1} \Sigma_{p} \gamma_{1} = \Sigma_{t}' \Sigma_{q}'^{-1};

-\]

-so that $\Sigma_{t}' \Sigma_{q}'^{-1}$ belongs to~$g$. But this is only possible if

-\[

-\Sigma_{t}' = \Sigma_{q}',

-\]

-which leads to

-\[

-\Sigma_{r} = \Sigma_{p},

-\]

-and

-\[

-\gamma_{2} = \gamma_{1}.

-\]

-The order of $\{G, G'\}$ is therefore the product of the orders of $G$

-and~$G'$, divided by the order of~$g$.

-

-If every operation of~$G$ is permutable with~$G'$, then $g$~must

-be a self-conjugate sub-group of~$G$. For $G$ and~$G'$ are transformed,

-each into itself, by any operation of~$G$; and therefore

-their common sub-group~$g$ must be transformed into itself by

-any operation of~$G$.

-

-Moreover those operations of~$G$, which are permutable with

-every operation of~$G'$, form a self-conjugate sub-group of~$G$.

-For if $T$~is an operation of~$G$, which is permutable with every

-operation~$S'$ of~$G'$, so that

-\[

-T^{-1} S' T = S',

-\]

-and if $S$~is any operation of~$G$, then

-\[

-S^{-1} T^{-1} S · S_{-1} S'S · S_{-1} TS = S^{-1} S'S,

-\]

-so that $S_{-1} TS$ is permutable with every operation of~$G'$. Hence

-every operation of~$G$, which is conjugate to~$T$, is permutable

-with every operation of~$G'$; and the operations of~$G$, which are

-permutable with every operation of~$G'$, therefore form a self-conjugate

-sub-group.

-

-If $G$ is a simple group, $g$~must consist of the identical

-operation only; and either all the operations of~$G$, or none of

-them, must be permutable with every operation of~$G'$.

-

-A special case is that in which the two groups $G'$ and~$G$ are

-respectively a self-conjugate sub-group~$\Gamma$ and any sub-group~$H$

-of some third group; for then every operation of~$H$ is permutable

-with~$\Gamma$. If $H$~is a cyclical sub-group generated by an

-operation~$S$ of order~$n$, and if $S^{m}$~is the lowest power of~$S$

-%% -----File: 059.png---Folio 43-------

-which occurs in~$\Gamma$, then $m$~must be a factor of~$n$. For if $m'$~is

-the greatest common factor of $m$ and~$n$, integers $x$ and~$y$ can be

-found such that

-\[

-xm + yn = m'.

-\]

-

-Now

-\[

-S^{m'} = S^{xm + yn} = S^{xm},

-\]

-and therefore $S^{m'}$ belongs to~$\Gamma$. Hence $m'$ cannot be less than~$m$,

-and therefore $m$~is a factor of~$n$. Moreover, since $\{S^{m}\}$~is

-a sub-group of~$\Gamma$, the order of~$\Gamma$ must be divisible by~$\dfrac{n}{m}$.

-Hence:---

-

-\begin{Theorem}[VIII.]

-If an operation~$S$, of order~$n$, is permutable

-with a group~$\Gamma$, and if $S^{m}$~is the lowest power of~$S$ which occurs

-in~$\Gamma$; then $m$~is a factor of~$n$, and $\dfrac{n}{m}$~is a factor of the order

-of~$\Gamma$.

-\end{Theorem}

-

-The operations of~$\{\Gamma, S\}$ can clearly be distributed in the

-sets

-\[

-\Gamma,\ \Gamma S,\ \Gamma S^{2},\ \dots,\ \Gamma S^{m-1};

-\]

-and no two of the operations $S$, $S^{2}$,~\dots, $S^{m-1}$ are conjugate in

-$\{\Gamma, S\}$.

-

-\Par{34.} A still more special case, but it is most important, is

-that in which the two groups are both of them self-conjugate

-sub-groups of some third group. If in this case the two groups

-are $G$ and~$H$, while $S$~and~$T$ are any operations of the two

-groups respectively, then

-\[

-S^{-1}HS = H,

-\]

-and

-\[

-T^{-1}GT = G;

-\]

-so that every operation of~$G$ is permutable with~$H$ and every

-operation of~$H$ is permutable with~$G$.

-

-Consider now the operation $S^{-1} T^{-1} ST$. Regarded as the

-product of $S^{-1}$ and~$T^{-1}ST$ it belongs to~$G$, and regarded as the

-product of $S^{-1}T^{-1}S$ and~$T$ it belongs to~$H$. Every operation of

-this form therefore belongs to the common group of $G$ and~$H$.

-If $G$ and~$H$ have no common operation except identity, then

-\[

-S^{-1} T^{-1} ST = 1,

-\]

-%% -----File: 060.png---Folio 44-------

-or

-\[

-ST = TS;

-\]

-and $S$~and~$T$ are permutable. Hence:---

-

-\begin{Theorem}[IX\protect\footnotemark.]

-If every operation of~$G$ transforms~$H$ into

-itself and every operation of~$H$ transforms~$G$ into itself, and

-if $G$~and~$H$ have no common operation except identity; then

-every operation of~$G$ is permutable with every operation of~$H$.

-\end{Theorem}

-%[** No comma after journal title]

-\footnotetext{Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XXII} (1883), p.~97.}%

-\index{Dyck@\Topic{Dyck}, quoted}%

-

-\begin{Corollary}

-If every operation of~$G$ transforms~$H$ into itself

-and every operation of~$H$ transforms~$G$ into itself, and if either

-$G$ or~$H$ is a simple group; then $G$ and~$H$ have no common

-operation except identity, and every operation of~$G$ is permutable

-with every operation of~$H$.

-\end{Corollary}

-

-For, by §~33, if $G$ and~$H$ had a common sub-group, it would

-be a self-conjugate sub-group of both of them; and neither of

-them could then be simple, contrary to hypothesis. Consequently,

-the only sub-group common to $G$ and~$H$ is the

-identical operation.

-

-\begin{Remark}

-\Par{35.} If $N$ and~$N'$ are the orders of two permutable groups $G$ and~$G'$,

-the $NN'$~equations expressing every operation of the form~$SS'$,

-where $S$~belongs to~$G$ and $S'$ to~$G'$, in the form~$\Sigma'\Sigma$, where $\Sigma$ and~$\Sigma'$

-belong to $G$ and $G'$ respectively, are never all independent. In

-particular, if

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{r}

-\]

-are a set of independent generating operations of~$G$, and

-\[

-m_{1},\ m_{2},\ \Dots,\ m_{r}

-\]

-their orders, it is clearly sufficient that each operation of the form

-\[

-S'S_{p}^{\alpha}\quad (\alpha = 1, 2, \Dots, m_{p} - 1;\ p = 1, 2, \Dots, r)

-\]

-should be capable of expression in the form~$\Sigma'\Sigma$.

-

-When, in fact, these conditions are satisfied, it is always possible

-by a series of steps to express any operation~$S'S$ in the form~$\Sigma\Sigma'$.

-\end{Remark}

-

-\begin{Ex}[1.] Shew that, in the group whose defining relations are

-\[

-A^{4} = 1,\quad B^{3} = 1,\quad (AB)^{2} = 1,

-\]

-the three operations $A^{2}$, $B^{-1} A^{2} B$, $BA^{2} B^{-1}$ are permutable and that

-they form a complete set of conjugate operations. Hence shew that

-$\{A^{2}, B\}$ is a self-conjugate sub-group, and that the order of the group

-is~$24$.

-\end{Ex}

-%% -----File: 061.png---Folio 45-------

-

-\begin{Ex}[2.] Shew that the cyclical group generated by the substitution

-$(1234567)$ is permutable with the group

-\[

-\{(267)(345),\ (23)(47)\};

-\]

-and that the order of the group resulting from their product is~$168$.

-\end{Ex}

-

-\begin{Ex}[3.] If $g_{1}$ and $g_{2}$ are the orders of the groups $G_{1}$ and~$G_{2}$, $\gamma$~the

-order of their greatest common sub-group and $g$~the order of

-$\{G_{1}, G_{2}\}$, shew that

-\[

-g\gamma \geq g_{1}g_{2},

-\]

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-and that, if $g\gamma = g_{1}g_{2}$, then $G_{1}$ and $G_{2}$ are permutable. \Attrib{Frobenius.}

-\end{Ex}

-

-\begin{Ex}[4.] If $G_{1}$ and $G_{2}$ are two sub-groups of~$G$ of orders $g_{1}$ and~$g_{2}$,

-and $S$~any operation of~$G$, prove that the number of distinct operations

-of~$G$ contained in the set~$S_{1}SS_{2}$, when for $S_{1}$ and $S_{2}$ are put in

-turn every pair of operations of $G_{1}$ and $G_{2}$ respectively, is~$\dfrac{g_{1}g_{2}}{\gamma}$; $\gamma$~being

-the order of the greatest sub-group common to $S^{-1}G_{1}S$ and~$G_{2}$.

-

-If $T$~is any other operation of~$G$, shew also that the sets $S_{1}SS_{2}$

-and $S_{1}TS_{2}$ are either identical or have no operation in common. \Attrib{Frobenius.}

-\end{Ex}

-

-\begin{Ex}[5.] If a group~$G$ of order~$mn$ has a sub-group~$H$ of order~$n$,

-and if $n$~has no prime factor which is less than~$m$, shew that $H$~must

-be a self-conjugate sub-group. \Attrib{Frobenius.}

-\end{Ex}

-%% -----File: 062.png---Folio 46-------

-

-

-\Chapter{IV.}{On Abelian Groups.}

-\index{Abelian group@\Topic{Abelian group}!definition of}%

-\index{Group@\Topic{Group}!Abelian}%

-

-\Par{36.} \First{We} shall now apply the general results, that have

-been obtained in the last chapter, to the study of two special

-classes of groups; in the present chapter we shall deal particularly

-with those groups whose operations are all permutable

-with each other.

-

-\begin{Definition}

-A group, whose operations are all permutable

-\index{Abel@\Topic{Abel}, quoted}%

-\index{Frobenius and Stickelberger@\Topic{Frobenius and Stickelberger}, quoted}%

-\index{Jordan@\Topic{Jordan}, quoted}%

-\index{Weber@\Topic{Weber}, quoted}%

-with each other, is called an \emph{Abelian}\footnote

-  {On Abelian groups, the reader may consult Frobenius and Stickelberger,

-  ``Ueber Gruppen vertauschbarer Elemente,'' \textit{Crelle's Journal}, Vol.~\VolNo{LXXXVI} (1879),

-  p.~217; and a very complete discussion in the second volume of Herr Weber's

-  recently published \textit{Lehrbuch der Algebra}. In the proof of the existence of a

-  set of independent generating operations (§~41) we have directly followed Herr

-  Weber.

-

-  The name ``Abelian group'' has been applied by M.~Jordan (\textit{Traité des substitutions~etc.}\

-  pp.~171 et~seq.)\ to an entirely different class of groups, whose

-  operations are not permutable. Most writers, we believe, have used the phrase

-  in the sense defined in the text.

-

-  The connection of Abel's name with groups of permutable operations is due

-  to his having been the first to investigate, with complete generality, the application

-  of such groups to the theory of equations, ``Mémoire sur une classe

-  particulière d'équations résolubles algébriquement,'' \textit{Crelle's Journal}, Vol.~\VolNo{IV}

-  (1829), p.~131; or Collected Works, 1881 edition, Vol.~\VolNo{I}, p.~478.}

-group.

-\end{Definition}

-

-It is to be expected (and it will be found) that the theory of

-Abelian groups is much simpler than that of groups in general;

-for the process of multiplication of the operations of such groups

-is commutative as well as associative.

-

-Every sub-group of an Abelian group is itself an Abelian

-group, since its operations are necessarily all permutable. For

-%% -----File: 063.png---Folio 47-------

-\index{Abelian group@\Topic{Abelian group}!sub-groups of}%

-the same reason, every operation and every sub-group of an

-Abelian group is self-conjugate both in the group itself and in

-any sub-group in which it is contained.

-

-If $G$ is an Abelian group and $H$~any sub-group of~$G$, then

-since $H$~is necessarily self-conjugate, there exists a factor-group~$\dfrac{G}{H}$,

-and this again must be an Abelian group. (The

-reader must not however infer that, if $H$ and $\dfrac{G}{H}$ are both

-Abelian, then $G$~is also Abelian. It is indeed clear that this

-is not necessarily the case.)

-

-\Par{37.} Let now $G$ be any Abelian group, and let $p^{m}$ be the

-highest power of a prime~$p$ that divides its order. We shall

-first shew that $G$~has a single sub-group of order~$p^{m}$, consisting

-of all the operations of~$G$ whose orders are powers of~$p$.

-

-If $S_{1}$,~$S_{2}$ are any two operations of~$G$, it follows from §~33,

-that, because $S_{1}$ and $S_{2}$ are permutable, the order of~$\{S_{1}, S_{2}\}$ is

-equal to or is a factor of the product of the orders of $S_{1}$~and~$S_{2}$.

-So again, if $S_{3}$~is an operation of~$G$ not contained in~$\{S_{1}, S_{2}\}$, the

-order of $\{S_{1}, S_{2}, S_{3}\}$ is equal to or is a factor of the product of

-the orders of $\{S_{1}, S_{2}\}$ and~$S_{3}$. Now by continually including a

-fresh operation, not contained in the group already arrived at,

-we must in this way after a finite number of steps arrive at the

-group~$G$, whose order is divisible by~$p$. Hence, among the

-operations $S_{1}$,~$S_{2}$, $S_{3}$,~\Dots, there must be at least one whose

-order is divisible by~$p$, and some power of this, say~$S$, will be

-an operation of order~$p$. Now, if $m$~is greater than unity, the

-order of the factor-group $\dfrac{G}{\{S\}}$, which is also Abelian, is divisible

-by~$p$, and therefore this factor-group must have an operation of

-order~$p$. Hence $G$~will (§~28) contain a sub-group of order~$p^{2}$.

-If $m$~is greater than~$2$, the same reasoning may be

-repeated to shew that $G$~has a sub-group of order~$p^{3}$, and so on.

-Hence, finally, $G$~has a sub-group of order~$p^{m}$. Let $H$~be this

-sub-group; and suppose if possible that $G$~contains an operation~$T$,

-whose order is a power of~$p$, which does not belong to~$H$.

-Then $\{H, T\}$ is a sub-group of~$G$, whose order is a power of~$p$,

-greater than~$p^{m}$; and this is impossible (§~22). The sub-group~$H$

-%% -----File: 064.png---Folio 48-------

-\index{Abelian group@\Topic{Abelian group}!sub-groups of}%

-must therefore contain all the operations of~$G$ whose orders

-are powers of~$p$. Hence:---

-

-\begin{Theorem}[I.]

-If $p^{m}$ is the highest power of a prime~$p$ that

-divides the order of an Abelian group~$G$, then $G$~contains a

-single sub-group of order~$p^{m}$, which consists of all the operations

-of~$G$ whose orders are powers of~$p$.

-\end{Theorem}

-

-\Par{38.} Let the order of~$G$ be

-\[

-N = p_{1}^{m_{1}} p_{2}^{m_{2}} \Dots p_{n}^{m_{n}},

-\]

-where $p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are distinct primes; and let

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{n}

-\]

-be the sub-groups of~$G$ of orders

-\[

-p_{1}^{m_{1}},\ p_{2}^{m_{2}},\ \Dots,\ p_{n}^{m_{n}}.

-\]

-

-Since the orders of $H_{1}$ and $H_{2}$ are relatively prime, they

-can have no common operation except identity; and therefore

-the order of $\{H_{1}, H_{2}\}$ is $p_{1}^{m_{1}} p_{2}^{m_{2}}$. This sub-group contains all

-the operations of~$G$ whose orders are relatively prime to

-$\dfrac{N}{p_{1}^{m_{1}} p_{2}^{m_{2}}}$. For if $T$~were an operation of~$G$ of order~$p_{1}^{\alpha} p_{2}^{\beta}$, not

-contained in $\{H_{1}, H_{2}\}$, then $\{H_{1}, H_{2}, T\}$ would be a sub-group

-of~$G$, of order $p_{1}^{n_{1}} p_{2}^{n_{2}}$, where $n_{1} > m_{1}$ if $\alpha > 0$ and $n_{2} > m_{2}$ if $\beta > 0$;

-and this is impossible (§~22).

-

-This process may clearly be continued to shew that, if

-$N = \mu\nu$, where $\mu$~and~$\nu$ are relatively prime, then $G$~contains a

-single sub-group of order~$\mu$, consisting of all the operations of~$G$

-whose orders are relatively prime to~$\nu$. Moreover $G$~itself is

-the direct product (§~31) of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

-

-\Par{39.} The first problem of pure group-theory that presents

-itself in connection with Abelian groups is the determination

-of all distinct Abelian groups of given order~$N$. Let $H_{1}$

-and $H_{1}'$ be two distinct Abelian groups of order~$p_{1}^{m_{1}}$, \ie\ two

-groups which are not simply isomorphic. Then two Abelian

-groups of order~$N$, whose sub-groups of order~$p_{1}^{m_{1}}$ are simply

-isomorphic with $H_{1}$ and $H_{1}'$ respectively, are necessarily distinct.

-Since then $G$~is the direct product of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$, the

-general problem for any composite order~$N$ will be completely

-solved when we have determined all distinct types of Abelian

-%% -----File: 065.png---Folio 49-------

-groups of order~$p_{1}^{m_{1}}$, $p_{2}^{m_{2}}$,~\Dots, $p_{n}^{m_{n}}$. We may therefore, for the

-purpose of this problem, confine our attention to those Abelian

-groups whose orders are powers of primes.

-

-\Par{40.} Suppose then that $G$~is an Abelian group whose order

-is the power of a prime. If among the operations of~$G$ we

-choose at random a set

-\[

-P,\ P',\ P'',\ \Dots

-\]

-from which the group can be generated, they will not in

-general be independent of each other.

-

-\begin{Remark}

-As an instance, we may take the group whose multiplication

-\index{Group@\Topic{Group}!multiplication table of}%

-\index{Multiplication table@\Topic{Multiplication table} of a group}%

-table is:---

-\[

-\begin{array}{c|*{8}{c}}

-\multicolumn{1}{c}{}

-      & 1     & P_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} \\

-\cline{2-9}

-\Strut

-1     & 1     & P_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} \\

-P_{1} & P_{1} & P_{2} & P_{3} & 1     & P_{5} & P_{6} & P_{7} & P_{4} \\

-P_{2} & P_{2} & P_{3} & 1     & P_{1} & P_{6} & P_{7} & P_{4} & P_{5} \\

-P_{3} & P_{3} & 1     & P_{1} & P_{2} & P_{7} & P_{4} & P_{5} & P_{6} \\

-P_{4} & P_{4} & P_{5} & P_{6} & P_{7} & 1     & P_{1} & P_{2} & P_{3} \\

-P_{5} & P_{5} & P_{6} & P_{7} & P_{4} & P_{1} & P_{2} & P_{3} & 1     \\

-P_{6} & P_{6} & P_{7} & P_{4} & P_{5} & P_{2} & P_{3} & 1     & P_{1} \\

-P_{7} & P_{7} & P_{4} & P_{5} & P_{6} & P_{3} & 1     & P_{1} & P_{2}

-\end{array}

-\]

-

-Here $P_{1}$ and $P_{5}$ are two operations from which every operation

-of the group may be generated: an inspection of the table will

-shew that they are connected by the relation

-\[

-P_{1}^{2} P_{5}^{2} = 1.

-\]

-

-On the other hand, if we choose $P_{1}$ and $P_{4}$ as generating

-operations, we find that every operation of the group can be expressed

-in the form

-\[

-P_{1}^{\alpha} P_{4}^{\beta},\quad (\alpha = 0, 1, 2, 3;\ \beta = 0, 1),

-\]

-while the only conditions to which the permutable operations $P_{1}$

-and $P_{4}$ are submitted are

-\[

-P_{1}^{4} = 1,\quad P_{4}^{2} = 1.

-\]

-\end{Remark}

-

-The question then arises as to whether $G$~can be generated

-by a set of permutable and independent operations, \ie\ by a

-set of permutable operations which are connected by no

-relations except those that give their orders. That this

-%% -----File: 066.png---Folio 50-------

-question is always to be answered in the affirmative may be

-proved in the following manner.

-

-\Par{41.} Let the order of~$G$ be~$p^{m}$; and let $P_{1}$ be an operation

-of~$G$ whose order~$p^{m_{1}}$ is not less than that of any other operation

-of the group. Then every operation of the group satisfies

-the equation

-\[

-S^{p^{m_{1}}} = 1.

-\]

-

-If $m_{1} = m$, the order of $\{P_{1}\}$ is equal to the order of~$G$; the

-latter is then a cyclical group generated by the operation~$P_{1}$.

-

-If $m_{1} < m$, $G$~must contain other operations besides those of~$\{P_{1}\}$.

-Denoting $\{P_{1}\}$ by~$G_{1}$, let $Q$ be any operation of~$G$ not

-contained in~$G_{1}$, and let $Q^{a}$ be the lowest power of~$Q$ that is

-contained in~$G_{1}$. Then (§~33) $a$~must be a power of~$p$; and

-when for $Q$ each operation of~$G$ that is not contained in~$G_{1}$ is

-taken in turn, $a$~must have some maximum value, say~$p^{m_{2}}$.

-Since no operation of~$G$ is of greater order than~$p^{m_{1}}$ it follows

-that $m_{2} \leq m_{1}$. We may suppose then $Q$~to be an operation of~$G$,

-such that $Q^{p^{m_{2}}}$ is the lowest power of~$Q$ which is contained

-in~$G_{1}$.

-

-Then

-\[

-Q^{p^{m_{2}}} = P_{1}^{\lambda_{1}},

-\]

-and

-\[

-Q^{p^{m_{1}}} = 1 = P_{1}^{\DPtypo{\lambda}{\lambda_{1}} p^{m_{1} - m_{2}}},

-\]

-so that $\lambda_{1}$~is divisible by~$p^{m_{2}}$ and may be expressed in the form~$xp^{m_{2}}$.

-The case $\lambda_{1} = 0$ forms no exception to this statement,

-since $\lambda_{1}$~is congruent to zero, $\pmod{p^{m_{1}}}$; but we actually take

-$x = 0$ in this case.

-

-If now we write

-\[

-Q P_{1}^{-x} = P_{2},

-\]

-then

-\[

-P_{2}^{p^{m_{2}}} = 1,

-\]

-and $P_{2}^{p^{m_{2}}}$ is the lowest power of~$P_{2}$ which is contained in~$G_{1}$.

-Let the sub-group $\{P_{1}, Q\}$ be denoted by~$G_{2}$. Then $G_{2}$~is

-generated by the two independent operations $P_{1}$ and $P_{2}$ of

-orders $p^{m_{1}}$ and~$p^{m_{2}}$; and every operation~$S$ of~$G$ is such that

-\[

-S^{p^{m_{2}}} = P_{1}^{\alpha_{1}},

-\]

-where $\alpha_{1}$~is divisible by~$p^{m_{2}}$.

-

-This process may now be continued step by step. That

-it will ultimately lead to a representation of~$G$ as generated

-by a set of independent operations may be shewn by induction.

-%% -----File: 067.png---Folio 51-------

-To this end, we will suppose that a sub-group~$G_{n}$ has been

-arrived at which is generated by the $n$ independent operations

-\[

-P_{1},\ P_{2},\ \Dots,\ P_{n},

-\]

-of orders

-\[

-p^{m_{1}},\ p^{m_{2}},\ \Dots,\ \DPtypo{p^{m}}{p^{m_{n}}},

-\]

-where

-\[

-m_{1} \geq m_{2} \geq \Dots \geq m_{n}.

-\]

-

-We will also suppose that, if $T$~is any operation of~$G$,

-\[

-T^{p^{m_{n}}} = P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{n-1}^{\alpha_{n-1}},

-\]

-where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{n-1}$ are all divisible by~$p^{m_{n}}$. In the special

-case $n = 2$ these suppositions have been justified.

-

-Let $Q'$ be any operation of~$G$, and let $Q'^{a}$ be the lowest

-power of~$Q'$ that occurs in~$G_{n}$. Then when for $Q'$ each operation

-of~$G$ that is not contained in~$G_{n}$ is taken in turn, $a$~will have

-a certain maximum value, say~$p^{m_{n+1}}$; and from the suppositions

-made above, $m_{n+1}$~is not greater than~$m_{n}$. We may therefore

-write

-\[

-Q'^{p^{m_{n+1}}}

-  = P_{1}^{\beta_{1}}

-    P_{2}^{\beta_{2}} \Dots

-    P_{n-1}^{\beta_{n-1}}

-    P_{n}^{\beta_{n}},

-\]

-and hence

-\[

-Q'^{p^{m_{n}}}

-  = P_{1}^{\beta_{1} p^{m_{n}-m_{n+1}}}

-    P_{2}^{\beta_{2} p^{m_{n}-m_{n+1}}} \Dots

-    P_{n-1}^{\beta_{n-1} p^{m_{n}-m_{n+1}}}

-    P_{n}^{\beta_{n} p^{m_{n}-m_{n+1}}}.

-\]

-

-Now $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ are independent; it follows therefore

-from the second assumption made above, that the

-exponents of $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ in this last equation are divisible

-by~$p^{m_{n}}$; and hence that $\beta_{1}$, $\beta_{2}$,~\Dots, $\beta_{n}$ are divisible by~$p^{m_{n+1}}$.

-

-If we now suppose that $Q'$ itself is chosen so that $Q'^{p^{m_{n+1}}}$ is

-the lowest power of~$Q'$ that occurs in~$G_{n}$, then

-\[

-Q'^{p^{m_{n+1}}}

-  = P_{1}^{x_{1} p^{m_{n+1}}}

-    P_{2}^{x_{2} p^{m_{n+1}}} \Dots

-    P_{n}^{x_{n} p^{m_{n+1}}};

-\]

-hence, if

-\[

-Q'P_{1}^{-x_{1}} P_{2}^{-x_{2}} \Dots P_{n}^{-x_{n}} = P_{n+1},

-\]

-then

-\[

-P_{n+1}^{p^{m_{n+1}}} = 1;

-\]

-and $P_{n+1}^{p^{m_{n+1}}}$ is the lowest power of~$P$ that occurs in~$G_{n}$. Hence

-finally $\{P_{1}, P_{2}, \Dots, P_{n}, Q'\}$ is generated by the $n + 1$ independent

-operations $P_{1}$, $P_{2}$,~\Dots, $P_{n+1}$ of orders $p^{m_{1}}$, $p^{m_{2}}$,~\Dots,

-$p^{m_{n+1}}$ ($m_{1} \geq m_{2} \geq \Dots \geq m_{n+1}$); and $S$~being any operation of~$G$,

-\[

-S^{p^{m_{n+1}}} = P_{1}^{\gamma_{1}} P_{2}^{\gamma_{2}} \Dots P_{n}^{\gamma_{n}},

-\]

-where $\gamma_{1}$, $\gamma_{2}$,~\Dots, $\gamma_{n}$ are divisible by~$p^{m_{n+1}}$.

-%% -----File: 068.png---Folio 52-------

-

-This completes the proof by induction, and we may now

-state the result in the form of the following theorem:---

-

-\begin{Theorem}[II.]

-An Abelian group of order~$p^{m}$, where $p$~is a

-\index{Abelian group@\Topic{Abelian group}!existence of independent generating operations of}%

-prime, can be generated by $r$ \($\leq m$\) independent operations $P_{1}$,

-$P_{2}$,~\Dots, $P_{r}$, of orders $p^{m_{1}}$, $p^{m_{2}}$,~\Dots, $p^{m_{r}}$, where

-\[

-m_{1} + m_{2} + \dots + m_{r} = m,

-\]

-and $m_{1} \geq m_{2} \geq \Dots \geq m_{r}$. Moreover if $\mu$~is any positive integer

-such that $m_{s} \geq \mu \geq m_{s+1}$, while $Q$~is any operation of the group, then

-\[

-Q^{p^{\mu}} = P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{s}^{\alpha_{s}},

-\]

-where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{s}$ are divisible by~$p^{\mu}$.

-\end{Theorem}

-

-The actual existence of the set of independent generating

-operations is demonstrated by the above inductive proof.

-The other inferences in the theorem may be established as

-follows. A set of independent permutable operations of orders

-$p^{m_{1}}$, $p^{m_{2}}$,~\Dots, $p^{m_{r}}$ generate a group of order $p^{m_{1} + m_{2} + \dots + m_{r}}$, and

-therefore

-\[

-m_{1} + m_{2} + \Dots + m_{r} = m.

-\]

-

-Since each number in this equation is a positive integer,

-the number of terms on the left-hand side cannot be greater

-than~$m$. Hence

-\[

-r \leq m;

-\]

-and, if $r = m$, every operation of the group except identity is of

-order~$p$.

-

-Finally, from the above inductive proof it follows that, if $Q$~is

-any operation of the group, then

-\[

-Q^{p^{m_{s}+1}} = P_{1}^{\beta_{1}} P_{2}^{\beta_{2}} \Dots P_{s}^{\beta_{s}},

-\]

-where $\beta_{1}$, $\beta_{2}$,~\Dots, $\beta_{s}$ are all divisible by~$p^{m_{s}+1}$.

-

-Hence

-\begin{align*}

-Q^{p^{\mu}}

-  &= P_{1}^{\beta_{1} p^{\mu - m_{s + 1}}}

-     P_{2}^{\beta_{2} p^{\mu - m_{s + 1}}} \Dots

-     P_{s}^{\DPtypo{\beta}{\beta_{s}} p^{\mu - m_{s + 1}}} \\

-  &= P_{1}^{\alpha_{1}}

-     P_{2}^{\alpha_{2}} \Dots

-     P_{s}^{\alpha_{s}},

-\end{align*}

-where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{s}$ are all divisible by~$p^{\mu}$.

-

-\Par{42.} It is clear, from the synthetic process by which it has

-been proved that an Abelian group of order~$p^{m}$ can be generated

-by a set of independent operations, that a considerable latitude

-exists in the choice of the actual generating operations; and

-the question arises as to the relations between the orders of the

-distinct sets of independent generating operations.

-

-The discussion of this question is facilitated by a consideration

-of certain special sub-groups of~$G$. If $A$ and $B$ are two

-%% -----File: 069.png---Folio 53-------

-operations of~$G$, and if the order of~$A$ is not less than that of~$B$,

-the order of~$AB$ is equal to, or is a factor of, the order of~$A$.

-Hence the totality of those operations of~$G$ whose orders do not

-exceed~$p^{\mu}$, or in other words of those operations which satisfy

-the relation

-\[

-S^{p^{\mu}} = 1,

-\]

-form a sub-group~$G_{\mu}$. The order of~$G_{\mu}$ clearly depends on the

-orders of the various operations of~$G$ and in no way on a special

-choice of generating operations. Now if

-\[

-P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{r}^{\alpha_{r}}

-\]

-belongs to~$G_{\mu}$, then

-\[

-P_{1}^{\alpha_{1}p^{\mu}} P_{2}^{\alpha_{2}p^{\mu}} \Dots P_{r}^{\alpha_{r}p^{\mu}} = 1.

-\]

-

-Hence if $m_{s+1}$ is the first of the series

-\[

-m_{1},\ m_{2},\ \Dots,\ m_{r},

-\]

-which is less than~$\mu$, then $\alpha_{s+1}$,~\Dots, $\alpha_{r}$ may have any values

-whatever; but $\alpha_{t}$ ($t = 1, 2, \Dots, s$) must be a multiple of~$p^{m_{t}-\mu}$.

-

-It follows from this that $G_{\mu}$~is generated by the $r$~independent

-operations

-\[

-P_{1}^{p^{m_{1}-\mu}},\

-P_{2}^{p^{m_{2}-\mu}},\ \Dots,\

-P_{s}^{p^{m_{s}-\mu}},\

-P_{s+1},\ \Dots,\ P_{r}.

-\]

-If then the order of~$G_{\mu}$ is~$p^{\nu}$, we have

-\[

-\nu = \mu s + \sum_{s+1}^{r} m_{t}.

-\]

-

-The order of~$G_{1}$, the sub-group formed of all operations of~$G$

-whose order is~$p$, is clearly~$p^{r}$.

-

-\Par{43.} Suppose now that by a fresh choice of independent

-generating operations, it were found that $G$ could be generated

-by the $r'$~independent operations

-\[

-P_{1}',\ P_{2}',\ \Dots,\ P_{r'}'

-\]

-of orders

-\[

-p^{m_{1}'},\ p^{m_{2}'},\ \Dots,\ \DPtypo{p^{m_{r'}}}{p^{m_{r'}'}},

-\]

-where

-\[

-m_{1}' \geq m_{2}' \geq \Dots \geq \DPtypo{m_{r'}}{m_{r'}'}.

-\]

-

-If $m_{s'+1}'$ is the first of this series which is less than~$\mu$, the

-order of~$G_{\mu}$ will be~$p^{\nu'}$, where

-\[

-\nu' = \mu s' + \sum_{s'+1}^{r'} m_{t}'.

-\]

-

-The order of~$G_{\mu}$ is independent of the choice of generating

-operations; so that for all values of~$\mu$

-\[

-\nu = \nu'.

-\]

-%% -----File: 070.png---Folio 54-------

-

-Hence, by taking $\mu = 1$,

-\index{Abelian group@\Topic{Abelian group}!invariance of the orders of a set of independent generating operations of}%

-\[

-r = r',

-\]

-or the number of independent generating operations is independent

-of their choice.

-

-If now

-\[

-m_{t} = m_{t}'\quad (t = s + 1, s + 2, \Dots, r),

-\]

-and

-\[

-m_{s} > m_{s}',

-\]

-and if we choose $\mu$ so that

-\[

-m_{s} \geq \mu > m_{s}';

-\]

-then

-\[

-\nu = \mu s + \sum_{s + 1}^{r} m_{t},

-\]

-and

-\[

-\nu' = \mu (s - 1) + m_{s}' + \sum_{s + 1}^{r} m_{t}.

-\]

-

-The condition

-\[

-\nu = \nu'

-\]

-gives

-\[

-\mu = m_{s}',

-\]

-in contradiction to the assumption just made.

-

-Similarly we can prove that the assumption $m_{s}' > m_{s}$ cannot

-be maintained; hence

-\[

-m_{s} = m_{s}';

-\]

-and therefore, however the independent generating operations

-of~$G$ are chosen, their number is always~$r$, and their orders are

-\[

-p^{m_{1}},\ p^{m_{2}},\ \Dots,\ p^{m_{r}}.

-\]

-

-\Par{44.} If $G'$ is a second Abelian group of order~$p^{m}$, simply

-isomorphic with~$G$, and if

-\[

-P_{1}',\ P_{2}',\ \Dots,\ \DPtypo{P_{r'}}{P_{r'}'}

-\]

-of orders

-\[

-p^{m_{1}'},\ p^{m_{2}'},\ \Dots,\ \DPtypo{p^{m_{r'}}}{p^{m_{r'}'}},

-\]

-where

-\[

-m_{1}' \geq m_{2}' \geq \Dots \geq \DPtypo{m_{r'}}{m_{r'}'},

-\]

-are a set of independent generating operations of~$G'$, exactly the

-same process as that of the last paragraph may be used to shew

-that

-\[

-r = r',

-\]

-and

-\[

-m_{s} = m_{s}'\quad (s = 1, 2, \Dots, r).

-\]

-

-In fact, since corresponding operations of two simply isomorphic

-groups have the same order, the order of~$G_{\mu}$ must be

-equal to the order of~$G_{\mu}'$; and this is the condition that has

-been used to obtain the result of the last paragraph.

-

-Two Abelian groups of order~$p^{m}$ cannot therefore be simply

-%% -----File: 071.png---Folio 55-------

-\index{Abelian group@\Topic{Abelian group}!sub-groups of|(}%

-\index{Abelian group@\Topic{Abelian group}!symbol for, of given type}%

-isomorphic unless the series of integers $m_{1}$, $m_{2}$,~\Dots, $m_{r}$ is the

-same for each. On the other hand when this condition is satisfied,

-it is clear that the two groups are simply isomorphic, since

-by taking $P_{s}$ and $P_{s}'$ ($s = 1, 2, \Dots, r$) as corresponding operations,

-the isomorphism is actually established.

-

-The number of distinct types of Abelian groups of order~$p^{m}$,

-where $p$~is a prime, \ie\ the number of such groups no one

-of which is simply isomorphic with any other, is therefore equal

-to the number of partitions of~$m$. When the prime~$p$ is given,

-each type of group may be conveniently, and without ambiguity,

-represented by the symbol of the corresponding partition.

-Thus the typical group~$G$ that we have been dealing with

-would be represented by the symbol $(m_{1}, m_{2}, \Dots, m_{r})$.

-

-\Par{45.} Having thus determined all distinct types of Abelian

-groups of order~$p^{m}$, a second general problem in this connection

-is the determination of all possible types of sub-group

-when the group itself is given. This will be facilitated by the

-consideration of a second special class of sub-groups in addition

-to the sub-groups~$G_{\mu}$ already dealt with.

-

-If $S$ and $S'$ are any two operations of~$G$, then

-\[

-S^{p^{\mu}} S'^{p^{\mu}} = (SS')^{p^{\mu}};

-\]

-and therefore the totality of the distinct operations obtained by

-raising every operation of~$G$ to the power~$p^{\mu}$ will form a

-sub-group~$H_{\mu}$.

-

-If

-\[

-m_{s} \geq \mu > m_{s + 1},

-\]

-then (Theorem~II, §~41)

-\[

-S^{p^{\mu}}

-  = P_{1}^{\alpha_{1} p^{\mu}}

-    P_{2}^{\alpha_{2} p^{\mu}} \Dots

-    P_{s}^{\alpha_{s} p^{\mu}},

-\]

-$S$~being any operation of~$G$. Hence $H_{\mu}$~is generated by the $s$~independent

-operations

-\[

-P_{1}^{p^{\mu}},\ P_{2}^{p^{\mu}},\ \Dots,\ P_{s}^{p^{\mu}};

-\]

-and the order of~$H_{\mu}$ is~$p^{\esum_{1}^{s} m_{t} - \mu s}$.

-

-\Par{46.} Let now $\Gamma$ of type $(n_{1}, n_{2}, \Dots, n_{s})$ be any sub-group of~$G$.

-The order of the group~$\Gamma_{1}$, formed of all the operations of~$\Gamma$

-which satisfy the equation

-\[

-S^{p} = 1,

-\]

-%% -----File: 072.png---Folio 56-------

-is~$p^{s}$ (§~42). This group must be identical with or be a sub-group

-of~$G_{1}$ whose order is~$p^{r}$. Hence

-\[

-s \leq r,

-\]

-\ie\ the number of independent generating operations of any

-sub-group of~$G$ is equal to or is less than the number of independent

-generating operations of $G$~itself.

-

-If $p^{n}$ is the order of~$\Gamma$, there must be one or more sub-groups

-of~$G$ of order~$p^{n + 1}$ which contain~$\Gamma$; for $\dfrac{G}{\Gamma}$~has operations

-of order~$p$. Hence $\Gamma$~must be contained in one or more

-sub-groups of~$G$ of order~$p^{m - 1}$. We may begin then by

-considering all possible types of sub-groups of~$G$ of order~$p^{m - 1}$.

-Let $g$ be such a sub-group and $(m_{1}', m_{2}', \Dots, m_{r'}')$ its type; and

-suppose that

-\[

-m_{t} = m_{t}'\quad (t = 1, 2, \Dots, k - 1),

-\]

-and

-\[

-m_{k} \neq m_{k}'.

-\]

-

-Then if

-\[

-m_{k} < m_{k}',

-\]

-the order of~$h_{m_{k}}$, the sub-group of~$g$ which results by raising all

-its operations to the power~$p^{m_{k}}$, is greater than the order of~$H_{m_{k}}$.

-This is impossible, since $h_{m_{k}}$~must coincide with~$H_{m_{k}}$ or with one

-of its sub-groups. Hence

-\[

-m_{k}' < m_{k}.

-\]

-

-Now

-\[

-\sum_{1}^{r'} m_{t}' = \sum_{1}^{r} m_{t} - 1;

-\]

-therefore

-\[

-m_{k}' = m_{k} - 1,

-\]

-and

-\[

-m_{t}' = m_{t}\quad (t = k + 1, \Dots, r).

-\]

-

-In the particular case in which $m_{r}$~is unity if we take $k$

-equal to~$r$, $m_{r}'$~is zero; and in this case $g$ would have $r - 1$

-generating operations.

-

-It is easy to see that sub-groups of order~$p^{m - 1}$ and of all

-the types just determined actually exist. For

-\[

-P_{1},\ P_{2},\ \Dots,\ P_{k - 1},\ P_{k^{p}},\ P_{k + 1},\ \Dots,\ P_{r}

-\]

-are a set of independent operations which generate a sub-group

-of the type $(m_{1}, m_{2}, \dots, m_{k - 1}, m_{k} - 1, m_{k + 1}, \dots, m_{r})$.

-%% -----File: 073.png---Folio 57-------

-

-\Par{47.} Returning to the general case, we will assume for all

-orders not less than~$p^{n}$ the existence of a sub-group~$\Gamma$ of type

-$(n_{1}, n_{2}, \Dots, n_{s})$, when the inequalities

-\[

-\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots\Add{,} r)

-\]

-are satisfied, where if $t > s$, $n_{t}$~is zero. When $n$ is equal to $m - 1$,

-the truth of this assumption has been established.

-

-If $\Gamma'$ of type $(n_{1}', n_{2}', \Dots, n_{s'}')$ is a sub-group of order~$p^{n - 1}$, it

-must be contained in some sub-group of order~$p^{n}$, say $\Gamma$ of type

-$(n_{1}, n_{1}, \Dots, n_{s})$. Then, by the preceding discussion,

-\[

-\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} n_{u}\quad (t = 1, 2, \Dots, s);

-\]

-and therefore

-\[

-\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r),

-\]

-so that the conditions assumed for sub-groups of order not less

-than~$p^{n}$ are necessary for sub-groups of order~$p^{n - 1}$. Moreover if

-the conditions

-\[

-\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r)

-\]

-are satisfied, and if $n_{k}'$ is the first of the series

-\[

-n_{1}',\ n_{2}',\ \Dots,

-\]

-which is less than the corresponding term of the series

-\[

-m_{1},\ m_{2},\ \Dots,

-\]

-we may take

-\begin{align*}

-n_{u} &= n_{u}',\quad u \neq k, \\

-n_{k} &= n_{k}' + 1;

-\end{align*}

-and then the conditions

-\[

-\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (\DPtypo{k}{t} = 1, 2, \Dots, r);

-\]

-are satisfied. But these conditions being satisfied it follows,

-from the assumption made, that $G$ contains a sub-group of order~$p^{n}$

-and type $(n_{1}, n_{2}, \Dots, n_{s})$ and therefore also a sub-group of

-order~$p^{n-1}$ and type $(n_{1}', n_{2}', \Dots, \DPtypo{n_{s'}}{n_{s'}'})$. If the conditions assumed

-are sufficient for the existence of a sub-group of order~$p^{n}$, they

-are thus proved to be sufficient for the existence of one of

-order~$p^{n-1}$. Hence since they are sufficient in the case of

-sub-groups of order~$p^{m-1}$, they are so generally.

-%% -----File: 074.png---Folio 58-------

-

-We may summarize the results obtained in the last three

-paragraphs as follows:---

-

-\begin{Theorem}[III.]

-The number of distinct types of Abelian

-groups of order~$p^{m}$, where $p$~is a prime, is equal to the number of

-partitions of~$m$; and each type may be completely represented

-by the symbol $(m_{1}, m_{2}, \Dots, m_{r})$ of the corresponding partition.

-If the numbers in the partition are written in descending order,

-a group of type $(m_{1}, m_{2}, \Dots, m_{r})$ will have a sub-group of type

-$(n_{1}, n_{2}, \Dots, n_{s})$, when the conditions

-\begin{gather*}

-s \leq r, \\

-\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r),

-\end{gather*}

-are satisfied; and the type of every sub-group must satisfy these

-conditions.

-\end{Theorem}

-

-\Par{48.} It will be seen later that the Abelian group of order~$p^{m}$

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$}%

-and type $(1, 1, 1, \Dots, \text{with $m$ units})$ is of special importance

-in the general theory, and we shall here discuss one or two of

-its simpler properties.

-

-Since the generating operations of the group are all of order~$p$,

-every operation except identity is of order~$p$; and therefore

-the type of any sub-group of order~$p^{s}$ is $(1, 1, 1, \Dots, \text{with $s$

-units})$. If the group be denoted by~$G$, every sub-group~$G_{\mu}$

-coincides with~$G$; while of the sub-groups~$H_{\mu}$, the first coincides

-with~$G$ and all the rest consist of the identical operation only.

-

-In choosing a set of independent generating operations, we

-may take for the first,~$P_{1}$, any one of the $p^{m} - 1$ operations of

-the group, other than identity. The sub-group~$\{P_{1}\}$ is of order~$p$;

-and therefore $G$ has $p^{m} - p$ operations which are not contained

-in~$\{P_{1}\}$. If we choose any one of these,~$P_{2}$, it is necessarily

-independent of~$P_{1}$, and may be taken as a second generating

-operation. The sub-group $\{P_{1}, P_{2}\}$ is of order~$p^{2}$ and type

-$(1, 1)$; and $G$ has $p^{m} - p^{2}$ operations which are not contained in

-this sub-group. If $P_{3}$ be any one of these, no power of~$P_{3}$

-other than identity is contained in $\{P_{1}, P_{2}\}$; and $P_{1}$,~$P_{2}$,~$P_{3}$ are

-therefore three independent operations which generate a sub-group

-of order~$p^{3}$. This process may clearly be continued till

-all $m$ generating operations have been chosen. If then the

-%% -----File: 075.png---Folio 59-------

-position which each generating operation occupies in the set of~$m$,

-when they are written in order, be taken into account, there are

-\[

-(p^{m} - 1) (p^{m} - p) (p^{m} - p^{2}) \Dots (p^{m} - p^{m - 1})

-\]

-distinct ways in which a set may be chosen. If on the other

-hand the sets of generating operations which consist of the

-same operations written in different orders be regarded as

-identical, the number of distinct sets is

-\[

-\frac{(p^{m} - 1) (p^{m} - p) \Dots (p^{m} - p^{m - 1})}{m!}.

-\]

-

-\Par{49.} No operation $P$ of the group can belong to two distinct

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!number of distinct ways of choosing a set of independent generating operations of}%

-sub-groups of order~$p$ except the identical operation. Hence

-since every sub-group of order~$p$ contains $p - 1$ operations

-besides identity, $G$~must contain $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of order~$p$.

-

-Let $N_{m, r}$ be the number of sub-groups of~$G$ of order~$p^{r}$, so

-that

-\[

-N_{m, 1} = \frac{p^{m} - 1}{p - 1}.

-\]

-

-There are, in~$G$, $p^{m} - p^{r}$ operations not contained in any given

-sub-group of order~$p^{r}$. If $P$~occurs among these operations, so

-also do $P^{2}$, $P^{3}$,~\Dots, $P^{p - 1}$. Hence there are $\dfrac{p^{m} - p^{r}}{p - 1}$ sub-groups of

-order~$p$ in~$G$ which are not contained in a given sub-group of

-order~$p^{r}$. Each of these may be combined with the given

-sub-group to give a sub-group of order~$p^{r + 1}$. When every

-sub-group of order~$p^{r}$ is treated in this way, every sub-group of

-order~$p^{r + 1}$ will be formed and each of them the same number,~$x$,

-of times. Hence

-\[

-xN_{m, r + 1} = N_{m, r} \frac{p^{m} - p^{r}}{p - 1}.

-\]

-

-Now a sub-group of order~$p^{r + 1}$ contains $N_{r + 1, r}$ sub-groups of

-order~$p^{r}$, and $\dfrac{p^{r + 1} - p^{r}}{p - 1}$ sub-groups of order~$p$ which are not

-contained in any given sub-group of order~$p^{r}$. Hence

-\[

-x = \frac{p^{r + 1} - p^{r}}{p - 1}  N_{r + 1, r},

-\]

-and therefore

-\[

-N_{m, r + 1} = \frac{N_{m, r}}{N_{r + 1\DPtypo{\,}{,} r}} · \frac{p^{m - r} - 1}{p - 1}.

-\]

-\index{Abelian group@\Topic{Abelian group}!sub-groups of|)}%

-%% -----File: 076.png---Folio 60-------

-

-We will now assume that

-\[

-N_{m, t} = \frac{(p^{m} - 1)(p^{m - 1} - 1) \Dots (p^{m - t + 1} - 1)}{(p - 1)(p^{2} - 1) \Dots (p^{t} - 1)},

-\]

-for all values of~$m$ and for values of $t$ not exceeding~$r$. This

-has been proved for $r = 1$. Then it follows, from the above

-relation, that

-\[

-N_{m, r + 1} = \frac{(p^{m} - 1)(p^{m - 1} - 1) \Dots (p^{m - r} - 1)}{(p - 1)(p^{2} - 1) \Dots (p^{r + 1} - 1)},

-\]

-that is to say, if the result is true for values of $t$ not exceeding~$r$,

-it is also true when $t = r + 1$. Hence the formula is true

-generally.

-

-It may be noticed that

-\[

-N_{m, t} = N_{m, m - t}.

-\]

-

-\Par{50.} \begin{Ex}[1.] Shew that a group whose operations except

-identity are all of order~$2$ is necessarily an Abelian group.

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!number of sub-groups of, whose order is given}%

-\end{Ex}

-

-\begin{Ex}[2.] Prove that in a group of order~$16$, whose operations

-except identity are all of order~$2$, the $15$~operations of order~$2$ may

-be divided into $5$~sets of $3$ each so that each set of~$3$ with identity

-forms a sub-group of order~$4$; and that this division into sets may

-be carried out in $\Erratum{7}{56}$ distinct ways.

-\end{Ex}

-

-\begin{Ex}[3.] If $G$ is an Abelian group and $H$ a sub-group of~$G$,

-shew that $G$~contains one or more sub-groups simply isomorphic

-with~$\dfrac{G}{H}$.

-\end{Ex}

-

-\begin{Ex}[4.] If the symbols in the successive rows of a determinant

-of $n$~rows are derived from those of the first row by performing on

-them the substitutions of a regular Abelian group of order~$n$, prove

-that the determinant is the product of $n$~linear factors.

-\Attrib{\textit{Messenger of Mathematics}, Vol.~\VolNo{XXIII} p.~112.}

-\end{Ex}

-

-\begin{Ex}[5.] Discuss the number of ways in which a set of

-independent generating operations of an Abelian group of order~$p^{m}$

-and given type may be chosen. Shew that, for a group of type

-$(m_{1}, m_{2}, \Dots, m_{r})$, where $m_{1} > m_{2} > \Dots > m_{r}$, the number of ways is

-of the form $p^{\alpha}(p - 1)^{r}$; and in particular that for a group of order

-$p^{\efrac{1}{2} n(n + 1)}$ and type $(n, n - 1, \Dots, 2, 1)$, the number of ways is

-$p^{\nu}(p - 1)^{n}$, where $\nu = \frac{1}{6} n(n + 1)(2n + 1) - n$.

-\end{Ex}

-

-\begin{Ex}[6.] Shew that for any Abelian group a set of independent

-generating operations

-\[

-S_{1},\ S_{2},\ \dots,\ S_{r - 1},\ S_{r},\ \dots,\ S_{n}

-\]

-can be chosen such that, for each value of~$r$, the order of~$S_{r}$ is equal

-to, or is a factor, of the order of~$S_{r - 1}$.

-\end{Ex}

-%% -----File: 077.png---Folio 61-------

-

-

-\Chapter{V.}{On Groups Whose Orders Are the Powers of

-Primes.}

-

-\Par{51.} \First{Having} in the last chapter dealt in some detail with

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime|EtSeq}%

-Abelian groups of order~$p^{m}$, where $p$~is a prime, we shall now investigate

-some of the more important properties of groups, which

-have the power of a prime for their order but are not necessarily

-Abelian. Besides illustrating and leading to many interesting

-applications of the general theorems of Chapter~III, the discussion

-of groups, whose order is the power of a prime, will be

-found in many ways to facilitate the subsequent discussion of

-other groups, whose order is not thus limited.

-

-\Par{52.} If $G$~is a group whose order is~$p^{m}$, where $p$~is a prime,

-the order of every sub-group of~$G$ must also be a power of~$p$; and

-therefore (§~25) the total number of operations of~$G$ which are

-conjugate with any given operation must be a power of~$p$. The

-identical operation of~$G$ is self-conjugate. Hence if the operations

-of~$G$, other than the identical operation, are distributed

-in conjugate sets containing $p^{\alpha}$, $p^{\beta}$,~\Dots\ operations, the total

-number of operations of~$G$ is $1 + p^{\alpha} + p^{\beta} + \Dots$; and therefore

-\[

-p^{m} = 1 + p^{\alpha} + p^{\beta} + \Dots\Add{.}

-\]

-

-This equation can only be true if $p^{r_{1}} - 1$ of the indices

-$\alpha$, $\beta$,~\Dots\ are zero, $r_{1}$~being some integer not less than unity.

-%% -----File: 078.png---Folio 62-------

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!always contain self-conjugate operations}%

-\index{Self-conjugate operation@\Topic{Self-conjugate operation}!of a group whose order is the power of a prime}%

-Hence $G$~must contain $p^{r_{1}}$ self-conjugate operations, which form

-(§~27) a self-conjugate sub-group\footnotemark.

-\footnotetext{Sylow, \textit{Math.\ Ann.}\ \VolNo{V}.\ (1872), p.~588.} %[** No comma or "Vol."]

-\index{Sylow@\Topic{Sylow}, quoted}%

-Hence:---

-

-\begin{Theorem}[I.]

-Every group whose order is the power of a

-prime contains self-conjugate operations, other than the identical

-operation; and no such group can be simple.

-\end{Theorem}

-

-\Par{53.} Let $H_{1}$ be the sub-group of~$G$ of order~$p^{r_{1}}$, formed of

-its self-conjugate operations. Then $\dfrac{G}{H_{1}}$~is a group of order~$p^{m - r_{1}}$,

-and it must contain self-conjugate operations other

-than identity, forming a self-conjugate sub-group of order~$p^{r_{2}}$.

-Corresponding to this self-conjugate sub-group of~$\dfrac{G}{H_{1}}$, $G$~must

-have a self-conjugate sub-group~$H_{2}$ of order~$p^{r_{1} + r_{2}}$, which

-contains~$H_{1}$. If $r_{1} + r_{2} < m$, $\dfrac{G}{H_{2}}$~is a group of order~$p^{m - r_{1} - r_{2}}$; and

-it again will have self-conjugate operations forming a sub-group

-of order~$p^{r_{3}}$. Corresponding to this, $G$~will have a self-conjugate

-sub-group~$H_{3}$ of order~$p^{r_{1} + r_{2} + r_{3}}$, which contains~$H_{2}$.

-If the process thus indicated is continued, till $G$~itself is arrived

-at, a series of groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{n},\ G

-\]

-is formed, each of which is self-conjugate in~$G$ and contains all

-that precede it\footnotemark.

-\footnotetext{Let $P$~be an operation or \DPchg{subgroup}{sub-group} of~$G$, and let $H$ be a \DPchg{subgroup}{sub-group} of~$G$

-  which contains~$P$. If every operation of~$H$ is permutable with~$P$, we shall

-  speak of $P$ as ``self-conjugate in~$H$.'' The phrase is to be regarded as

-  emphasizing the relation of~$P$ to a sub-group of~$G$ which contains it; and not

-  as implying that $H$ is the greatest, or the only, \DPchg{subgroup}{sub-group} of~$G$, within which $P$

-  is a self-conjugate operation or \DPchg{subgroup}{sub-group}.}

-Moreover $\dfrac{H_{r + 1}}{H_{r}}$~is the group formed of the

-self-conjugate operations of~$\dfrac{G}{H_{r}}$, and it is therefore an Abelian

-group.

-

-The series of sub-groups thus arrived at is of considerable

-importance. From the mode of formation, it is clear that a

-second group~$G'$ of order~$p^{m}$ cannot be simply isomorphic with~$G$,

-unless $\dfrac{H_{r + 1}}{H_{r}}$ and $\dfrac{H_{r + 1}'}{H_{r}'}$ are simply isomorphic for each value

-%% -----File: 079.png---Folio 63-------

-of~$r$. It by no means however follows that, when this latter

-condition is satisfied, $G$~and $G'$ are always simply isomorphic.

-This is indeed not necessarily the case; instances

-to the contrary will be found among the groups of order

-$p^{3}$ and $p^{4}$ in §§~69--72. While in general there is no limitation

-on the type of any Abelian factor-group $\dfrac{H_{r + 1}}{H_{r}}$ formed from

-two consecutive terms of the series, it is to be noted that

-the last factor-group $\dfrac{G}{H_{n}}$ cannot be cyclical\footnotemark,

-\footnotetext{Young, \textit{American Journal}, vol.~\VolNo{XV} (1893), p.~132.}%

-\index{Young@\Topic{Young}, quoted}%

-a restriction that

-can be established as follows. If $\dfrac{G}{H_{n}}$ were cyclical, then, because

-$\dfrac{H_{n}}{H_{n - 1}}$ is the group formed of the self-conjugate operations of

-$\dfrac{G}{H_{n - 1}}$, the operations of $\dfrac{G}{H_{n - 1}}$ could be arranged in the sets

-\[

-\dfrac{H_{n}}{H_{n - 1}}, \quad

-P\dfrac{H_{n}}{H_{n - 1}}, \quad

-P^{2} \dfrac{H_{n}}{H_{n - 1}}, \quad

-\dots,\quad

-P^{p^{\alpha - 1}} \dfrac{H_{n}}{H_{n - 1}},

-\]

-$p^{\alpha}$ being the order of $\dfrac{G}{H_{n}}$ and $P^{p^{\alpha}}$~being an operation of~$\dfrac{H_{n}}{H_{n - 1}}$.

-But since $P$~is permutable with its own powers and with every

-operation of~$\dfrac{H_{n}}{H_{n - 1}}$, these operations would form an Abelian

-group. Now the self-conjugate operations of~$\dfrac{G}{H_{n - 1}}$ form the

-group $\dfrac{H_{n}}{H_{n - 1}}$, and therefore $\dfrac{G}{H_{n - 1}}$ cannot be Abelian. Hence

-the assumption, that $\dfrac{G}{H_{n}}$~is cyclical, is incorrect.

-

-\begin{Remark}

-One immediate consequence of this result is that a group~$G$ of

-order~$p^{2}$ is necessarily Abelian. For if it were not, its self-conjugate

-operations would form a sub-group~$H$ of order~$p$, and $\dfrac{G}{H}$~being of

-order~$p$ would be cyclical.

-

-A second consequence is that if $G$, of order~$p^{m}$, is not Abelian,

-the order of the sub-group~$H_{1}$, formed of the self-conjugate operations

-of~$G$, cannot exceed $p^{m - 2}$.

-\end{Remark}

-%% -----File: 080.png---Folio 64-------

-

-\Par{54.} It was proved in the last chapter that an Abelian

-group has one or more sub-groups the order of which is any

-given factor of the order of the group itself. Hence, since $\dfrac{H_{r + 1}}{H_{r}}$

-is Abelian, $H_{r + 1}$~must have one or more sub-groups which contain~$H_{r}$,

-of any given order greater than that of~$H_{r}$ and less than its

-own.

-

-In particular, if $p^{s}$ and $p^{t}$ are the orders of $H_{r}$ and $H_{r + 1}$,

-then $\dfrac{H_{r + 1}}{H_{r}}$ must have a series of sub-groups of orders

-\[

-p^{t - s - 1},\ p^{t - s - 2},\ \Dots,\ p,\ 1,

-\]

-each of which contains those that follow it, while each one is of

-necessity self-conjugate in~$\dfrac{G}{H_{r}}$. Hence between $H_{r}$ and~$H_{r + 1}$, a

-series of groups $h_{s + 1}$, $h_{s + 2}$,~\Dots, $h_{t - 1}$ of orders

-$p^{s + 1}$, $p^{s + 2}$,~\Dots, $p^{t - 1}$

-can be chosen, so that each of the series

-\[

-H_{r},\ h_{s + 1},\ h_{s + 2},\ \Dots,\ h_{t - 1},\ H_{r + 1}

-\]

-is self-conjugate in~$G$ and contains all that precede it. We can

-therefore always form, in at least one way, a series of sub-groups

-of~$G$ of orders

-\[

-1,\ p,\ p^{2},\ \Dots,\ p^{m - 1},\ p^{m},

-\]

-such that each is self-conjugate in~$G$ and contains all that

-precede it.

-

-It will be shewn in the next paragraph that every sub-group

-of order~$p^{s}$ of~$G$ is contained self-conjugately in some

-sub-group of order~$p^{s + 1}$, and that every sub-group of order~$p^{s}$ of

-a group of order~$p^{s + 1}$ is self-conjugate. Assuming this result, it

-follows at once that a series of sub-groups of orders

-\[

-1,\ p,\ p^{2},\ \Dots,\ p^{m - 1},\ p^{m}

-\]

-can always be formed, which shall contain any one given

-sub-group and which shall be such that each group of the

-series contains the previous one self-conjugately. It will not

-however, in general, be the case, as in the previous series, that

-each group of the series is self-conjugate in~$G$.

-

-\Par{55.} Let $G_{s}$ be any sub-group of~$G$ of order~$p^{s}$, and let $H_{r + 1}$

-be the first of the series of sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{n}

-\]

-%% -----File: 081.png---Folio 65-------

-which is not contained in~$G_{s}$. Then $\dfrac{G_{s}}{H_{r}}$~is a sub-group of $\dfrac{G}{H_{r}}$

-which does not contain all the operations of~$\dfrac{H_{r + 1}}{H_{r}}$. Every

-operation of $\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate in~$\dfrac{G}{H_{r}}$; and therefore $\dfrac{G_{s}}{H_{r}}$ is

-self-conjugate in $\left\{\dfrac{G_{s}}{H_{r}}, \dfrac{H_{r + 1}}{H_{r}}\right\}$, a group of order greater than its

-own. Hence $G_{s}$~must be contained self-conjugately in some

-group $G_{s + t}$ of order~$p^{s + t}$, where $t$~is not less than unity.

-Moreover since $\dfrac{G_{s + t}}{G_{s}}$ must contain operations of order~$p$, there

-must be one or more groups of order~$p^{s + 1}$ which contain~$G_{s}$

-self-conjugately. Hence:---

-

-\begin{Theorem}[II.]

-If $G_{s}$ of order~$p^{s}$ is a sub-group of~$G$, which

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!every sub-group of, is contained self-conjugately in a greater sub-group}%

-is of order~$p^{m}$, then $G$~must contain a sub-group of order~$p^{s + t}$,

-$t \nless 1$, within which $G_{s}$~is self-conjugate. In particular, every

-sub-group of order~$p^{m - 1}$ of~$G$ is a self-conjugate sub-group\footnotemark.

-\end{Theorem}

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),

-  p.~173: Burnside, ``Notes on the theory of groups of finite order,'' \textit{Proc.\ London

-  Mathematical Society}, Vol.~\VolNo{XXVI} (1895), p.~209.}

-

-Suppose now that $G_{s + t}$, of order~$p^{s + t}$, is the greatest sub-group

-of~$G$ which contains a given sub-group~$G_{s}$, of order~$p^{s}$,

-self-conjugately; so that $G_{s}$~is one of $p^{m - s - t}$ conjugate sub-groups.

-Suppose also that $G_{s + t + u}$ of order~$p^{s + t + u}$, ($u \nless 1$), is

-the greatest sub-group of~$G$ that contains $G_{s + t}$ self-conjugately.

-Every operation of~$G_{s + t + u}$ transforms $G_{s + t}$ into itself; and no

-operation of $G_{s + t + u}$ that is not contained in $G_{s + t}$ transforms $G_{s}$

-into itself. Hence, in~$G_{s + t + u}$, $G_{s}$~is one of $p^{u}$~conjugate sub-groups

-and each of these is self-conjugate in~$G_{s + t}$.

-

-The $p^{m - s - t}$ sub-groups conjugate with~$G_{s}$ may therefore be

-divided into $p^{m - s - t - u}$ sets of $p^{u}$~each, ($u \nless 1$), such that any

-operation of one of the sets transforms each sub-group of that

-set into itself.

-

-Similarly if $G_{r}$, of order~$p^{r}$, is the greatest sub-group of~$G$

-that contains a given operation $P$ self-conjugately, and if $G_{r + s}$,

-$s \nless 1$, is the greatest sub-group that contains $G_{r}$ self-conjugately,

-then $G_{r}$~must contain self-conjugately $p^{s}$~operations of the

-%% -----File: 082.png---Folio 66-------

-conjugate set to which $P$ belongs, and therefore any two of

-these $p^{s}$~operations are permutable. Hence the $p^{m - r}$ conjugate

-operations of the set to which $P$ belongs can be divided into

-$p^{m - r - s}$ sets of $p^{s}$~each, ($s \nless 1$), such that all the operations of

-any one set are permutable with each other. In particular, if

-$P$~is one of a set of $p$ conjugate operations, all the operations of

-the set are permutable.

-

-\Par{56.} If $P$ is an operation of~$G$, of order~$p^{n}$, which is conjugate

-to one of its own powers~$P^{\alpha}$, there must be some other

-operation~$Q$ such that

-\[

-Q^{-1}PQ = P^{\alpha}.

-\]

-

-From this equation it follows that

-\[

-Q^{-1}P^{x}Q = (Q^{-1}PQ)^{x} = P^{x \alpha}.

-\]

-

-Again

-\begin{alignat*}{2}

-Q^{-2}PQ^{2} &= Q^{-1}P^{\alpha}Q &&= P^{\alpha^{2}}, \\

-Q^{-3}PQ^{3} &= Q^{-1}P^{\alpha^{2}}Q &&= P^{\alpha^{3}},

-\end{alignat*}

-and so on. Hence

-\[

-Q^{-y}P^{x}Q^{y} = P^{x \alpha^{y}}.

-\]

-

-If $Q^{\beta}$ is the lowest power of~$Q$ which is permutable with~$P$,

-then in~$\{P, Q\}$\DPnote{* comma?} $P$~will be one of $\beta$ conjugate operations, and

-therefore $\beta$~must be a power of~$p$, say~$p^{r}$, less than~$p^{n}$. Now

-\[

-Q^{-p^{r}}PQ^{p^{r}} = P^{\alpha^{p^{r}}};

-\]

-and therefore

-\[

-\alpha^{p^{r}} \equiv 1\ (\mod p^{n}),

-\]

-while

-\[

-\alpha^{p^{r - 1}} \nequiv 1\ (\mod p^{n}).

-\]

-

-First, we will suppose that $p$~is an odd prime. Then since

-\[

-x^{p^{r}} = x\ (\mod p),

-\]

-whatever integer $x$ may be, we may assume that $\alpha = 1 + kp^{s}$,

-where $k$~is not a multiple of~$p$; and then

-\begin{gather*}

-\alpha^{p^{r}} = 1 + kp^{r + s} + \Dots \\

-\alpha^{p^{r - 1}} = 1 + kp^{r + s - 1} + \Dots

-\end{gather*}

-

-Hence

-\[

-r + s = n,

-\]

-and

-\[

-\alpha \equiv 1\ (\mod p^{n - r}).

-\]

-

-In particular, we see that no operation of order~$p$ can be

-conjugate to one of its powers. Hence if $P$~and~$P'$ are two

-%% -----File: 083.png---Folio 67-------

-conjugate operations of order~$p$, $\{P\}$~and $\{P'\}$ have no operation

-in common except identity. Also, if $\{P\}$ be a self-conjugate

-sub-group of order~$p$, each of its operations is self-conjugate.

-

-If $p$ is~$2$, we must take

-\[

-\alpha = ±1 + k2^{s},

-\]

-where $k$~is odd, and we are led by the same process to the

-result

-\[

-\alpha \equiv ±1\ (\mod 2^{n - r}).

-\]

-

-\Par{57.} If $P$ is an operation of~$G$ which belongs to the sub-group~$H_{r + 1}$

-(§~53) and to no previous sub-group in the set

-\[

-1,\ H_{1},\ H_{2},\ \Dots,\ H_{n},\ G,

-\]

-then $\{P, H_{r}\}$ is a sub-group of~$H_{r + 1}$; and therefore every

-operation of $\dfrac{\{P, H_{r}\}}{H_{r}}$ is self-conjugate in~$\dfrac{G}{H_{r}}$. The set of

-operations~$PH_{r}$ is therefore transformed into itself by every

-operation of~$G$, and hence every operation that is conjugate to~$P$

-is contained in the set~$PH_{r}$. Suppose now, if possible, that

-all the operations conjugate with~$P$ are contained in the set~$PH_{r - 1}$.

-Then every operation of~$G$ transforms this set into

-itself, and therefore every operation of $\dfrac{\{P, H_{r - 1}\}}{H_{r - 1}}$ is self-conjugate

-in~$\dfrac{G}{H_{r - 1}}$. This however cannot be the case, since by supposition

-$P$~does not belong to~$H_{r}$. Hence among the operations

-conjugate to~$P$, there must be some which belong to~$PH_{r}$

-and not to~$PH_{r - 1}$. In particular, if $P$~is one of $p$ conjugate

-operations, no operation conjugate to~$P$ can belong to the set~$PH_{r - 1}$.

-For there must be an operation~$Q$ such that $Q^{-1}PQ$

-belongs to $PH_{r}$ and not to~$PH_{r - 1}$; and this being the case, no

-one of the operations $Q^{-x}PQ^{x}$, ($x = 1, 2, \Dots, p - 1$), can belong

-to~$PH_{r - 1}$. But these operations with~$P$ constitute the conjugate

-set.

-

-It follows from the above reasoning that, if $P$~belongs to~$H_{r + 1}$,

-then each operation of the set $P^{-\alpha} Q^{-1} P^{\alpha}Q$, where $Q$~is any

-operation of~$G$, belongs to~$H_{r}$. If $Q$~belongs to~$H_{s + 1}$, ($s < r$), the

-operation $P^{-\alpha} Q^{-1} P^{\alpha}Q$, regarded as the product of $P^{-\alpha} Q^{-1} P^{\alpha}$ and~$Q$,

-must similarly belong to~$H_{s}$. Hence unless every operation

-%% -----File: 084.png---Folio 68-------

-of~$H_{s + 1}$ is permutable with~$P$, there must be operations conjugate

-to~$P$, which belong to the set~$PH_{s}$. Moreover those

-operations of~$G$ which transform $P$ into operations of the set~$PH_{s}$

-form a sub-group. For if

-\[

-S^{-1}PS = Ph_{s},

-\]

-and

-\[

-S'^{-1}PS' = Ph_{s}',

-\]

-then

-\[

-S'^{-1} S^{-1} PSS' = Ph_{s}' · S'^{-1} h_{s}S,

-\]

-and when both $h_{s}$ and $h_{s}'$ belong to~$H_{s}$, so also does $h_{s}' S'^{-1} h_{s}S'$.

-

-\begin{Remark}

-\Par{58.} In illustration of the foregoing paragraph, we will consider

-a group~$G$ of order~$p^{r + s}$\Erratum{}{ where $p$~is an odd prime} in which the group~$H_{1}$, is of type~$(r)$, while

-$\dfrac{G}{H_{1}}$~is an Abelian group of type~$(1, 1, \Dots \text{with $s$ units})$.

-

-Let $Q$~be an operation of order~$p^{r}$ that generates~$H_{1}$; and let

-$P_{1}$, $P_{2}$,~\Dots, $P_{s}$ be $s$~operations no one of which is a power of any

-other, such that, with $Q$, they generate~$G$.

-

-If $P_{1}$ and $P_{2}$ are not permutable, then

-\[

-P_{2}^{-1} P_{1} P_{2} = P_{1} Q^{\alpha},

-\]

-and

-\[

-P_{2}^{-p} P_{1} P_{2}^{p} = P_{1} Q^{p\alpha}.

-\]

-

-Now $P_{2}^{p}$ belongs to~$H_{1}$, and therefore

-\[

-Q^{p\alpha} = 1,

-\]

-so that

-\[

-\alpha = kp^{r-1}.

-\]

-

-Hence the only operations which are conjugate to~$P_{1}$ are

-\[

-P_{1},\ P_{1} Q^{p^{r - 1}},\ P_{1} Q^{2p^{r - 1}},\ \Dots,\ P_{1} Q^{(p-1)p^{r - 1}};

-\]

-and $P_{1}$ is one of a set of $p$ conjugate operations, which are transformed

-among themselves by~$P_{2}$. Similarly $P_{2}$~is one of a set of $p$

-conjugate operations which are transformed among themselves by~$P_{1}$.

-Hence both $P_{1}$ and $P_{2}$ are permutable with each of the

-operations $P_{3}$, $P_{4}$,~\Dots, $P_{s}$. Again, $P_{3}$~is not self-conjugate, and

-there must therefore be another, say~$P_{4}$, of the set $P_{3}$, $P_{4}$,~\Dots, $P_{s}$

-which, with its powers, transforms $P_{3}$ into a set of $p$ conjugate

-operations. Then $P_{3}$~will similarly transform~$P_{4}$; and $P_{3}$,~$P_{4}$ will

-be permutable with each of the set $P_{5}$,~\Dots, $P_{s}$, $P_{1}$,~$P_{2}$. Hence

-finally the set of $s$ operations $P_{1}$, $P_{2}$,~\Dots, $P_{s}$ must be divisible

-into sets of two, such that each pair are permutable with all the

-remaining operations, but are not permutable with each other. The

-group can therefore only exist if $s$~is even\footnotemark.

-\footnotetext{Young, ``On groups whose order is the power of a prime,'' \textit{American

-  Journal of Mathematics}, Vol.~\VolNo{XV} (1893), p.~171.}

-\index{Young@\Topic{Young}, quoted}%

-%% -----File: 085.png---Folio 69-------

-

-The sub-group $\{P_{1}, P_{2}, Q\}$ is a group which satisfies the same

-conditions as~$G$, when $s = 2$. Its order is~$p^{r + 2}$, and it contains a

-self-conjugate operation of order~$p^{r}$. Now we shall see in §§~65,~66

-that such a group is necessarily of one of two types\footnotemark,

-\footnotetext{Let

-  \begin{gather*}

-  S_{1}^{\alpha_{1}} = 1,\quad S_{2}^{\alpha_{2}} = 1,\ \Dots,\quad S_{n}^{\alpha_{n}} = 1, \\

-  f_{1}(S_{i}) = 1,\quad f_{2}(S_{i}) = 1,\ \Dots,\quad f_{k}(S_{i}) = 1,

-  \end{gather*}

-  where $f_{j}(S_{i})$ is an abbreviation for an expression of the form $S_{p}^{a} S_{q}^{b} \Dots S_{r}^{c}$, be a

-  set of relations, such as was considered in §~18, which completely specify a

-  group. We may then, without altering the sense in which the word ``type''

-  has been used (§§~19,~44), speak of \emph{the type of group} defined by these relations.

-  It is essential however that the relations should completely specify the group,

-  as otherwise they will define more than one type. For instance, it is clear, from

-  §~56, that the equations

-  \[

-  Q^{p} = 1,\quad P^{p^{2}} = 1,\quad Q^{-1}PQ = P^{\alpha},

-  \]

-  where $p$~is a given prime, but $\alpha$~is not given, will define more than one type of

-  group. Any data in fact, which completely specify a group, may be said to

-  define a type of group. Thus in dealing with Abelian groups of order~$p^{m}$, a type

-  is defined by a partition of~$m$.}%

-the generating

-operations of which satisfy one or the other of the sets of equations

-\[

-Q^{p^{r}} = 1,\quad

-P_{1}^{p} = 1,\quad

-P_{2}^{p} = 1,\quad

-P_{2}^{-1} P_{1}P_{2} = P_{1} Q^{p^{r - 1}}:

-\]

-or

-\[

-Q^{p^{r}} = 1,\quad

-P_{1}^{p} = Q,\quad

-P_{2}^{p} = 1,\quad

-P_{2}^{-1} P_{1}P_{2} = P_{1} Q^{p^{r - 1}}.

-\]

-

-The same is true for the sub-groups $\{P_{3}, P_{4}, Q\}$, etc.; and all the

-operations of any one of these sub-groups are permutable with each

-operation of $\{P_{1}, P_{2}, Q\}$.

-

-Hence finally, since the equations to be satisfied by each pair of

-operations, such as $P_{1}$~and~$P_{2}$, may be chosen in two distinct ways,

-there are in all $2^{\efrac{1}{2}s}$ distinct types of group of order~$p^{r + s}$, for which

-$H_{1}$~is a cyclical group of order~$p^{r}$, and $\dfrac{G}{H_{1}}$~is an Abelian group of type

-$(1, 1, \Dots\DPchg{\,}{,} \text{to $s$ units})$.

-\end{Remark}

-

-\Par{59.} It has been seen in §~55 that every sub-group~$G'$ of

-order~$p^{m - 1}$ of a group~$G$ of order~$p^{m}$ is self-conjugate. Suppose

-now that $G$~contains two such sub-groups $G'$~and~$G''$. Then

-since $G'$ and $G''$ are permutable with each other, while the

-order of~$\{G', G''\}$ is~$p^{m}$, the order of the greatest group~$g'$

-common to them must (§~33) be~$p^{m - 2}$; and since $g'$~is the

-greatest common sub-group of two self-conjugate sub-groups of~$G$,

-it must itself be a self-conjugate sub-group of~$G$. The factor

-group $\dfrac{G}{g'}$ of order~$p^{2}$ contains the two distinct sub-groups $\dfrac{G'}{g'}$

-and $\dfrac{G''}{g'}$, which are of order~$p$ and permutable with each other.

-%% -----File: 086.png---Folio 70-------

-Hence $\dfrac{G}{g'}$ must be an Abelian group of type $(1, 1)$, and

-it therefore contains (§~49) $p + 1$ sub-groups of order~$p$. Hence,

-besides~$G'$, $G$~must contain $p$~other sub-groups of order~$p^{m-1}$

-which have in common with~$G'$ the sub-group~$g'$. If the $p + 1$

-sub-groups thus obtained do not exhaust the sub-groups of~$G$

-of order~$p^{m - 1}$, let $G'''$ be a new one. Then, as before, $G'$ and $G'''$

-must have a common sub-group~$g''$, of order~$p^{m - 2}$, which is

-self-conjugate in~$G$. If $g''$ were the same as~$g'$, $\dfrac{G'''}{g'}$~would be

-contained in~$\dfrac{G}{g'}$, which by supposition is not the case. It may

-now be shewn as above that there are, in addition to~$G'$, $p$~sub-groups

-of order~$p^{m - 1}$ which have in common with~$G'$ the

-group~$g''$. These are therefore necessarily distinct from those

-before obtained. This process may clearly be repeated till all

-the sub-groups of order~$p^{m - 1}$ are exhausted. Hence finally, if

-the number of sub-groups of~$G$, of order~$p^{m - 1}$ be $r_{m - 1}$, we have

-$r_{m - 1} \equiv 1\ (\mod p)$.

-

-\Par{60.} The self-conjugate operations of a group~$G$ of order~$p^{m}$,

-whose orders are~$p$, form with identity a self-conjugate

-sub-group whose order is some power of~$p$; and therefore their

-number must be congruent to $-1, \pmod{p}$. On the other hand,

-if $P$~is any operation of~$G$ of order~$p$ which is not self-conjugate,

-the number of operations in the conjugate set to which $P$

-belongs is a power of~$p$. Hence the total number of operations

-of~$G$, of order~$p$, is congruent to $-1, \pmod{p}$. Now if $r_{1}$~is the

-total number of sub-groups of~$G$ of order~$p$, the number of

-operations of order~$p$ is $r_{1}(p - 1)$, since no two of these sub-groups

-can have a common operation, except identity. It

-follows that

-\[

-r_{1}(p - 1) \equiv -1,\ (\mod p),

-\]

-and therefore

-\[

-r_{1} \equiv 1,\ (\mod p).

-\]

-

-If now $G_{s}$ is any sub-group of~$G$ of order~$p^{s}$, and if $G_{s + t}$~is the

-greatest sub-group of~$G$ in which $G_{s}$ is contained self-conjugately,

-then every sub-group of~$G$ which contains $G_{s}$ self-conjugately is

-contained in~$G_{s + t}$. But every sub-group of order~$p^{s + 1}$, which

-contains~$G_{s}$, contains $G_{s}$ self-conjugately; and therefore every

-%% -----File: 087.png---Folio 71-------

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of sub-groups of given order is congruent to unity, $\pmod{p}$}%

-sub-group of order~$p^{s + 1}$, which contains~$G_{s}$, is itself contained in~$G_{s + t}$.

-By the preceding result, the number of sub-groups of

-$\dfrac{G_{s + t}}{G_{s}}$ of order~$p$ is congruent to unity, $\pmod{p}$. Hence the

-number of sub-groups of~$G$ of order~$p^{s + 1}$, which contain~$G_{s}$

-of order~$p^{s}$, is congruent to unity,~$\pmod{p}$.

-

-\Par{61.} Let now $r_{s}$ represent the total number of sub-groups of

-order~$p^{s}$ contained in a group~$G$ of order~$p^{m}$. If any one of

-them is contained in $a_{x}$~sub-groups of order~$p^{s + 1}$, and if any one

-of the sub-groups of order~$p^{s + 1}$ contains $b_{y}$~sub-groups of order~$p^{s}$\DPchg{;}{,}

-then

-\[

-\sum_{x = 1}^{x = r} a_{x} = \sum_{y = 1}^{y = r_{s + 1}} b_{y};

-\]

-for the numbers on either side of this equation are both equal

-to the number of sub-groups of order~$p^{s + 1}$, when each of the

-latter is reckoned once for every sub-group of order~$p^{s}$ that it

-contains. It has however been shewn, in the two preceding

-paragraphs, that for all values of $x$~and~$y$

-\[

-a_{x} \equiv 1,\quad b_{y} \equiv 1\ (\mod p).

-\]

-

-Hence

-\[

-r_{s} \equiv r_{s + 1}\ (\mod p).

-\]

-

-Now it has just been proved that

-\[

-r_{1} \equiv 1 \quad \text{and}\quad r_{m - 1} \equiv 1\ (\mod p);

-\]

-and therefore finally, for all values of~$s$,

-\[

-r_{s} \equiv 1\ (\mod p).

-\]

-We may state the results thus obtained as follows:---

-

-\begin{Theorem}[III.]

-The number of sub-groups of any given

-%[** TN: "mod" italicized in the original]

-order~$p^{s}$ of a group of order~$p^{m}$ is congruent to unity,~$\pmod{p}$\footnotemark.

-\end{Theorem}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, ``Verallgemeinerung des Sylow'schen Satzes,'' \textit{Berliner Sitzungsberichte},

-  (1895), p.~989.}

-

-\begin{Corollary}

-The number of self-conjugate sub-groups of

-order~$p^{s}$ of a group of order~$p^{m}$ is congruent to unity, \DPtypo{mod~$p$}{$\pmod{p}$}.

-\end{Corollary}

-

-This is an immediate consequence of the theorem, since the

-number of sub-groups in any conjugate set is a power of~$p$.

-

-\Par{62.} Having shewn that the number of sub-groups of~$G$ of

-order~$p^{s}$ is of the form $1 + kp$, we may now discuss under what

-%% -----File: 088.png---Folio 72-------

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!case in which there is only one sub-group of a given order|(}%

-circumstances it is possible for $k$ to be zero, so that $G$ then

-contains only one sub-group~$G_{s}$ of order~$p^{s}$.

-

-If this is the case, and if $P$~is any operation of~$G$ not contained

-in~$G_{s}$, the order of~$P$ must be not less than~$p^{s + 1}$; for if

-it were less, $G$~would have some sub-group of order~$p^{s}$ containing~$P$

-and this would necessarily be different from~$G_{s}$. If the order

-of~$P$ is~$p^{s + t}$, then $\{P^{p^{t}}\}$~is a cyclical sub-group of order~$p^{s}$; and it

-must coincide with~$G_{s}$. Hence, if $G_{s}$ is the only sub-group of

-order~$p^{s}$, it must be cyclical.

-

-Suppose now that $G$~contains operations of order~$p^{r}$ ($r > s$),

-but no operations of order~$p^{r + 1}$; and let $P$ be an operation of~$G$

-of order~$p^{r}$. Then $\{P\}$ must be contained self-conjugately in a

-non-cyclical sub-group of order~$p^{r + 1}$.

-

-We will take first the case in which $p$~is an odd prime.

-Then (§~55) $G$~must contain an operation~$P'$ which does not

-belong to~$\{P\}$, such that

-\[

-P'^{-1} PP' = P^{\alpha},\quad P'^{p} = P^{\beta}.

-\]

-If $\alpha$ were unity, $\{P, P'\}$ would be an Abelian group of order~$p^{r + 1}$

-containing no operation of order~$p^{r + 1}$. Its type would

-therefore be $(r, 1)$, and it would necessarily contain an operation

-of order~$p$ not occurring in~$\{P\}$. It has been shewn that this is

-impossible if $G_{s}$ is the only sub-group of order~$p^{s}$, and therefore

-$\alpha$~cannot be unity.

-

-We may then without loss of generality (§~56) assume that

-\[

-\alpha = 1 + p^{r - 1}.

-\]

-Moreover if $\beta$ were not divisible by~$p$, the order of~$P'$ would

-be~$p^{r + 1}$, contrary to supposition. Hence we must have the

-relations

-\[

-P'^{-1} PP' = P^{1 + p^{r - 1}},\quad P'^{p} = P^{\gamma p}.

-\]

-

-By successive applications of the first of these equations, we

-get

-\[

-P'^{-y} P^{x} P'^{y} = P^{x(1 + yp^{r - 1})},

-\]

-for all values of $x$ and~$y$; and from this it immediately follows

-that

-\begin{align*}

-(P^{x}P')^{p} &= P'^{p} P^{x\{p + \efrac{1}{2} p(p + 1)p^{r - 1}\}} \\

-              &= P^{(x + \gamma)p}.

-\end{align*}

-%% -----File: 089.png---Folio 73-------

-

-Hence if

-\[

-x = p^{r - 1} - \gamma,

-\]

-the order of $P^{x}P'$, an operation not contained in~$\{P\}$, is~$p$. This

-is impossible if $G_{s}$ is the only sub-group of order~$p^{s}$. If then

-$r < m$, $G$~must contain operations of order greater than~$p^{r}$; and

-$G$~is therefore a cyclical group. Hence:---

-

-\begin{Theorem}[IV.]

-If $G$, of order~$p^{m}$, where $p$~is an odd prime,

-contains only one sub-group of order~$p^{s}$, $G$~must be cyclical.

-\end{Theorem}

-

-\Par{63.} When $p = 2$, the result is not so simple.

-

-Let $Q$ be an operation that transforms $\{P\}$ into itself, and

-suppose that the lowest power of~$Q$ that occurs in~$\{P\}$ is~$Q^{4}$.

-Then $\{P, Q\}$ must be defined by

-\[

-Q^{-1}PQ = P^{\alpha},\quad Q^{4} = P^{\beta},

-\]

-where

-\[

-\alpha = ±1 + k2^{r - 2}.

-\]

-Moreover $\beta$ must be a multiple of~$4$, as otherwise the order of~$Q$

-would be greater than~$2^{r}$. A simple calculation now gives

-\[

-(P^{x}Q^{2})^{2} = P^{\beta + x(1 + \alpha^{2})};

-\]

-and since

-\[

-\beta \equiv 0\ (\mod 4),

-\]

-and

-\[

-\alpha^{2} \equiv 1\ (\mod 2^{r - 1}),

-\]

-$x$ can always be chosen so that

-\[

-\beta + x(1 + \alpha^{2}) \equiv 0\ (\mod 2^{r}).

-\]

-When $x$ is thus chosen, $P^{x}Q^{2}$~is an operation of order~$2$ which is

-not contained in~$\{P\}$. But this is inconsistent with $G_{s}$ being

-the only sub-group of order~$2^{s}$; hence $\{P\}$~must contain~$Q^{2}$.

-

-If now $\{P\}$ is not a self-conjugate sub-group, there must

-(§~55) be some operation~$P'$ of order~$p^{r}$, conjugate with~$P$ and

-not contained in~$\{P\}$, which transforms $\{P\}$ into itself; and

-then $\{P, P'\}$ is defined by

-\[

-P'^{-1}PP' = P^{\alpha},\quad P'^{2} = P^{2\beta},

-\]

-where $\alpha$ is $-1$ or $±1 + 2^{r - 1}$, and $\beta$~is odd.

-

-If

-\[

-\alpha = 1 + 2^{r - 1},

-\]

-then

-\[

-(P^{x}P')^{2} = P^{2\{\beta + x(1 + 2^{r - 2})\}};

-\]

-and $x$ can be chosen so that $P^{x}P'$~is of order~$2$. Hence this

-case cannot occur.

-%% -----File: 090.png---Folio 74-------

-

-If $\alpha$ is either $-1$ or $-1 + 2^{r - 1}$, the defining relations of

-$\{P, P'\}$ are not self-consistent unless $r$~is~$2$; for they lead to

-\[

-P'^{-1}P^{2\beta}P' = P^{-2\beta},

-\]

-and

-\[

-P'^{-1}P^{2\beta}P' = P^{2\beta}.

-\]

-

-If $r$ is~$2$, and if $P''$~is an operation that transforms $P$ into~$P'$,

-so that

-\[

-P''^{-1} PP'' = P',

-\]

-then

-\[

-(PP'')^{2} = PP''^{2}P' = P^{-1}P',

-\]

-since

-\[

-P^{2} = P'^{2} = P''^{2}.

-\]

-

-Hence $PP''$ would be of order~$8$; and therefore this case

-cannot occur.

-

-Hence, finally, $\{P\}$~must be self-conjugate.

-

-If $r + 1 < m$, $\{P\}$~must be transformed into itself by some

-operation~$Q'$ not contained in~$\{P,\ Q\}$, so that

-\[

-Q^{-1}PQ = P^{-1 + k2^{r - 1}},

-\]

-and

-\[

-Q'^{-1}PQ' = P^{-1 + k'2^{r - 1}},

-\]

-where $k$ and $k'$ are each either zero or unity. If both are zero

-or both unity, $QQ'$~is permutable with~$P$; and if one is zero and

-the other unity,

-\[

-(QQ')^{-1} P(QQ') = P^{1 + 2^{r - 1}}.

-\]

-In either case, the group contains an operation of order~$2$ that

-does not belong to~$\{P\}$, and this is inadmissible. Hence, lastly,

-$r$~must not be less than~$m - 1$.

-

-Suppose now that $r$~is equal to~$m - 1$, and that of the

-operations of~$G$, not contained in~$\{P\}$, $Q$~has as small an order

-as possible, say $2^t$ ($t \nless s + 1$). Then $Q^{2}$ of order~$2^{t - 1}$ is contained

-in~$\{P\}$; without loss of generality we may assume

-\[

-Q^{2} = P^{2^{m - t}}.

-\]

-

-Moreover

-\[

-Q^{-1}PQ = P^{-1}\quad \text{or}\quad P^{-1 + 2^{m - 2}},

-\]

-and hence

-\[

-Q^{-1} P^{2^{m - t}}Q = P^{-2^{m - t}}.

-\]

-Now $Q$ is permutable with~$P^{2^{m - t}}$, one of its own powers, and

-therefore $P^{2^{m - t}}$~is an operation of order~$2$. Hence $t$~is~$2$ and $s$~is

-unity. If then $s$~is greater than unity, $G$~must contain

-%% -----File: 091.png---Folio 75-------

-operations of order~$2^{m}$ and must therefore be cyclical. Moreover

-if $s$ is unity, the relations

-\[

-Q^{-1}PQ = P^{-1 + 2^{m - 2}}, \quad Q^{2} = P^{2^{m - 2}},

-\]

-lead to

-\[

-(PQ)^{2} = 1,

-\]

-and they are therefore inadmissible.

-

-On the other hand, the relations

-\[

-Q^{-1}PQ = P^{-1}, \quad Q^{2} = P^{2^{m - 2}},

-\]

-give

-\[

-(P^{x}Q)^{2} = P^{2^{m - 2}};

-\]

-and every operation of~$G$, not contained in~$\{P\}$, is of order~$4$.

-Also these relations are clearly self-consistent, and they define

-a group of order~$2^{m}$. Hence finally:---

-

-\begin{Theorem}[V.]

-If a group~$G$, of order~$2^{m}$, has a single

-sub-group of order~$2^{s}$, \($s > 1$\), it must be cyclical; if it has a

-single sub-group of order~$2$, it is either cyclical or of the type

-defined by

-\[

-P^{2^{m - 1}} = 1,\quad Q^{2} = P^{2^{m - 2}},\quad Q^{-1}PQ = P^{-1}.

-\]

-\end{Theorem}

-

-\Par{64.} We shall now proceed to discuss, in application of the

-foregoing theorems and for the importance of the results themselves,

-the various types of groups of order~$p^{m}$ which contain

-self-conjugate cyclical sub-groups of orders $p^{m - 1}$ and $p^{m - 2}$

-respectively. It is clear from Theorem~V that the case $p = 2$

-requires independent investigation; we shall only deal in detail

-with the case in which $p$~is an odd prime, and shall state the

-results for the case when $p = 2$.

-

-The types of Abelian groups of order~$p^{m}$ which contain

-operations of order~$p^{m - 2}$ are those corresponding to the symbols

-$(m)$, $(m - 1, 1)$, $(m - 2, 2)$, and $(m - 2, 1, 1)$. We will assume

-that the groups which we consider in the following paragraphs

-are not Abelian.

-

-\Par{65.} We will first consider a group $G$, of order~$p^{m}$, which

-contains an operation~$P$ of order~$p^{m - 1}$. The cyclical sub-group~$\{P\}$

-is self-conjugate and contains a single sub-group $\{P^{p^{m - 2}}\}$ of

-order~$p$. By Theorem~IV, since $G$~is not cyclical, it must

-contain an operation~$Q'$, of order~$p$, which does not occur in~$\{P\}$.

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!case in which there is only one sub-group of a given order|)}%

-%% -----File: 092.png---Folio 76-------

-Since $\{P\}$ is self-conjugate and the group is not Abelian, $Q'$~must

-transform $P$ into one of its own powers. Hence

-\[

-Q'^{-1}PQ' = P^{\alpha},

-\]

-and since $Q'^{p}$ is permutable with~$P$ it follows, from §~56, that

-\[

-\alpha = 1 + kp^{m - 2}.

-\]

-

-Since the group is not Abelian, $k$~cannot be zero; but it may

-have any value from $1$ to~$p - 1$. If now

-\[

-kx \equiv 1\ (\mod p),

-\]

-then

-\[

-Q'^{-x} PQ'^{x} = P^{1 + p^{m - 2}};

-\]

-and therefore, writing $Q$ for~$Q'^{x}$, the group is defined by

-\[

-P^{p^{m - 1}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p{m - 2}}.

-\]

-

-These relations are clearly self-consistent, and they define a

-group of order~$p^{m}$.

-

-There is therefore a single type of non-Abelian group of

-order~$p^{m}$ which contains operations of order~$p^{m - 1}$, because, for

-any such group, a pair of generating operations may be chosen

-which satisfy the above relations.

-

-From the relation

-\[

-Q^{-1}PQ = P^{1 + p^{m - 2}},

-\]

-it follows by repetition and multiplication that

-\[

-Q^{-y} P^{x} Q^{y} = P^{x(1 + yp^{m - 2})},

-\]

-and therefore that

-\[

-(P^{x} Q^{y})^{z} = P^{xz\{1 + \efrac{1}{2}(z + 1)yp^{m - 2}\}(1 - yzp^{m - 2})} Q^{yz},

-\]

-and

-\[

-(P^{x} Q^{y})^{p} = P^{xp}.

-\]

-

-Hence $G$ contains $p$~cyclical sub-groups of order~$p^{m - 1}$ of

-which $P$ and~$PQ^{y}$ ($y = 1, 2, \Dots, p - 1$) may be taken as the

-generating operations. Since $Q$ and $P^{p}$ are permutable, $G$~also

-contains an Abelian non-cyclical sub-group $\{Q, P^{p}\}$ of

-order~$p^{m - 1}$ It is easy to verify that the $1 + p$ sub-groups thus

-obtained exhaust the sub-groups of order~$p^{m - 1}$; and that, for any

-other order~$p^{s}$, there are also exactly $p + 1$ sub-groups of which

-$p$~are cyclical and one is Abelian of type~$(s - 1, 1)$.

-

-\begin{Remark}

-The reader will find it an instructive exercise to verify the

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$|(}%

-results of the corresponding case where $p$~is~$2$; they may be stated

-thus. There are four distinct types of non-Abelian group of order

-%% -----File: 093.png---Folio 77-------

-$2^{m}$, which contain operations of order~$2^{m - 1}$, when $m > 3$. Of these,

-one is the type given in Theorem~V, and the remaining three are

-defined by

-\begin{alignat*}{3}

-P^{2^{m - 1}} &= 1,\quad &Q^{2} &= 1,\quad &QPQ &= P^{1 + 2^{m - 2}}; \\

-P^{2^{m - 1}} &= 1,      &Q^{2} &= 1,      &QPQ &= P^{-1 + 2^{m - 2}}; \\

-P^{2^{m - 1}} &= 1,      &Q^{2} &= 1,      &QPQ &= P^{-1}.

-\end{alignat*}

-

-When $m = 3$, there are only two distinct types. In this case, the

-second and the fourth of the above groups are identical, and the

-third is Abelian.

-\end{Remark}

-

-\Par{66.} Suppose next that $G$, a group of order~$p^{m}$, has a

-self-conjugate cyclical sub-group~$\{P\}$ of order~$p^{m - 2}$, and that

-no operation of~$G$ is of higher order than~$p^{m - 2}$. We may

-at once distinguish two cases for separate discussion; viz.\

-\Inum{(i)}~that in which $P$~is a self-conjugate operation, \Inum{(ii)}~that in

-which $P$~is not self-conjugate.

-

-Taking the first case, there can be no operation~$Q'$ in~$G$ such

-that $Q'^{p^{2}}$~is the lowest power of~$Q'$ contained in~$\{P\}$, for if there

-were, $\{Q', P\}$ would be Abelian and, its order being~$p^{m}$, it would

-necessarily coincide with~$G$. Hence any operation~$Q'$, not

-contained in~$\{P\}$, generates with~$P$ an Abelian group of type

-$(m - 2, 1)$, and we may choose $P$~and~$Q$ as independent

-generators of this sub-group, the order of~$Q$ being~$p$. If now

-$R'$~is any operation of~$G$ not contained in~$\{Q, P\}$, $R'^{p}$~must

-occur in this sub-group, and therefore

-\[

-R'^{p} = P^{\alpha} Q^{\beta}.

-\]

-

-If $\beta$ were not zero, $R^{p^{2}}$~would be the lowest power of~$R$

-that occurs in~$\{P\}$, and we have just seen that this cannot be

-the case. Hence

-\[

-R'^{p} = P^{\alpha},

-\]

-and $\{R', P\}$ is again an Abelian group of type $(m - 2, 1)$. If

-$P$~and~$R$ are independent generators of this group, the latter

-cannot occur in~$\{Q, P\}$. Now since $Q$~is not self-conjugate,

-\[

-R^{-1}QR = QP^{\beta};

-\]

-and since $R^{p}$ is permutable with~$Q$

-\[

-P^{p\beta} = 1,

-\]

-so that

-\[

-\beta \equiv 0\ (\mod p^{m - 3}).

-\]

-

-Hence

-\[

-R^{-1}QR = QP^{kp^{m - 3}},

-\]

-%% -----File: 094.png---Folio 78-------

-where $k$~is not a multiple of~$p$. If finally, $P^{k}$~be taken as a

-generating operation in the place of~$P$, the group is defined by

-\begin{gather*}

-P^{p^{m - 2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP^{p^{m - 3}}, \\

-PQ = QP,\quad PR = RP.

-\end{gather*}

-There is therefore a single type of group of order~$p^{m}$, which

-contains a self-conjugate operation of order~$p^{m - 2}$ and no operation

-of order~$p^{m - 1}$.

-

-\Par{67.} Taking next the second case, we suppose that $G$

-contains a cyclical self-conjugate sub-group~$\{P\}$ of order~$p^{m - 2}$,

-but no self-conjugate operation of this order.

-

-If $G$ contains an operation~$Q'$ such that $Q'^{p^{2}}$~is the lowest

-power of~$Q'$ occurring in~$\{P\}$, it follows (§~56), since $\{P\}$~is

-self-conjugate, that

-\[

-Q'^{-1}PQ' = P^{1 + kp^{m - 4}}.

-\]

-

-Moreover, since the order of~$Q'$ cannot exceed~$p^{m - 2}$, we have

-\[

-Q'^{p^{2}} = P^{\alpha p^{2}}.

-\]

-

-A simple calculation, similar to that of §~65, will now shew

-that

-\[

-(P^{x}Q')^{p^{2}} = P^{(x + \alpha)p^{2}},

-\]

-while

-\[

-(P^{x}Q')^{p} = Q'^{p} P^{x \{p + \efrac{1}{2}k(p + 1)p^{m - 3}\}}.

-\]

-

-Hence $P^{-\alpha}Q'$ is an operation of order~$p^{2}$, none of whose

-powers, except identity, is contained in~$\{P\}$. If this operation

-be represented by~$Q$, $G$~is defined by

-\[

-P^{p^{m - 2}} = 1,\quad Q^{p^{2}} = 1,\quad Q^{-1}PQ = P^{1 + kp^{m - 4}}.

-\]

-

-If $k$ is a multiple of~$p^{2}$, the group is Abelian.

-

-If $k$ is~$k'p$, and if $Q^{\alpha}$~is taken for a new generating operation,

-where $\alpha$~is chosen so that

-\[

-\alpha k' \equiv 1\ (\mod p),

-\]

-we have a single type defined by

-\[

-P^{p^{m - 2}} = 1,\quad Q^{p^{2}} = 1,\quad Q^{-1}PQ = P^{1 + p^{m - 3}}.

-\]

-

-If $k$ is not a multiple of~$p$, we may take it equal to $k_{1} + k_{2}p$,

-where $k_{1}$~and~$k_{2}$ are both less than~$p$, and $k_{1}$~is not zero. (This

-case cannot occur if $m < 5$.) Then

-\[

-Q^{-x_{1} -x_{2}p} PQ^{x_{1} + x_{2}p}

-  = P^{1 + k_{1}x_{1}p^{m - 4} + (k_{1}x_{2} + k_{2}x_{1})p^{m - 3}},

-\]

-%% -----File: 095.png---Folio 79-------

-when $m > 5$; while if $m = 5$, the coefficient of $p^{m - 3}$, viz.\ of~$p^{2}$, in

-the index of~$P$ must be increased by $\frac{1}{2} x_{1}(x_{1} - 1)k_{1}^{2}$.

-

-If now we choose $x_{1}$ so that

-\[

-k_{1}x_{1} = 1 + y_{1}p,

-\]

-and then $x_{2}$ so that

-\[

-y_{1} + k_{1}x_{2} + k_{2}x_{1} \equiv 0\ (\mod p),

-\]

-(with the suitable modification when $m = 5$), then $Q^{x_{1} + x_{2}p}$

-transforms $P$ into $P^{1 + p^{m - 4}}$. We therefore again in this case

-get a single type of group, which, if $Q^{x_{1} + x_{2}p}$ is represented by~$R$,

-is defined by

-\[

-P^{p^{m - 2}} = 1,\quad R^{p^{2}} = 1,\quad R^{-1}PR = P^{1 + p^{m - 4}}.

-\]

-As already stated, this type exists only when $m > 4$.

-

-If every operation of~$G$ is such that its $p$th~power occurs in~$\{P\}$,

-$P$~must be one of a set of $p$ conjugate operations; for if $P$~is

-one of $p^{2}$ conjugate operations, there must be an operation~$Q$

-which transforms $P$ into $P^{1 + kp^{m - 4}}$, where $k$~is not a multiple of~$p$,

-and then $Q^{p^{2}}$ is the lowest power of~$Q$ which is permutable

-with~$P$. Since $P$ is one of $p$ conjugate operations, it must be

-self-conjugate in an Abelian group $\{Q, P\}$ of type $(m - 2, 1)$.

-Again, since $P$ and $P^{1 + p^{m - 3}}$ are conjugate operations, $P$~must be

-contained in a group of order~$p^{m - 1}$, defined by

-\[

-P^{p^{m - 2}} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p^{m - 3}}.

-\]

-

-If $Q$ is not a self-conjugate operation, it must be one of $p$

-conjugate operations and (§~57) these must be

-\[

-Q, \quad QP^{p^{m - 3}}, \quad QP^{2p^{m - 3}},\ \Dots,\quad QP^{(p - 1)p^{m - 3}}.

-\]

-

-Now if

-\[

-R^{-1}QR = QP^{\alpha p^{m - 3}},

-\]

-then

-\[

-R^{-1} P^{-\alpha} QR = P^{-\alpha}Q.

-\]

-Hence, if $Q$ is not self-conjugate, $P^{-\alpha}Q$~is a self-conjugate

-operation of order~$p^{m - 2}$: and the group is that determined

-in the last paragraph. Therefore $Q$~must be self-conjugate,

-and the group is defined by

-\begin{gather*}

-P^{p^{m - 2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p^{m - 3}}, \\

-PQ = QP,\quad RQ = QR.

-\end{gather*}

-%% -----File: 096.png---Folio 80-------

-

-It is clear from these relations that the group thus arrived

-at is the direct product of the groups $\{P, R\}$ and~$\{Q\}$\footnotemark.

-\footnotetext{In each of the five distinct cases to which we have been led in the discussion

-  contained in §§~65--67, we have arrived at a set of defining relations, containing

-  no indeterminate symbols, such that in each case a set of generating operations

-  can be chosen to satisfy these relations. To justify the statement, in each

-  particular case, that such a set of relations gives a distinct type of group, it is

-  finally necessary to verify that the relations actually define a group of order~$p^{m}$.

-  In the cases dealt with in the text, this verification is implicitly contained in the

-  process by which the relations have been arrived at. We have therefore omitted

-  the direct verification, which moreover is extremely simple. We shall similarly

-  omit the corresponding verifications in the discussion of groups of orders $p^{3}$ and~$p^{4}$,

-  as in none of these cases does it present any difficulty.

-

-  To illustrate the necessity of such a verification in general, we may consider

-  a simple case. The relations

-  \[

-  P^{3} = 1,\quad Q^{9} = 1,\quad P^{-1}QP = Q^{\alpha},

-  \]

-  where $\alpha$~is any given integer, certainly define a group whose order is equal to or

-  is a factor of~$27$, since they indicate that $\{P\}$~and~$\{Q\}$ are permutable. They will

-  not however give a type of group of order~$27$, unless $\alpha$~is $1$,~$4$ or~$7$. For instance,

-  if $\alpha = 5$, the relations involve

-  \[

-  Q = P^{-3}QP^{3} = Q^{8},\quad\text{or}\quad Q^{7} = 1.

-  \]

-  Hence

-  \[

-  Q = Q^{4 · 7 - 3 · 9} = 1,

-  \]

-  and the relations hold only for a group of order~$3$.

-

-  Again, if $\alpha = 3$, the relations give

-  \[

-  P^{-1}Q^{3}P = Q^{9} = 1,\quad\text{or}\quad Q^{3} = 1,

-  \]

-  and as before they define a group of order~$3$.}

-

-\begin{Remark}

-It is to be expected, from the result of the corresponding case at

-the end of §~65, that the number of distinct types when $p = 2$ is

-much greater than when $p$~is an odd prime. There are, in fact,

-when $m > 5$, fourteen distinct types of groups of order~$2^{m}$, which

-contain a self-conjugate cyclical sub-group of order~$2^{m - 2}$ and no

-operation of order~$2^{m - 1}$. They may be classified as follows.

-

-Suppose first that the group has a self-conjugate operation of

-order~$2^{m - 2}$. There is then a single type defined by the relations

-\[

-\Inum{(i)}\quad A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad C^{2} = 1,\quad CBC = BA^{2^{m - 3}}.

-\]

-

-Suppose next that the group~$G$ has no self-conjugate operation

-of order~$2^{m - 2}$, and let $\{A\}$~be a self-conjugate cyclical sub-group of

-order~$2^{m - 2}$. If $\dfrac{G}{\{A\}}$~is cyclical, there are, when $m > 5$, five distinct

-types. The common defining relations of these are

-\[

-A^{2^{m - 2}} = 1,\quad B^{4} = 1,\quad B^{-1}AB = A^{\alpha};

-\]

-and the five distinct types are

-\begin{gather*}

-\Inum{(ii)}\quad \alpha = -1,\qquad

-\Inum{(iii)}\quad \alpha = 1 + 2^{m - 3},\qquad

-\Inum{(iv)}\quad \alpha = -1 + 2^{m - 3}, \\

-\Inum{(v)}\quad \alpha = 1 + 2^{m - 4},\qquad

-\Inum{(vi)}\quad \alpha = -1 + 2^{m - 4}.

-\end{gather*}

-%% -----File: 097.png---Folio 81-------

-If $m = 5$, then \Inum{(iv)} and \Inum{(v)} are identical, and \Inum{(vi)}~is Abelian; so that

-there are only three distinct types. If $m = 4$, there is a single type;

-it is given by~\Inum{(ii)}.

-

-When $\dfrac{G}{\{A\}}$ is not cyclical, the square of every operation of~$G$ is

-contained in~$\{A\}$. If all the self-conjugate operations of~$G$ are not

-contained in~$\{A\}$, there must be a self-conjugate operation~$B$, of order~$2$,

-which does not occur in~$\{A\}$. If $C$~is any operation of~$G$, not

-contained in~$\{A, B\}$, then $\{A, C\}$ is a self-conjugate sub-group of

-order~$2^{m - 1}$, which has no operation except identity in common with~$\{B\}$.

-Hence $G$~is a direct product of a group of order~$2$ and a group

-of order~$2^{m - 1}$. There are therefore, for this case, four types \Inum{(vii)},

-\Inum{(viii)}, \Inum{(ix)}, \Inum{(x)}, when $m > 4$, corresponding to the four groups of

-order~$2^{m - 1}$ of §~65. If $m = 4$, there are two types.

-

-Next, let all the self-conjugate operations of~$G$ be contained in~$\{A\}$;

-and suppose that $A$~is one of two conjugate operations. Then

-$G$~must contain an Abelian sub-group of type $(m - 2, 1)$, in which $A$

-occurs; and it may be shewn that, when $m > 4$, there are two types

-defined by the relations

-\[

-\lintertext{\Inum{(xi)} and \Inum{(xii)}}

-\begin{gathered}

-A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad BAB = A,\quad C^{2} = 1, \\

-CBC = BA^{2^{m - 3}},\quad CAC = A^{-1}\quad \text{or}\quad A^{-1 + 2^{m - 3}}.

-\end{gathered}

-\]

-When $m = 4$, there is, for this case, no type.

-

-Lastly, suppose that $A$ is one of four conjugate operations. Then

-$G$~must contain sub-groups of order~$2^{m - 1}$, of the second and the

-third types of §~65, and a sub-group of order~$2^{m - 1}$ of either the

-first or fourth type (\lc). In this last case, there are two distinct

-types defined by

-\[

-\lintertext{\Inum{(xiii)} and \Inum{(xiv)}}

-\begin{aligned}

-&A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad BAB = A^{1 + 2^{m - 3}}, \\

-&\rlap{$C^{2} = 1,\quad CAC = A^{-1 + 2^{m - 3}},\quad CBC = B\quad \text{or}\quad BA^{2^{m - 3}}$.}

-\end{aligned}

-\]

-These two types exist only when $m > 4$.

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$|)}%

-

-\Par{68.} We shall now, as a final illustration, determine and

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime|(}%

-tabulate all types of groups of orders $p^{2}$,~$p^{3}$ and~$p^{4}$. It has been

-already seen that when $p = 2$ the discussion must, in part at least,

-be distinct from that for an odd prime; for the sake of brevity

-we shall not deal in detail with this case, but shall state the results

-only and leave their verification as an exercise to the reader.

-

-It has been shewn (§~53) that all groups of order~$p^{2}$ are Abelian;

-and hence the only distinct types are those represented by $(2)$ and

-$(1, 1)$.

-

-For Abelian groups of order~$p^{3}$, the distinct types are $(3)$, $(2, 1)$

-and $(1, 1, 1)$.

-%% -----File: 098.png---Folio 82-------

-

-If a non-Abelian group of order~$p^{3}$ contains an operation of

-order~$p^{2}$, the sub-group it generates is self-conjugate; hence (§~65)

-in this case there is a single type of group defined by

-\[

-P^{p^{2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}.

-\]

-

-If there is no operation of order~$p^{2}$, then since there must be a

-self-conjugate operation of order~$p$, the group comes under the head

-discussed in §~66; there is again a single type of group defined by

-\begin{gather*}

-P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP, \\

-R^{-1}PR = P,\quad Q^{-1}PQ = P.

-\end{gather*}

-

-These two types exhaust all the possibilities for non-Abelian

-groups of order~$p^{3}$.

-

-\Par{69.} For Abelian groups of order~$p^{4}$, the possible distinct types

-are $(4)$, $(3, 1)$, $(2, 2)$, $(2, 1, 1)$ and $(1, 1, 1, 1)$.

-

-For non-Abelian groups of order~$p^{4}$ which contain operations of

-order~$p^{3}$ there is a single type, namely that given in §~65 when $m$~is

-put equal to~$4$.

-

-For non-Abelian groups which contain a self-conjugate cyclical

-sub-group of order~$p^{2}$ and no operation of order~$p^{3}$, there are three

-distinct types, obtained by writing $4$ for~$m$ in the group of §~66 and

-in the first and the last groups of §~67. The defining relations of

-these need not be here repeated, as they will be given in the summarizing

-table (§~73).

-

-It remains now to determine all distinct types of groups of

-order~$p^{4}$, which contain no operation of order~$p^{3}$ and no self-conjugate

-cyclical sub-group of order~$p^{2}$. We shall first deal with groups

-which contain operations of order~$p^{2}$.

-

-Let $S$ be an operation of order~$p^{2}$ in a group~$G$ of order~$p^{4}$. The

-cyclical sub-group~$\{S\}$ must be self-conjugate in a non-cyclical sub-group

-$\{S, T\}$ of order~$p^{3}$, defined by

-\[

-S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S^{1 + kp}.

-\]

-

-If $R$ is any operation of~$G$, not contained in~$\{S, T\}$, then since

-$\{S\}$~is not self-conjugate, we must have (§~57)

-\[

-R^{-1}SR = S^{1 + \alpha p} T^{\beta},

-\]

-and therefore

-\[

-R^{-1} S^{p}R = S^{p}.

-\]

-

-Now

-\[

-T^{-1}S^{p}T = S^{p},

-\]

-and therefore the $p$th~power of every operation of order~$p^{2}$ in~$G$ is a

-self-conjugate operation.

-

-First let us suppose that $G$~contains other self-conjugate

-operations besides those of~$\{S^{p}\}$. Every such operation must occur

-%% -----File: 099.png---Folio 83-------

-in the group that contains $\{S\}$ self-conjugately; hence in this case

-$T$~must be self-conjugate.

-

-We now therefore have

-\begin{gather*}

-S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S, \\

-R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = T, \\

-R^{p} = S^{\gamma p}T^{\delta}.

-\end{gather*}

-

-These equations give

-\[

-(S^{x}R)^{p} = R^{p}S^{xp} = S^{(x + \gamma)p}T^{\delta}.

-\]

-

-Hence, if $\delta = 0$, $S^{-\gamma}R$~is an operation of order~$p$. Denoting this

-by~$R'$ and $S^{\alpha p} T^{\beta}$ by~$T'$, the group is defined by

-\begin{gather*}

-S^{p^{2}} = 1,\quad T'^{p} = 1,\quad R'^{p} = 1,\quad R'^{-1}SR' = ST', \\

-T'^{-1}ST' = S,\quad R'^{-1}T'R' = T'.

-\end{gather*}

-

-On the other hand, if $\delta$ is not zero, $S^{\efrac{\delta\alpha}{\beta} - \gamma}R$ is an operation of

-order~$p^{2}$ such that $R$ transforms it into a power of itself. This is

-contrary to the supposition that the group contains no cyclical self-conjugate

-sub-group of order~$p^{2}$. Hence $\delta$~cannot be different from

-zero: we therefore have only one type of group.

-

-\Par{70.} Next, let $\{S^{p}\}$ contain all the self-conjugate operations of~$G$;

-and as before, let $\{S, T\}$ be the group that contains $\{S\}$ self-conjugately.

-If $G$ contains an operation~$S'$ of order~$p^{2}$ which

-does not occur in~$\{S, T\}$, there must also be a non-cyclical sub-group

-$\{S', T'\}$ of order~$p^{3}$ which contains $\{S'\}$ self-conjugately.

-Now $\{S, T\}$ and $\{S', T'\}$ must have a common sub-group of order~$p^{2}$;

-since this is self-conjugate in~$G$, it cannot be cyclical. The only

-non-cyclical sub-groups of orders\DPnote{** [sic]}~$p^{2}$ that $\{S, T\}$ and $\{S', T'\}$ contain

-are $\{S^{p}, T\}$ and $\{S'^{p}, T'\}$. Hence these must be identical, and

-therefore $T$~must occur in~$\{S', T'\}$. If now $\{S, T\}$ and $\{S', T'\}$ were

-both Abelian, $T$~would be permutable both with~$S$ and with~$S'$, and

-would therefore, contrary to supposition, be a self-conjugate

-operation. Hence either \Inum{(i)}~$G$~must contain a non-Abelian group

-$\{S, T\}$ of order~$p^{3}$, in which $S$ is an operation of order~$p^{2}$; or \Inum{(ii)}~the

-Abelian group $\{S, T\}$, in which $S$ is an operation of order~$p^{2}$, must

-contain all the operations of~$G$ of order~$p^{2}$.

-

-In the case~\Inum{(i)}, the group is defined by

-\begin{gather*}

-S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S^{1 + p}, \\

-R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = S^{\gamma p}T, \\

-R^{p} = S^{\delta p}.

-\end{gather*}

-%% -----File: 100.png---Folio 84-------

-

-These equations give

-\[

-(S^{x}R)^{p} = R^{p} S^{xp\{1 + \efrac{1}{6} (p - 1)p(p + 1)\beta (\gamma - 2)\}},

-\]

-and therefore, except when $p = 3$, $S^{-\delta}R$~is an operation of order~$p$.

-Hence, if $p > 3$, we may, without loss of generality, put zero for~$\delta$.

-In this case a simple calculation gives

-\[

-R^{-x}SR^{x} = S^{1 + \alpha_{x} p}T^{\beta_{x}},\quad

-R^{-x}TR^{x} = S^{\gamma_{x} p}T,

-\]

-where

-\[

-\alpha_{x} = \tfrac{1}{2} \beta\gamma x(x - 1) + x\alpha,\quad

-\beta_{x} = x\beta,\quad

-\gamma_{x} = x\gamma.

-\]

-

-Hence if $R^{x} = R'$, $S^{y} = S'$, $S^{zp}T = T'$, where $x$,~$y$ and~$z$ are not

-multiples of~$p$, then

-\begin{gather*}

-S'^{p^{2}} = 1,\quad T'^{p} = 1,\quad T'^{-1}S'T' = S'^{1 + p}, \\

-R'^{-1}S'R' = S'^{1 + (\alpha_{x} - z\beta_{x})p} T'^{y\beta}, \\

-R'^{-1}T'R' = S'^{\efrac{\gamma_{x}p}{y}}T',\quad R'^{p} = 1.

-\end{gather*}

-

-If now we take

-\[

-z \equiv \frac{\alpha_{x}}{\beta_{x}},\quad y \equiv \frac{1}{\beta_{x}}\ (\mod p),

-\]

-then

-\[

-R'^{-1}S'R' = S'T',\quad R'^{-1}T'R' = S'^{x^{2}\beta\gamma p}T'.

-\]

-

-Dropping the accents, the defining relations now become

-\begin{gather*}

-S^{p^{2}} = 1,\quad T^{p} = 1,\quad R^{p} = 1,\quad T^{-1}ST = S^{1 + p}, \\

-R^{-1}SR = ST,\quad R^{-1}TR = S^{\alpha p}T;

-\end{gather*}

-where $\alpha$ is either zero, unity, or any given non-residue. Since the

-sub-group $\{R, T, S^{p}\}$ contains all the operations of order~$p$ of the

-group, it follows at once that these three cases give three distinct

-types.

-

-When $p = 3$, it will be found that the defining relations again

-give three distinct types. The operation~$R$ may always be chosen so

-that $\alpha$~and~$\gamma$ are zero. If it is so chosen, the three types correspond

-to the values $1$,~$0$; $-1$,~$1$; and $-1$,~$-1$; of $\beta$~and~$\delta$.

-

-In case~\Inum{(ii)}, the group is defined by

-\begin{gather*}

-S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S, \\

-R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = S^{\gamma p}T, \\

-R^{p} = 1;

-\end{gather*}

-with the condition that all operations of~$G$, not contained in~$\{S, T\}$,

-are of order~$p$.

-

-The formulæ for $R^{-x}SR^{x}$ and $R^{-x}TR^{x}$ enable us to calculate

-directly the power of any given operation of~$G$. Thus they give

-\[

-(S^{x}R)^{p} = S^{px\{1 + \efrac{1}{6} \beta\gamma(p + 1)p(p - 1) + \efrac{1}{2} \alpha p(p + 1)\}}.

-\]

-%% -----File: 101.png---Folio 85-------

-

-If $p > 3$, this gives

-\[

-(S^{x}R)^{p} = S^{px},

-\]

-so that, if $x$ is not a multiple of~$p$, the order of~$S^{x}R$ is~$p^{2}$. Hence

-the type of group under consideration can only occur when $p = 3$.

-

-In this case

-\[

-(S^{x}R)^{3} = S^{3x(1 + \beta \gamma)}.

-\]

-

-Hence if $p = 3$ and $\beta\gamma \equiv -1\ (\mod 3)$, we obtain a new type. A

-reduction similar to that in the previous case may now be effected;

-and taking unity for~$x$, the group is defined by

-\begin{gather*}

-S^{9} = 1,\quad T^{3} = 1,\quad R^{3} = 1,\quad T^{-1}ST = S, \\

-R^{-1}SR = ST,\quad R^{-1}TR = S^{-3}T.

-\end{gather*}

-

-\Par{71.} It only remains to determine the distinct types which

-contain no operation of order~$p^{2}$.

-

-Suppose first that the self-conjugate operations of~$G$ form a

-group of order~$p^{2}$. This must be generated by two independent

-operations $P$~and~$Q$ of order~$p$.

-

-If now $R$ is any other operation of the group, $\{P, Q, R\}$ must be

-an Abelian group of type $(1, 1, 1)$. If again $S$~is any operation not

-contained in $\{P, Q, R\}$, it cannot be permutable with~$R$; for if it

-were, $R$~would be self-conjugate. There must therefore be a relation

-of the form (§~57)

-\[

-S^{-1}RS = RP^{\alpha}Q^{\beta}.

-\]

-

-Since any operation of $\{P, Q\}$ may be taken for one of its

-generating operations, we may take $P^{\alpha}Q^{\beta}$ or $P'$ for one. If then $Q'$

-is an independent operation of~$\{P, Q\}$, $G$~will be defined by

-\[

-P'^{p} = 1,\quad Q'^{p} = 1,\quad R^{p} = 1,\quad S^{p} = 1,\quad S^{-1}RS = RP',

-\]

-in addition to the relations expressing that $P'$~and~$Q'$ are self-conjugate.

-There is then in this case a single type. That all the

-operations of~$G$ in this case are actually of order~$p$ follows from the

-obvious fact that every sub-group of order~$p^{3}$ is either Abelian or of

-the second type of non-Abelian groups of order~$p^{3}$.

-

-\Par{72.} Suppose, secondly, that the self-conjugate operations of~$G$

-form a sub-group of order~$p$, generated by~$P$. There must then be

-some operation~$Q$ which belongs to a set of $p$ conjugate operations;

-for if every operation of~$G$ which is not self-conjugate were one of

-a set of $p^{2}$ conjugate operations, the total number of operations in

-the group would be congruent to $p\ (\mod p^{2})$. It follows that $Q$~must

-be self-conjugate in a group of order~$p^{3}$; and since $P$~is also

-self-conjugate in this group, it must be Abelian. Let $P$,~$Q$,~$R$ be

-generators of this group and $S$~any operation of~$G$ not contained in

-%% -----File: 102.png---Folio 86-------

-it. We may now assume that $Q$~belongs to the sub-group~$H_{2}$

-(§~53), and therefore that

-\[

-S^{-1}QS = QP^{\alpha},

-\]

-while

-\[

-S^{-1}RS = RQ^{\beta}P^{\gamma}.

-\]

-

-If $\beta$ were zero, $Q^{\efrac{1}{\alpha}} R^{-\efrac{1}{\gamma}}$ would be a self-conjugate operation not

-contained in~$\{P\}$; and therefore $\beta$~must be different from zero. We

-may now put

-\[

-Q^{\beta} P^{\gamma} = Q',\quad P^{\alpha\beta} = P';

-\]

-and the group is then defined by the relations

-\begin{gather*}

-%[** TN: Set on one line in the original]

-P'^{p} = 1,\quad

-Q'^{p} = 1,\quad

-R^{p} = 1,\quad

-S^{p} = 1,\\

-S^{-1}RS = RQ',\quad

-S^{-1}Q'S = Q'P',

-\end{gather*}

-together with the relations expressing that $P$ is self-conjugate.

-There is thus again in this case, at most, a single type. It remains

-to determine whether the operations are all actually of order~$p$.

-

-The defining relations give

-\[

-S^{-1} P^{\alpha_{x}} Q^{\beta_{x}} R^{\gamma_{x}}S

-  = P^{\alpha_{x + 1}} Q^{\beta_{x + 1}} R^{\gamma_{x + 1}},

-\]

-where

-\[

-\alpha_{x + 1} = \alpha_{x} + \beta_{x},\quad

-\beta_{x + 1} = \beta_{x} + \gamma_{x},\quad

-\gamma_{x + 1} = \gamma_{x};

-\]

-and therefore

-\[

-S^{-x} P^{\alpha} Q^{\beta} R^{\gamma} S^{x}

-  = P^{\alpha_{x}} Q^{\beta_{x}} R^{\gamma_{x}},

-\]

-where

-\[

-\alpha_{x} = \alpha + x\beta + \tfrac{1}{2} x(x - 1)\gamma,\quad

-\beta_{x} = \beta + x\gamma,\quad

-\gamma_{x} = \gamma.

-\]

-

-Hence

-\begin{align*}

-(P^{\alpha} Q^{\beta} R^{\gamma}S)^{p}

-  &= P^{\esum_{1}^{p} \alpha_{x}} Q^{\esum_{1}^{p}\beta_{x}} R^{\esum_{1}^{p} \gamma_{x}} \\

-  &= P^{p\alpha + \efrac{1}{2} p(p - 1)\beta + \efrac{1}{6} (p + 1)p(p - 1)\gamma}

-     Q^{p\beta  + \efrac{1}{2} p(p - 1) \gamma} R^{p\gamma}.

-\end{align*}

-

-If $p$ is a greater prime than~$3$, the indices of $P$,~$Q$,~$R$ are all

-multiples of~$p$; hence $P^{\alpha} Q^{\beta} R^{\gamma}S$ is of order~$p$, $S$~being any operation

-not contained in~$G$. If however $p = 3$, then

-\[

-(P^{\alpha} Q^{\beta} R^{\gamma}S)^{3} = P^{\gamma},

-\]

-so that, if $\gamma$ is not a multiple of~$3$, $P^{\alpha} Q^{\beta} R^{\gamma}S$ is an operation of

-order~$9$. Hence this last type of group exists as a distinct type for

-all primes greater than~$3$; but for $p = 3$, it is not distinct from one

-of the previous types containing operations of order~$9$.

-\end{Remark}

-

-\Par{73.} In tabulating, as follows, the types of group thus

-obtained, we give with each group~$G$ a symbol of the form

-\[

-(a, b, \Dots)(a', b', \Dots)(a'', b'', \Dots) \Dots,

-\]

-indicating the types of $H_{1}$, $\dfrac{H_{2}}{H_{1}}$, $\dfrac{H_{3}}{H_{2}}$,~\Dots, where

-\[

-H_{1},\ H_{2},\ H_{3},\ \Dots,\ H_{n},\ G

-\]

-is the series of self-conjugate sub-groups defined in §~53. This

-%% -----File: 103.png---Folio 87-------

-symbol is to be read from the left so that $(a, b, \Dots)$ is the

-type of~$H_{1}$.

-

-Moreover in each group there is no operation of higher

-order than that denoted by~$P$.

-

-

-\Subsection{\textit{Table of groups of order $p^{n}$, $p$ an odd prime}\footnotemark.}

-\footnotetext{On groups of orders $p^{3}$ and~$p^{4}$, the reader may consult, in addition to

-  Young's memoir already referred to, Hölder, ``Die Gruppen der Ordnungen

-  $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.},\DPnote{** no "Vol."} \VolNo{XLIII}, (1893), in particular, pp.~371--410.}%

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime}%

-\index{Holder@\Topic{Hölder}, quoted}%

-\index{Young@\Topic{Young}, quoted}%

-

-%[** TN: Layout reformatted]

-\Class{\Inum{I.}}{$n = 2$, two types.}

-

-\Item{(i)} $(2)$;\qquad \Inum{(ii)}\quad $(1, 1)$.

-

-\Class{\Inum{II.}}{$n = 3$, five types.}

-

-\Item{(i)} $(3)$;\qquad \Inum{(ii)} $(2, 1)$;\qquad \Inum{(iii)} $(1, 1, 1)$;

-

-\Item{(iv)} $P^{p^{2}} = 1$,\quad $Q^{p} = 1$,\quad $Q^{-1}PQ = P^{1 + p}$,\quad $(1) (11)$;

-

-\Item{(v)} $\begin{gathered}[t]

-P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP, \\

-R^{-1}PR = P,\quad Q^{-1}PQ = P,\quad (1) (11).

-\end{gathered}$

-

-\Class{\Inum{III.}}{$n = 4$, fifteen types.}

-

-\Item{(i)} $(4)$;\qquad

-\Inum{(ii)}\quad $(3, 1)$;\qquad

-\Inum{(iii)}\quad $(2, 2)$;

-

-\Item{(iv)} $(2, 1, 1)$;\qquad

-\Inum{(v)}\quad $(1, 1, 1, 1)$;

-

-\Item{(vi)} $P^{p^{3}} = 1$,\quad $Q^{p} = 1$,\quad $Q^{-1}PQ = P^{1 + p^{2}}$,\quad $(2) (11)$;

-

-\Item{(vii)} $\begin{gathered}[t]

-P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP^{p}, \\

-Q^{-1}PQ = P,\quad R^{-1}PR = P,\quad (2) (11);

-\end{gathered}$

-

-\Item{(viii)} $P^{p^{2}} = 1$,\quad $Q^{p^{2}} = 1$,\quad $Q^{-1}PQ = P^{1 + p}$,\quad $(11)(11)$;

-

-\Item{(ix)} $\begin{gathered}[t]

-P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p}, \\

-P^{-1}QP = Q,\quad R^{-1}QR = Q,\quad (11) (11),

-\end{gathered}$ \\

-this group~\Inum{(ix)} being the direct product of $\{Q\}$ and~$\{P, R\}$;

-

-\Item{(x)} $\begin{gathered}[t]

-P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = PQ, \\

-Q^{-1}PQ = P,\quad R^{-1}QR = Q,\quad (11) (11);

-\end{gathered}$

-%% -----File: 104.png---Folio 88-------

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime}%

-

-\Item[]{(xi),} \Inum{(xii)}, and \Inum{(xiii)} $p > 3$,

-\begin{gather*}

-P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}, \\

-R^{-1}PR = PQ,\quad R^{-1}QR = P^{\alpha p}Q,\quad (1) (1) (11),

-\end{gather*}

-where for~\Inum{(xi)} $\alpha = 0$, for~\Inum{(xii)} $\alpha = 1$, for~\Inum{(xiii)} $\alpha =$~any non-residue,

-$(\mod p)$;

-

-\Item[]{(xi),} \Inum{(xii)}, and \Inum{(xiii)} $p = 3$,

-\begin{gather*}

-P^{9} = 1,\quad Q^{3} = 1,\quad R^{3} = P^{3 \alpha},\quad Q^{-1}PQ = P^{4}, \\

-R^{-1}PR = PQ^{\beta},\quad R^{-1}QR = Q,\quad (1) (1) (11),

-\end{gather*}

-where for~\Inum{(xi)} $\alpha = 0$, $\beta = 1$, for~\Inum{(xii)} $\alpha = 1$, $\beta = -1$, for~\Inum{(xiii)}

-$\alpha = -1$, $\beta = -1$.

-

-\Item{(xiv)} $\begin{gathered}[t]

-P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad S^{p} = 1,\quad S^{-1}RS = RP, \\

-S^{-1}QS = Q,\quad S^{-1}PS = P,\quad R^{-1}QR = Q, \\

-R^{-1}PR = P,\quad Q^{-1}PQ = P,\quad (11) (11),

-\end{gathered}$ \\

-this group~\Inum{(xiv)} being the direct product of $\{Q\}$ and $\{P, R, S\}$;

-

-\Item{(xv)} $p > 3$,

-\begin{gather*}

-P^{p} = 1,\ Q^{p} = 1,\ R^{p} = 1,\ S^{p} = 1,\ S^{-1}RS = RQ,\  S^{-1}QS = QP, \\

-S^{-1}PS = P,\ R^{-1}QR = Q,\ R^{-1}PR = P, \\

-Q^{-1}PQ = P,\ (1) (1) (11);

-\end{gather*}

-

-\Item{(xv)} $p = 3$,

-\begin{gather*}

-P^{9} = 1,\ Q^{3} = 1,\ R^{3} = 1,\ Q^{-1}PQ = P,\ R^{-1}PR = PQ, \\

-R^{-1}QR = P^{-3}Q,\ (1) (1) (11).

-\end{gather*}

-

-\Par{74.} To complete the list, we add, as was promised in §~68,

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!table of distinct types of orders $8$ and~$16$}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$8$ or~$16$}%

-the types of non-Abelian groups of orders $2^{3}$ and~$2^{4}$; the

-possible types of Abelian groups being the same as for an

-odd prime.

-

-\Class{}{\textit{Non-Abelian groups of order~$2^{3}$}; two types.}

-

-\Item[]{(i)} identical with \Inum{II}~\Inum{(iv)}, writing $2$ for~$p$;

-

-\Item[]{(ii)} $P^{4} = 1$,\quad $Q^{4} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $Q^{2} = P^{2}$,\quad $(1)(11)$.

-

-\Class{}{\textit{Non-Abelian groups of order~$2^{4}$}; nine types.}

-

-\Item[]{(i)}, \Inum{(ii)}, \Inum{(iii)}, \Inum{(iv)} and \Inum{(v)} identical with \Inum{III}~\Inum{(vi)}, \Inum{(vii)},

-\Inum{(viii)}, \Inum{(ix)} and~\Inum{(x)}, writing $2$ for~$p$;

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime|)}%

-%% -----File: 105.png---Folio 89-------

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%

-\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!table of distinct types of orders $8$ and~$16$}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$8$ or~$16$}%

-

-\Item{(vi)} $\begin{gathered}[t]

-P^{4} = 1,\quad Q^{4} = 1,\quad R^{2} = 1,\quad Q^{-1}PQ = P^{-1}, \\

-Q^{2} = P^{2},\quad R^{-1}QR = Q,\quad R^{-1}PR = P,\quad (11) (11),

-\end{gathered}$ \\

-this group~\Inum{(vi)} being the direct product of $\{R\}$ and $\{P, Q\}$;

-

-\Item{(vii)} $P^{8} = 1$,\quad $Q^{2} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $(1) (1) (11)$;

-

-\Item{(viii)} $P^{8} = 1$,\quad $Q^{2} = 1$,\quad $Q^{-1}PQ = P^{3}$,\quad $(1) (1) (11)$;

-

-\Item{(ix)} $P^{8} = 1$,\quad $Q^{4} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $Q^{2} = P^{4}$,\quad $(1) (1) (11)$.

-

-\begin{Remark}

-\Par{75.} In the following examples, $G$~is a non-Abelian group whose

-order is the power of a prime~$p$; and the sub-groups $H_{1}$, $H_{2}$,~\Dots, $H_{n}$

-referred to are the series of self-conjugate sub-groups of §~53.

-\end{Remark}

-

-\begin{Ex}[1.] If $P$ and $Q$ are any two operations of~$G$, the totality of

-operations of the form $P^{-1}Q^{-1}PQ$ generate a self-conjugate sub-group

-identical with or contained in~$H_{n}$.

-\end{Ex}

-

-\begin{Ex}[2.] Shew that, if every sub-group of~$G$ is Abelian, $\dfrac{G}{H_{1}}$~is an

-Abelian group of type $(1, 1)$; and any two operations of~$G$, neither

-of which belongs to~$H_{1}$ and neither of which is a power of the other,

-generate~$G$.

-\end{Ex}

-

-\begin{Ex}[3.] If $\dfrac{G}{H_{n}}$, $\dfrac{H_{n}}{H{n-1}}$, $\dfrac{H_{n-1}}{H_{n-2}}$,~\Dots\ are of types $(1, 1)$, $(1)$,

-$(1)$,~\Dots, shew that $G$~can be generated by two operations.

-\end{Ex}

-

-\begin{Ex}[4.] If $G$, of order~$p^{m}$, is not Abelian, and if every sub-group

-of~$G$ is self-conjugate, shew that $p$~must be~$2$. \Attrib{Dedekind.}

-\index{Dedekind@\Topic{Dedekind}, quoted}%

-\end{Ex}

-

-\begin{Ex}[5.] If $G$ is of order~$p^{m}$, and $H_{1}$ of type $(1, 1, \Dots\DPchg{\,}{,} \text{with $m - 2$

-units})$, and if $m > 5$, $G$~is the direct product of two groups. If

-$m = 5$, there is one type for which $G$ is not a direct product, viz.\

-\begin{gather*}

-Q_{1}^{p^{2}} = 1,\quad

-Q_{2}^{p^{2}} = 1,\quad

-P^{p} = 1,\quad

-Q_{2}^{-1}Q_{1}Q_{2} = Q_{1}P, \\

-%

-Q_{1}^{-1}PQ_{1} = P,\quad

-Q_{2}^{-1}PQ_{2} = P.

-\end{gather*}

-\end{Ex}

-

-\begin{Ex}[6.] A group~$G$, of order~$p^{m}$, ($m > 4$), where $p$~is an odd

-prime, contains an operation~$P$ of order~$p^{m-2}$, and no cyclical self-conjugate

-sub-group of order~$p^{m-2}$. Shew that, if $P$ is one of a set

-of $p^{2}$ conjugate operations, the defining relations of the group are of

-the form

-\begin{gather*}

-P^{p^{m-2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p^{m - 3}},\quad R^{p} = P^{\alpha p} Q^{\beta}, \\

-R^{-1}QR = Q,\quad R^{-1}PR = P^{1 + \beta p^{m - 4}}Q;

-\end{gather*}

-and that, if $P$ is one of a set of $p$ conjugate operations, the defining

-relations are of the form

-\begin{gather*}

-P^{p^{m-2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P,\quad R^{p} = Q^{\alpha}, \\

-R^{-1}QR = P^{\beta p^{m - 3}}Q,\quad R^{-1}PR = PQ.

-\end{gather*}

-Determine in each case the number of distinct types.

-\end{Ex}

-%% -----File: 106.png---Folio 90-------

-

-

-\Chapter{VI.}{On Sylow's Theorem.}

-\index{Sylow@\Topic{Sylow}, quoted}%

-

-\Par{76.} \First{It} has been proved (§~22) that the order of any sub-group

-of a group~$G$ is a factor of the order of~$G$; and it

-results at once from the investigation of §§~45--47 that, in

-an Abelian group, there is always at least one sub-group

-whose order is any given factor of the order of the group.

-The latter result is not however generally true for groups

-which are not Abelian; and for factors of the order of a group

-which contain more than one distinct prime, no general law is

-known as to the existence or non-existence of corresponding

-sub-groups. If however $p^{m}$, where $p$~is a prime, divides the

-order of the group, it may be shewn that the group will always

-contain a sub-group of order~$p^{m}$. The special form of this

-theorem, that a group whose order is divisible by a prime~$p$

-contains operations of order~$p$, is due originally to Cauchy\footnotemark.

-\index{Cauchy@\Topic{Cauchy}, quoted}%

-\footnotetext{Cauchy, \textit{Exercises d'analyse},~\VolNo{III}, (1844), p.~250.}%

-The more general result was first established by Sylow. He

-has shewn\footnote

-  {Sylow, Théorèmes sur les groupes de substitutions, \textit{Math.\ Ann.},\DPnote{** No "Vol." here, below} \VolNo{V} (1872)

-  pp.~584 et~seq. Compare also Frobenius, Neuer Beweis des Sylow'schen Satzes,

-  \textit{Crelle}, \VolNo{C.}\ (1886), p.~179.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-that, if $p^{\alpha}$~is the highest power of a prime~$p$ which

-divides the order of a group, the group contains a single conjugate

-set of $kp + 1$ sub-groups of order~$p^{\alpha}$.

-

-\begin{Remark}

-For the special case $p = 2$, the following proof of Cauchy's

-theorem is perhaps worth giving for its simplicity. Arrange the

-operations of a group of even order in pairs $S$ and $S^{-1}$ of inverse

-operations. If at any stage no further pairs can be formed, the

-remaining operations must all, except the identical operation, be

-%% -----File: 107.png---Folio 91-------

-of order~$2$. The number of these remaining operations, which

-include the identical operation, is even. Hence there must be at

-least one operation of order~$2$, and the total number of such operations

-is odd.

-\end{Remark}

-

-We shall devote the present chapter to the proof of Sylow's

-theorem; and a consideration of some of its more immediate

-consequences. These constitute, as will be seen later on, a most

-important set of results.

-

-\Par{77.} We shall divide the proof of Sylow's theorem into two

-parts. First we shew that, if $p^{\alpha}$~is the highest power of a prime~$p$

-which divides the order of a group, the group must have a

-sub-group of order~$p^{\alpha}$; and secondly that the sub-groups of

-order~$p^{\alpha}$ form a single conjugate set and that their number is

-congruent to unity, $\pmod{p}$.

-

-\begin{Lemma}

-If $p^{\alpha}$ is the highest power of a prime~$p$ by which

-the order of a group~$G$ is divisible, $G$~must contain a sub-group

-whose order is divisible by~$p^{\alpha}$.

-\end{Lemma}

-

-If the group~$G$ is Abelian, this has been already proved in

-§~37. Suppose then that $G$ of order~$N$ ($= p^{\alpha}m$, where $m$~is

-prime to~$p$) is not Abelian; and let $G'$ of order~$N'$ be the sub-group

-of~$G$ formed of its self-conjugate operations. If $S$~is any

-operation of~$G$ which is not self-conjugate, and if $n$~is the order

-of the greatest sub-group within which $S$~is permutable, $S$~forms

-one of a set of $\dfrac{N}{n}$ conjugate operations; hence, by equating

-the order of the group to the sum of the numbers of

-operations contained in the different conjugate sets, we obtain

-the equation

-\[

-N = N' + \sum \frac{N}{n},

-\]

-where the sign of summation is extended to all the different

-conjugate sets of those operations which are not self-conjugate.

-

-If $N'$ is not divisible by~$p$, or in other words if $G$ contains

-no self-conjugate operations of order~$p$, this equation requires

-that at least one of the symbols~$n$ should be divisible by~$p^{\alpha}$;

-therefore in this case the lemma is true.

-

-If $N'$ is divisible by~$p^{\alpha'}$, the group~$G'$, being Abelian, must

-%% -----File: 108.png---Folio 92-------

-contain a sub-group~$g'$ of order~$p^{\alpha'}$; and this sub-group is self-conjugate

-in~$G$. Consider now the group $\dfrac{G}{g'}$ of order~$p^{\alpha - \alpha'}m$.

-If it has no self-conjugate operations of order~$p$, it must

-(by the result just obtained) contain a sub-group whose order

-is divisible by~$p^{\alpha - \alpha'}$; and hence $G$~contains a sub-group whose

-order is divisible by~$p^{\alpha}$. If, on the other hand, $\dfrac{G}{g'}$~has a sub-group

-of self-conjugate operations of order~$p^{\alpha''}$, then $G$~must

-contain a self-conjugate sub-group of order~$p^{\alpha' + \alpha''}$, say~$g''$.

-We may now repeat the same reasoning with the group $\dfrac{G}{g''}$ of

-order~$p^{\alpha - \alpha' - \alpha''}m$. In this way we must in any case be ultimately

-led to a sub-group of~$G$ whose order is divisible by~$p^{\alpha}$. The

-lemma is therefore established.

-

-We have as an immediate inference:---

-\index{Sylow's theorem@\Topic{Sylow's theorem}|(}%

-

-\begin{Theorem}[I.]

-If $p^{\alpha}$~is the highest power of a prime~$p$ which

-divides the order of a group, the group contains a sub-group of

-order~$p^{\alpha}$.

-\end{Theorem}

-

-For the group contains a sub-group whose order is divisible

-by~$p^{\alpha}$; this sub-group must itself contain a sub-group whose

-order is divisible by~$p^{\alpha}$, and so on. Hence at last a sub-group

-must be arrived at whose order is~$p^{\alpha}$.

-

-\begin{Corollary}

-Since it has been seen in §~53 that a group of

-order~$p^{\alpha}$ contains sub-groups of every order~$p^{\beta}$ ($\beta < \alpha$), it follows

-that, if the order of a group is divisible by~$p^{\beta}$, the group must

-contain a sub-group of order~$p^{\beta}$. In particular, a group whose

-order is divisible by a prime~$p$ has operations of order~$p$.

-\end{Corollary}

-

-\Par{78.} \begin{Theorem}[II.]

-If $p^{\alpha}$~is the highest power of a prime~$p$

-which divides the order of a group~$G$, the sub-groups of~$G$ of

-order~$p^{\alpha}$ form a single conjugate set, and their number is congruent

-to unity, $\pmod{p}$.

-\end{Theorem}

-

-If $H$~is a sub-group of~$G$ of order~$p^{\alpha}$, the only operations of~$G$,

-which are permutable with~$H$ and have powers of~$p$ for

-their orders, are the operations of $H$ itself. For if $P$ is an

-operation of order~$p^{r}$, not contained in~$H$ and permutable with

-it, and if $p^{s}$ is the order of the greatest group common to~$\{P\}$

-%% -----File: 109.png---Folio 93-------

-and~$H$, the order of $\{H, P\}$ is~$p^{\alpha + r + s}$. But $G$ can have no such

-sub-group, since $p^{\alpha + r - s}$ is not a factor of the order of~$G$.

-

-Suppose now that $H'$~is any sub-group conjugate to~$H$;

-and let $p^{\beta}$ be the order of the group~$h$ common to $H$ and~$H'$.

-When $H'$~is transformed by all the operations of~$H$, the operations

-of~$h$ are the only ones which transform $H'$ into itself.

-Hence the operations of~$H$ can be divided into $p^{\alpha - \beta}$ sets of $p^{\beta}$

-each, such that the operations of each set transform $H'$ into a

-distinct sub-group. In this way, $p^{\alpha - \beta}$~sub-groups are obtained

-distinct from each other and from~$H$ and conjugate to~$H$. If

-these sub-groups do not exhaust the set of sub-groups conjugate

-to~$H$, let $H''$ be a new one. From $H''$ another set of $p^{\alpha - \beta'}$

-sub-groups can be formed, distinct from each other and from~$H$

-and conjugate to~$H$. Moreover no sub-group of this latter set

-can coincide with one of the previous set. For if

-\[

-P_{1}^{-1} H''P_{1} = P_{2}^{-1} H'P_{2},

-\]

-where $P_{1}$ and $P_{2}$ are operations of~$H$, then

-\[

-H'' = P_{3}^{-1} H'P_{3},

-\]

-where $P_{3}$ ($= P_{2}P_{1}^{-1}$) is an operation of~$H$; and this is contrary to

-the supposition that $H''$~is different from each group of the

-previous set. By continuing this process, it may be shewn that

-the number of sub-groups in the conjugate set containing~$H$ is

-\[

-1 + p^{\alpha - \beta} + p^{\alpha - \beta'} + \Dots,

-\]

-where no one of the indices $\alpha - \beta$, $\alpha - \beta'$,~\Dots\ can be less than

-unity. The number of sub-groups in the conjugate set containing~$H$

-is therefore congruent to unity, $\pmod{p}$.

-

-If now $G$ contains another sub-group~$H_{1}$, of order~$p^{\alpha}$, it must

-belong to a different conjugate set. The number of sub-groups

-in the new set may be shewn, as above, to be congruent

-to unity, $\pmod{p}$. But on transforming $H_{1}$ by the operations of~$H$,

-a set of $p^{\alpha - \gamma}$ conjugate sub-groups is obtained, where $p^{\gamma}$~is

-the order of the sub-group common to $H$ and~$H_{1}$. A further

-sub-group of the set, if it exists, gives rise to $p^{\alpha - \gamma'}$ additional

-conjugate sub-groups, distinct from each other and from the

-previous $p^{\alpha - \gamma}$. Proceeding thus we shew that the number of

-sub-groups in the conjugate set is a multiple of~$p$; and as it

-cannot be at once a multiple of~$p$ and congruent to unity, $\pmod{p}$,

-%% -----File: 110.png---Folio 94-------

-the set does not exist. The sub-groups of order~$p^{\alpha}$ therefore

-form a single conjugate set and their number is congruent to

-unity, $\pmod{p}$.

-

-\begin{Corollary}[I.]

-If $p^{\alpha}m$~is the order of the greatest group~$I$,

-within which the group~$H$ of order~$p^{\alpha}$ is contained self-conjugately,

-the order of the group~$G$ must be of the form

-\[

-p^{\alpha}m(1 + kp).

-\]

-\end{Corollary}

-

-\begin{Corollary}[II.]

-The number of groups of order~$p^{\alpha}$ contained

-in~$G$, \ie\ the factor $1 + kp$ in the preceding expression for the

-order of the group, can be expressed in the form

-\[

-1 + k_{1}p + k_{2}p^{2} + \dots + k_{\alpha}p^{\alpha},

-\]

-where $k_{r}p^{r}$ is the number of groups having with a given group~$H$

-of the set greatest common sub-groups of order~$p^{\alpha - r}$.

-\end{Corollary}

-

-This follows immediately from the arrangement of the set

-of groups given in the proof of the theorem. Thus each of the

-$p^{\alpha - \beta}$ groups, obtained on transforming $H'$ by the operations of~$H$,

-has in common with $H$ a greatest common sub-group of

-order~$p^{\beta}$. It may of course happen that any one or more of the

-numbers $k_{1}$, $k_{2}$,~\dots, $k_{\alpha}$ is zero. If no two sub-groups of the set

-have a common sub-group whose order is greater than~$p^{r}$, then

-$k_{1}$, $k_{2}$,~\dots, $k_{\alpha - r - 1}$ all vanish; and the number of sub-groups in the

-set is congruent to unity, $\pmod{p^{\alpha - r}}$. Conversely, if $p^{s}$~is the

-highest power of~$p$ that divides~$kp$, some two sub-groups of the

-set must have a common sub-group whose order is not less than

-$p^{\alpha - s}$; for if there were no such common sub-groups, the number

-of sub-groups in the set would be congruent to unity, $\pmod{p^{s + 1}}$.

-

-\begin{Corollary}[III.]

-Every sub-group of~$G$ whose order is~$p^{\beta}$,

-($\beta < \alpha$), must be contained in one or more sub-groups of order~$p^{\alpha}$.

-\end{Corollary}

-

-For if the sub-group of order~$p^{\beta}$ is contained in no sub-group

-of order~$p^{\beta + 1}$, the only operations whose orders are powers of~$p$

-that transform it into itself are its own. In this case, the

-preceding method may be used to shew that the number of

-sub-groups in the conjugate set to which the given sub-group

-belongs must be congruent to unity, $\pmod{p}$. But this is

-impossible, as the number of such sub-groups must, on the

-assumption made, be a multiple of $p^{\alpha - \beta}$. The sub-group of

-%% -----File: 111.png---Folio 95-------

-order~$p^{\beta}$ is therefore contained in one of order~$p^{\beta + 1}$, and hence

-repeating the same reasoning in one of order~$p^{\alpha}$.

-

-\Par{79.} We shall refer to Theorems I and II together as Sylow's

-theorem. In discussing in this and the following paragraphs

-some of the results that follow from Sylow's theorem, we shall

-adhere to the notation that has been used in establishing the

-theorem itself. Thus $p^{\alpha}$ will always denote the highest power

-of a prime~$p$ which divides the order of~$G$; the sub-groups of~$G$

-of order~$p^{\alpha}$ will be denoted by $H$, $H_{1}$,~\dots, and the greatest sub-groups

-of~$G$ that contain these self-conjugately by $I$, $I_{1}$,~\dots.

-These latter form a single conjugate set of sub-groups of~$G$,

-whose orders are~$p^{\alpha}m$, the order of $G$ itself being $p^{\alpha}m(1 + kp)$.

-Moreover the number of groups in this conjugate set is~$1 + kp$.

-

-Suppose now that $S$ is any operation of~$G$ whose order is a

-power of~$p$. When the $1 + kp$ sub-groups

-\[

-H,\ H_{1},\ H_{2},\ \dots,\ H_{kp}

-\]

-are transformed by~$S$, each one that contains~$S$ is transformed

-into itself, while the remainder are interchanged in sets, the

-number in any set being a power of~$p$. Hence the number of

-these groups which contain~$S$ must be congruent to unity,

-$\pmod{p}$. In precisely the same way, it may be shewn that the

-number of sub-groups of order~$p^{\alpha}$, which contain a given sub-group

-of order~$p^{\beta}$, is congruent to unity, $\pmod{p}$.

-\index{Sylow's theorem@\Topic{Sylow's theorem}|)}%

-

-A sub-group (or operation), having a power of~$p$ for its order

-and occurring in $1 + lp$ sub-groups of order~$p^{\alpha}$, will not necessarily

-be one of the same number of conjugate sub-groups (or

-operations) in each of these $1 + lp$ sub-groups. If then $h$~is a sub-group,

-of order~$p^{r}$, that occurs in $1 + lp$ sub-groups of order~$p^{\alpha}$,

-we may choose one of these, say~$H$, in which $h$~is one of as small

-a number as possible of conjugate sub-groups. Let this number

-be $p^{\alpha - r - s}$, so that, in~$H$, $h$~is self-conjugate in a group of order~$p^{r + s}$.

-Then in~$G$, the order of the greatest group~$i$, in which

-$h$~is self-conjugate, must be $p^{r + s}n$, where $n$~is relatively prime to~$p$.

-The number of sub-groups of~$i$ of order~$p^{r + s}$ must be

-congruent to unity, $\pmod{p}$. No two of these groups of order~$p^{r + s}$

-can occur in the same group of order~$p^{\alpha}$; for if they did,

-they would generate a group of order~$p^{r + s + t}$, ($t \nless 1$), and this is

-%% -----File: 112.png---Folio 96-------

-impossible, $p^{r + s}$~being the highest power of~$p$ that divides the

-order of~$i$. Moreover the number of groups of order~$p^{\alpha}$ in which

-any one of these groups of order~$p^{r + s}$ enters is congruent to

-unity, $\pmod{p}$. Hence the number of groups of order~$p^{\alpha}$, in

-which $h$~is one of $p^{\alpha - r - s}$ (the least possible number) conjugate

-sub-groups, is congruent to unity, $\pmod{p}$. Suppose now that,

-in~$H'$, $h$~is one of $p^{\alpha - r - s'}$ ($s' < s$) conjugate sub-groups; so that

-in~$H'$ it is self-conjugate in a sub-group of order~$p^{r + s'}$ and

-in no greater sub-group. Then the highest power of $p$ that

-divides the order of the group common to $I'$ and $i$ is~$p^{r + s'}$; and

-therefore, when $H'$ is transformed by all the operations of~$i$, the

-number of groups of order~$p^{\alpha}$ formed is a multiple of~$p^{s - s'}$. In

-each of these groups, $h$~is one of $p^{\alpha - r - s'}$ conjugate sub-groups.

-If this does not exhaust all the groups of order~$p^{\alpha}$ in which $h$ is

-one of $p^{\alpha - r - s'}$ conjugate sub-groups, let $H''$ be another. Then

-from this another set of groups, whose number is a multiple of~$p^{s - s'}$,

-may be formed, which are distinct from each other and

-from the previous set, such that in each of them $h$~is one of

-$p^{\alpha - r - s'}$ conjugate sub-groups. This process may clearly be

-continued till all such groups are exhausted. Hence the

-number of groups of order~$p^{\alpha}$, in which $h$ is one of $p^{\alpha - r - s'}$ ($s' < s$)

-conjugate sub-groups, is a multiple of~$p^{s - s'}$. If $h$~enters as one

-of $p^{\alpha - r - s'}$ conjugate sub-groups in~$H'$, a sub-group conjugate

-to~$h$ must enter as one of $p^{\alpha - r - s'}$ conjugate sub-groups in~$H$.

-Hence if $p^{r + s}$, $p^{r + s'}$, $p^{r + s''}$,~\dots\ are the orders of the greatest

-sub-groups that contain $h$ self-conjugately in the various sub-groups

-of order~$p^{\alpha}$ in which it appears, then $H$~must contain

-sub-groups of the conjugate set (in~$G$) to which $h$ belongs in

-conjugate sets of $p^{r + s}$, $p^{r + s'}$, $p^{r + s''}$,~\dots\ only. The total number~$x$

-of such sub-groups contained in~$H$ is clearly connected with

-the total number $1 + lp$ of sub-groups of order~$p^{\alpha}$, in which $h$~appears,

-by the relation

-\[

-x(1 + kp) = y(1 + lp),

-\]

-$y$ being the number of sub-groups in the conjugate set. For

-every sub-group of order~$p^{\alpha}$ will contain $x$ of the set: and the

-two sides of the equation represent two distinct ways of reckoning

-all the sub-groups of the set $h$ contained in the sub-groups

-of the set~$H$, when all repetitions are counted.

-%% -----File: 113.png---Folio 97-------

-

-It is to be noticed that, with the above notation,

-\[

-y = \frac{p^{\alpha - r - s} m(1 + kp)}{n};

-\]

-and therefore

-\[

-x = \frac{p^{\alpha - r - s} m(1 + lp)}{n}.

-\]

-

-\Par{80.} \begin{Theorem}[III.]

-\index{Sylow's theorem@\Topic{Sylow's theorem}!some direct consequences of|(}%

-Let $p^{\alpha}$ be the highest power of a prime~$p$

-which divides the order of a group~$G$, and let $H$ be a sub-group

-of~$G$ of order~$p^{\alpha}$. Let $h$ be a sub-group common to $H$ and some

-other sub-group of order~$p^{\alpha}$, such that no sub-group, which contains~$h$

-and is of greater order, is common to any two sub-groups of

-order~$p^{\alpha}$. Then there must be some operation of~$G$, of order

-prime to~$p$, which is permutable with~$h$ and not with~$H$\footnotemark.

-\end{Theorem}

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),

-  p.~176: and Burnside, ``Notes on the theory of groups of finite order,'' \textit{Proc.\

-  London Mathematical Society}, Vol.~\VolNo{XXVI} (1895), p.~209.}

-

-Suppose that $H$ and $H'$ are two groups of order~$p^{\alpha}$ to

-which $h$ is common; and let $h_{1}$ and $h_{1}'$ be sub-groups of $H$ and~$H'$,

-of greater order than~$h$, in which $h$~is self-conjugate. If $h_{1}$

-and $h_{1}'$ generate a group whose order is a power of~$p$, it must

-occur in some group~$H''$ of order~$p^{\alpha}$; and then $H$ and $H''$ have

-a common group~$h_{1}$, which contains~$h$ and is of greater order.

-This is contrary to supposition, and therefore the order of the

-group generated by $h_{1}$ and $h_{1}'$ is not a power of~$p$. Hence $h$ is

-permutable with some operation whose order is prime to~$p$.

-

-Let $p^{r}$ be the order of~$h$, and $p^{r + s}n$ be the order of the

-greatest sub-group~$i$ of~$G$ that contains $h$ self-conjugately. If $i$~contained

-a self-conjugate sub-group of order~$p^{r + s}$, $h_{1}$~and $h_{1}'$

-would be sub-groups of it and they would generate a group

-whose order is a power of~$p$. This is not the case, and therefore

-$i$~must contain $1 + k'p$ sub-groups of order~$p^{r + s}$, so that we may

-write $m'(1 + k'p)$ for~$n$; and then, in~$i$, a sub-group of order~$p^{r + s}$

-is self-conjugate in a sub-group of order~$p^{r + s}m'$. No sub-group

-of~$i$ of order~$p^{r + t}$ ($t > 0$) can occur in more than one sub-group

-of order~$p^{\alpha}$; and the $1 + k'p$ sub-groups of~$i$ of order~$p^{r + s}$ belong

-therefore to $1 + k'p$ distinct sub-groups of order~$p^{\alpha}$. Moreover

-$h$~occurs in no sub-groups of order~$p^{\alpha}$ other than these $1 + k'p$.

-For if $h$ occurred in another sub-group~$H_{1}$, it would in this

-sub-group be self-conjugate in a group of order~$p^{r + s'}$ ($s' > 0$); and

-%% -----File: 114.png---Folio 98-------

-this group would occur in~$i$. This group would then be common

-to two sub-groups of order~$p^{\alpha}$, contrary to supposition.

-

-An operation of~$i$, which transforms one of its sub-groups

-of order~$p^{r + s}$ into another, must transform the sub-group of order~$p^{\alpha}$

-containing the one into that containing the other. Hence $i$

-must contain operations which are not permutable with~$H$.

-The greatest common sub-group of $i$~and~$I$ is that sub-group

-of~$i$ of order~$p^{r + s}m'$ which contains the sub-group of order~$p^{r + s}$

-belonging to $H$ self-conjugately. For every operation, that

-transforms this sub-group of order~$p^{r + s}$ into itself, must transform

-$H$ into itself; and no operation can transform $H$ into itself

-which transforms this sub-group of order~$p^{r + s}$ into another.

-

-\Par{81.} Let $P$ be an operation, or sub-group, which is self-conjugate

-in~$H$; and let $Q$~be another operation, or sub-group,

-of~$H$, which is conjugate to~$P$ in~$G$, but not conjugate to~$P$

-in~$I$. Suppose first that, if possible, $Q$~is self-conjugate in~$H$.

-There must be an operation~$S$ which transforms $P$ into~$Q$ and

-$H$ into some other sub-group~$H'$, so that

-\begin{align*}

-S^{-1}PS &= Q, \\

-S^{-1}HS &= H'.

-\end{align*}

-

-Now in the sub-group which contains $Q$ self-conjugately,

-the sub-groups of order~$p^{\alpha}$ form a single conjugate set, and

-$H$~must occur among them. Hence this sub-group must contain

-an operation~$T$ such that

-\[

-T^{-1}QT = Q,

-\]

-and

-\[

-T^{-1}H'T = H.

-\]

-It follows that

-\begin{align*}

-T^{-1}S^{-1}PST &= Q, \\

-T^{-1}S^{-1}HST &= H,

-\end{align*}

-or that, contrary to supposition, $P$~and~$Q$ are conjugate in~$I$.

-Hence:---

-

-\begin{Theorem}[IV.]

-Let $G$ and $H$ be defined as in the previous

-theorem, and let $I$ be the greatest sub-group of~$G$ which contains

-$H$ self-conjugately. Then if $P$ and $Q$ are two self-conjugate

-operations or sub-groups of~$H$, which are not conjugate in~$I$,

-they are not conjugate in~$G$.

-\end{Theorem}

-%% -----File: 115.png---Folio 99-------

-

-\begin{Corollary}

-If $H$ is Abelian, no two operations of~$H$ which

-are not conjugate in~$I$ can be conjugate in~$G$. Hence the

-number of distinct sets of conjugate operations in~$G$, which

-have powers of~$p$ for their orders, is the same as the number of

-such sets in~$I$.

-\end{Corollary}

-

-\Par{82.} Suppose next that $Q$ is not self-conjugate in~$H$.

-Then every operation that transforms $Q$ into~$P$ must transform

-$H$ into a sub-group of order~$p^{\alpha}$ in which $P$~is not self-conjugate.

-Of the sub-groups of order~$p^{\alpha}$, to which $P$ belongs and in which

-$P$~is not self-conjugate, choose $H'$ so that, in~$H'$, $P$~forms one

-of as small a number of conjugate operations or sub-groups as

-possible. Let $g$ be the greatest sub-group of~$H'$ that contains

-$P$ self-conjugately. Among the sub-groups of order~$p^{\alpha}$ that

-contain $P$ self-conjugately, there must be one or more to which $g$

-belongs. Let $H$ be one of these; and suppose that $h$~and~$h'$

-are the greatest sub-groups of $H$ and $H'$ respectively that

-contain $g$ self-conjugately. The orders of both $h$ and $h'$ must

-(Theorem~II, §~55) be greater than the order of~$g$; and in consequence

-of the assumption made with respect to~$H'$, every

-sub-group, having a power of~$p$ for its order and containing~$h$,

-must contain $P$ self-conjugately.

-

-Now consider the sub-group $\{h, h'\}$. Since it does not

-contain $P$ self-conjugately, its order cannot be a power of~$p$.

-Also if $p^{\beta}$ is the highest power of~$p$ that divides its order, it

-must contain more than one sub-group of order~$p^{\beta}$. For any

-sub-group of order~$p^{\beta}$, to which $h$ belongs, contains $P$ self-conjugately;

-and any sub-group of order~$p^{\beta}$, to which $h'$ belongs,

-does not. Suppose now that $S$~is an operation of~$\{h, h'\}$, having

-its order prime to~$p$ and transforming a sub-group of~$\{h, h'\}$ of

-order~$p^{\beta}$, to which $h$ belongs, into one to which $h'$ belongs.

-Then $S$~cannot be permutable with~$P$; for if it were, $P$~would

-be self-conjugate in each of these sub-groups of order~$p^{\beta}$.

-

-When $P$ is an operation, we may reason in the same way

-with respect to~$\{P\}$.

-

-Since $g$ is self-conjugate in both $h$ and~$h'$, $S$~must transform

-$g$ into itself. Now $P$ is self-conjugate in~$g$, and therefore

-%% -----File: 116.png---Folio 100-------

-$S^{-r}PS^{r}$ is also self-conjugate in~$g$ for all values of~$r$. If then

-$S^{t}$~is the first power of~$S$ which is permutable with~$P$, the

-series of groups $P$, $S^{-1}PS$,~\Dots, $S^{-t+1} PS^{t-1}$ are all distinct and

-each is a self-conjugate sub-group of~$g$. Every group in this

-series is therefore permutable with every other. Hence:---

-

-\begin{Theorem}[V.]

-If $G$ and $H$ are defined as in the two preceding

-theorems, and if $P$ is a self-conjugate sub-group or operation of~$H$,

-then either \Inum{(i)}~$P$~must be self-conjugate in every sub-group

-of~$G$, of order~$p^{\alpha}$, in which it enters, or \Inum{(ii)}~there must be an

-operation~$S$, of order~$q$ prime to~$p$, such that the set of sub-groups

-$S^{-r} \{P\} S^{r}$ \($r = 0, 1, \Dots, q - 1$\) are all distinct and permutable

-with each other.

-\end{Theorem}

-

-\begin{Remark}

-\Par{83.} In illustration of Sylow's theorem and its consequences, we

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$pq$}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$pq$, where $p$~and $q$ are different primes}%

-will now consider certain special types of group; and we will deal

-first with a \emph{group whose order is the product of two different primes}.

-

-If $p_{1}$ and $p_{2}$ ($p_{1} < p_{2}$) are distinct primes, a group of order~$p_{1}p_{2}$

-must, by Sylow's theorem, have a self-conjugate sub-group of order~$p_{2}$.

-If $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, the group must also

-have a self-conjugate sub-group of order~$p_{1}$. The two self-conjugate

-sub-groups of orders $p_{1}$ and $p_{2}$ can have no common operation

-except the identical operation; and therefore (Theorem~IX, §~34)

-every operation of one must be permutable with every operation of

-the other. Hence if $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, a group

-of order~$p_{1}p_{2}$ must be Abelian, and therefore also cyclical.

-

-If $p_{2}$ is congruent to unity, $\pmod{p_{1}}$, there may be either $1$ or $p_{2}$

-conjugate sub-groups of order~$p_{1}$.

-

-If there is one, it is self-conjugate and the group is cyclical.

-

-If there are $p_{2}$ sub-groups of order~$p_{1}$, let $P_{1}$ denote an operation

-which generates one of them. Then if $P_{2}$ is an operation of order~$p_{2}$,

-\[

-P_{1}^{-1} \{P_{2}\} P_{1} = \{P_{2}\};

-\]

-therefore

-\[

-P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},

-\]

-so that

-\[

-P_{2} = P_{1}^{-p_{1}} P_{2} P_{1}^{p_{1}} = P_{2}^{\alpha^{p_{1}}}.

-\]

-Hence

-\[

-\alpha^{p_{1}} \equiv 1\ (\mod p_{2}).

-\]

-If $\alpha$ were unity, $P_{2}$~would be permutable with~$P_{1}$, and $\{P_{1}\}$~would not

-be one of $p_{2}$ conjugate sub-groups. Hence $\alpha$~must be a primitive

-root of the above congruence.

-

-A group of order~$p_{1}p_{2}$, which has $p_{2}$ conjugate sub-groups of order~$p_{1}$,

-must therefore, when it exists, be defined by the relations

-\[

-P_{1}^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},

-\]

-\index{Sylow's theorem@\Topic{Sylow's theorem}!some direct consequences of|)}%

-%% -----File: 117.png---Folio 101-------

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$pq$, where $p$~and $q$ are different primes}%

-where $\alpha$~is a primitive root of the congruence\footnote

-  {It should be noticed that, if $\alpha$ is not a root of the congruence, the relations

-  define a group of order~$p_{1}$. Thus from

-  \[

-  P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},

-  \]

-  we have

-  \[

-  P_{1}^{-p_{1}} P_{2} P_{1}^{p_{1}} = P_{2}^{\alpha^{p_{1}}}.

-  \]

-  Hence

-  \[

-  P_{2}^{\alpha^{p_{1} - 1}} = 1,\quad \text{and}\quad P_{2}^{p_{2}} = 1,

-  \]

-  so that $P_{2} = 1$, and the group reduces to~$\{P_{1}\}$.}

-\[

-\alpha^{p_{1}} \equiv 1\ (\mod p_{2}).

-\]

-

-It follows from §~33 that the order of the group defined by these

-relations cannot exceed~$p_{1} p_{2}$.  But also from the given relations it

-is clearly impossible to deduce new relations of the form

-\[

-P_{1}^{x} P_{2}^{y} = P_{1}^{x'} P_{2}^{y'},

-\]

-where $x$ and $x'$ are less than~$p_{1}$, and $y$ and $y'$ are less than~$p_{2}$; hence

-the order cannot be less than~$p_{1} p_{2}$.

-

-Again, let $\beta$ be a primitive root of the congruence, distinct from~$\alpha$,

-so that

-\[

-\beta^{x} \equiv \alpha\ (\mod p_{2}).

-\]

-

-Then if, in the group defined by

-\[

-P_{1}^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}^{-1} P_{2} P_{1} = P_{2}^{\beta},

-\]

-we represent $P_{1}^{x}$ by~$P_{1}'$, the defining relations become

-\[

-P_{1}'^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}'^{-1} P_{2} P_{1}' = P_{2}^{\alpha};

-\]

-and the group is simply isomorphic with the previous group.

-

-Hence finally, when $p_{2}$~is congruent to unity, $\pmod{p_{1}}$, there is a

-single type of group of order~$p_{1} p_{2}$, which contains $p_{2}$ conjugate

-sub-groups of order~$p_{1}$.

-

-\Par{84.} We will next deal with the problem of determining all

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$24$|(}%

-distinct types of group of order~$24$.

-

-A group of order~$24$ must contain either $1$ or $3$~sub-groups of

-order~$8$, and either $1$ or $4$~sub-groups of order~$3$. If it has one

-sub-group of order~$8$ and one sub-group of order~$3$, the group must,

-since each of these sub-groups is self-conjugate, be their direct

-product. We have seen (§§~68, 74) that there are five distinct types

-of group of order~$8$; there are therefore five distinct types of group

-of order~$24$, which are obtained by taking the direct product of any

-group of order~$8$ and a group of order~$3$.

-

-If there are $3$~groups of order~$8$, some two of them must

-(Theorem~II, Cor.~II, §~78) have a common sub-group of order~$4$;

-and (Theorem~III, §~80) this common sub-group must be a self-conjugate

-sub-group of the group of order~$24$. Moreover if, in

-this case, a sub-group of order~$8$ is Abelian, each operation of the

-%% -----File: 118.png---Folio 102-------

-self-conjugate sub-group of order~$4$ must (Theorem~IV, Cor.\ §~81)

-be a self-conjugate operation of the group of order~$24$.

-

-With the aid of these general considerations, it now is easy to

-determine for each type of group of order~$8$, the possible types of

-group of order~$24$, in addition to the five types already obtained.

-

-\Inum{(i)} Suppose a group of order~$8$ to be cyclical, and let $A$ be an

-operation that generates it. If $\{A\}$~is self-conjugate and $B$~is an

-operation of order~$3$, then

-\[

-B^{-1}AB = A^{\alpha},

-\]

-and therefore

-\[

-B^{-3}AB^{3} = A^{\alpha^{3}}.

-\]

-Hence

-\[

-\alpha^{3} \equiv 1\ (\mod 8),

-\]

-and therefore

-\[

-\alpha \equiv 1\ (\mod 8);

-\]

-so that $A$ and~$B$ are permutable. This is one of the types already

-obtained. Hence for a new type, $\{A\}$~cannot be self-conjugate, and

-$A^{2}$~must be a self-conjugate operation; $B$~is therefore one of two

-conjugate operations, while $\{B\}$~is self-conjugate. Hence the only

-possible new type in this case is given by

-\[

-A^{-1}BA = B^{-1}.

-\]

-

-\Inum{(ii)} Next, let a group of order~$8$ be an Abelian group defined by

-\[

-A^{4} = 1,\quad B^{2} = 1,\quad AB = BA.

-\]

-

-If this is self-conjugate, then, by considerations similar to those

-of the preceding case, we infer that the group is the direct product

-of groups of orders $8$~and~$3$. Hence there is not in this case a new

-type.

-

-If the group of order~$8$ is not self-conjugate, the self-conjugate

-group of order~$4$ may be either $\{A\}$ or $\{A^{2}, B\}$. In either case, if $C$~is

-an operation of order~$3$, it must be one of two conjugate operations

-while $\{C\}$~is self-conjugate. Hence there are two new types respectively

-given by

-\[

-C^{3} = 1,\quad BCB = C^{-1},\quad A^{-1}CA = C;

-\]

-and

-\[

-C^{3} = 1,\quad A^{-1}CA = C^{-1},\quad BCB = C.

-\]

-

-\Inum{(iii)} Let a group of order~$8$ be an Abelian group defined by

-\[

-A^{2} = 1,\quad B^{2} = 1,\quad C^{2} = 1,\quad  AB = BA,\quad BC = CB,\quad CA = AB.

-\]

-

-If it is self-conjugate, and if the group of order~$24$ is not the

-direct product of groups of orders $8$~and~$3$, an operation~$D$ of order~$3$

-must transform the $7$~operations of order~$2$ among themselves; and

-it must therefore be permutable with one of them. Now the relations

-\[

-D^{-1}AD = A,\quad D^{-1}BD = AB,

-\]

-are not self-consistent, because they give

-\[

-D^{-2}BD^{2} = B.

-\]

-%% -----File: 119.png---Folio 103-------

-Hence, since the group of order~$8$ is generated by $A$,~$B$ and any

-other operation of order~$2$ except~$AB$, we may assume, without loss

-of generality, that

-\[

-D^{-1}AD = A,\quad  D^{-1}BD = C,\quad  D^{-1}CD = A^{x}B^{y}C^{z}.

-\]

-These relations give

-\[

-B = D^{-3}BD^{3} = D^{-1}A^{x}B^{y}C^{z}D = A^{x(1 + z)}B^{yz}C^{y + z^{2}},

-\]

-and therefore

-\[

-y = z = 1.

-\]

-

-Now if

-\[

-D^{-1}CD = ABC,

-\]

-and if

-\[

-AB = B',\quad AC = C',

-\]

-then

-\[

-D^{-1}B'D = C',\quad  D^{-1}C'D = B'C';

-\]

-so that the two alternatives $x = 0$ and $x = 1$ lead to simply

-isomorphic groups.

-

-Hence there is in this case a single type. It is the direct

-product of $\{A\}$ and $\{D, B, C\}$, where

-\[

-D^{-1}BD = C,\quad D^{-1}CD = BC.

-\]

-

-If the group of order~$8$ is not self-conjugate, the self-conjugate

-group of order~$4$ may be taken to be~$\{A, B\}$; and $D$~being an operation

-of order~$3$, there is a single new type given by

-\[

-D^{3} = 1,\quad CDC = D^{-1},\quad ADA = D,\quad BDB = D.

-\]

-

-\Inum{(iv)} Let a group of order~$8$ be a non-Abelian group defined by

-\[

-A^{4} = 1,\quad B^{4} = 1,\quad A^{2} = B^{2},\quad B^{-1}AB = A^{-1};

-\]

-and let $C$ be an operation of order~$3$. If the group of order~$8$ is

-self-conjugate, and the group of order~$24$ is not a direct product of

-groups of orders $8$~and~$3$, $C$~must transform the $3$~sub-groups of

-order~$4$, $\{A\}$,~$\{B\}$ and~$\{AB\}$, among themselves. Hence we may take

-\[

-C^{-1}AC = B,

-\]

-and

-\[

-C^{-1}BC = AB\quad \text{or}\quad (AB)^{3}.

-\]

-If $C$ transforms $B$ into~$(AB)^{3}$, then

-\[

-C^{-3}AC^{3} = A^{-1},

-\]

-and $C$ cannot be an operation of order~$3$. Hence in this case there

-is only one new type, given by

-\[

-C^{3} = 1,\quad C^{-1}AC = B,\quad C^{-1}BC = AB.

-\]

-

-If the sub-group of order~$8$ is not self-conjugate, the self-conjugate

-sub-group of order~$4$ is cyclical, and each of its operations must be

-permutable with~$C$. Hence again we get a single new type, given by

-\[

-C^{3} = 1,\quad A^{-1}CA = C,\quad B^{-1}CB = C^{-1}.

-\]

-

-\Inum{(v)} Lastly, let a sub-group of order~$8$ be a non-Abelian group

-defined by

-\[

-A^{4} = 1,\quad B^{2} = 1,\quad BAB = A^{-1}.

-\]

-%% -----File: 120.png---Folio 104-------

-

-This contains one cyclical and two non-cyclical sub-groups of

-order~$4$. If it is self-conjugate, the group of order~$24$ must therefore

-be the direct product of groups of orders $8$~and~$3$; and there is no

-new type.

-

-If the sub-group of order~$8$ is not self-conjugate, and the self-conjugate

-sub-group of order~$4$ is the cyclical group~$\{A\}$, then $A$

-must be permutable with an operation~$C$ of order~$3$, and there is a

-single new type given by

-\[

-C^{3} = 1,\quad A^{-1}CA = C,\quad B^{-1}CB = C^{-1}.

-\]

-

-If the self-conjugate sub-group of order~$4$ is not cyclical, it may

-be taken to be $\{1, A^{2}, B, A^{2}B\}$. If $C$ is permutable with each

-operation of this sub-group, there is a single type given by

-\[

-C^{3}= 1,\quad A^{-1}CA = C^{-1},\quad B^{-1}CB = C.

-\]

-

-If $C$ is not permutable with every operation of the self-conjugate

-sub-group, it must transform $A^{2}$,~$B$,~$A^{2}B$ among themselves and we

-may take

-\[

-C^{-1}A^{2}C = B,\quad C^{-1}BC = A^{2}B.

-\]

-

-Now $\{C, A^{2}, B\}$ is self-conjugate, and therefore $A$ must transform~$C$

-into another operation of order~$3$ contained in this sub-group.

-Hence

-\[

-A^{-1}CA = C^{x}A^{2y}B^{z}.

-\]

-

-The only values of $x$,~$y$,~$z$ which are consistent with the previous

-relation

-\[

-A^{2}CA^{2} = CA^{2}B,

-\]

-are

-\[

-x = 2,\quad y = z = 1.

-\]

-

-The last new type is therefore defined by

-\begin{gather*}

-A^{4} = 1,\quad B^{2} = 1,\quad BAB = A^{-1}, \\

-C^{3} = 1,\quad C^{-1}A^{2}C = B,\quad C^{-1}BC = A^{2}B, \\

-A^{-1}CA = C^{2}A^{2}B.

-\end{gather*}

-

-When $B$ is eliminated between these relations, it will be found

-that the only independent relations remaining are

-\[

-A^{4} = 1,\quad C^{3} = 1,\quad (AC)^{2} = 1.

-\]

-

-It is a good exercise to verify that these form a complete set of

-defining relations for the group. (Compare Ex.~1, §~35.)

-

-There are therefore, in all, fifteen distinct types of group of order~$24$.

-The last of these is the only type, which has neither a self-conjugate

-sub-group of order~$8$, nor one of order~$3$. The reader

-should satisfy himself, as an exercise, that, in the ten cases where

-the group is not a direct product of groups of orders $8$~and~$3$, the

-defining relations which we have given are self-consistent. This is

-of course an essential part of the investigation, and it may, in

-more complicated cases, involve some little difficulty. We have

-omitted the verification here, where it is very easy, for the sake of

-brevity.

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$24$|)}%

-%% -----File: 121.png---Folio 105-------

-

-It is to be noticed that the last type obtained gives an example,

-and indeed the simplest possible, of Theorem~V, §~82. Thus in

-$\{A, B\}$ of order~$8$, $A^{2}$~is a self-conjugate operation and $B$~is not. In

-the group of order~$24$, the operations $A^{2}$ and $B$ are conjugate; and

-$C$~is an operation, of order prime to~$2$, such that $A^{2}$, $C^{-1}A^{2}C$, $C^{-2}A^{2}C^{2}$

-generate three mutually permutable sub-groups.

-

-A discussion similar to that of the present section (but simpler,

-since in each case the number of types is smaller), will verify the

-following table\footnotemark:---

-\footnotetext{Miller, \textit{Comptes Rendus}, \VolNo{CXXII} (1896), p.~370.}

-\index{Miller@\Topic{Miller}, quoted}%

-\[

-\begin{array}{l|*{11}{c|}}

-\text{Order} & 6 & 10 & 12 & 14 & 15 & 18 & 20 & 22 & 26 & 28 & 30 \\

-\hline

-\Strut

-\text{Number} & 2 & 2 & 5 & 2 & 1 & 5 & 5 & 2 & 2 & 4 & 4 \\

-\hline

-\end{array}

-\]

-

-This table, taken with the results of Chapter~V, gives the number

-of distinct types of groups for all orders less than~$32$.

-

-\Par{85.} As a second example, we will discuss the various distinct

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$60$|(}%

-types of group of order~$60$.

-

-A group of order~$60$ must, by Sylow's theorem, contain either $1$

-or $6$ cyclical sub-groups of order~$5$.

-

-We will first suppose that a group~$G$ of order~$60$ contains a

-single cyclical sub-group of order~$5$, which is necessarily self-conjugate.

-There will then be four operations of order~$5$ in~$G$; and

-we may deal with two sub-cases according as these operations are or

-are not self-conjugate.

-

-\Inum{(i)} Suppose that each operation of order~$5$ is self-conjugate.

-There must then be either $1$ or $3$ sub-groups of order~$4$. If there is

-only one, it must be permutable with an operation of order~$3$;

-and then $G$~contains a sub-group of order~$12$. If there are three,

-it follows, by Theorem~II, Cor.~II (§~78), that some pair of them

-must have a common sub-group of order~$2$. But (Theorem~III, §~80)

-this sub-group of order~$2$ must be permutable with some operation

-of order prime to~$2$, which is not permutable with a sub-group of

-order~$4$. This operation must be of order~$3$; hence in this case also

-there must be a sub-group of order~$12$. Thus then the group is,

-with either alternative, the direct product of a group of order~$5$ and

-a group of order~$12$. Now there are five distinct types of group of

-order~$12$; there are therefore five distinct types of group of order~$60$

-which contain self-conjugate operations of order~$5$.

-

-\Inum{(ii)} Suppose that $S$ is an operation of order~$5$ which is not self-conjugate.

-If $T$~is an operation which is not permutable with~$S$,

-then

-\[

-T^{-1}ST = S^{\alpha},

-\]

-%% -----File: 122.png---Folio 106-------

-where $\alpha$~is not unity. Also, if $T^{x}$ is the lowest power of~$T$ which is

-permutable with~$S$, then

-\[

-T^{-x}ST^{x} = S^{\alpha^{x}} = S,

-\]

-and therefore

-\[

-\alpha^{x} \equiv 1\ (\mod 5).

-\]

-

-It follows that $x$ must be either $2$ or~$4$. If $x$~is~$2$ for every

-operation~$T$ which is not permutable with~$S$, then $S$ and $S^{4}$ form a

-complete set of conjugate operations; as also do $S^{2}$ and~$S^{3}$. If $x$~is~$4$

-for any operation~$T$, the four operations $S$, $S^{2}$, $S^{3}$, $S^{4}$ form a

-single conjugate set.

-

-First, let $S$ be one of two conjugate operations; it must then be

-self-conjugate in a sub-group of order~$30$, and by Sylow's theorem

-this sub-group must contain a single sub-group of order~$3$. It will

-therefore be given by

-\begin{alignat*}{2}

-&\Inum{($\alpha$)}\quad & A^{2} &= 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ABA = B, \\

-\lintertext{or}

-&\Inum{($\beta$)}\quad & A^{2} &= 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ABA = B^{2},

-\end{alignat*}

-according as $B$ is or is not a self-conjugate operation; $S$~in either

-case being permutable with both $A$ and~$B$.

-

-If the sub-groups of order~$4$ are cyclical, $G$~must contain an

-operation~$A_{1}$ of order~$4$ whose square is~$A$; and $A_{1}$~must transform $S$

-into its inverse and $\{B\}$ into itself. The latter condition clearly

-cannot be satisfied if the self-conjugate sub-group of order~$30$ is of

-type~\Inum{($\beta$)}. Hence we have two types given by

-\begin{gather*}

-A_{1}^{4} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S, \\

-A_{1}^{-1}SA_{1} = S^{4},\quad \text{and}\quad A_{1}^{-1}BA_{1} = B\quad \text{or}\quad B^{2}.

-\end{gather*}

-

-If the sub-groups of order~$4$ are not cyclical, $G$~must contain an

-operation~$A'$ of order~$2$, which is permutable with~$A$; and $A'$~transforms

-$S$ into its inverse and $\{B\}$~into itself. In this case, if the self-conjugate

-sub-group is of type~\Inum{($\alpha$)}, there are two types given by

-\begin{gather*}

-A'^{2} = 1,\quad A^{2} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ASA = S,\quad B^{-1}SB = S, \\

-A'AA' = A,\quad A'SA' = S^{4},\quad ABA = B,\quad A'BA' = B\quad \text{or}\quad B^{2}.

-\end{gather*}

-

-If the self-conjugate sub-group is of type~\Inum{($\beta$)}, there is a single

-type in which the last two of the preceding equations are replaced

-by

-\[

-ABA = B^{2},\quad A'BA' = B.

-\]

-

-Secondly, let the operations of order~$5$ form a single conjugate

-set. The sub-groups of order~$4$ must then be cyclical since $G$~contains

-an operation~$T$, such that $T^{4}$~is the lowest power of~$T$ which

-is permutable with~$S$. Also $S$~is permutable in a sub-group of order~$15$.

-This sub-group must be self-conjugate; and therefore $G$~contains

-a self-conjugate sub-group of order~$3$. Let

-\[

-B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S

-\]

-define the self-conjugate sub-group of order~$15$; and let $A_{1}$ be an

-%% -----File: 123.png---Folio 107-------

-operation of order~$4$, none of whose powers is permutable with~$S$.

-We may then take

-\[

-A_{1}^{-1}SA_{1} = S^{2};

-\]

-since $\{B\}$ is self-conjugate, $A_{1}$~must transform this sub-group into

-itself. There are therefore two types given by

-\[

-A_{1}^{4} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S,\quad A_{1}^{-1}SA_{1} = S^{2},

-\]

-and

-\[

-A_{1}^{-1}BA_{1} = B\quad \text{or\quad} B^{2}.

-\]

-Hence there are in all twelve distinct groups of order~$60$, each of

-which has a self-conjugate sub-group of order~$5$.

-

-\Inum{(iii)} Next, suppose that $G$~contains $6$~conjugate sub-groups of

-order~$5$. No operation of order~$3$ can be permutable with an

-operation of order~$5$, and therefore by Sylow's theorem there must

-be $10$~conjugate sub-groups of order~$3$. Hence $G$~contains $24$~operations

-of order~$5$ and $20$~operations of order~$3$. If any one

-operation of order~$5$ were permutable with an operation of order~$2$,

-all its powers would be permutable with the same operation, and

-therefore, since the sub-groups of order~$5$ form a single conjugate

-set, every operation of order~$5$ would be permutable with an

-operation of order~$2$. The group would then contain at least $24$~operations

-of order~$10$. This is clearly impossible, since the sum of

-the numbers of operations of orders $3$,~$5$ and~$10$ would be greater

-than the order of the group. Hence the sub-group of order~$10$,

-which contains self-conjugately a sub-group of order~$5$, must be of

-the type

-\[

-A^{2} = 1,\quad S^{5} = 1,\quad ASA = S^{4}.

-\]

-

-In a similar way, we shew that a sub-group of order~$6$, which

-contains self-conjugately a sub-group of order~$3$, is of the type

-\[

-A^{2} = 1,\quad B^{3} = 1,\quad ABA = B^{-1}.

-\]

-

-Since no operation of order $3$~or~$5$ is permutable with an operation

-of order~$2$, it follows (Theorem~III, §~80) that no two sub-groups of

-order~$4$ can have a common operation other than identity. Hence

-there must be $5$~sub-groups of order~$4$; for if there were $3$~or~$15$,

-some of them would necessarily have common operations. Each

-sub-group of order~$4$ is therefore contained self-conjugately in a

-sub-group of order~$12$. Such a sub-group of order~$12$ can contain no

-self-conjugate operation of order~$2$, since $G$~contains no operation of

-order~$6$. Hence the sub-groups of order~$4$ are non-cyclical, and the

-$3$~operations of order~$2$ in any sub-group of order~$4$ are conjugate

-operations in the sub-group of order~$12$ containing it. This sub-group

-must therefore be of the type

-\[

-B^{3} = 1,\quad B^{-1}A_{1}B = A_{2},\quad B^{-1}A_{2}B = A_{1}A_{2},

-\]

-where $A_{1}$ and $A_{2}$ are two permutable operations of order~$2$.

-

-The $5$~sub-groups of order~$4$ contain therefore $15$~distinct operations

-of order~$2$; and these form a conjugate set. We have already

-%% -----File: 124.png---Folio 108-------

-seen that the $20$~operations of order~$3$ form a conjugate set, and

-that the $24$~operations of order~$5$ form two conjugate sets of $12$

-each. Hence the $60$~operations of the group are distributed in $5$~conjugate

-sets, containing respectively $1$,~$12$, $12$,~$15$ and $20$~operations.

-It follows at once (§~27, \PageRef{p.}{35}) that the group, when it exists,

-is simple.

-

-A sub-group of order~$12$, the existence of which has been proved,

-must be one of $5$~conjugate sub-groups; and, since the group is

-simple, no operation can transform each of these into itself. Hence

-if the $5$~conjugate sub-groups

-\[

-H_{1},\ H_{2},\ H_{3},\ H_{4},\ H_{5}

-\]

-are transformed by any operation of the group into

-\[

-H_{1}',\ H_{2}',\ H_{3}',\ H_{4}',\ H_{5}',

-\]

-and if we regard

-\[

-%[** TN: Left delimiter missing in the original]

-\Sub{H_{1}, H_{2}, H_{3}, H_{4}, H_{5}}

-    {H_{1}', H_{2}', H_{3}', H_{4}', H_{5}'}

-\]

-as a substitution performed on $5$~symbols, the group is simply

-isomorphic with a substitution group of $5$~symbols. In other words,

-the group can be represented as a group of substitutions of $5$~symbols.

-Now there are just $60$~even substitutions of $5$~symbols; and it is

-easy to verify that the group they form satisfy all the conditions

-above determined. Moreover it will be formally proved in Chapter~VIII,

-and it is indeed almost obvious, that no group of substitutions

-can be simple if it contains odd substitutions. Hence finally, there

-is one and only one type of group of order~$60$ which contains $6$~sub-groups

-of order~$5$.

-\end{Remark}

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$60$|)}%

-

-\begin{Ex}[1.] If $p$,~$q$,~$r$ are distinct primes, shew that a group of order~$p^{2}qr$,

-which contains a self-conjugate operation of order~$r$, must be

-the direct product of two groups of orders $p^{2}q$ and $r$ respectively.

-\end{Ex}

-

-\begin{Ex}[2.] Shew that there is a single type of group of order~$84$

-which contains $28$~sub-groups of order~$3$; and determine its defining

-relations.

-\end{Ex}

-

-\begin{Ex}[3.] If $p^{\alpha}$ ($\alpha > 1$) is the highest power of~$p$ which divides the

-order of~$G$, and if $1 + kp$ be the number of sub-groups of~$G$ of order~$p^{\alpha}$,

-shew that \Inum{(i)}~if $1 + kp < p^{2}$, \Inum{(ii)}~if a group of order~$p^{x}$ is cyclical

-and $1 + kp < p^{\alpha}$, $G$~is composite.

-\index{Maillet@\Topic{Maillet}, quoted}%

-\Attrib{Maillet, \textit{Comptes Rendus}, \VolNo{CXVIII}, (1894), p.~1188.}

-\end{Ex}

-

-\Par{86.} A remarkable and important extension of Sylow's

-theorem has recently been given by Herr Frobenius\footnotemark.

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius: \textit{Berliner Sitzungsberichte} (1895), p.~988.}%

-In

-the theorem, as stated above, $p^{\alpha}$~is the highest power of a

-%% -----File: 125.png---Folio 109-------

-prime~$p$ that divides the order of a group. Herr Frobenius

-shews that, if $p^{k}$~is any power of a prime~$p$ that divides the

-order of a group, the number of sub-groups of order~$p^{k}$ is

-congruent to unity, $\pmod{p}$. These groups do not however, in

-general, form a single conjugate set.

-

-For a group whose order is a power of~$p$ higher than~$p^{k}$, this

-result has been already proved in §~61. Suppose now that

-\[

-G_{k},\ G_{k}',\ G_{k}'',\ \Dots,

-\]

-are the sub-groups of order~$p^{k}$ contained in a group~$G$, and that

-$p^{\alpha}$~is the highest power of~$p$ dividing the order of~$G$. It has

-been seen, in the proof of Sylow's theorem, that $G$~contains at

-least one sub-group~$G_{\alpha}$ of order~$p^{\alpha}$. The above set of sub-groups

-of order~$p^{k}$ may then be divided into two classes, those

-namely which are contained in~$G_{\alpha}$, and those which are not.

-The result of §~61 shews that the number of the sub-groups

-contained in the first of these classes is congruent to unity,

-$\pmod{p}$; and if $G_{\alpha}$ is a self-conjugate sub-group of~$G$, all the

-sub-groups must be contained in this class. If $G_{\alpha}$ is not

-self-conjugate in~$G$, and if $G_{k}$, any one of the sub-groups

-belonging to the second class, be transformed by all the

-operations of~$G_{\alpha}$, a set of $p^{x}$ sub-groups of order~$p^{k}$ will result.

-

-Now it is easy to see that $x$~is not less than unity, and that

-every one of these $p^{x}$~sub-groups belongs to the second class.

-For suppose first that $x$~is zero, so that $G_{k}$~is transformed into

-itself by every operation of~$G_{\alpha}$. Then $\{G_{\alpha}, G_{k}\}$ is a sub-group

-of~$G$ whose order is a power of~$p$; and since $G_{k}$~is not contained

-in~$G_{\alpha}$, the order of this sub-group is not less than $p^{\alpha + 1}$. This is

-impossible, since $p^{\alpha + 1}$~is not a factor of the order of~$G$. Secondly,

-if $P^{-1}G_{k}P$ were contained in~$G_{\alpha}$, $P$~being some operation of~$G_{\alpha}$,

-then $G_{k}$~would be contained in~$PG_{\alpha}P^{-1}$ or in~$G_{\alpha}$, contrary

-to supposition. If the sub-groups of the second class are

-not thus exhausted, and if $G_{k}'$ is a new one, we may on transforming~$G_{k}'$

-by the operations of~$G_{\alpha}$ form a fresh set of $p^{x'}$~sub-groups

-of the second class which are distinct from each

-other and from the previous set; and this process may be

-continued till the second class is exhausted. The number of

-sub-groups in the second class is therefore a multiple of~$p$.

-Hence:---

-%% -----File: 126.png---Folio 110-------

-

-\begin{Theorem}[VI.]

-If $p^{k}$, where $p$~is prime, divides the order of

-\index{Sylow's theorem@\Topic{Sylow's theorem}!extension of}%

-a group~$G$, the number of sub-groups of~$G$ of order~$p^{k}$ is congruent

-to unity, $\pmod{p}$.

-\end{Theorem}

-

-\Par{87.} If $p$~is a prime which divides the order of a group~$G$,

-it immediately follows from the foregoing theorem that the

-number of operations of~$G$, which satisfy the relation

-\[

-S^{p} = 1,

-\]

-is a multiple of~$p$. For the number of sub-groups of~$G$ of order~$p$

-is $k_{1}p + 1$, and no two of these sub-groups can contain a

-common operation except identity. There are therefore

-\[

-(k_{1}p + 1)(p - 1)

-\]

-distinct operations of order~$p$ in~$G$; these, together with the

-identical operation, which also satisfies

-\[

-S^{p} = 1,

-\]

-give in all $(k_{1}p - k_{1} + 1)p$ operations.

-

-This result has been generalized by Herr Frobenius\footnote

-  {``Verallgemeinerung des Sylow'schen Satzes,'' \textit{Berliner Sitzungsberichte}

-  (1895), pp.~984,~985.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-in the

-following form:---

-\index{Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$|(}%

-

-\begin{Theorem}[VII.]

-If $n$~is a factor of the order~$N$ of a group~$G$,

-the number of operations of~$G$, including identity, whose

-orders are factors of~$n$, is a multiple of~$n$.

-\end{Theorem}

-

-It is easy to verify the truth of this theorem directly for

-small values of~$N$; and we may therefore assume it true for

-every group whose order is less than that of the given group~$G$.

-Again, when $n$~is equal to~$N$, the theorem is obviously true.

-If then, on the assumption that the theorem is true for all

-factors of~$N$ which are greater than~$n$, we shew that it is true

-for~$n$, the general truth of the theorem will follow by induction.

-

-If $p$ is any factor of~$\dfrac{N}{n}$, we assume that the number of

-operations of~$G$ whose orders divide~$np$ is a multiple of~$np$, and

-therefore also of~$n$; and we have to shew that the number of

-operations of~$G$, whose orders divide~$np$ and do not divide~$n$, is

-%% -----File: 127.png---Folio 111-------

-also a multiple of~$n$. Let this set of operations be denoted by~$A$.

-If $p^{\lambda - 1}$ is the highest power of~$p$ that divides~$n$, the order of

-every operation of the set~$A$ must be equal to or be a multiple

-of~$p^{\lambda}$. Let $P$ be one of these operations and $m$~its order. Then,

-if $\phi(m)$ is the number of integers less than and prime to~$m$,

-the cyclical sub-group~$\{P\}$ contains $\phi(m)$~operations of order~$m$,

-and each of these belongs to the set~$A$. If these do not exhaust

-the set, let $P'$ of order~$m'$ be another operation belonging to it.

-Then no one of the $\phi(m')$~operations of order~$m'$, contained in

-the cyclical sub-group~$\{P'\}$, can be identical with any of the

-preceding set of $\phi(m)$~operations, since $P'$~itself is not contained

-in that set; at the same time, the new set of $\phi(m')$~operations

-all belong to~$A$. This process may be continued till

-the set~$A$ is exhausted. Now $m$, $m'$,~\Dots\ are all divisible by~$p^{\lambda}$,

-and therefore $\phi(m)$, $\phi(m')$,~\Dots\ are all divisible by~$p^{\lambda - 1}(p - 1)$.

-Hence the number of operations in the set~$A$ is a multiple

-of~$p^{\lambda - 1}$.

-

-If now $n = p^{\lambda - 1}s$, where by supposition $s$~is relatively prime

-to~$p$, it remains to shew that the number of operations in the

-set~$A$ is a multiple of~$s$. For this purpose, let $P$ be any

-operation of~$G$ of order~$p^{\lambda}$; and let those operations of~$G$, which

-are permutable with~$P$, form a sub-group~$H$ of order~$p^{\lambda}r$. The

-number of operations of~$H$ whose orders divide~$s$ is the same as

-the number whose orders divide~$t$, where $t$~is the greatest common

-measure of $r$~and~$s$. Now the order of~$\dfrac{H}{\{P\}}$ is less than~$N$, and

-therefore we may assume that the number of operations of~$\dfrac{H}{\{P\}}$

-whose orders divide~$t$ is a multiple of~$t$, say~$kt$. $H$~therefore

-contains $kt$~operations of the form~$PT$, where $P$ and $T$ are

-permutable and the order of~$T$ divides~$s$. Now $P$ is one of

-a set of $\dfrac{N}{p^{\lambda}r}$ conjugate operations in~$G$; and corresponding to

-each of these, there is a similar set of $kt$~operations. Moreover

-no two of these operations can be identical; for we have

-seen (§~16) that, if $m$~and~$n$ are relatively prime, an operation

-of order~$mn$ of a group~$G$, can be expressed in only one way

-as the product of two permutable operations of~$G$ of orders $m$~and~$n$.

-%% -----File: 128.png---Folio 112-------

-

-The complete set of $\dfrac{N}{p^{\lambda} r} kt$ operations of the form~$PT$ belongs

-to~$A$; if $A$ is not thus exhausted, its remaining operations can

-be divided into similar sets. Now $N$ is divisible by both $r$

-and $s$, and therefore by their least common multiple~$\dfrac{rs}{t}$. Hence

-$\dfrac{Nkt}{p^{\lambda} r}$ is divisible by~$s$, and therefore also the number of operations

-in the set~$A$ is divisible by~$s$. The number of operations

-in~$A$, being divisible both by~$p^{\lambda - 1}$ and by~$s$, is therefore divisible

-by~$n$. Hence finally, the number of operations of~$G$, whose

-orders divide~$n$, is a multiple of~$n$; and the theorem is proved.

-

-\begin{Corollary}[I.]

-Let $N_{n}$ be the number of operations of~$G$

-whose orders divide~$n$, and suppose that

-\[

-N_{n} = n.

-\]

-Let $p$~be the smallest prime factor of~$n$, and $p^{\alpha}$~the highest

-power of~$p$ that divides~$n$, so that

-\[

-n = p^{\alpha} n_{1},

-\]

-where every prime factor of~$n_{1}$ is greater than~$p$. If the order

-of~$G$ is divisible by a higher power of~$p$ than~$p^{\alpha}$, then

-\[

-N_{pn} = kpn,

-\]

-where $k$~is an integer. Now if the order~$m$ of any operation

-divides~$pn$ and does not divide~$n$, then $m$~must be a

-multiple of~$p^{\alpha + 1}$; and therefore among the operations, whose

-orders divide~$n$, there must be operations whose orders are

-equal to or are multiples of~$p^{\alpha}$. Hence

-\[

-N_{p^{\alpha - 1} n_{1}} = \lambda p^{\alpha - 1} n_{1},

-\]

-where

-\[

-\lambda < p.

-\]

-Now $N_{p^{\alpha} n_{1}} - N_{p^{\alpha - 1} n_{1}}$ is the number of operations whose orders

-are factors of~$n$ and multiples of~$p^{\alpha}$; and it has been shewn, in

-the proof of the theorem, that this number is a multiple of~$p^{\alpha - 1} (p - 1)$.

-Hence

-\[

-(p - \lambda) p^{\alpha - 1} n_{1} = \mu p^{\alpha - 1} (p - 1).

-\]

-Since every prime factor of~$n_{1}$ is greater than~$p$, this equation

-requires that $\lambda$~should be unity; therefore

-\[

-N_{p^{\alpha - 1} n_{1}} = p^{\alpha - 1} n_{1}.

-\]

-\index{Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$|)}%

-%% -----File: 129.png---Folio 113-------

-

-This process may be repeated to shew that, for each value

-of~$\beta$ which is not greater than~$\alpha$, we have

-\[

-N_{p^{\alpha - \beta} n_{1}} = p^{\alpha - \beta}n_{1},

-\]

-so that, finally,

-\[

-N_{n_{1}} = n_{1}.

-\]

-

-Moreover it is easy to see that the reasoning holds when

-$p^{\alpha}$~is the highest power of~$p$ that divides the order of~$G$, provided

-that then the sub-groups of~$G$, of order~$p^{\alpha}$, are cyclical.

-

-Suppose now that

-\[

-n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}},

-\]

-where $p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are primes in ascending order; and either

-that, for each value of~$r$ from $1$ to~$n - 1$, $p_{r}^{\alpha_{r}}$~is not the highest

-power of~$p_{r}$ which divides the order of~$G$: or that, if $p_{r}^{\alpha_{r}}$~is the

-highest power of~$p_{r}$ that divides the order of~$G$, the sub-groups

-of order~$p_{r}^{\alpha_{r}}$ are cyclical.

-

-Then we may prove as above, first, that $G$~contains operations

-of each of the orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n - 1}^{\alpha_{n - 1}}$; and secondly

-that, for each value of~$r$ from $2$ to~$n$,

-\[

-N_{p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}}

- = p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}.

-\]

-

-The equation

-\[

-N_{p_{n}^{\alpha_{n}}} = p_{n}^{\alpha_{n}}

-\]

-implies that $G$ has a self-conjugate sub-group of order~$p_{n}^{\alpha_{n}}$; for

-$G$~has a sub-group of this order, and if $G$ had more than one, it

-would necessarily contain more than $p_{n}^{\alpha_{n}}$~operations whose

-orders divide~$p_{n}^{\alpha_{n}}$.

-

-Again, since $G$ contains a self-conjugate sub-group of order~$p_{n}^{\alpha_{n}}$

-and also operations of order~$p_{n - 1}^{\alpha_{n - 1}}$, it must contain a sub-group of

-order $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$. If it had more than one sub-group of this order,

-it would contain more than $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$ operations whose orders

-divide $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$.  But since

-\[

-N_{p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}}

- = p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}},

-\]

-this is impossible. Hence $G$~contains a single sub-group of

-%% -----File: 130.png---Folio 114-------

-order $p_{n - 1}^{\alpha_{n -1}}p_{n}^{\alpha_{n}}$, which is necessarily self-conjugate. In the

-same way we shew that, for each order

-\[

-p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}\quad (r = n - 1, n - 2, \Dots, 1),

-\]

-$G$~contains a single sub-group.

-

-Finally then, under the conditions stated above, the equation

-\[

-N_{n} = n

-\]

-involves the property that $G$~has a self-conjugate sub-group of

-order~$n$ and no other sub-group of the same order.

-\end{Corollary}

-

-\begin{Corollary}[II\protect\footnotemark.]

-\footnotetext{Frobenius, \textit{Berliner Sitzungsberichte}, (1895), p.~987.}%

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-If $m$~and~$n$ are relatively prime factors of

-the order of~$G$, and if the numbers of operations of~$G$ whose

-orders divide $m$~and~$n$ respectively are equal to $m$~and~$n$; then

-every operation whose order is a factor of~$m$ is permutable with

-every operation whose order is a factor of~$n$, and the number of

-operations of~$G$ whose orders divide~$mn$ is equal to~$mn$.

-\end{Corollary}

-

-Every operation of~$G$ whose order divides~$mn$ can be expressed

-as the product of two permutable operations whose

-orders divide $m$~and $n$ respectively; and therefore the number

-of operations whose orders divide~$mn$ cannot be greater than~$mn$.

-On the other hand, it follows from the theorem that the

-number of such operations cannot be less than~$mn$; and therefore

-every operation whose order divides~$m$ must be permutable

-with every operation whose order divides~$n$.

-

-\begin{Corollary} [III\protect\footnotemark.]

-\footnotetext{Frobenius: ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),

-  p.~170.}%

-If a group of order~$mn$, where $m$~and~$n$ are

-relatively prime, contains a self-conjugate sub-group of order~$n$,

-the group contains exactly $n$ operations whose orders divide~$n$.

-\end{Corollary}

-

-If the group~$G$ has an operation~$S$ whose order divides~$n$,

-and if $S$ is not contained in the self-conjugate sub-group~$H$ of

-order~$n$, $\{S, H\}$~would be a sub-group of~$G$, whose order is

-greater than~$n$ and at the same time contains no factor in

-common with~$m$. This is impossible, and therefore $H$~must

-contain all the operations of~$G$ whose orders divide~$n$.

-%% -----File: 131.png---Folio 115-------

-

-\begin{Corollary} [IV\protect\footnotemark.]

-\footnotetext{Frobenius, \textit{loc.~cit.}\ p.~170.}%

-If $G$~has a self-conjugate sub-group~$H$ of

-order~$mn$, where $m$ and $n$ are relatively prime, and if $H$ has a

-self-conjugate sub-group~$K$ of order~$n$, then $K$~is a self-conjugate

-sub-group of~$G$.

-\end{Corollary}

-

-For by the preceding Corollary, $H$~contains exactly $n$~operations

-whose orders divide~$n$, namely the operations forming the

-sub-group~$K$; and every operation that transforms $H$ into itself

-must interchange these $n$~operations among themselves. Hence

-every operation of~$G$ is permutable with~$K$.

-

-\begin{Ex}[1.]

-Shew that if, the conditions of Corollary~II being satisfied,

-the order of~$G$ is~$mn$, then either $G$~is the direct product of two

-groups of orders $m$~and~$n$, or $G$~contains an Abelian self-conjugate

-sub-group.

-\end{Ex}

-

-\begin{Ex}[2.]

-If $p$~and~$q$ are distinct primes, there cannot be more than

-one type of group of order~$p^{\alpha}q^{\beta}$ which contains no operation of order~$pq$.

-\end{Ex}

-

-\Par{88.} We have seen in §~53 that a group~$G$, whose order is

-the power of a prime, contains a series of self-conjugate sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{n - 1},\ H_{n},\ G,

-\]

-such that in $\dfrac{G}{H_{r}}$ every operation of~$\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate. We

-shall conclude the present chapter by shewing that any group

-which has such a series of self-conjugate sub-groups is the

-direct product of two or more groups whose orders are powers

-of primes.

-

-\begin{Theorem}[VIII.]

-If a group~$G$, of order~$p^{\alpha}q^{\beta} \Dots r^{\gamma}$, where

-$p$, $q$,~\Dots, $r$ are distinct primes, has a series of self-conjugate

-sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{n - 1},\ H_{n},\ G,

-\]

-such that in $\dfrac{G}{H_{r}}$ every operation of~$\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate, then

-$G$~is the direct product of groups of order~$p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$.

-\end{Theorem}

-

-Suppose, if possible, that $p$~divides the order of~$H_{2}$ and does

-not divide the order of~$H_{1}$. If $P$ is an operation of order~$p$

-%% -----File: 132.png---Folio 116-------

-contained in~$H_{2}$, $\{P, H_{1}\}$~is a self-conjugate sub-group of~$G$;

-and every operation conjugate to~$P$ is contained in the set~$PH_{1}$.

-But the only operation of this set, whose order is~$p$, is~$P$. Hence

-$P$~must be a self-conjugate operation, contrary to the supposition

-that has been made. Hence if the order of~$H_{1}$ is not

-divisible by~$p$, neither is the order of~$H_{2}$. Suppose, next, that

-$p^{x}$~is the highest power of~$p$ that divides the orders of both $H_{r}$

-and~$H_{r - 1}$. Then the order of the sub-group $\dfrac{H_{r}}{H_{r - 1}}$ of~$\dfrac{G}{H_{r - 1}}$,

-formed of the self-conjugate operations of the latter, is not

-divisible by~$p$; and therefore the order of~$\dfrac{H_{r + 1}}{H_{r - 1}}$ is not divisible

-by~$p$. Hence $p^{x}$~is the highest power of~$p$ that divides the

-order of~$H_{r + 1}$. This reasoning may be repeated to shew that

-$p^{x}$~is the highest power of~$p$ that divides the order of each of the

-groups $H_{r + 2}$, $H_{r + 3}$,~\Dots. Hence $x$~must be equal to~$\alpha$; and

-therefore the order of~$H_{1}$ must be divisible by each of the

-primes $p$, $q$,~\Dots,~$r$.

-

-Suppose now that, for each prime~$p$ which divides the order

-of~$G$, every operation of~$H_{r}$, whose order is a power of~$p$, is

-permutable with every operation of~$G$ whose order is relatively

-prime to~$p$. Let $P$ be any operation, whose order is a power of~$p$,

-belonging to~$H_{r + 1}$ and not to~$H_{r}$; and let $Q$ be any operation

-of~$G$ whose order is relatively prime to~$p$. If $Q$~is not permutable

-with~$P$, then

-\[

-Q^{-1}PQ = Ph_{r},

-\]

-where $h_{r}$~is some operation of~$H_{r}$. The order of~$h_{r}$ must be a

-power of~$p$. For let $h_{r} = h_{r}' h_{r}''$, where the order of~$h_{r}'$ is a power

-of~$p$ and the order of~$h_{r}''$ is relatively prime to~$p$. Then, from

-the supposition made with regard to the sub-group~$H_{r}$, the

-operation~$Ph_{r}$ is the product of the permutable operations $Ph_{r}'$

-and~$h_{r}''$. But, since the order of~$Ph_{r}$ is a power of~$p$, this

-is impossible unless $h_{r}''$~is identity. If the order of~$h_{r}$ is~$p^{\beta}$,

-then

-\[

-Q^{-p^{\beta}} PQ^{p^{\beta}} = Ph_{r}^{p^{\beta}} = P;

-\]

-and this equation implies that $Q$ is permutable with~$P$, since

-$p^{\beta}$~and the order of~$Q$ are relatively prime. Hence if the

-%% -----File: 133.png---Folio 117-------

-supposition that has been made holds for~$H_{r}$, it also holds for~$H_{r + 1}$.

-But it certainly holds for~$H_{1}$, and therefore it is true for~$G$.

-Hence every operation of~$G$ whose order is a power of~$p$ is

-permutable with every operation of~$G$ whose order is relatively

-prime to~$p$. The group therefore contains self-conjugate sub-groups

-of each of the orders $p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$; and it follows, from

-the definition of §~31, that $G$~is the direct product of these

-groups.

-

-\begin{Remark}

-We add here two examples in further illustration of the applications

-of Sylow's theorem.

-\end{Remark}

-

-\begin{Ex}[1.] If $p$~is a prime, greater than~$3$, shew that the number of

-distinct types of group of order~$6p$ is $6$~or~$4$, according as $p$~is congruent

-to $1$~or $5$, $\pmod{6}$.

-\end{Ex}

-

-\begin{Ex}[2.] If $p$~is a prime, greater than~$5$, shew that the number of

-distinct types of group of order~$12p$ is $18$,~$12$, $15$ or~$10$, according

-as $p$~is congruent to $1$,~$5$, $7$ or $11$, $\pmod{12}$.

-\end{Ex}

-%% -----File: 134.png---Folio 118-------

-

-

-\Chapter{VII.}{On the Composition-Series of Group.}

-\index{Composition factors@\Topic{Composition factors}!definition of}%

-\index{Composition-series@\Topic{Composition-series}!definition of}%

-\index{Factor groups@\Topic{Factor groups}!set of, for a given group}%

-\index{Factor groups@\Topic{Factor groups}!invariance of}%

-

-\Par{89.} \First{Let} $G_{1}$~be a maximum self-conjugate sub-group (§~27)

-of a given group~$G$, $G_{2}$~a maximum self-conjugate sub-group of~$G_{1}$,

-and so on. Since $G$ is a group of finite order, we must, after

-a finite number of sub-groups, arrive in this way at a sub-group

-$G_{n - 1}$, whose only self-conjugate sub-group is that formed of the

-identical operation alone, so that $G_{n - 1}$ is a simple group.

-

-\begin{Definitions} The series of groups

-\[

-G,\ G_{1},\ G_{2},\ \Dots,\ G_{n - 1},\ 1,

-\]

-obtained in the manner just described is called a \emph{composition-series}

-of~$G$.

-

-The set of groups

-\[

-\frac{G}{G_{1}},\ \frac{G_{1}}{G_{2}},\ \Dots,\ \frac{G_{n - 2}}{G_{n - 1}},\ G_{n - 1},

-\]

-is called a set of \emph{factor-groups} of~$G$, and the orders of these

-groups are said to form a set of \emph{composition-factors} of~$G$.

-\end{Definitions}

-

-Each of the set of factor-groups is necessarily (§~30) a

-simple group.

-

-The set of groups forming a composition-series of $G$ is not,

-in general, unique. Thus $G$ may have more than one maximum

-self-conjugate sub-group, in which case the second term in the

-series may be taken different from~$G_{1}$. Moreover the groups

-succeeding $G_{1}$ are not all necessarily self-conjugate in~$G$; and

-when some of them are not so, we obtain a new composition-series

-%% -----File: 135.png---Folio 119-------

-on transforming the whole set by a suitably chosen

-operation of~$G$. That the new set thus obtained is again a

-composition-series is obvious; for if $G_{r + 1}$ is a maximum self-conjugate

-sub-group of~$G_{r}$, so also is $S^{-1} G_{r + 1} S$ of~$S^{-1} G_{r} S$. We

-proceed to prove that, if a group has two different composition-series,

-the number of terms in them is the same and the

-factor-groups derived from them are identical except as regards

-the sequence in which they occur.

-

-This result, which is of great importance in the subsequent

-theory, is due to Herr Hölder\footnotemark;

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{``Zurückführung einer beliebigen algebraischen Gleichung auf eine Kette

-  von Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV}, (1889), p.~33.}%

-and the proof we here give does

-not differ materially from his.

-

-The less general result, that, however the composition-series

-may be chosen, the composition-factors are always the same

-except as regards their sequence, had been proved by M.~Jordan\footnote

-  {``Traité des substitutions,'' (1870), p.~42.}

-\index{Jordan@\Topic{Jordan}, quoted}%

-some years before the date of Herr Hölder's memoir.

-

-\Par{90.} \begin{Theorem}[I.]

-If $H$ is any self-conjugate sub-group of a

-group~$G$; and if $K$,~$K'$ are two self-conjugate sub-groups of~$G$ contained

-in~$H$, such that there is no self-conjugate sub-group of~$G$

-contained in~$H$ and containing either $K$ or $K'$ except $H$,~$K$ and

-$K'$ themselves; and if $L$ is the greatest common sub-group of $K$

-and~$K'$, so that $L$ is necessarily self conjugate in~$G$; then the

-groups $\dfrac{H}{K}$ and $\dfrac{K'}{L}$ are simply isomorphic, as also are the groups

-$\dfrac{H}{K'}$ and~$\dfrac{K}{L}$.

-\end{Theorem}

-

-Since $K$ and $K'$ are self-conjugate sub-groups of~$G$ contained

-in~$H$, $\{K, K'\}$~must also be a self-conjugate sub-group of~$G$

-contained in~$H$; and since, by supposition, there is in~$H$ no self-conjugate

-sub-group of~$G$ other than $H$ itself, which contains

-either $K$ or~$K'$, $\{K, K'\}$~must coincide with~$H$. Hence (§~33)

-the product of the orders of $K$ and $K'$ is equal to the product

-of the orders of $H$ and~$L$.

-

-If the order of~$\dfrac{K}{L}$ is~$m$, the operations of~$K$ may be divided

-into the $m$~sets

-\[

-L,\ S_{1}L,\ S_{2}L,\ \Dots,\ S_{m - 1}L,

-\]

-%% -----File: 136.png---Folio 120-------

-such that any operation of one set multiplied by any operation

-of a second gives some operation of a definite third set, and the

-group $\dfrac{K}{L}$ is defined by the laws according to which the sets

-combine.

-

-Consider now the $m$~sets of operations

-\[

-K',\ S_{1}K',\ S_{2}K',\ \Dots,\ S_{m - 1}K'.

-\]

-

-No two operations of any one set can be identical. If

-operations from two different sets are the same, say

-\[

-S_{p}k_{1}' = S_{q}k_{2}',

-\]

-where $k_{1}'$ and $k_{2}'$ are operations of~$K'$, then

-\[

-S_{q}^{-1} S_{p} = k_{2}' k_{1}'^{-1},

-\]

-some operation of~$K'$. But $S_{q}^{-1}S_{p}$ is an operation of~$K$; hence,

-as it belongs both to $K$ and~$K'$, it must belong to~$L$, so that

-\[

-S_{p} = S_{q}l,

-\]

-where $l$ is some operation of~$L$. This however contradicts the

-supposition that the operations $S_{p}L$ and $S_{q}L$ are all distinct. It

-follows that the operations of the above $m$~sets are all distinct.

-

-Now they all belong to the group~$H$; and their number,

-being the order of~$K'$ multiplied by the order of~$\dfrac{K}{L}$, is equal to

-the order of~$H$. Hence in respect of the self-conjugate sub-group~$K'$,

-which $H$ contains, the operations of the group~$H$ can

-be divided into the sets

-\[

-K',\ S_{1}K',\ S_{2}K',\ \Dots,\ S_{m - 1}K',

-\]

-and the group $\dfrac{H}{K'}$ is defined by the laws according to which

-these sets combine. But if

-\[

-S_{p}L · S_{q}L = S_{r}L,

-\]

-then necessarily

-\[

-S_{p}K' · S_{q}K' = S_{r}K'.

-\]

-Hence the groups $\dfrac{H}{K'}$ and $\dfrac{K}{L}$ are simply isomorphic. In precisely

-the same way it is shewn that $\dfrac{H}{K}$ and $\dfrac{K'}{L}$ are simply

-isomorphic.

-%% -----File: 137.png---Folio 121-------

-

-\begin{Corollary}

-If $H$ coincides with~$G$, $K$~and $K'$ are maximum

-self-conjugate sub-groups of~$G$. Hence if $K$ and $K'$ are maximum

-self-conjugate sub-groups of~$G$, and if $L$ is the greatest group

-common to $K$ and~$K'$, then $\dfrac{G}{K}$ and $\dfrac{K'}{L}$ are simply isomorphic; as

-also are $\dfrac{G}{K'}$ and~$\dfrac{K}{L}$.

-\end{Corollary}

-

-Now $\dfrac{G}{K}$ and $\dfrac{G}{K'}$ are simple groups; and therefore, $\dfrac{K}{L}$ and $\dfrac{K'}{L}$

-being simple groups, $L$~must be a maximum self-conjugate

-sub-group of both $K$ and~$K'$.

-

-\Par{91.} We may now at once proceed to prove by a process of

-induction the properties of the composition-series of a group

-stated at the end of §~89. Let us suppose that, for groups

-whose orders do not exceed a given number~$n$, it is already

-known that any two composition-series contain the same

-number of groups and that the factor-groups defined by them

-are the same except as regards their sequence. If $G$, a group

-whose order does not exceed~$2n$, has more than one composition-series,

-let two such series be

-\begin{align*}

-&G,\ G_{1},\ G_{2},\Dots,\ 1; \\

-\lintertext{and}

-&G,\ G_{1}',\ G_{2}',\Dots,\ 1.

-\end{align*}

-If $H$ is the greatest common sub-group of $G_{1}$ and~$G_{1}'$, and if

-\[

-H,\ I,\ J,\ \Dots,\ 1

-\]

-is a composition-series of~$H$, then, by the Corollary in the

-preceding paragraph,

-\begin{align*}

-G,\ G_{1},\ H,\ I,\ &J,\Dots,\ 1, \\

-\lintertext{and}

-G,\ G_{1}',\ H,\ I,\ &J,\Dots,\ 1

-\end{align*}

-are two composition-series of~$G$ which contain the same number

-of terms and give the same factor-groups. For it has there

-been shewn that $\dfrac{G}{G_{1}}$ and $\dfrac{G_{1}'}{H}$ are simply isomorphic; as also are

-$\dfrac{G}{G_{1}'}$ and $\dfrac{G_{1}}{H}$. Now the order of~$G_{1}$, being a factor of the order of~$G$,

-cannot exceed~$n$. Hence the two composition-series

-\begin{alignat*}{3}

-&G_{1},&&\ G_{2}, &&\Dots,\ 1, \\

-\lintertext{and}

-&G_{1},&&\ H,\ I, &&\Dots,\ 1,

-\end{alignat*}

-%% -----File: 138.png---Folio 122-------

-\index{Composition-series@\Topic{Composition-series}!invariance of}%

-by supposition contain the same number of groups and give the

-same factor-groups; and the same is true of the two composition-series

-\begin{alignat*}{2}

-&G_{1}',\ G_{2}',&&\Dots,\ 1, \\

-\lintertext{and}

-&G_{1}',\ H,\ I,&&\Dots,\ 1.

-\end{alignat*}

-Hence finally, the two original series are seen, by comparing

-them with the two new series that have been formed, to have

-the same number of groups and to lead to the same factor-groups.

-The property therefore, if true for groups whose order

-does not exceed~$n$, is true also for groups whose order does not

-exceed~$2n$. Now the simplest group, which has more than one

-composition-series, is that defined by

-\[

-A^{2} = 1,\quad B^{2} = 1,\quad AB = BA.

-\]

-For this group there are three distinct composition-series, viz.\

-\begin{alignat*}{2}

-&\{A,\, B\},\ \{A\}, &&1; \\

-&\{A,\, B\},\ \{B\}, &&1; \\

-\lintertext{and}

-&\{A,\, B\},\ \{AB\},\ &&1:

-\end{alignat*}

-and for these the theorem is obviously true. It is therefore

-true generally. Hence:---

-

-\begin{Theorem}[II.]

-Any two composition-series of a group consist

-of the same number of sub-groups, and lead to two sets of factor-groups

-which, except as regards the sequence in which they occur,

-are identical with each other.

-\end{Theorem}

-

-The definite set of simple groups, which we thus arrive at

-from whatever composition-series we may start, are essential

-constituents of the group: the group is said to be \emph{compounded}

-from them. The reader must not, however, conclude either that

-the group is defined by its set of factor-groups, or that it necessarily

-contains a sub-group simply isomorphic with any given

-one of them.

-

-\Par{92.} It has been already pointed out that the groups in a

-composition-series of~$G$ are not necessarily, all of them, self-conjugate

-groups of~$G$.

-

-Suppose now that a series of groups

-\[

-G,\ H_{1},\ H_{2},\ \Dots,\ H_{m - 1},\ 1

-\]

-%% -----File: 139.png---Folio 123-------

-are chosen so that each one is a self-conjugate sub-group of~$G$,

-while there is no self-conjugate sub-group of~$G$ contained in any

-one group of the series and containing the next group.

-

-\begin{Definition}

-The series of groups, obtained in the manner

-\index{Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}!definition of}%

-\index{Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}!invariance of}%

-just described, is called a \emph{chief composition-series}, or a \emph{chief-series}

-of~$G$.

-\end{Definition}

-

-It should be noticed that such a series is not necessarily

-obtained by dropping out from a composition-series those of its

-groups which are not self-conjugate in the original group.

-Thus it follows immediately, from the results of §~54, that the

-composition-series of a group whose order is the power of a

-prime can be chosen, either \Inum{(i)}~so that every group of the series

-is a self-conjugate sub-group, or \Inum{(ii)}~so as to contain any given

-sub-group, self-conjugate or not.

-

-A chief composition-series of a group is not necessarily

-unique; and when a group has more than one, the following

-theorem, exactly analogous to Theorem~II, holds:---

-

-\begin{Theorem}[III.]

-Any two chief-composition-series of a group

-consist of the same number of terms and lead to two sets of

-factor-groups, which, except as regards the sequence in which they

-occur, are identical with each other.

-\end{Theorem}

-

-The formal proof of this theorem would be a mere repetition

-of the proof of §~91, Theorem~I itself being used to start from

-instead of its Corollary; it is therefore omitted.

-

-Although it is not always possible to pass from a composition-series

-to a chief-series, the process of forming a composition-series

-on the basis of a given chief-series can always be carried

-out. Thus if, in a chief-series, $H_{r + 1}$~is not a maximum self-conjugate

-sub-group of~$H_{r}$, the latter group must have a

-maximum self-conjugate sub-group $G_{r, 1}$ which contains~$H_{r + 1}$.

-If $H_{r + 1}$ is not a maximum self-conjugate sub-group of~$G_{r, 1}$, then

-such a group, $G_{r, 2}$, may be found still containing~$H_{r + 1}$; and this

-process may be continued till we arrive at a group~$G_{r, s - 1}$, of

-which $H_{r + 1}$~is a maximum self-conjugate sub-group. A similar

-process may be carried out for each pair of consecutive terms

-in the chief-series; the resulting series so obtained is a composition-series

-of the original group.

-%% -----File: 140.png---Folio 124-------

-

-\Par{93.} The factor-groups $\dfrac{H_{r}}{H_{r + 1}}$ arising from a chief-series are

-not necessarily simple groups. If between $H_{r}$ and $H_{r + 1}$ no

-groups of a corresponding composition-series occur, the group

-$\dfrac{H_{r}}{H_{r + 1}}$ is simple; but when there are such intermediate groups,

-$\dfrac{H_{r}}{H_{r + 1}}$ cannot be simple. We proceed to discuss the nature of

-this group in the latter case.

-

-Let $G$ be multiply isomorphic with~$G'$, so that the self-conjugate

-sub-group $H_{r + 1}$ of~$G$ corresponds to the identical

-operation of~$G'$. Also let

-\[

-H_{1}',\ H_{2}',\ \Dots,\ H_{p}',\ H'_{p + 1},\ \Dots,\ H_{r}',\ 1

-\]

-be the sub-groups of~$G'$ which correspond to the sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{p},\ H_{p + 1},\ \Dots,\ H_{r},\ H_{r + 1}

-\]

-of~$G$. Since $H_{p}$ contains~$H_{p + 1}$, $H_{p}'$~must contain~$H'_{p + 1}$; and

-since $H_{p + 1}$ is self-conjugate in~$G$, $H'_{p + 1}$~is self-conjugate in~$G'$.

-Also if $G'$ had a self-conjugate sub-group contained in~$H_{p}'$ and

-containing~$H'_{p + 1}$, $G$~would have a self-conjugate sub-group

-contained in~$H_{p}$ and containing~$H_{p + 1}$. This is not the case,

-and therefore

-\[

-G',\ H_{1}',\ H_{2}',\ \Dots,\ H_{r}',\ 1

-\]

-is a chief-series of~$G'$. Hence $\dfrac{H_{r}}{H_{r + 1}}$ is simply isomorphic with~$H_{r}'$,

-the last group but one in the chief-series of~$G'$.

-

-\begin{Definition}

-If $\Gamma$ is a self-conjugate sub-group of~$G$, and

-\index{Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}!definition of}%

-if $G$~has no self-conjugate sub-group, contained in~$\Gamma$, whose

-order is less than that of~$\Gamma$, then $\Gamma$~is called a \emph{minimum} self-conjugate

-sub-group of~$G$.

-\end{Definition}

-

-Making use of the phrase thus defined, the discussion of the

-factor-groups $\dfrac{H_{r}}{H_{r + 1}}$ of a chief-series is the same as that of the

-minimum self-conjugate sub-groups of a given group.

-

-\Par{94.} To simplify the notation as much as possible, let $I$ be

-a minimum self-conjugate sub-group of~$G$; and, if $I$ is not a

-simple group, suppose that in a composition-series of~$G$ the

-%% -----File: 141.png---Folio 125-------

-term following~$I$ is~$g_{1}$. Since $g_{1}$~cannot be self-conjugate in~$G$,

-it must be one of a set of conjugate sub-groups

-\[

-%[** TN: Item tag set like an equation number in the original; only instance]

-\Inum{($\alpha$)}\quad g_{1},\ g_{2},\ \Dots,\ g_{n}.

-\]

-

-Now, if $g_{r} = S^{-1}g_{1}S$, then as $g_{1}$~is a maximum self-conjugate

-sub-group of~$I$, $S^{-1}g_{1}S$~or $g_{r}$ is a maximum self-conjugate

-sub-group of $S^{-1}IS$ or~$I$; and hence every one of the above set

-of conjugate sub-groups is a maximum self-conjugate sub-group

-of~$I$. If $g_{rs}$ represents the greatest sub-group common to $g_{r}$ and~$g_{s}$,

-then, by Theorem~I of the present chapter (§~90),

-\begin{gather*}

-\Dots,\ I,\ g_{1},\ g_{1r},\ \Dots \\

-\Dots,\ I,\ g_{r},\ g_{1r},\ \Dots

-\end{gather*}

-are composition-series of~$G$. Hence $\dfrac{g_{1}}{g_{1r}}$  is simply isomorphic

-with~$\dfrac{I}{g_{r}}$. Now $\dfrac{I}{g_{r}}$ and $\dfrac{SIS^{-1}}{Sg_{r}S^{-1}}$, or~$\dfrac{I}{g_{1}}$ are simply isomorphic;

-therefore $\dfrac{g_{1}}{g_{1r}}$ and $\dfrac{I}{g_{1}}$ are simply isomorphic. Similarly we may

-shew that, whatever $r$~and~$s$ may be, $\dfrac{g_{r}}{g_{rs}}$~and $\dfrac{I}{g_{1}}$ are simply

-isomorphic.

-

-If $g_{1r}$ and $g_{1s}$ are the same group, whatever $r$~and~$s$

-may be, this group is common to the whole set of conjugate

-groups~\Inum{($\alpha$)}. If these groups have any common sub-group,

-except identity, it is (Theorem~V, §~27) a self-conjugate sub-group

-of~$G$, and this self-conjugate sub-group would be contained

-in~$I$, contrary to supposition. Hence if $g_{1r}$ and $g_{1s}$ are

-the same group, for all values of $r$~and~$s$, this group consists of

-the identical operation alone, and a composition-series is

-\[

-\Dots,\ I,\ g_{1},\ 1.

-\]

-

-If $g_{1r}$ and $g_{1s}$ are distinct and if $g_{1rs}$ is their greatest common

-sub-group, then

-\begin{gather*}

-\Dots,\ I,\ g_{1},\ g_{1r},\ g_{1rs},\ \Dots \\

-\Dots,\ I,\ g_{1},\ g_{1s},\ g_{1rs},\ \Dots

-\end{gather*}

-are two composition-series, and $\dfrac{g_{1r}}{g_{1rs}}$~is simply isomorphic with with~$\dfrac{I}{g_{1}}$

-for all values of $r$~and~$s$.

-%% -----File: 142.png---Folio 126-------

-

-This reasoning may be repeated. If $g_{1rs}$ and $g_{1rt}$ are the

-same group whatever $s$~and $t$ may be, they must each be the

-identical operation. If not, $g_{1rst}$~is another term of the composition-series,

-and $\dfrac{g_{1rs}}{g_{1rst}}$~is simply isomorphic with~$\dfrac{I}{g_{1}}$.

-

-Hence however the composition-series from $I$ onwards be

-constructed, the corresponding factor-groups are all simply

-isomorphic with each other. Moreover every group after $I$ in

-the composition-series is self-conjugate in~$I$. For $g_{1}$ and $g_{r}$

-being self-conjugate sub-groups of~$I$, so also is their greatest

-common sub-group~$g_{1r}$; and $g_{1r}$,~$g_{1s}$ being self-conjugate sub-groups

-of~$I$, $g_{1rs}$~is also self-conjugate; and so on. Let the

-composition-series thus arrived at be now written

-\[

-\Dots,\ I,\ \gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s-1},\ 1;

-\]

-where, as has been proved, $\dfrac{I}{\gamma_{1}}$, $\dfrac{\gamma_{1}}{\gamma_{2}}$,~\Dots, $\gamma_{s-1}$, are simply isomorphic.

-

-The final group~$\gamma_{s-1}$ must be one of a conjugate set of, say,

-$\nu$~groups in~$G$, no one of which has any operation except

-identity in common with any other. Since each of these $\nu$~groups

-is self-conjugate in~$I$, and since no two of them have a

-common operation except identity, it follows by Theorem~IX,

-§~34, that every operation of any one of them is permutable

-with every operation of the remaining $\nu - 1$. The group generated

-by the $\nu$ groups conjugate to~$\gamma_{s-1}$, being self-conjugate in~$G$

-and contained in~$I$, must coincide with~$I$. Now, if $\gamma_{s-1}^{1}$ and

-$\gamma_{s-1}^{2}$ are any two of this set of $\nu$~groups, $\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$~is their

-direct product, and it is a self-conjugate sub-group of~$I$. If

-$s > 2$, $\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$ does not coincide with~$I$; and therefore there

-must be another sub-group~$\gamma_{s-1}^{3}$, of the set to which $\gamma_{s-1}$

-belongs, which is not contained in~$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$. Since both

-the latter group and $\gamma_{s-1}^{3}$ are self-conjugate in~$I$, while $\gamma_{s-1}^{3}$

-is a simple group, no operation of~$\gamma_{s-1}^{3}$ except identity can be

-contained in~$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$. Hence $\gamma_{s-1}^{1}$,~$\gamma_{s-1}^{2}$ and~$\gamma_{s-1}^{3}$ are independent,

-\ie\ no operation of one of these groups can be expressed

-in terms of operations of the other two. The group

-$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}, \gamma_{s-1}^{3}\}$ is the direct product of $\gamma_{s-1}^{1}$,~$\gamma_{s-1}^{2}$ and~$\gamma_{s-1}^{3}$,

-and it is self-conjugate in~$I$. If $s > 3$, the same reasoning

-%% -----File: 143.png---Folio 127-------

-may be repeated. Finally, from the set of $\nu$~groups conjugate

-to~$\gamma_{s - 1}$, it must be possible to choose $s$~independent groups

-\[

-\gamma_{s - 1}^{1},\ \gamma_{s - 1}^{2},\ \Dots,\ \gamma_{s - 1}^{s},

-\]

-such that no operation of any one of them can be expressed in

-terms of the operations of the remaining $s - 1$; and $I$~will then

-be the direct product of these $s$~groups. Hence:---

-

-\begin{Theorem}[IV.]

-If between two consecutive terms $H_{r}$ and $H_{r+1}$

-in the chief-composition-series of a group there occur the groups

-$G_{r, 1}$, $G_{r, 2}$,~\Dots, $G_{r, s - 1}$ of a composition-series; then \Inum{(i)}~the factor groups

-\[

-\frac{H_{r}}{G_{r, 1}},\ \frac{G_{r, 1}}{G_{r, 2}},\ \Dots,\ \frac{G_{r, s - 1}}{H_{r + 1}}

-\]

-are all simply isomorphic, and \Inum{(ii)}~$\dfrac{H_{r}}{H_{r + 1}}$ is the direct product of

-$s$~groups of the type~$\dfrac{H_{r}}{G_{r, 1}}$.

-\end{Theorem}

-

-\begin{Corollary}

-If the order of $\dfrac{H_{r}}{H_{r+1}}$ is a power,~$p^{s}$, of a prime,

-$\dfrac{H_{r}}{H_{r + 1}}$ must be an Abelian group of type $(1, 1, \Dots\DPchg{\,}{,} \text{to $s$ units})$.

-\end{Corollary}

-

-\Par{95.} A chief-series of a group~$G$ can always be constructed

-\index{Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}!is a simple group or the direct product of simply isomorphic simple groups}%

-which shall contain among its terms any given self-conjugate

-sub-group of~$G$. For if $\Gamma$ is a self-conjugate sub-group of~$G$,

-and if $\dfrac{G}{\Gamma}$~is simple, we may take $\Gamma$ for the group which follows~$G$

-in the chief-series. If on the other hand $\dfrac{G}{\Gamma}$~is not simple, it

-must contain a minimum self-conjugate sub-group. Then $\Gamma_{1}$,

-the corresponding self-conjugate sub-group of~$G$, contains~$\Gamma$;

-and if there were a self-conjugate sub-group of~$G$ contained in~$\Gamma_{1}$,

-and containing~$\Gamma$, the self-conjugate sub-group of~$\dfrac{G}{\Gamma}$, which

-corresponds to~$\Gamma_{1}$, would not be a minimum self-conjugate sub-group.

-We may now repeat the same process with~$\Gamma_{1}$, and so

-on; the sub-groups thus introduced will, with $G$~and~$\Gamma$,

-clearly form the part of a chief-series extending from $G$ to~$\Gamma$.

-The series may be continued from~$\Gamma$, till we arrive at the

-identical operation, in the usual way.

-%% -----File: 144.png---Folio 128-------

-

-\begin{Remark}

-\Par{96.} It will perhaps assist the reader if we illustrate the foregoing

-\index{Composition-series@\Topic{Composition-series}!examples of}%

-theory by one or two simple examples. We take first a group

-of order~$12$, defined by the relations

-\begin{gather*}

-A^{2} = 1,\quad B^{2} = 1,\quad AB = BA, \\

-R^{3} = 1,\quad R^{-1}AR = B,\quad R^{-1}BR = AB\footnotemark.

-\end{gather*}

-\footnotetext{The reader will notice that $B$~can be eliminated from these relations, and

-  that the group can be defined by $A^{2} = 1$, $R^{3} = 1$, $(AR)^{3} = 1$. The structure of the

-  group however is given, at a glance, by the equations in the text.}%

-From the last two equations, it follows that

-\[

-R^{-1}ABR = A,

-\]

-and therefore $R$~transforms the sub-group $\{A, B\}$ of order~$4$ into

-itself; so that this sub-group is self-conjugate, and the order of the

-group is $12$ as stated. The self-conjugate sub-group $\{A, B\}$ thus

-determined is clearly a maximum self-conjugate sub-group. Also it is

-the only one. For if there were another its order would be~$6$, and

-it would contain all the operations of order~$3$ in the group. Now

-since $\{R\}$~is only permutable with its own operations, the group

-contains $4$~sub-groups of order~$3$, and therefore there can be no

-self-conjugate sub-group of order~$6$. The three cyclical sub-groups

-$\{A\}$,~$\{B\}$ and~$\{AB\}$ of order~$2$ are transformed into each other by~$R$,

-and therefore no one of them is self-conjugate.

-

-Hence the only chief-series is

-\[

-\{R, A, B\},\quad \{A, B\},\quad 1,

-\]

-and there are three composition-series, viz.\

-\begin{alignat*}{4}

-&\{R, A, B\}, &&\{A, B\}, &&\{A\}, &&1; \\

-&\{R, A, B\}, &&\{A, B\}, &&\{B\}, &&1; \\

-\lintertext{and}

-&\{R, A, B\},\quad &&\{A, B\},\quad &&\{AB\},\quad &&1.

-\end{alignat*}

-

-The orders of the factor-groups in the chief-series are $3$ and~$2^{2}$,

-and the group of order~$2^{2}$ is, as it should be, an Abelian group whose

-operations are all of order~$2$. The composition-factors are $3$,~$2$,~$2$ in

-the order written.

-

-\Par{97.} As a rather less simple instance, we will now take a group

-generated by four permutable independent operations $A$,~$B$, $P$,~$Q$, of

-orders $2$,~$2$, $3$,~$3$ respectively and an operation~$R$ of order~$3$, for

-which

-\[

-R^{-1}AR = B,\quad R^{-1}BR = AB,\quad R^{-1}PR = P,\quad R^{-1}QR = QP\footnotemark.

-\]

-

-\footnotetext{Here again the group can clearly be defined in terms of $A$,~$Q$ and~$R$.}%

-The sub-group $\{A, B, P, Q\}$, of order~$36$, is clearly a maximum

-self-conjugate sub-group, and therefore the order of the group is~$108$.

-Since $A$,~$B$ and~$AB$ are conjugate operations, every self-conjugate

-sub-group that contains~$A$ must contain~$B$; and since $Q$~and $QP$ are

-%% -----File: 145.png---Folio 129-------

-conjugate, every self-conjugate sub-group that contains~$Q$ must

-contain~$P$. Hence the only other possible maximum self-conjugate

-sub-groups are those of the form $\{A, B, P, RQ^{\alpha}\}$; and since

-\[

-Q^{-1} RQ^{\alpha}Q = RQ^{\alpha} P^{-1},

-\]

-these groups actually are self-conjugate. The same reasoning shews

-that the only maximum self-conjugate sub-group of $\{A, B, P, Q\}$ or

-of $\{A, B, P, RQ^{\alpha}\}$, which is self-conjugate in the original group, is

-$\{A, B, P\}$; and the only maximum self-conjugate sub-groups of the

-latter, which are self-conjugate in the original group, are $\{A, B\}$ and~$\{P\}$.

-Hence all the chief-series of the group are given by

-\[

-\{R, A, B, P, Q\},\quad

-\begin{gathered}

-\{A, B, P, Q\}, \\

-\text{or} \\

-\{A, B, P, RQ^{\alpha}\},

-\end{gathered} \quad

-\{A, B, P\}, \quad

-\begin{gathered}

-\{A, B\}, \\

-\text{or} \\

-\{P\},

-\end{gathered}

-\quad 1.

-\]

-Since $\{A, B, P, Q\}$ is an Abelian group, all of its sub-groups are

-self-conjugate. Hence if $G_{1}$,~$G_{2}$, and~$G_{3}$ are \emph{any} maximum sub-groups

-of $\{A, B, P, Q\}$, $G_{1}$~and $G_{2}$ respectively, then

-\[

-\{R, A, B, P, Q\},\quad \{A, B, P, Q\},\quad G_{1},\quad G_{2},\quad G_{3},\quad 1

-\]

-is a composition-series.

-

-Again, since $A$~and~$B$ are conjugate in $\{A, B, P, RQ^{\alpha}\}$, the only

-maximum self-conjugate sub-groups of this group are those of the

-form $\{A, B, P^{x}(RQ^{\alpha})^{y}\}$. If $y$~is zero, this sub-group is Abelian; and

-we may take for the next term in the composition series any

-maximum sub-group~$g_{2}$ of this Abelian sub-group, and for the last

-term but one any sub-group~$g_{3}$ of~$g_{2}$. But if $y$~is not zero,

-$\{A, B, P^{x}(RQ^{\alpha})^{y}\}$ can only be followed by~$\{A, B\}$. Hence the

-remaining composition-series are of the forms:---

-\[

-\{R, A, B, P, Q\},\quad \{A, B, P, RQ^{\alpha}\},\quad \{A, B, P\},\quad g_{2},\quad g_{3},\quad 1,

-\]

-and

-\begin{gather*}

-\{R, A, B, P, Q\},\quad \{A, B, P, RQ^{\alpha}\},\quad \{A, B, P^{x}(RQ^{\alpha})^{y}\}, \\

-\{A, B\},\quad \{A\}\ \text{or}\ \{B\}\ \text{or}\ \{AB\},\quad 1.

-\end{gather*}

-It should be noticed that if, in the last of these series, we drop out

-the terms which are not self-conjugate in the original group, here

-the third and fifth terms, we do not arrive at a chief-series. This

-illustrates a remark made in §~92.

-\end{Remark}

-\index{Composition-series@\Topic{Composition-series}!examples of}%

-

-\Par{98.} \begin{Theorem}[V.]

-If $H$~is a sub-group of~$G$, each composition-factor

-of~$H$ must be equal to or be a factor of some

-composition-factor of~$G$.

-\end{Theorem}

-

-If $G$ is simple, its only composition-factor is equal to its

-order: the theorem in this case is obvious.

-%% -----File: 146.png---Folio 130-------

-

-If $G$ is not simple, let $G_{r + 1}$ be the first term in a composition-series

-of~$G$ which does not contain~$H$; and let $G_{r}$ be

-the term preceding~$G_{r + 1}$. If $H_{1}$ is the greatest common sub-group

-of $G_{r + 1}$ and~$H$, then $H_{1}$~is a self-conjugate sub-group

-of~$H$. For every operation of~$H$ transforms both $H$ and $G_{r + 1}$

-into themselves; and therefore every operation of $H$ transforms~$H_{1}$,

-the greatest common sub-group of $H$ and~$G_{r + 1}$, into itself.

-Now the order of $\{H, G_{r + 1}\}$ is equal to the product of the

-orders of $H$ and $G_{r + 1}$ divided by the order of~$H_{1}$; and $\{H, G_{r + 1}\}$

-is contained in~$G_{r}$. Hence the order of $\dfrac{H}{H_{1}}$ is equal to or is

-a factor of the order of~$\dfrac{G_{r}}{G_{r + 1}}$. If then a composition-series of~$H$

-be taken, containing the term~$H_{1}$, the orders of the factor-groups,

-formed by those terms of the series terminating with~$H_{1}$,

-are equal to or are factors of the order of~$\dfrac{G_{r}}{G_{r + 1}}$. The same

-reasoning may now be used for $H_{1}$ that has been applied to~$H$;

-and the theorem is therefore true.

-

-\begin{Corollary}

-If all the composition-factors of~$G$ are primes,

-so also are the composition-factors of every sub-group of~$G$.

-\end{Corollary}

-

-\Par{99.} \begin{Definition} A group, all whose composition-factors

-\index{Group@\Topic{Group}!soluble}%

-\index{Soluble groups@\Topic{Soluble groups}!definition of}%

-are primes, is called a \emph{soluble} group.

-\end{Definition}

-

-The Corollary of the last theorem may be stated in the

-form:---if a group is soluble, so also are all its sub-groups.

-

-A soluble group of order $p^{\alpha} q^{\beta} \Dots r^{\gamma}$, where $p$, $q$,~\Dots, $r$

-are distinct primes, has $\alpha + \beta + \Dots + \gamma$ composition-factors;

-these are capable of $\dfrac{(\alpha + \beta + \Dots + \gamma)!}{\alpha!\, \beta!\, \Dots\, \gamma!}$ distinct arrangements.

-

-For a specified group the composition-factors may, as we

-have already seen, occur in two or more distinct arrangements;

-but it is immediately obvious that two groups of the same

-order cannot be of the same type, \ie\ simply isomorphic, unless

-the distinct arrangements, of which the composition-factors are

-capable, are the same for both. A first step therefore towards

-the enumeration of all distinct types of soluble groups of a

-given order, will be to classify them according to the distinct

-%% -----File: 147.png---Folio 131-------

-\index{Soluble groups@\Topic{Soluble groups}!properties of}%

-arrangements of which the composition-factors are capable; for

-no two groups belonging to different classes can be of the same

-type.

-

-The case, in which the composition-factors are capable of all

-possible arrangements, is one which will always occur. Taking

-in this case $\beta$~$q$'s followed by $\alpha$~$p$'s for the last $\alpha + \beta$ composition-factors,

-the group contains a sub-group~$G'$ of order~$p^{\alpha} q^{\beta}$. In

-the composition-series of this group, with the composition-factors

-taken as proposed, there is a sub-group~$H$ of order~$p^{\alpha}$

-contained self-conjugately in a sub-group~$H_{1}$ of order~$p^{\alpha}q$. This

-sub-group~$H_{1}$ is contained self-conjugately in a group~$H_{2}$ of

-order~$p^{\alpha}q^{2}$. Hence (Theorem~VII, Cor.~IV, §~87) $H$~is

-contained self-conjugately in~$H_{2}$. Again, $H_{2}$~is contained self-conjugately

-in a group~$H_{3}$ of order~$p^{\alpha}q^{3}$, and therefore again $H$~is

-self-conjugate in~$H_{3}$. Proceeding thus, we shew that $H$~is

-self-conjugate in~$G'$. It follows that $n$, the number of

-conjugate sub-groups of order~$p^{\alpha}$ contained in the group, is

-not a multiple of~$q$. Now $q$~may be any one of the distinct

-primes other than~$p$ that divide the order of the group. Hence

-finally the group contains a self-conjugate sub-group of order~$p^{\alpha}$.

-In the same way we shew that it contains self-conjugate

-sub-groups of orders $q^{\beta}$,~\Dots, $r^{\gamma}$. The group must therefore

-be the direct product of groups whose orders are $p^{\alpha}$, $q^{\beta}$,~\Dots,~$r^{\gamma}$.

-

-Hence:---

-

-\begin{Theorem}[VI.]

-A soluble group, the composition-factors

-of which may be taken in any order, is the direct product of

-groups whose orders are powers of primes.

-\end{Theorem}

-

-\Par{100.} \begin{Theorem}[VII.]

-If $G$ is a soluble group of order~$p^{\alpha} m$,

-where $p$~is prime and does not divide~$m$, and if every operation

-of~$G$ whose order is a power of~$p$ is permutable with every

-operation whose order is relatively prime to~$p$; then $G$~is the

-direct product of two groups of orders $p^{\alpha}$ and~$m$.

-\end{Theorem}

-

-Let $H$ be a sub-group of~$G$ of order~$p^{\alpha}$. This sub-group,

-from the conditions of the theorem, is necessarily self-conjugate.

-Let a chief-series of~$G$ be formed which contains~$H$, say

-\[

-G,\ \Dots,\ H',\ H,\ \Dots,\ 1.

-\]

-%% -----File: 148.png---Folio 132-------

-

-The order of~$H'$ must be of the form~$p^{\alpha}q^{\beta}$; and $H'$~must

-contain a single sub-group~$h$ of order~$q^{\beta}$. This sub-group~$h$ is

-self-conjugate in~$G$, since $H'$~is self-conjugate in~$G$; and $h$~is the

-only sub-group of order~$q^{\beta}$ contained in~$H'$.

-

-Consider now the group~$\dfrac{G}{h}$. Every operation of this group

-whose order is a power of~$p$ is permutable with every operation

-whose order is relatively prime to~$p$. Hence we may repeat

-the same reasoning to shew that $\dfrac{G}{h}$~contains a self-conjugate

-sub-group of order~$r^{\gamma}$ where $r$~is a prime distinct from~$p$, but

-possibly the same as~$q$. It follows that $G$~has a self-conjugate

-sub-group~$h_{1}$ of order~$q^{\beta}r^{\gamma}$. We may now repeat the same

-reasoning with~$\dfrac{G}{h_{1}}$; and in this way we must at last reach

-a self-conjugate sub-group of~$G$ of order~$m$. Hence, since $G$~contains

-self-conjugate sub-groups of orders $p^{\alpha}$ and~$m$, which

-are relatively prime, $G$~must be the direct product of these

-sub-groups.

-

-It should be noticed that the above reasoning does not

-necessarily hold if $G$ is not soluble; for then the order

-of~$H'$ may be of the form~$p^{\alpha} \mu^{\beta}$, where~$\mu$, the order of a simple

-group, contains more than one prime factor. In that case it

-would not be necessarily true that $H'$~contains a group of order~$\mu^{\beta}$.

-

-\begin{Remark}

-\Par{101.} Though the actual determination of all types of soluble

-\index{Soluble groups@\Topic{Soluble groups}!properties of}%

-groups of a given order more properly forms part of the subject of

-Chapter~XI, we will here, as a further illustration of the subject of

-the present Chapter, deal with the problem for groups whose orders

-are of the form~$p^{2}q$, $p$~and~$q$ being distinct primes.

-

-Every such group must be soluble. In fact, if $p > q$, the group

-must contain a self-conjugate sub-group of order~$p^{2}$; and if $p < q$, there

-must be a self-conjugate sub-group of order~$q$ unless $p = 2$, $q = 3$;

-while in this case if there are $4$~sub-groups of order~$3$, there must be

-a self-conjugate sub-group of order~$4$. These statements all follow

-immediately from Sylow's theorem.

-

-The Abelian groups of order~$p^{2}q$ may be specified immediately;

-and therefore in what follows we will assume that the group is not

-Abelian. There are $3$~possible arrangements for the composition-factors,

-viz.\

-\[

-p,\ p,\ q; \quad p,\ q,\ p; \quad q,\ p,\ p.

-\]

-%% -----File: 149.png---Folio 133-------

-

-If the factors are capable of all three arrangements, the group is

-the direct product of groups of orders $p^{2}$ and~$q$; it is therefore an

-Abelian group.

-

-If the two arrangements $p$,~$p$,~$q$ and $q$,~$p$,~$p$ are possible, there are

-self-conjugate sub-groups of orders $p^{2}$ and~$q$; the group again is

-Abelian, and all three arrangements are possible.

-

-There are now five other possibilities.

-

-\Class{\Inum{I.}}{$p$,~$p$,~$q$ and $p$,~$q$,~$p$, the only possible arrangements.}

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$p^{2}q$|(}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}q$, where $p$~and $q$ are different primes|(}%

-

-There must be here a sub-group of order~$pq$, containing self-conjugate

-sub-groups of orders $p$ and $q$ and therefore Abelian. Let

-this be generated by operations $P$~and~$Q$, of orders $p$~and~$q$. Since the

-group has sub-groups of order~$p^{2}$, there must be operations of orders

-$p$ or~$p^{2}$, not contained in the sub-group of order~$pq$, and permutable

-with~$P$. Let $R$ be such an operation, so that $R^{p}$~belongs to the

-sub-group of order~$pq$. $R$~cannot be permutable with~$Q$, as the

-group would be then Abelian; hence

-\[

-R^{-1}QR = Q^{\alpha},

-\]

-so that

-\[

-R^{-p} QR^{p} = Q^{\alpha^{p}},

-\]

-and

-\[

-\alpha^{p} \equiv 1\ (\mod q).

-\]

-This case can therefore only occur if $p$~is a factor of~$q - 1$. There are

-two distinct types, according as $R$ is of order~$p$ or of order~$p^{2}$; \ie\

-according as the sub-groups of order~$p^{2}$ are non-cyclical or cyclical.

-If $\alpha$~and~$\beta$ are any two distinct primitive roots of the congruence

-\[

-\alpha^{p} \equiv 1\ (\mod q),

-\]

-the relations

-\[

-R^{-1}QR = Q^{\alpha},

-\]

-and

-\[

-R^{-1}QR = Q^{\beta},

-\]

-do not lead to distinct types, since the latter may be reduced to the

-form by replacing $R$ by~$R^{x}$, where

-\[

-\beta^{x} \equiv \alpha\ (\mod q).

-\]

-

-The two types are respectively defined by the relations

-\begin{gather*}

-Q^{q} = 1,\quad P^{p} = 1,\quad R^{p} = 1,\quad P^{-1}QP = Q, \\

-R^{-1}PR = P,\quad R^{-1}QR = Q^{\alpha};

-\end{gather*}

-and

-\[

-Q^{q} = 1,\quad R^{p^{2}} = 1,\quad R^{-1}QR = Q^{\alpha}.

-\]

-

-In each case, $\alpha$~is a primitive root of the congruence $\alpha^{p} \equiv 1\ (\mod q)$.

-

-\Class{\Inum{II.}}{$p$,~$q$,~$p$ and $q$,~$p$,~$p$, the only possible arrangements.}

-

-There must be a self-conjugate sub-group of order~$pq$, in which

-the sub-group of order~$q$ is not self-conjugate, and a self-conjugate

-sub-group of order~$p^{2}$. The sub-group of order~$pq$ must be given by

-\begin{gather*}

-P^{p} = 1,\quad Q^{q} = 1,\quad Q^{-1}PQ = P^{\alpha}, \\

-\alpha^{q} \equiv 1\ (\mod p);

-\end{gather*}

-%% -----File: 150.png---Folio 134-------

-so that in this case $q$~must be a factor of~$p - 1$. If the sub-group of

-order~$p^{2}$ is not cyclical, there must be an operation~$R$ of order~$p$, not

-contained in the sub-group of order~$pq$. Any such operation must

-be permutable with~$P$. Moreover since the sub-group of order~$pq$ is

-self-conjugate and contains only $p$~sub-groups of order~$q$, the

-sub-group~$\{Q\}$ must be permutable with some operation of order~$p$.

-Hence we may assume that $R$~is permutable with~$\{Q\}$, and, since

-$p > q$, with~$Q$.

-

-We thus obtain a single type defined by

-\begin{alignat*}{3}

-P^{p} &= 1,      &R^{p} &= 1,       &       PR &= RP, \\

-Q^{q} &= 1,\quad &   QR &= RQ,\quad & Q^{-1}PQ &= P^{\alpha}.

-\end{alignat*}

-

-If the sub-group of order~$p^{2}$ is cyclical, all the operations, which

-have powers of~$p$ for their orders and are not contained in the sub-group

-of order~$pq$, must be of order~$p^{2}$. There can therefore be no

-operation of order~$p$, which is permutable with~$\{Q\}$; therefore there

-is no corresponding type.

-

-\Class{\Inum{III.}}{$p$,~$p$,~$q$, the only possible arrangement.}

-

-There must be a self-conjugate sub-group of order~$pq$, which has

-no self-conjugate sub-group of order~$p$; it is therefore defined by

-\begin{gather*}

-P^{p} = 1,\quad Q^{q} = 1, \quad P^{-1}QP = Q^{\alpha}, \\

-\alpha^{p} \equiv 1\ (\mod q);

-\end{gather*}

-so that here $p$~must be a factor of~$q - 1$.

-

-If the sub-groups of order~$p^{2}$ are not cyclical, there must be an

-operation~$R'$ of order~$p$, not contained in this sub-group and permutable

-with~$P$. Hence

-\[

-R'^{-1}QR' = Q^{\beta};

-\]

-and if

-\[

-\beta \equiv \alpha^{x}\ (\mod q),

-\]

-then $R'P^{-x}$ is an operation of order~$p$, which is not contained in the

-sub-group of order~$pq$ and is permutable with~$Q$. It is therefore a

-self-conjugate operation of order~$p$. Hence $p$,~$q$,~$p$ is a possible

-arrangement of the composition-factors, and there is in this case

-no type.

-

-If the sub-groups of order~$p^{2}$ are cyclical, there must be an operation~$R$

-of order~$p^{2}$, such that

-\[

-R^{p} = P.

-\]

-Hence

-\[

-R^{-1}QR = Q^{\beta},

-\]

-where $\beta$ is a primitive root of the congruence

-\[

-\beta^{p^{2}} = 1\ (\mod q).

-\]

-This case then can only occur when $p^{2}$~is a factor of~$q - 1$; and we

-again have a single type defined by

-\[

-R^{p^{2}} = 1, \quad Q^{q} = 1,\quad R^{-1}QR = Q^{\beta}.

-\]

-%% -----File: 151.png---Folio 135-------

-

-\Class{\Inum{IV.}}{$p$,~$q$,~$p$, the only possible arrangement.}

-

-Here the self-conjugate sub-group of order~$pq$ must be given by

-\begin{gather*}

-P^{p} = 1,\quad Q^{q} = 1,\quad Q^{-1}PQ = P^{\alpha}, \\

-\alpha^{q} \equiv 1\ (\mod p),

-\end{gather*}

-and $q$ must be a factor of~$p - 1$. As in~\Inum{II}, there must be an

-operation~$R$ of order~$p$, permutable with~$\{Q\}$ and therefore with~$Q$;

-and since $R$~transforms $\{P, Q\}$ into itself, it must be permutable

-with~$P$. This however makes the sub-group $\{P, R\}$ self-conjugate,

-which requires $q$,~$p$,~$p$ to be a possible arrangement of the composition-factors.

-Hence there is no type corresponding to this case.

-

-\Class{\Inum{V.}}{$q$,~$p$,~$p$, the only possible arrangement.}

-

-If the sub-group of order~$p^{2}$ is cyclical, and is generated by~$P$,

-while $Q$~is an operation of order~$q$, we must have

-\[

-Q^{-1}PQ = P^{\alpha},

-\]

-where

-\[

-\alpha^{q} \equiv 1\ (\mod p^{2}).

-\]

-Here $q$ must be a factor of~$p - 1$; since the congruence has just

-$q - 1$ primitive roots, there is a single type of group.

-

-If the sub-group of order~$p^{2}$ is not cyclical, it can be generated

-by two permutable operations $P_{1}$ and $P_{2}$ of order~$p$. Since a

-sub-group of order~$q$ is not self-conjugate, either $p$ or $p^{2}$ must be

-congruent to unity $(\mod q)$; and therefore $q$~must be a factor of

-either $p - 1$ or~$p + 1$.

-

-Suppose first, that $q$ is a factor of~$p - 1$.

-

-There are $p + 1$ sub-groups of order~$p$. When these are

-transformed by any operation~$Q$ of order~$q$, those which are not

-permutable with~$Q$ must be interchanged in sets of~$q$. Hence at

-least two of these sub-groups must be permutable with~$Q$, and we

-may take the generating operations of two such sub-groups for $P_{1}$

-and~$P_{2}$. Therefore

-\[

-Q^{-1}P_{1}Q = P_{1}^{\alpha},\quad Q^{-1}P_{2}Q = P_{2}^{\beta}.

-\]

-

-Now if either $\alpha$ or~$\beta$, say~$\beta$, were unity, then $\{Q, P_{1}\}$ would be a

-self-conjugate sub-group and $p$,~$q$,~$p$ would be a possible arrangement

-of the composition-factors. Hence neither $\alpha$ nor~$\beta$ can be unity, and

-we may take

-\[

-Q^{-1}P_{1}Q = P_{1}^{\alpha},\quad Q^{-1}P_{2}Q = P_{2}^{\alpha^{x}},

-\]

-where $\alpha$ is a primitive root of

-\[

-\alpha^{q} \equiv 1\ (\mod p),

-\]

-and $x$ is not zero.

-

-It remains to determine how many distinct types these equations

-contain. When $q = 2$, the only possible value of~$x$ is unity; and there

-is a single type. When $q$~is an odd prime, and we take

-\[

-Q^{y} = Q',\quad P_{1} = P_{2}',\quad P_{2} = P_{1}',\quad xy \equiv 1\ (\mod q),

-\]

-%% -----File: 152.png---Folio 136-------

-the equations become

-\[

-Q'^{-1} P_{1}' Q' = P_{1}'^{\alpha},\quad Q'^{-1} P_{2}' Q' = P_{2}'^{\alpha^{y}},

-\]

-and therefore the values $x$~and~$y$ of the index of~$\alpha$, where

-\[

-xy \equiv 1\ (\mod q),

-\]

-give the same type. Now the only way, in which the two equations

-can be altered into two equations of the same form, is by replacing~$Q$

-by some other operation of the group whose order is~$q$ and by

-either interchanging $P_{1}$ and~$P_{2}$ or leaving each of them unchanged.

-Moreover the other operations of the group whose orders are~$q$ are

-those of the form $Q^{l} P_{1}^{m} P_{2}^{n}$, where $l$ is~not zero, and this operation

-transforms $P_{1}$ and~$P_{2}$ in the same way as~$Q^{l}$. Hence finally, the

-values $x$~and~$y$ of the index will only give groups of the same type

-when

-\[

-xy \equiv 1\ (\mod q).

-\]

-There are therefore $\frac{1}{2} (q + 1)$ distinct types, when $q$~is an odd prime;

-they are given by the above equations.

-

-Suppose next, that $q$~is a factor of~$p + 1$.

-

-Any operation~$Q$, of order~$q$, will transform the sub-groups of

-order~$p$, with which it is not permutable, so as to interchange them

-in sets of~$q$; and hence if it is permutable with any sub-group, it

-must be permutable with $q$ at least. This, by the last case, is

-clearly impossible, and hence $Q$~is permutable with no sub-group of

-order~$p$. We may therefore, by suitably choosing the generating

-operations of the group of order~$p^{2}$, assume that

-\[

-Q^{-1} P_{1} Q = P_{2},\quad Q^{-1} P_{2} Q = P_{1}^{\alpha} P_{2} ^{\beta}.

-\]

-

-If now

-\[

-Q^{-x - 1} P_{1} Q^{x + 1} = P_{1}^{\alpha_{x}} P_{2}^{\beta_{x}},

-\]

-then

-\[

-\alpha_{x + 1} \equiv \alpha \beta_{x},\quad

-\beta_{x + 1}  \equiv \alpha_{x} + \beta \beta_{x}\ (\mod p),

-\]

-and therefore

-\[

-\beta_{x + 1} - \beta \beta_{x} - \alpha \beta _{x - 1} \equiv 0\ (\mod p).

-\]

-Hence if $\iota_{1}$ and $\iota_{2}$ are the roots of the congruence

-\[

-\iota^{2} - \beta \iota - \alpha \equiv 0\ (\mod p),

-\]

-then

-\[

-\beta_{x} \equiv \frac{\iota_{2}^{x + 1} - \iota_{1}^{x + 1}}{\iota_{2} - \iota_{1}}.

-\]

-

-Now since $Q^{q}$ is the lowest power of~$Q$ that is permutable with~$P_{1}$,

-$\beta_{q - 1}$~must be the first term of the series $\beta_{1}$, $\beta_{2}$,~\dots\ which vanishes.

-Hence $q$~is the least value of~$z$ for which

-\[

-\iota_{2}^{z} \equiv \iota_{1}^{z},

-\]

-and therefore the congruence

-\[

-\iota^{2} - \beta \iota - \alpha \equiv 0

-\]

-is irreducible. Moreover $\alpha_{q - 1}$ must be congruent to unity, and

-therefore

-\[

-1 \equiv -\iota_{1}\iota_{2} \frac{\iota_{2}^{q - 1} - \iota_{1}^{q - 1}}{\iota_{2} - \iota_{1}} \equiv \iota_{1}^{q}.

-\]

-%% -----File: 153.png---Folio 137-------

-

-From the quadratic congruence satisfied by~$\iota$, it follows that

-\[

-\alpha \equiv -\iota^{p + 1} \equiv -1,\quad \beta \equiv \iota^{p} + \iota\ (\mod p);

-\]

-and thence

-\[

-\alpha_{x} \equiv -\frac{\iota^{px} - \iota^{x}}{\iota^{p} - \iota},\quad

-\beta_{x}  \equiv  \frac{\iota^{p(x + 1)} - \iota^{x + 1}}{\iota^{p} - \iota}.

-\]

-

-Finally, we may shew that, when $q$~is a factor of~$p + 1$, the

-equations

-\begin{gather*}

-P_{1}^{p} = 1,\quad P_{2}^{p} = 1,\quad Q^{q} = 1,\quad P_{1}P_{2} = P_{2}P_{1}, \\

-Q^{-1}P_{1}Q = P_{2},\quad  Q^{-1}P_{2}Q = P_{1}^{-1}P_{2}^{\iota^{p} + \iota},

-\end{gather*}

-where $\iota$ is a primitive root of the congruence

-\[

-\iota^{q} \equiv 1\ (\mod p),

-\]

-define a single type of group, whatever primitive root of the congruence

-is taken for~$\iota$.

-

-Thus from the given equations it follows that

-\begin{align*}

-Q^{-x}P_{1}Q^{x} &= P_{1}^{\alpha_{x - 1}}P_{2}^{\beta_{x - 1}} = P_{3},\text{ say,} \\

-\intertext{and}

-Q^{-x}P_{2}'Q^{x} &= (P_{1}^{\alpha_{x - 1}}P_{2}^{\beta_{x - 1}})^{\alpha_{x - 1}} (P_{1}^{\alpha_{x}}P_{2}^{\beta_{x}})^{\beta_{x - 1}}, \\

-&= P_{1}^{\alpha_{x}\beta_{x - 1} - \alpha_{x - 1}\beta_{x}} P_{3}^{\alpha_{x - 1} + \beta_{x}} \\

-&= P_{1}^{-1}P_{3}^{\iota^{px} + \iota^{x}}.

-\end{align*}

-If then we take $P_{1}$,~$P_{3}$ and~$Q^{x}$ as generating operations in the place

-of $P_{1}$,~$P_{2}$ and~$Q$, the defining relations are reproduced with $\iota^{x}$ in the

-place of~$\iota$. The relations therefore define a single type of group\footnotemark.

-\index{Cole and Glover@\Topic{Cole} and \Topic{Glover}, quoted}%

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{On groups whose order is of the form~$p^{2}q$ the reader may consult; Hölder,

-  ``Die Gruppen der Ordnungen $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.}\ \DPnote{** No "Vol."}\VolNo{XLIII} (1893), in

-  particular pp.~335--360; and Cole and Glover, ``On groups whose orders are

-  products of three prime factors,'' \textit{Amer.\ Journal}, \VolNo{XV} (1893), pp.~202--214.}%

-

-We have, for the sake of brevity, in each case omitted the

-verification that the defining relations actually give a group of

-order~$p^{2}q$. This presents no difficulty, even for the last type; for

-the previous types it is immediately obvious.

-\end{Remark}

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$p^{2}q$|)}%

-\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}q$, where $p$~and $q$ are different primes|)}%

-%% -----File: 154.png---Folio 138-------

-

-

-\Chapter{VIII.}{On Substitution-Groups: Transitive and

-Intransitive Groups.}

-

-\Par{102.} \First{It} has been proved, in the theorem in §~20, that every

-group is capable of being represented as a group of substitutions

-performed on a number of symbols equal to the order of the

-group. For applications to Algebra, and in particular to the

-Theory of Equations, the presentation of a group as a group of

-substitutions is of special importance; and we shall now proceed

-to consider the more important properties of this special mode

-of representing groups\footnotemark.

-\footnotetext{When it is necessary to call attention directly to the fact that the group

-  we are dealing with is supposed to be presented as a group of substitutions, the

-  group will be spoken of as a \emph{substitution-group}.}

-

-\begin{Definition}

-When a group is represented by means of

-\index{Degree@\Topic{Degree} of a substitution group!definition of}%

-\index{Substitution group@\Topic{Substitution group}!degree of, definition of}%

-substitutions performed on a finite set of $n$ distinct symbols,

-the integer~$n$ is called the \emph{degree} of the group.

-\end{Definition}

-

-It is obvious, by a consideration of simple cases, that a

-group can always be represented in different forms as a group

-of substitutions, the number of symbols which are permuted

-in two forms not being necessarily the same; examples have

-already been given in Chapter~II\@. The ``degree of a group''

-is therefore only an abbreviation of ``the degree of a special

-representation of the group as a substitution-group.''

-

-The $n!$ substitutions, including the identical substitution,

-that can be performed upon $n$~distinct symbols, clearly form a

-%% -----File: 155.png---Folio 139-------

-group; for they satisfy the conditions of the definition (§~12).

-Moreover they form the greatest group of substitutions that can

-be performed on the $n$~symbols, because every possible substitution

-occurs among them. When a group then is spoken of as of

-degree~$n$, it is implicitly being regarded as a sub-group of this

-most general group of order~$n!$ which can be represented by

-substitutions of the $n$~symbols; and therefore (Theorem~I,

-§~22) the order of a substitution-group of degree~$n$ must be

-a factor of~$n!$.

-

-\Par{103.} It has been seen in §~11 that any substitution performed

-on $n$~symbols can be represented in various ways as

-the product of transpositions; but that the number of transpositions

-entering in any such representation of the substitution

-is either always even or always odd. In particular, the identical

-substitution can only be represented by an even number of

-transpositions. Hence if $S$~and~$S'$ are any two even (§~11)

-substitutions of $n$~symbols, and $T$~any substitution at all of $n$~symbols,

-then $SS'$ and $T^{-1}ST$ are even substitutions. The

-even substitutions therefore form a self-conjugate sub-group~$H$

-of the group~$G$ of all substitutions.

-

-If now $T$ is any odd substitution, the set of substitutions~$TH$

-are all odd and all distinct. Moreover they give all the

-odd substitutions; for if $T'$~is any odd substitution distinct

-from~$T$, then $T^{-1}T'$~is an even substitution and must be

-contained in~$H$. Hence the number of even substitutions is

-equal to the number of odd substitutions: and the order of~$G$

-is twice that of~$H$.

-

-\begin{Definitions}

-The group of order~$n!$ which consists of all

-\index{Alternating group@\Topic{Alternating group}!definition of}%

-\index{Group@\Topic{Group}!symmetric}%

-\index{Group@\Topic{Group}!alternating}%

-\index{Symmetric group@\Topic{Symmetric group}!definition of}%

-the substitutions that can be performed on $n$ symbols is called

-the \emph{symmetric} group of degree~$n$.

-

-The group of order~$\frac{1}{2} n!$ which consists of all the even

-substitutions of $n$~symbols is called the \emph{alternating}\footnote

-  {The symmetric group has been so called because the only functions of the

-  $n$~symbols which are unaltered by all the substitutions of the group are the

-  symmetric functions.

-

-  All the substitutions of the alternating group leave the square root of the

-  discriminant unaltered (§~11).}

-group of

-degree~$n$.

-\end{Definitions}

-%% -----File: 156.png---Folio 140-------

-

-If the substitutions of a group of degree~$n$ are not all even,

-the preceding reasoning may be repeated to shew that its even

-substitutions form a self-conjugate sub-group whose order is

-half the order of the group; and this sub-group is a sub-group

-of the alternating group of the $n$~symbols.

-

-\Par{104.} \begin{Definition}

-A substitution-group is called \emph{transitive}

-\index{Degree@\Topic{Degree} of a substitution group!is a factor of the order, if the group is transitive}%

-\index{Intransitive groups@\Topic{Intransitive groups}!definition of}%

-\index{Substitution group@\Topic{Substitution group}!transitive and intransitive, definition of}%

-\index{Substitution group@\Topic{Substitution group}!degree of transitive, is a factor of its order}%

-\index{Transitive group@\Topic{Transitive group}!definition of}%

-when, by means of its substitutions, a given symbol~$a_{1}$ can be

-changed into every other symbol $a_{2}$, $a_{3}$,~\Dots, $a_{n}$ operated on

-by the group. When it has not this property, the group is

-called \emph{intransitive}.

-\end{Definition}

-

-A transitive group contains substitutions changing any one

-symbol into any other. For if $S$~and~$T$ respectively change~$a_{1}$

-into $a_{s}$~and~$a_{t}$, then $S^{-1}T$~changes $a_{s}$ into~$a_{t}$.

-

-\begin{Theorem}[I.]

-The substitutions of a transitive group~$G$, which

-leave a given symbol~$a_{1}$ unchanged, form a sub-group; and the

-number of substitutions, which change $a_{1}$ into any other symbol~$a_{r}$,

-is equal to the order of this sub-group.

-\end{Theorem}

-

-The substitutions which leave $a_{1}$ unchanged must form a

-sub-group~$H$ of~$G$; for if $S$~and~$S'$ both leave~$a_{1}$ unchanged, so

-also does~$SS'$.

-

-Let the operations of~$G$ be divided into the sets

-\[

-H,\ HS_{1},\ HS_{2},\ \Dots,\ HS_{m - 1}.

-\]

-

-No operation of the set~$HS_{p}$ leaves $a_{1}$ unchanged; and each

-operation of the set~$HS_{p}$ changes $a_{1}$ into~$a_{p}$, if $S_{p}$~does so. If

-the operations of any other set~$HS_{q}$ also changed $a_{1}$ into~$a_{p}$, then

-$S_{p}S_{q}^{-1}$ would leave $a_{1}$ unchanged and would belong to~$H$, which

-it does not. Hence each set changes~$a_{1}$ into a distinct symbol.

-The number of sets must therefore be equal to the number of

-symbols, while from their formation each set contains the same

-number of substitutions. If $N$~is the order and $n$ the degree

-of the transitive group~$G$, then $\dfrac{N}{n}$~is the order of the sub-group

-that leaves any symbol~$a_{1}$ unchanged; and there are $\dfrac{N}{n}$

-substitutions changing $a_{1}$ into any other given symbol~$a_{p}$.

-

-\begin{Corollary}

-The order of a transitive group must be

-divisible by its degree.

-\end{Corollary}

-%% -----File: 157.png---Folio 141-------

-

-Every group conjugate to~$H$ leaves one symbol unchanged.

-\index{Substitution group@\Topic{Substitution group}!transitive, whose substitutions, except identity, displace all or all but one of the symbols|(}%

-For if $S$ changes $a_{1}$ into~$a_{p}$, then $S^{-1}HS$~leaves $a_{p}$ unchanged.

-The sub-groups which leave the different symbols unchanged

-form therefore a conjugate set.

-

-{\Loosen A transitive group of degree~$n$ and order~$mn$ has, as we

-have just seen, $m - 1$ substitutions other than identity which

-leave a given symbol~$a_{1}$ unchanged. Hence there must be at

-least $mn - 1 - n(m - 1)$, \ie\ $n - 1$, substitutions in the group

-that displace every symbol. If the $m - 1$ substitutions, other

-than identity, that leave $a_{1}$ unchanged are all distinct from the

-$m - 1$ that leave $a_{p}$ unchanged, whatever other symbol~$a_{p}$ may

-be, $n- 1$~will be the actual number of substitutions that

-displace all the symbols; and no operation other than identity

-will displace less than $n - 1$ symbols. If however the sub-groups

-that leave $a_{1}$~and~$a_{p}$ unchanged have other substitutions

-besides identity in common, these substitutions must displace

-less than $n - 1$ symbols; and there will be more than $n - 1$

-substitutions which displace all the symbols.}

-

-\begin{Ex} If the substitutions of two transitive groups of degree~$n$

-which displace all the symbols are the same, the groups can only

-differ in the substitutions that keep just one symbol unchanged.

-\index{Netto@\Topic{Netto}, quoted}%

-\Attrib{Netto.}

-\end{Ex}

-

-\begin{Remark}

-\Par{105.} We have seen that every group can be represented as a

-substitution-group whose degree is equal to its order. A reference

-to the proof of this theorem (§~20) will shew that such a substitution-group

-is transitive, and that the identical substitution is the only

-one which leaves any symbol unchanged.

-

-We will now consider some of the properties of a transitive

-group of degree~$n$, whose operations, except identity, displace all or

-all but one of the $n$~symbols. It has just been seen that such a group

-has exactly $n - 1$ operations which displace all the $n$~symbols.

-If these $n - 1$ operations, with identity, form a sub-group, the sub-group

-must clearly be self-conjugate.

-

-Suppose now that $nm$ is the order of the group. Then the order of

-the sub-group, that leaves one symbol~$a_{1}$ unchanged, is~$m$; and if $a_{2}$~is

-any other of the $n$~symbols, no two operations of this sub-group

-can change $a_{2}$ into the same symbol. For if $S$,~$S'$ were two operations

-of the sub-group both changing $a_{2}$ into~$a_{r}$, then $SS'^{-1}$~would be

-an operation, distinct from identity, which would keep both $a_{1}$ and

-$a_{2}$ unchanged. Let now $P$ be any operation that displaces all the

-symbols. Then the set of $m$~operations $S^{-1}PS$, where for $S$ is put in

-%% -----File: 158.png---Folio 142-------

-turn each operation of the sub-group that keeps $a_{1}$ unchanged, are all

-distinct; for each of them changes $a_{1}$ into a different symbol. If this

-set does not exhaust the operation conjugate to~$P$, and if $P_{1}$~is another

-such operation, then the set of $m$~operations $S^{-1}P_{1}S$ are all distinct

-from each other and from those of the previous set. This process

-may be continued till the operations conjugate to~$P$ are exhausted.

-The number of operations conjugate to~$P$ is therefore a multiple

-of~$m$. Since $P$~itself does not belong to any one of the $n$~conjugate

-sub-groups that each keep one symbol fixed, no operation conjugate

-to~$P$ can belong to any of them. Hence each of the $km$~operations

-of the conjugate set to which $P$ belongs displaces all the symbols.

-The $n - 1$ operations that displace all the symbols can therefore be

-divided into sets, so that the number in each set is a multiple of~$m$;

-and hence $m$~must be a factor of~$n - 1$.

-

-Suppose now that $p$ is any prime factor of~$n$, and that $p^{\alpha}$~is the

-highest power of~$p$ which divides~$n$. If $P$~is an operation whose

-order is a power of~$p$, and if $p^{\beta}\mu$~is the order of the greatest sub-group~$h$

-that contains $P$ self-conjugately, then $P$~is one of $\dfrac{nm}{p^{\beta}\mu}$

-conjugate operations. Now (§~87) the sub-group~$h$ contains~$k\mu$ ($k \nless 1$)

-operations whose orders are relatively prime to~$p$; and therefore

-there are $k\mu$~operations of the form~$PQ$, where $Q$~is permutable with~$P$

-and the order of~$Q$ is relatively prime to the order of~$P$. If $P'$~is

-any operation conjugate to~$P$, there are similarly $k\mu$~operations of

-the form~$P'Q'$; and (§~16) no one of these operations can be

-identical with any one of those of the previous set. The group

-therefore will contain $\dfrac{nmk}{p^{\beta}}$ distinct operations, which are conjugate

-to the various operations of the set~$PQ$. Moreover since $P$~displaces

-all the symbols, each one of these operations must displace all the

-symbols.

-

-If then the group has $r$~distinct sets of conjugate operations

-whose orders are powers of~$p$, the number of operations, whose orders

-are divisible by~$p$, is equal to

-\[

-\sum_{s=1}^{s=r} \frac{nmk_{s}}{p^{\beta_{s}}}.

-\]

-

-Also the number of operations, which displace all the symbols

-and the orders of which are not divisible by~$p$, is of the form $\dfrac{nk_{0}}{p^{\alpha}} - 1$

-(§~87).

-

-Hence finally

-\begin{align*}

-m \sum_{s=1}^{s=r} \frac{k_{s}}{p^{\beta_{s}}}

-  &= \frac{1}{n} \left\{n - 1 - \left(\frac{nk_{0}}{p^{\alpha}} - 1\right)\right\} \\

-  &= \frac{p^{\alpha} - k_{0}}{p^{\alpha}}.

-\end{align*}

-%% -----File: 159.png---Folio 143-------

-The greatest possible value of~$m$ will correspond to the suppositions

-\[

-r = 1,\quad k_{1} = 1,\quad k_{0} = 1;

-\]

-and these give

-\[

-m = p^{\alpha} - 1.

-\]

-

-Hence if $n = p^{\alpha} q^{\beta} \dots r^{\gamma}$, where $p$, $q$,~\dots, $r$ are distinct primes, $m$~cannot

-be greater than the least of the numbers

-\[

-p^{\alpha} - 1,\quad q^{\beta} - 1,\quad \dots,\quad r^{\gamma} - 1.

-\]

-

-Certain particular cases may be specially noticed. First, a group

-of degree~$n$ and order~$n(n - 1)$, whose operations other than

-identity displace all or all but one of the symbols, can exist only

-when $n$~is the power of a prime\footnotemark.

-\index{Jordan@\Topic{Jordan}, quoted}%

-\footnotetext{Jordan, ``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér.\

-  Vol.~\VolNo{XVII} (1872), p.~355.}%

-Groups which satisfy these

-conditions will be discussed in §~112.

-

-Similarly, a group of degree~$n$ and order~$nm$, where $m$~is not less

-than~$\sqrt{n}$, whose operations other than identity displace all or all but

-one of the symbols, can only exist when $n$~is the power of a prime.

-

-If $n$ is equal to twice an odd number, a transitive group of

-degree~$n$, none of whose operations except identity leave two symbols

-unchanged, must be of order~$n$.

-

-Lastly we may shew that, if $m$~is even, a group of degree~$n$ and

-order~$nm$, none of whose operations except identity leave two symbols

-unchanged, must contain a self-conjugate Abelian sub-group of order

-and degree~$n$.

-

-A sub-group that keeps one symbol fixed must, if $m$~is even,

-contain an operation of order~$2$. If it contained $r$~such operations,

-the group would contain~$nr$; and each of these could be expressed

-as the product of $\frac{1}{2}(n - 1)$ independent transpositions. Now from $n$~symbols

-$\frac{1}{2}n(n - 1)$ transpositions can be formed. If then $r$ were

-greater than~$1$, among the operations of order~$2$ that keep one

-symbol fixed there would be pairs of operations with a common

-transposition; and the product of two such operations would be an

-operation, distinct from identity, which would keep two symbols at

-least fixed. This is impossible; therefore $r$~must be unity. Now let

-\[

-A_{1},\ A_{2},\ \Dots,\ A_{n}

-\]

-be the $n$~operations of order~$2$ belonging to the group. Since no

-two of these operations contain a common transposition,

-\[

-A_{1}A_{r},\ A_{2}A_{r},\ \Dots,\ A_{r - 1}A_{r},\ A_{r + 1}A_{r},\ \Dots, A_{n}A_{r}

-\]

-are the $n - 1$ operations which displace all the symbols. These

-operations may also be expressed in the form

-\[

-A_{r}A_{1},\ A_{r}A_{2},\ \Dots,\ A_{r}A_{r - 1},\ A_{r}A_{r + 1},\ \Dots, A_{r}A_{n};

-\]

-and since

-\[

-A_{p}A_{r} · A_{r}A_{q} = A_{p}A_{q},

-\]

-%% -----File: 160.png---Folio 144-------

-the product of any two of these operations is either identity or

-another operation which displaces all the symbols. Hence the $n - 1$

-operations which displace all the symbols, with identity, form a self-conjugate

-sub-group. Now

-\[

-A_{r} · A_{p}A_{r} · A_{r} = A_{r}A_{p},

-\]

-so that $A_{r}$~transforms every operation of this sub-group into its

-inverse. Hence

-\[

-A_{r}A_{p} · A_{q}A_{r} = A_{q}A_{p} = A_{q}A_{r} · A_{r}A_{q};

-\]

-\ie\ every two operations of this sub-group are permutable, and the

-sub-group is therefore Abelian.

-\end{Remark}

-

-\Par{106.} If

-\[

-S = (a_{1}a_{2} \Dots a_{i})(a_{i + 1}a_{i + 2} \Dots a_{j})\Dots,

-\]

-and

-\[

-T = \Sub{a_{1}, a_{2}, \Dots, a_{n}}{b_{1}, b_{2}, \Dots, b_{n}},

-\]

-are any two substitutions of a group, then (§~10)

-\[

-T^{-1}ST = (b_{1}b_{2} \Dots b_{i})(b_{i + 1}b_{i + 2} \Dots b_{j})\Dots.

-\]

-

-Hence every substitution of the group, which is conjugate

-\index{Self-conjugate operation@\Topic{Self-conjugate operation}!of a transitive substitution group must be a regular substitution}%

-\index{Substitution group@\Topic{Substitution group}!conjugate substitutions of, are similar}%

-to~$S$, is also similar to~$S$. It does not necessarily or generally

-follow that two similar substitutions of a group are conjugate.

-That this is true however of the symmetric group is obvious,

-for then the substitution~$T$ may be chosen so as to replace the

-$n$~symbols by any permutation of them whatever.

-

-A self-conjugate substitution of a transitive group of degree~$n$

-must be a regular substitution (§~9) changing all the $n$~symbols.

-For if it did not change all the $n$~symbols, it would belong to

-one of the sub-groups that keep a symbol unchanged. Hence,

-since it is a self-conjugate substitution, it would belong to each

-sub-group that keeps a symbol unchanged, which is impossible

-unless it is the identical substitution. Again if it were not

-regular, one of its powers would keep two or more symbols

-unchanged, and this cannot be the case since every power of a

-self-conjugate substitution must be self-conjugate. On the

-other hand, a self-conjugate sub-group of a transitive group

-need not contain any substitution which displaces all the

-symbols. Thus if

-\begin{align*}

-S &= (12)(34), \\

-T &= (135)(246),

-\end{align*}

-then $\{S, T\}$ is a transitive group of degree~$6$. The only

-substitutions conjugate to~$S$ are

-\[

-T^{-1}ST = (34)(56) \quad \text{and} \quad TST^{-1} = (12)(56);

-\]

-\index{Substitution group@\Topic{Substitution group}!transitive, whose substitutions, except identity, displace all or all but one of the symbols|)}%

-%% -----File: 161.png---Folio 145-------

-and these, with $S$ and identity, form a self-conjugate sub-group

-of order~$4$, none of whose substitutions displace more than $4$

-symbols. The form of a self-conjugate sub-group of a transitive

-group will be considered in greater detail in the next Chapter.

-

-\Par{107.} Since a self-conjugate substitution of a transitive

-\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%

-substitution-group~$G$ of degree~$n$ must be a regular substitution

-which displaces all the symbols, the self-conjugate sub-group~$H$

-of~$G$ which consists of all its self-conjugate operations must

-have $n$ or some submultiple of~$n$ for its order. For if $S$ and $S'$

-are two self-conjugate substitutions of~$G$, so also is~$S^{-1}S'$; and

-therefore $S$ and $S'$ cannot both change $a$ into~$b$. The order of~$H$

-therefore cannot exceed~$n$; and if the order is~$n'$, the

-substitutions of~$H$ must interchange the $n$~symbols in sets of~$n'$,

-so that $n'$~is a factor of~$n$. Let now $S$, some substitution

-performed on the $n$~symbols of the transitive substitution-group~$G$

-of degree~$n$, be permutable with every substitution of~$G$.

-Then $S$ is a self-conjugate operation of the transitive

-substitution-group $\{S, G\}$ of degree~$n$, and it is therefore a

-regular substitution in all the $n$~symbols. The totality of the

-substitutions~$S$, which are permutable with every substitution

-of~$G$, form a group (not necessarily Abelian); and the order of

-this group is~$n$ or a factor of~$n$.

-

-The special case, in which $G$ is a group whose order~$n$ is

-equal to its degree, is of sufficient importance to merit particular

-attention. If

-\[

-S_{1} (= 1),\ S_{2},\ \Dots,\ S_{n}

-\]

-are the operations of~$G$, it has been seen (§~20) that the

-substitution-group can be expressed as consisting of the $n$~substitutions

-\[

-\Sub{\PadTo{S_{1} S_{x}}{S_{1}}, \PadTo{S_{2} S_{x}}{S_{2}}, \Dots, \PadTo{S_{n} S_{x}}{S_{n}}}

-    {S_{1} S_{x}, S_{2} S_{x}, \Dots, S_{n}S_{x}},\quad (x = 1, 2, \Dots, n),

-\]

-performed upon the symbols of the operations of the group.

-

-Now if pre-multiplication be used in the place of post-multiplication,

-it may be verified exactly as in §~20 that the $n$~substitutions

-\[

-\Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \PadTo{S_{x} S_{2}}{S_{2}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}

-    {S_{x} S_{1}, S_{x} S_{2}, \Dots, S_{x} S_{n}},\quad (x = 1, 2, \Dots, n)

-\]

-%% -----File: 162.png---Folio 146-------

-form a group~$G'$ simply isomorphic with~$G$; and the substitution

-of~$G'$ which is given corresponds to the operation~$S_{x}^{-1}$ of~$G$.

-Moreover

-\begin{multline*}

-  \Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}

-      {S_{x} S_{1}, \Dots, S_{x} S_{n}}^{-1}

-  \Sub{\PadTo{S_{1} S_{y}}{S_{1}}, \Dots, \PadTo{S_{n} S_{y}}{S_{n}}}

-      {S_{1} S_{y}, \Dots, S_{n} S_{y}}

-  \Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}

-      {S_{x} S_{1}, \Dots, S_{x} S_{n}} \\

-  \begin{aligned}

-    &= \Sub{S_{x} S_{1}, \Dots, S_{x} S_{n}}

-           {S_{1} S_{y}, \Dots, S_{1} S_{n}}

-       \Sub{\PadTo{S_{x} S_{1} S_{y}}{S_{1} S_{y}}, \Dots,

-            \PadTo{S_{x} S_{n} S_{y}}{S_{n} S_{y}}}

-           {S_{x} S_{1} S_{y}, \Dots, S_{x} S_{n} S_{y}} \\

-    &= \Sub{\PadTo{S_{x} S_{1} S_{y}}{S_{x} S_{1}}, \Dots,

-            \PadTo{S_{x} S_{n} S_{y}}{S_{x} S_{n}}}

-           {S_{x} S_{1} S_{y}, \Dots, S_{x} S_{n} S_{y}}

-      = \Sub{\PadTo{S_{1} S_{y}}{S_{1}}, \Dots, \PadTo{S_{n} S_{y}}{S_{n}}}

-            {S_{1} S_{y}, \Dots, S_{n} S_{y}},

-  \end{aligned}

-\end{multline*}

-so that every operation of~$G'$ is permutable with every operation

-of~$G$, while $G$~and $G'$ can be expressed as transitive

-substitution-groups in the same $n$~symbols. Hence:---

-

-\begin{Theorem}[II.]

-Those substitutions of $n$~symbols which are

-\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%

-\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a group, whose degree is equal to its order, form a simply isomorphic group}%

-permutable with every substitution of a substitution-group~$G$ of

-order~$n$, transitive in the $n$~symbols, form a group~$G'$ of order

-and degree~$n$, simply isomorphic with~$G$\footnotemark.

-\end{Theorem}

-\footnotetext{Jordan, \textit{Traité des Substitutions} (1870), p.~60.}%

-\index{Jordan@\Topic{Jordan}, quoted}%

-

-It has, in fact, been seen that there is a substitution-group~$G'$

-of order and degree~$n$, every one of whose substitutions is

-permutable with every substitution of~$G$; and also that the

-order of any such group cannot exceed~$n$.

-

-If $S_{x}$ is a self-conjugate substitution of~$G$, the substitutions

-\[

-\Sub{\PadTo{S_{1} S_{x}}{S_{1}}, \Dots, \PadTo{S_{n} S_{x}}{S_{n}}}

-      {S_{1} S_{x}, \Dots, S_{n} S_{x}}

-\quad \text{and} \quad

-\Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}

-      {S_{x} S_{1}, \Dots, S_{x} S_{n}}

-\]

-are the same. Hence $G$~and $G'$ have for their greatest common

-sub-group, that which is constituted by the self-conjugate

-substitutions of either; and if $n'$ is the order of this sub-group,

-the order of $\{G, G'\}$ is~$\dfrac{n^{2}}{n'}$. In particular, if $G$ is Abelian, $G$ and

-$G'$ coincide; and if $G$ has no self-conjugate operation except

-identity, $\{G, G'\}$~is the direct product of $G$ and~$G'$.

-\index{Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$}%

-

-The sub-group of $\{G, G'\}$ which leaves one symbol, say~$S_{1}$,

-unchanged, is formed of the distinct substitutions of the set

-\[

-\Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},

-     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}

-    {S_{x}^{-1} S_{1} S_{x},

-     S_{x}^{-1} S_{2} S_{x}, \Dots,

-     S_{x}^{-1} S_{n} S_{x}},\quad (x = 1, 2, \Dots, n).

-\]

-%% -----File: 163.png---Folio 147-------

-

-When $G$ has no self-conjugate operation except identity,

-\index{Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$}%

-the order of this sub-group is~$n$, and it is simply isomorphic

-with~$G$. In fact, in this case the order of $\{G, G'\}$, a transitive

-group of degree~$n$, is~$n^{2}$, and therefore the order of a sub-group

-that keeps one symbol unchanged is~$n$. Again

-\begin{multline*}

-\Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},

-     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}

-    {S_{x}^{-1} S_{1} S_{x},

-     S_{x}^{-1} S_{2} S_{x}, \Dots,

-     S_{x}^{-1} S_{n} S_{x}}

-\Sub{\PadTo{S_{y}^{-1} S_{1} S_{y}}{S_{1}},

-     \PadTo{S_{y}^{-1} S_{2} S_{y}}{S_{2}}, \Dots,

-     \PadTo{S_{y}^{-1} S_{n} S_{y}}{S_{n}}}

-    {S_{y}^{-1} S_{1} S_{y},

-     S_{y}^{-1} S_{2} S_{y}, \Dots,

-     S_{y}^{-1} S_{n} S_{y}} \\

-%

-\begin{aligned}

-&= \Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},

-     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}

-    {S_{x}^{-1} S_{1} S_{x},

-     S_{x}^{-1} S_{2} S_{x}, \Dots,

-     S_{x}^{-1} S_{n} S_{x}} \\

-&\qquad\qquad

-  \Sub{\PadTo{S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y}}{S_{x}^{-1} S_{1} S_{x}},

-       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}}{S_{x}^{-1} S_{2} S_{x}}, \Dots,

-       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}}{S_{x}^{-1} S_{n} S_{x}}}

-      {S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y},

-       S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}, \Dots,

-       S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}} \\

-%

-&= \Sub{\PadTo{S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y}}{S_{1}},

-       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}}{S_{2}}, \Dots,

-       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}}{S_{n}}}

-      {S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y},

-       S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}, \Dots,

-       S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}};

-\end{aligned}

-\end{multline*}

-thus giving a direct verification that the sub-group is isomorphic

-with the group whose operations are

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}.

-\]

-

-When $G$ contains self-conjugate operations, it will be

-multiply isomorphic with the sub-group~$K_{1}$ of~$\{G, G'\}$ which

-keeps the symbol $S_{1}$ fixed; and if $g$ is the group constituted by

-the self-conjugate operations of~$G$ (or~$G'$), then $K_{1}$~is simply

-isomorphic with~$\dfrac{G}{g}$.

-

-If $K_{1}$ is not a maximum sub-group of~$\{G, G'\}$, let $I$ be a

-greater sub-group containing~$K_{1}$. Then $I$ and $G'$ (or~$G$) must

-contain common substitutions. For every substitution of

-$\{G, G'\}$ is of the form

-\[

-\Sub{\PadTo{S_{y} S_{1} S_{x}}{S_{1}},

-     \PadTo{S_{y} S_{2} S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{y} S_{n} S_{x}}{S_{n}}}

-    {S_{y} S_{1} S_{x}, S_{y} S_{2} S_{x}, \Dots, S_{y} S_{n} S_{x}};

-\]

-and if this substitution belongs to~$I$, then

-\[

-%[** TN: Set on two lines in the original]

-\Sub{\PadTo{S_{y} S_{1} S_{x}}{S_{1}},

-     \PadTo{S_{y} S_{2} S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{y} S_{n} S_{x}}{S_{n}}}

-    {S_{y} S_{1} S_{x}, S_{y} S_{2} S_{x}, \Dots, S_{y} S_{n} S_{x}}

-\Sub{\PadTo{S_{x} S_{1} S_{x}^{-1}}{S_{1}},

-     \PadTo{S_{x} S_{2} S_{x}^{-1}}{S_{2}}, \Dots,

-     \PadTo{S_{x} S_{n} S_{x}^{-1}}{S_{n}}}

-    {S_{x} S_{1} S_{x}^{-1}, S_{x} S_{2} S_{x}^{-1}, \Dots, S_{x} S_{n} S_{x}^{-1}},

-\]

-or

-\[

-\Sub{\PadTo{S_{x} S_{y} S_{1}}{S_{1}},

-     \PadTo{S_{x} S_{y} S_{2}}{S_{2}}, \Dots,

-     \PadTo{S_{x} S_{y} S_{n}}{S_{n}}}

-    {S_{x} S_{y} S_{1}, S_{x} S_{y} S_{2}, \Dots, S_{x} S_{y} S_{n}},

-\]

-%% -----File: 164.png---Folio 148-------

-is a substitution of~$G'$ which belongs to~$I$. Moreover, since $G'$

-is a self-conjugate sub-group of~$\{G, G'\}$, the substitutions of~$G'$

-which belong to~$I$ form a self-conjugate sub-group of~$I$: this

-sub-group we will call~$H'$.

-

-Now every substitution of the group can be represented as

-the product of a substitution of~$K_{1}$ by a substitution of~$G$: and

-therefore all the sub-groups conjugate to~$I$ will be obtained on

-transforming $I$ by the operations of~$G$. Hence, because every

-substitution of~$G$ transforms $H'$ into itself, $H'$~is common to

-the complete set of conjugate sub-groups to which $I$ belongs;

-and $H'$~is therefore a self-conjugate sub-group of~$\{G, G'\}$.

-Finally then, $K_{1}$~is a maximum sub-group of~$\{G, G'\}$, if and only

-if $G$~is a simple group.

-

-\Par{108.} \begin{Definition}

-A substitution-group, that contains one

-\index{Substitution group@\Topic{Substitution group}!multiply transitive, definition of}%

-or more substitutions changing $k$ given symbols $a_{1}$, $a_{2}$,~\dots, $a_{k}$

-into any other $k$~symbols, is called \emph{$k$-ply transitive}.

-\end{Definition}

-

-Such a group clearly contains substitutions changing any

-set of $k$~symbols into any other set of~$k$; and the order of the

-sub-group keeping any~$j$ ($\ngtr k$) symbols unchanged is independent

-of the particular $j$~symbols chosen.

-

-\begin{Theorem}[III.]

-The order of a $k$-ply transitive group of

-\index{Substitution group@\Topic{Substitution group}!order of a $k$-ply transitive, whose degree is~$n$, is a multiple of $n(n - 1)\Dots (n - k + 1)$}%

-degree~$n$ is $n(n - 1) \Dots (n - k + 1)m$, where $m$~is the order of

-the sub-group that leaves any $k$~symbols unchanged. This

-sub-group is contained self-conjugately in a sub-group of order~$k!m$.

-\end{Theorem}

-

-If $N$ is the order of the group, the order of the sub-group

-which keeps one symbol fixed is~$\dfrac{N}{n}$, by Theorem~I (§~104).

-Now this sub-group is a transitive group of degree~$n - 1$; and

-therefore the order of the sub-group that keeps two symbols

-unchanged is~$\dfrac{N}{n(n - 1)}$. If $k > 2$, this sub-group again is a

-transitive group of degree~$n - 2$; and so on. Proceeding thus,

-the order of the sub-group which keeps $k$~symbols unchanged

-is seen to be

-\[

-\frac{N}{n(n - 1) \Dots (n - k + 1)},

-\]

-which proves the first part of the theorem.

-%% -----File: 165.png---Folio 149-------

-

-Let $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ be the $k$~symbols which are left unchanged

-by a sub-group~$H$ of order~$m$. Since the group is

-$k$-ply transitive, it must contain substitutions of the form

-\[

-\Sub{a_{1}, a_{2}, \Dots, a_{k}, \PadTo[l]{b'}{b}, \PadTo[l]{c'}{c}, \Dots}

-    {a_{1}', a_{2}', \Dots, a_{k}', b', c', \Dots},

-\]

-where $a_{1}'$, $a_{2}'$,~\Dots, $a_{k}'$ are the same $k$~symbols as $a_{1}$, $a_{2}$,~\Dots, $a_{k}$

-arranged in any other sequence. Also every substitution of this

-form is permutable with~$H$, since it interchanges among themselves

-the symbols left unchanged by~$H$. Further, if $S_{1}$ and

-$S_{2}$ are any two substitutions of this form, $S_{1}^{-1} S_{2}$~will belong to~$H$

-if, and only if, $S_{1}$~and $S_{2}$ give the same permutation of the

-symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$. Hence finally, since $k!$~distinct

-substitutions can be performed on the $k$~symbols, the order of

-the sub-group that contains $H$ self-conjugately is~$k!m$.

-

-If $m$ is unity, the identical substitution is the only one that

-keeps any $k$~symbols fixed, and there is just one substitution

-that changes $k$~symbols into any other~$k$. In the same case, the

-group contains substitutions which displace $n - k + 1$ symbols

-only, and there are none, except the identical substitution,

-which displace less.

-

-If $m > 1$, the group will contain $m - 1$ substitutions besides

-identity, which leave unchanged any $k$ given symbols, and

-therefore displace $n - k$ symbols at most.

-

-It follows from §~105 that a $k$-ply transitive group of degree~$n$

-and order $n(n - 1) \Dots (n - k + 1)$ can exist only if $n - k + 2$

-is the power of a prime. For such a group must contain

-sub-groups of order $(n - k + 2)(n - k + 1)$, which keep $k - 2$

-symbols unchanged and are doubly transitive in the remaining

-$n - k + 2$. When $k$~is~$n$, the group is the symmetric group; and

-when $k$~is~$n - 2$, we shall see (in §~110) that the group is the

-alternating group. If $k$ is less than~$n - 2$, M.~Jordan\footnote

-  {``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér., Vol.~\VolNo{XVII}

-  (1872), pp.~357--363.}

-\index{Jordan@\Topic{Jordan}, quoted}%

-\PageLabel{149}%

-has

-shewn that, with two exceptions for $n = 11$ and $n = 12$, the

-value of~$k$ cannot exceed~$3$. The actual existence of triply

-transitive groups of order $(p^{n} + 1) p^{n} (p^{n} - 1)$, for all prime

-values of~$p$, will be established in §~113.

-%% -----File: 166.png---Folio 150-------

-

-\Par{109.} A $k$-ply transitive group, of degree~$n$ and order~$N$,

-\index{Substitution group@\Topic{Substitution group}!construction of multiply transitive}%

-is not generally contained in some $(k + 1)$-ply transitive

-group of degree~$n + 1$ and order~$N(n + 1)$. To determine

-whether this is the case for any given group, M.~Jordan\footnote

-  {\textit{Traité des Substitutions} (1870), pp.~31,~32.}

-\index{Jordan@\Topic{Jordan}, quoted}%

-has

-suggested the following tentative process, which for moderate

-values of~$n$ is always practicable.

-

-Let $G$ be a transitive group of order~$N$ in the $n$~symbols

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{n};

-\]

-and suppose that $G$~is that sub-group of the transitive group~$\Gamma$

-in the $n + 1$ symbols

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{n},\ a_{n + 1},

-\]

-which leaves the symbol $a_{n + 1}$ unchanged. Then $\Gamma$ must be at

-least doubly transitive, and it therefore contains a substitution

-of order~$2$ which interchanges the two symbols $a_{1}$~and~$a_{n + 1}$.

-Let $A$ be such a substitution. Then

-\[

-\Gamma = \{G, A\};

-\]

-for $\{G, A\}$ is contained in~$\Gamma$, and its order cannot be less than

-the order of~$\Gamma$. Also if $S$ is any substitution of~$G$\Erratum{}{ which displaces~$a_{1}$}, two other

-substitutions $S'$ and $S''$ of~$G$ can always be found such that

-\[

-ASA = S'AS''.

-\]

-

-In fact, if $ASA$ changes $a_{r}$ into~$a_{n + 1}$, and if $S'$~changes $a_{r}$

-into~$a_{1}$, then $AS'^{-1} ASA$ leaves $a_{n + 1}$ unchanged, and it therefore

-belongs to~$G$.

-

-Conversely, if $A$~is any operation of order~$2$ which changes

-$a_{1}$ into $a_{n + 1}$ and permutes the remaining~$a$'s among themselves;

-and if, whatever substitution of~$G$ is represented by~$S$,\Erratum{ either $ASA$~belongs to~$G$ or} two other

-substitutions of~$G$ can be found such that

-\[

-ASA = S'AS'';

-\]

-then $\{G, A\}$ is a group with the required properties. In fact,

-when these conditions are satisfied, every substitution of the

-group $\{G, A\}$ can be expressed in one of the two forms

-\[

-S_{1},\ S_{2}AS_{3},

-\]

-%% -----File: 167.png---Folio 151-------

-where $S_{1}$,~$S_{2}$,~$S_{3}$ are substitutions of~$G$. For instance, the

-substitution

-\begin{align*}

-AS_{p}AS_{q}AS_{r}

-  &= S_{p}'AS_{p}'' · S_{q}AS_{r}, \\

-  &= S_{p}'AS_{t}AS_{r}, \quad\text{if}\quad S_{p}''S_{q} = S_{t}, \\

-  &= S_{p}'S_{t}'AS_{t}''S_{r},

-\end{align*}

-which is of the second form.

-\Erratum{}{The reduction is here carried out on supposition that $S_{p}$~and~$S_{t}$ displace~$a_{1}$.

-The modification, when this is not the case, is obvious.}

-

-Moreover every operation of the form $S_{2}AS_{3}$ displaces~$a_{n + 1}$,

-and therefore the sub-group of~$\{G, A\}$ which leaves $a_{n + 1}$ unchanged

-is~$G$.

-

-It is clearly sufficient that the conditions

-\[

-ASA = S'AS''

-\]

-should be satisfied, when each of a set of independent generating

-operations of~$G$ is taken for~$S$.

-

-\begin{Ex} Construct a doubly transitive group of degree~$12$ of which

-the sub-group that keeps one symbol unchanged is

-\[

-\{(123456789ab),\ (256a4)(39b87)\}.

-\]

-\end{Ex}

-

-\Par{110.} Let

-\[

-S = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{j - 1}a_{j})

-    (a_{j + 1} \Dots a_{k - 1}a_{k} \Dots) \Dots

-\]

-be a substitution of a $k$-ply transitive group displacing $s$ ($> k$)

-symbols. If $j < k - 1$, take

-\[

-T = \Sub{a_{1}, a_{2}, \Dots, a_{k - 1}, a_{k}, \Dots}

-        {a_{1}, a_{2}, \Dots, a_{k - 1}, b_{k}, \Dots},

-\]

-where $b_{k}$~is some other symbol occurring in~$S$. Since the

-group is $k$-ply transitive, it must contain a substitution such as~$T$.

-Now

-\[

-T^{-1}ST = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{j - 1}a_{j})

-           (a_{j + 1} \Dots a_{k - 1}b_{k} \Dots) \Dots;

-\]

-and this is certainly not identical with~$S$, so that $T^{-1}STS^{-1}$

-cannot be the identical substitution. Moreover $a_{1}$, $a_{2}$,~\Dots, $a_{k - 2}$

-are not affected by $T^{-1}STS^{-1}$; and therefore this substitution

-will displace at most $2s - 2k + 2$ symbols.

-

-If $j = k - 1$, take

-\[

-T = \Sub{a_{1}, a_{2}, \Dots, a_{k - 1}, a_{k}, \Dots}

-        {a_{1}, a_{2}, \Dots, a_{k - 1}, c_{k}, \Dots},

-\]

-where $c_{k}$ is a symbol that does not occur in~$S$. Then

-\[

-T^{-1}ST = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{k - 2}a_{k - 1})(c_{k} \Dots),

-\]

-%% -----File: 168.png---Folio 152-------

-and this cannot coincide with~$S$. Now in this case, $a_{1}$, $a_{2}$,~\Dots, $a_{k-1}$

-are not affected by $T^{-1}STS^{-1}$; and therefore this operation

-will again displace at most $2s - 2k + 2$ symbols.

-

-If then

-\[

-2s - 2k + 2 < s,

-\]

-or

-\[

-s < 2k - 2,

-\]

-the group must contain a substitution affecting fewer symbols

-than~$S$. This process may be repeated till we arrive at a

-substitution

-\[

-\Sigma = (\alpha_{1}\alpha_{2} \Dots \alpha_{i})

-         (\alpha_{i+1} \Dots \alpha_{j}) \Dots

-         (\alpha_{j+1} \Dots \alpha_{k}),

-\]

-which affects exactly $k$~symbols; and if this substitution be

-transformed by

-\[

-T = \Sub{\alpha_{1}, \alpha_{2}, \Dots, \alpha_{k-1}, \alpha_{k}, \Dots}

-        {\alpha_{1}, \alpha_{2}, \Dots, \alpha_{k-1},  \beta_{k}, \Dots},

-\]

-then

-\[

-T^{-1} \Sigma T = (\alpha_{1}\alpha_{2} \dots \alpha_{i})

-                  (\alpha_{i+1} \dots \alpha_{j}) \dots

-                  (\alpha_{j+1} \dots a_{k-1} \beta_{k}),

-\]

-and

-\[

-\Sigma^{-1} T^{-1} \Sigma T = (\alpha_{k} \beta_{k} \alpha_{j+1}).

-\]

-

-Thus in the case under consideration the group contains

-one, and therefore every, circular substitution of three symbols;

-and hence (§~11) it must contain every even substitution. It

-is therefore either the alternating or the symmetric group. If

-then a $k$-ply transitive group of degree~$n$ does not contain the

-alternating group of $n$~symbols, no one of its substitutions,

-except identity, must displace fewer than $2k - 2$ symbols.

-It has been shewn that such a group contains substitutions

-displacing not more than $n - k + 1$ symbols; and therefore, for

-a $k$-ply transitive group of degree~$n$, other than the alternating

-or the symmetric group, the inequality

-\[

-n - k + 1  \nless 2k- 2,

-\]

-or

-\PageLabel{152}%

-\[

-k \ngtr \tfrac{1}{3} n + 1,

-\]

-must hold. Hence:---

-

-\begin{Theorem}[IV.]

-A group of degree~$n$, which does not contain

-\index{Substitution group@\Topic{Substitution group}!limit to the degree of transitivity of}%

-the alternating group of $n$~symbols, cannot be more than

-$(\frac{1}{3}n + 1)$-ply transitive.

-\end{Theorem}

-

-The symmetric group is $n$-ply transitive; and, since of the

-two substitutions

-\[

-\Sub{a_{1}, a_{2}, \Dots, a_{n-2}, a_{n-1}, a_{n}}

-    {b_{1}, b_{2}, \Dots, b_{n-2}, b_{n-1}, b_{n}}

-\quad\text{and}\quad

-\Sub{a_{1}, a_{2}, \Dots, a_{n-2}, a_{n-1}, a_{n}}

-    {b_{1}, b_{2}, \Dots, b_{n-2}, b_{n}, b_{n-1}},

-\]

-%% -----File: 169.png---Folio 153-------

-one is evidently even and the other odd, the alternating group

-is $(n - 2)$-ply transitive. The discussion just given shews that

-no other group of degree~$n$ can be more than $(\frac{1}{3}n + 1)$-ply

-transitive\footnotemark.

-\index{Bochert@\Topic{Bochert}, quoted}%

-\index{Jordan@\Topic{Jordan}, quoted}%

-\footnotetext{For a further discussion of the limits of transitivity of a substitution-group,

-  compare Jordan, \textit{Traité des Substitutions}, pp.~76--87; and Bochert,

-  \textit{Math.\ Ann.},\DPnote{** No "Vol."} \VolNo{xxix}, (1886) pp.~27--49; \VolNo{XXXIII}, (1888) pp.~573--583.}%

-

-\Par{111.} The process used in the preceding paragraph may be

-applied to shew that, unless $n = 4$, the alternating group of $n$~symbols

-is simple. It has just been shewn that the alternating

-group is $(n - 2)$-ply transitive. Therefore, if $S$ is a substitution

-of the alternating group displacing fewer than $n - 1$~symbols, a

-substitution $T^{-1}ST$ can certainly be found such that $S^{-1}T^{-1}ST$

-is a circular substitution of three symbols. In this case, the

-self-conjugate group generated by~$S$ and its conjugate substitutions

-contains all the circular substitutions of three symbols,

-and therefore it coincides with the alternating group itself. If

-$S$~displaces $n - 1$ symbols, then $T^{-1}ST$ can be taken so that

-$S^{-1}T^{-1}ST$ displaces not more than $2(n - 1) - 2(n - 2) + 2$, or

-$4$~symbols; and if $S$~displaces $n$~symbols, $S^{-1}T^{-1}ST$~can be found

-to displace not more than $2n - 2(n - 2) + 2$, or $6$~symbols.

-

-It is therefore only necessary to consider the case $n = 5$, when

-$S$~displaces $n - 1$ symbols; and the cases $n = 4$, $5$,~$6$, when $S$~displaces

-$n$~symbols; in all other cases, the group generated by~$S$

-and its conjugate substitutions must contain circular substitutions

-of $3$~symbols.

-

-When $n = 5$, and $S$ is an even substitution displacing $4$~symbols,

-we may take

-\[

-S = (12) (34).

-\]

-If

-\[

-T = (12) (35),

-\]

-then

-\[

-T^{-1}ST = (12) (45),

-\]

-and

-\[

-S^{-1}T^{-1}ST = (345).

-\]

-Hence, in this case again, we are led to the alternating group

-itself.

-%% -----File: 170.png---Folio 154-------

-

-When $n = 6$, and $S$~is an even substitution displacing all

-the symbols, we may take

-\[

-S = (12) (3456),

-\]

-or

-\[

-S' = (123) (456).

-\]

-If now

-\[

-T = (12) (3645),

-\]

-then

-\[

-S^{-1}T^{-1}ST = (356),

-\]

-and

-\[

-S'^{-1}T^{-1}S'T = (14263);

-\]

-and, in either case, we are led to the alternating group.

-

-When $n = 5$, and $S$~is an even substitution displacing all

-the symbols, we may put

-\[

-S = (12345).

-\]

-If

-\[

-T = (345),

-\]

-then

-\[

-S^{-1}T^{-1}ST = (134);

-\]

-and again the alternating group is generated.

-

-When $n = 4$, and $S$~is an even substitution displacing all the

-symbols, we may take

-\[

-S = (12) (34).

-\]

-Here the only two substitutions conjugate to~$S$ are clearly

-$(13) (24)$ and $(14) (23)$, which are permutable with each other

-and with~$S$. Hence the alternating group of $4$~symbols, which

-is of order~$12$, has a self-conjugate sub-group of order~$4$.

-

-Finally when $n = 3$, the alternating group, being the group

-$\{(123)\}$, is a simple cyclical group of order~$3$. Hence:---

-

-\begin{Theorem}[V.]

-The alternating group of $n$~symbols is a simple

-\index{Alternating group@\Topic{Alternating group}!is simple, except for degree~$4$}%

-group except when $n = 4$.

-\end{Theorem}

-

-\begin{Remark}

-\Par{112.} It has been seen in §~108 that the order of a doubly transitive

-group of degree~$n$ is equal to or is a multiple of~$n(n - 1)$. If it

-is equal to this number, every substitution of the group, except

-identity, must displace either all or all but one of the symbols; for

-a sub-group of order~$n - 1$ which keeps one symbol fixed is transitive

-in the remaining $n - 1$ symbols, and therefore all its substitutions,

-except identity, displace all the $n - 1$ symbols.

-

-Now it has been shewn in §~105 that a transitive group of degree~$n$

-and order~$n(n - 1)$, whose operations displace all or all but one of

-the symbols can exist only if $n$~is the power of a prime~$p$. The

-$n - 1$ operations displacing all the symbols are the only operations

-of the group whose orders are powers of~$p$; and therefore with

-%% -----File: 171.png---Folio 155-------

-\index{Doubly transitive groups@\Topic{Doubly transitive groups}!of degree~$n$ and order~$n(n - 1)$|(}%

-\index{Substitution group@\Topic{Substitution group}!doubly transitive, of degree~$n$ and order~$n(n - 1)$|(}%

-identity they form a self-conjugate sub-group of order~$n$. Moreover

-it also follows from §~105 that the $n - 1$ operations of this sub-group

-other than identity form a single conjugate set. Hence this sub-group

-must be Abelian, and all its operations are of order~$p$.

-

-Suppose first that $n$~is a prime~$p$, and that $P$~is any operation of

-the group of order~$p$. If $\alpha$~is a primitive root of~$p$, there must be

-an operation~$S$ in the group such that

-\[

-S^{-1}PS = P^{\alpha};

-\]

-then $S^{p-1}$ is the lowest power of~$S$ which is permutable with~$P$.

-Now $S$~must belong to a sub-group of order~$p - 1$ that keeps one

-symbol fixed, and we have just seen that the order of~$\{S\}$ is not less

-than~$p - 1$. The sub-group of order~$p - 1$ is therefore cyclical, and

-\[

-S^{p-1} = 1.

-\]

-Hence the group, if it exists, must be defined by

-\[

-P^{p} = 1,\quad S^{p-1} = 1,\quad S^{-1}PS = P^{\alpha}.

-\]

-

-It is an immediate result of a theorem, which will be proved in the

-next chapter~(§~123), that this group can be actually represented as

-a transitive substitution-group of degree~$p$; this may be also verified

-directly as follows.

-

-Let

-\[

-P = (a_{1}a_{2} \dots a_{p}),

-\]

-so that

-\[

-P^{\alpha} = (a_{1}a_{\alpha + 1}a_{2\alpha + 1} \dots a_{(p - 1)\alpha + 1}),

-\]

-where the suffixes are to be reduced $(\mod p)$; and suppose that $S$~is

-a substitution that keeps unchanged. Then since

-\[

-S^{-1}PS = P^{\alpha},

-\]

-$S$ must change $a_{2}$ into~$a_{\alpha + 1}$, $a_{3}$ into~$a_{2 \alpha + 1}$, and generally, $a_{r}$ into

-$a_{(r - 1)\alpha + 1}$. Hence

-\[

-S = (a_{2} a_{\alpha + 1} a_{\alpha^{2} + 1} \dots) \dots;

-\]

-and since $\alpha$~is a primitive root of~$p$, there is only a single cycle; so

-that

-\[

-S = (a_{2} a_{\alpha + 1} a_{\alpha^{2} + 1} \dots a_{\alpha^{p - 2} + 1}).

-\]

-

-The substitutions $P$ and~$S$ thus constructed actually generate a

-doubly transitive substitution-group of degree~$p$ and order~$p(p - 1)$.

-

-Without making a complete investigation of the case in which $n$~is

-the power of a prime, we go on to shew that, $p$~being any prime,

-there is always a doubly transitive group of degree~$p^{m}$ and order

-$p^{m}(p^{m} - 1)$, in which a sub-group of order~$p^{m} - 1$ is cyclical\footnotemark.

-\index{Mathieu@\Topic{Mathieu}, quoted}%

-\footnotetext{On the subject of this and the following paragraph, the reader  should

-\PageLabel{155}%

-  consult the memoirs by Mathieu, \textit{Liouville's Journal}, \Ord{2}{me} Sér., t.~\VolNo{V} (1860),

-  pp.~9--42; \textit{ib.}\ t.~\VolNo{VI} (1861), pp.~241--323; where the groups here considered were

-  first shewn to exist.}

-%% -----File: 172.png---Folio 156-------

-

-Let $i$ be a primitive root of the congruence

-\[

-i^{p^{m} - 1} \equiv 1\ (\mod p),

-\]

-so that the distinct roots of the congruence are

-\[

-i,\ i^{2},\ i^{3},\ \Dots,\ i^{p^{m} - 1}.

-\]

-Every rational function of~$i$ with real integral coefficients satisfies

-the same congruence; and therefore every such function is congruent

-$(\mod p)$ to some power of~$i$ not exceeding the $(p^{m} - 1)$th.

-

-Consider now a set of transformations of the form

-\[

-x' \equiv \alpha x + \beta\ (\mod p),

-\]

-where $\alpha$ is a power of~$i$, and $\beta$~is either a power of~$i$ or zero. Two

-such transformations, performed successively, give another transformation

-of the same form; and since $\alpha$~cannot be zero, the inverse

-of each transformation is another definite transformation; so that

-the totality of transformations of this form constitute a group.

-Moreover

-\[

-\begin{aligned}

-x' &\equiv \alpha x + \beta, \\

-\lintertext{and}

-x' &\equiv \alpha' x + \beta'

-\end{aligned}

-\ (\mod p),

-\]

-are not the same transformation unless

-\[

-\alpha \equiv \alpha' \quad\text{and}\quad \beta \equiv \beta'\ (\mod p).

-\]

-{\Loosen Hence, since $\alpha$ can take $p^{m} - 1$ distinct values and $\beta$ can take $p^{m}$

-distinct values, the order of the group, formed of the totality of

-these transformations, is $p^{m}(p^{m} - 1)$.}

-

-The transformations for which $\alpha$ is unity clearly form a sub-group.

-If $S$~and~$T$ represent

-\[

-x' \equiv \alpha x + \beta \quad\text{and}\quad x' \equiv x + \gamma

-\]

-respectively, $S^{-1}TS$ represents

-\[

-x' \equiv x + \alpha \gamma.

-\]

-Hence the transformations for which $\alpha$ is unity form a self-conjugate

-sub-group whose order is~$p^{m}$. Every two transformations of this

-sub-group are clearly permutable; and the order of each of them

-except identity is~$p$.

-

-Again, the transformations for which $\beta$~is zero form a sub-group.

-Since every one of them is a power of the transformation

-\[

-x' \equiv ix,

-\]

-this sub-group is a cyclical sub-group of order~$p^{m} - 1$. If the transformation

-just written be denoted by~$I$, then $S^{-1}IS$ is

-\[

-x' \equiv ix + \beta (1 - i).

-\]

-Hence the only operations permutable with~$\{I\}$ are its own operations,

-and therefore $\{I\}$~is one of $p^{m}$ conjugate sub-groups.

-%% -----File: 173.png---Folio 157-------

-

-The set of transformations

-\[

-x' \equiv \alpha x + \beta

-\]

-therefore forms a group of order~$p^{m}(p^{m} - 1)$. This group contains a

-self-conjugate Abelian sub-group of order~$p^{m}$ and type $(1, 1, \dots, 1)$,

-and $p^{m}$~conjugate cyclical sub-groups of order~$p^{m} - 1$, none of whose

-operations are permutable with any of the operations of the self-conjugate

-sub-group.

-

-Now if the operation

-\[

-x' \equiv \alpha x + \beta

-\]

-be performed on each term of the series

-\[

-0,\ i,\ i^{2},\ \Dots,\ i^{p^{m}-1},

-\]

-it will, since every rational integral function of~$i$ with real integral

-coefficients is congruent $(\mod p)$ to some power of~$i$, change the

-term into another of the same series; and since the congruence

-\[

-\alpha i^{x} + \beta \equiv \alpha i^{y} + \beta

-\]

-gives

-\[

-x \equiv y\ (\mod p^{m-1}),

-\]

-no two terms of the series can thus be transformed into the same

-term. Moreover the only operation that leaves every term of the

-series unchanged is clearly the identical operation.

-

-To each operation of the form

-\[

-x' \equiv \alpha x + \beta

-\]

-therefore will correspond a single substitution performed on the $p^{m}$

-symbols just written, so that to the product of two operations will

-correspond the product of the two homologous substitutions. The

-group is therefore simply isomorphic with a substitution group of

-degree~$p^{m}$. Moreover since the linear congruence

-\[

-x \equiv \alpha x + \beta\ (\mod p)

-\]

-has only a single solution when $\alpha$ is different from unity, and none

-when $\alpha$~is unity, every substitution except identity must displace all

-or all but one of the symbols. The substitution group is therefore

-doubly transitive\footnotemark.

-\index{Burnside@\Topic{Burnside}, quoted}%

-\footnotetext{The author has shewn (\textit{Messenger of Mathematics}, Vol.~\VolNo{XXV} (1896), pp.~147--153)

-  that the type of group considered in the text is the only type of doubly

-  transitive group of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$ when $m = 3$; and that, when

-  $m = 2$ and $p > 3$, the same is true. When $m = 2$ and $p = 3$, there is one other type.}

-\end{Remark}

-

-\begin{Ex}[1.]

-Apply the method just explained to the actual construction

-of a doubly transitive group of degree~$8$ and order~$56$.

-\end{Ex}

-

-\begin{Ex}[2.] Shew that the equations

-\[

-A^{2} = 1,\quad S^{2^{m}-1} = 1,\quad AS^{-1}AS = S^{-n}AS^{n},

-\]

-where $n$~is such that a primitive root of the congruence,

-\[

-i^{2^{m}-1} - 1 \equiv 0\ (\mod 2),

-\]

-\index{Doubly transitive groups@\Topic{Doubly transitive groups}!of degree~$n$ and order~$n(n - 1)$|)}%

-\index{Substitution group@\Topic{Substitution group}!doubly transitive, of degree~$n$ and order~$n(n - 1)$|)}%

-%% -----File: 174.png---Folio 158-------

-satisfies the congruence

-\[

-i^{n} + i + 1 \equiv 0\ (\mod 2),

-\]

-suffice to define a group which can be expressed as a doubly transitive

-group of degree~$2^{m}$ and order~$2^{m}(2^{m} - 1)$.

-\Attrib{\textit{Messenger of Mathematics}, Vol.~\VolNo{XXV}, p.~189.}

-\end{Ex}

-

-\begin{Remark}

-\Par{113.} A slight extension of the method of the preceding paragraph

-\index{Substitution group@\Topic{Substitution group}!triply transitive, of degree~$n$ and order~$n(n - 1)(n - 2)$}%

-will enable us to shew that, for every prime~$p$, it is possible to

-construct a triply transitive group of degree~$p^{m} + 1$ and order

-\index{Triply transitive@\Topic{Triply transitive} groups of degree~$n$ and order $n(n - 1)(n - 2)$}%

-$(p^{m} + 1)p^{m}(p^{m} - 1)$. The analysis of this group will form the

-subject of investigation in Chapter~XIV; here we shall only demonstrate

-its existence.

-

-In the place of the operations of the last paragraph, we now

-consider those of the form

-\[

-x' \equiv \frac{\alpha x + \beta}{\gamma x + \delta}\ (\mod p),

-\]

-where again $\alpha$, $\beta$, $\gamma$, $\delta$ are powers of~$i$, limited now by the condition

-that $\alpha\delta - \beta\gamma$ is not congruent to zero $(\mod p)$. When this relation

-is satisfied, the set of operations again clearly form a group. Moreover

-if we represent $\dfrac{i^{x}}{0}$ by~$\infty$ for all values of~$x$, any operation of this

-group, when carried out on the set of quantities

-\[

-\infty,\ 0,\ i,\ i^{2},\ \Dots,\ i^{p^{m} - 1},

-\]

-will change each of them into another of the set; while no operation

-except identity will leave each symbol of the set unchanged. Hence

-the group can be represented as a substitution group of degree

-$p^{m} + 1$.

-

-Now

-\[

-\frac{x' - i^{a'}}{x'- i^{b'}}\, \frac{i^{c'} - i^{b'}}{i^{c'} - i^{a'}} \equiv

-\frac{x - i^{a}}{x - i^{b}}\, \frac{i^{c} - i^{b}}{i^{c} - i^{a}}

-\]

-is an operation of the above form, which changes the three symbols

-$i^{a}$,~$i^{b}$,~$i^{c}$ into $i^{a'}$,~$i^{b'}$,~$i^{c'}$ respectively; and it is easy to modify this form

-so that it holds when $0$~or~$\infty$ occurs in the place of~$i^{a}$, etc. Hence

-the substitution group is triply transitive, since it contains an

-operation transforming any three of the $p^{m} + 1$ symbols into any

-other three.

-

-On the other hand, if the typical operation keeps the symbol~$x$

-unchanged, then

-\[

-\gamma x^{2} + (\delta - \alpha) x - \beta \equiv 0\ (\mod p),

-\]

-and this congruence cannot have more than two roots among the

-set of $p^{m} + 1$ symbols. Hence no substitution of the group, except

-identity, keeps more than two symbols fixed.

-

-Finally then, since the group is triply transitive and since it

-contains no operation, except identity, that keeps more than two

-symbols fixed, its order must (§~108) be $(p^{m} + 1) p^{m} (p^{m} - 1)$.

-\end{Remark}

-%% -----File: 175.png---Folio 159-------

-

-\Par{114.} An intransitive substitution group, as defined in §~104,

-is one which does not contain substitutions changing $a_{1}$ into

-each of the other symbols $a_{2}$, $a_{3}$,~\Dots, $a_{n}$ operated on by the

-group. Let us suppose that the substitutions of such a group

-change $a_{1}$ into $a_{2}$, $a_{3}$,~\Dots, $a_{k}$ only. Then all the substitutions

-of the group must interchange these $k$~symbols among themselves;

-for if the group contained a substitution changing $a_{2}$

-into~$a_{k + 1}$, then the product of any substitution changing $a_{1}$ into~$a_{2}$

-by this latter substitution would change $a_{1}$ into~$a_{k + 1}$. Hence

-the $n$~symbols operated on by the group can be divided into a

-number of sets, such that the substitutions of the group change

-the symbols of any one set transitively among themselves, but

-do not interchange the symbols of two distinct sets. It follows

-immediately that the order of the group must be a common

-multiple of the numbers of symbols in the different sets.

-

-Suppose now that $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ is a set of symbols which

-are interchanged transitively by all the substitutions of a group~$G$

-of degree~$n$. If for a time we neglect the effect of the

-substitutions of~$G$ on the remaining $n - k$ symbols, the group~$G$

-will reduce to a transitive group~$H$ of degree~$k$. The group~$G$

-is isomorphic with the group~$H$; for if we take as the substitutions

-of~$G$, that correspond to a given substitution of~$H$, those

-which produce the same permutation of the symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$

-as that produced by the substitution of~$H$, then to the product

-of any two substitutions of~$G$ will correspond the product of

-the corresponding two substitutions of~$H$. The isomorphism

-thus shewn to exist may be simple or multiple. In the former

-case, the order of~$H$ is the same as that of~$G$; in the

-latter case, the substitutions of~$G$ which correspond to the

-identical substitution of~$H$, \ie\ those substitutions of~$G$ which

-change none of the symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ form a self-conjugate

-sub-group.

-

-We will consider in particular an intransitive group~$G$

-\index{Intransitive groups@\Topic{Intransitive groups}!properties of|(}%

-\index{Intransitive groups@\Topic{Intransitive groups}!transitive sets of symbols in}%

-which interchanges the symbols in \emph{two} transitive sets; these

-we will refer to as the $\alpha$'s and the~$\beta$'s. Let $G_{\alpha}$~and $G_{\beta}$ be the

-two groups transitive in the $\alpha$'s and $\beta$'s respectively, to which

-$G$~reduces when we alternately leave out of account the effect

-of the substitutions on the $\beta$'s and the~$\alpha$'s. Also let $g_{\alpha}$ and~$g_{\beta}$

-%% -----File: 176.png---Folio 160-------

-be the self-conjugate sub-groups of~$G$, which keep respectively

-all the $\beta$'s and all the $\alpha$'s unchanged; and denote the group

-$\{g_{\alpha}, g_{\beta}\}$ by~$g$. This last group~$g$, which is the direct product

-of $g_{\alpha}$ and~$g_{\beta}$, is self-conjugate in~$G$, since it is generated by the

-two self-conjugate groups $g_{\alpha}$ and~$g_{\beta}$. Now $g_{\alpha}$~is self-conjugate

-not only in~$G$ but also in~$G_{\alpha}$; for $G_{\alpha}$~permutes the~$\alpha$'s in the

-same way that $G$ does, while any substitution of~$g_{\alpha}$, not

-affecting the~$\beta$'s, is necessarily permutable with every substitution

-performed on the~$\beta$'s. The group~$G_{\alpha}$ is simply

-isomorphic with the group~$\dfrac{G}{g_{\beta}}$, and $G_{\beta}$ with~$\dfrac{G}{g_{\alpha}}$; hence, using $n_{H}$

-to denote the order of a group~$H$,

-\[

-n_{G} = n_{G_{\alpha}} n_{g_{\beta}} = n_{G_{\beta}} n_{g_{\alpha}}.

-\]

-Let the substitutions of~$G$ be now divided into sets in respect

-of the self-conjugate sub-group~$g$, so that

-\[

-G = g,\ Sg,\ Tg,\ \Dots.

-\]

-

-The group $\dfrac{G}{g}$ is defined by the laws according to which

-these sets combine among themselves, the sets being such that,

-if any substitution of one set be multiplied by any substitution

-of a second set, the resulting substitution will belong to a

-definite third set.

-

-If we now neglect the effect of the substitutions on the

-symbols~$\beta$, the group~$G$ reduces to~$G_{\alpha}$ and $g$~reduces to~$g_{\alpha}$, and

-hence

-\[

-G_{\alpha} = g_{\alpha},\ S_{\alpha} g_{\alpha},\ T_{\alpha} g_{\alpha},\ \Dots,

-\]

-where $S_{\alpha}$, $T_{\alpha}$,~\Dots\ represent the substitutions $S$, $T$,~\Dots, so far

-as they affect the~$\alpha$'s. Moreover the substitutions in the different

-sets into which $G_{\alpha}$ is thus divided must be all distinct since,

-by the preceding relations between the orders of the groups,

-their number is just equal to the order of~$G_{\alpha}$. Hence $\dfrac{G_{\alpha}}{g_{\alpha}}$ is

-defined by the laws according to which these sets of substitutions

-combine. But if

-\[

-Sg · Tg = Ug,

-\]

-then necessarily

-\[

-S_{\alpha} g_{\alpha} · T_{\alpha} g_{\alpha} = U_{\alpha} g_{\alpha},

-\]

-%% -----File: 177.png---Folio 161-------

-and therefore, finally, the three groups $\dfrac{G}{g}$,~$\dfrac{G_{\alpha}}{g_{\alpha}}$, and~$\dfrac{G_{\beta}}{g_{\beta}}$ are simply

-isomorphic.

-

-\Par{115.} The relation of simple isomorphism between $\dfrac{G_{\alpha}}{g_{\alpha}}$ and $\dfrac{G_{\beta}}{g_{\beta}}$

-thus arrived at establishes between the groups $G_{\alpha}$ and $G_{\beta}$ an

-isomorphism of the most general kind (§~32).

-

-To every operation of~$G_{\alpha}$ correspond $n_{g_{\beta}}$~operations of~$G_{\beta}$,

-and to every operation of~$G_{\beta}$ correspond $n_{g_{\alpha}}$~operations

-of~$G_{\alpha}$; so that to the product of any two operations of~$G_{\alpha}$ (or~$G_{\beta}$)

-there corresponds a definite set of $n_{g_{\beta}}$~operations of~$G_{\beta}$ (or~$n_{g_{\alpha}}$

-operations of~$G_{\alpha}$).

-

-Returning now to the intransitive group~$G$, its genesis from

-the two transitive groups $G_{\alpha}$ and~$G_{\beta}$, with which it is isomorphic,

-may be represented as follows. The $n_{g_{\alpha}}$ to $n_{g_{\beta}}$ correspondence,

-such as has just been described, having been established

-between the groups $G_{\alpha}$ and~$G_{\beta}$, each substitution of~$G_{\alpha}$ is

-multiplied by the $n_{g_{\beta}}$~substitutions that correspond to it in~$G_{\beta}$.

-The set of $n_{G_{\alpha}}n_{g_{\beta}}$ substitutions so obtained form a group, for

-\[

-S_{\alpha} S_{\beta} · S_{\alpha}' S_{\beta}'

-  = S_{\alpha} S_{\alpha}' · S_{\beta} S_{\beta}'

-  = S_{\alpha}'' S_{\beta}'',

-\]

-where, if $S_{\beta}$,~$S_{\beta}'$ are substitutions corresponding to $S_{\alpha}$,~$S_{\alpha}'$, then

-$S_{\beta}''$~is a substitution corresponding to~$S_{\alpha}''$. Moreover, this

-group may be equally well generated by multiplying every one

-of the substitutions of~$G_{\beta}$ by the $n_{g_{\alpha}}$~corresponding substitutions

-of~$G_{\alpha}$; and by a reference to the representations of $G$,~$G_{\alpha}$ and~$G_{\beta}$,

-as divided into sets of substitutions given above, it is immediately

-obvious that all these substitutions occur in~$G$. Hence,

-as their number is equal to the order of~$G$, the group thus

-formed coincides with~$G$.

-

-\Par{116.} The general result for any intransitive group, the

-simplest case of which has been considered in the two last

-paragraphs, may be stated in the following form:---

-

-\begin{Theorem}[VI.]

-If $G$ is an intransitive group of degree~$n$

-which permutes the $n$~symbols in $s$~transitive sets, and if \Inum{(i)}~$G_{r}$~is

-what $G$~becomes when the substitutions of~$G$ are performed on

-%% -----File: 178.png---Folio 162-------

-the $r$th~set of symbols only, \Inum{(ii)}~$\Gamma_{r}$~is what $G$ becomes when the

-substitutions of~$G$ are performed on all the sets except the~$r$th,

-\Inum{(iii)}~$g_{r}$~is that sub-group of~$G$ which changes the symbols of the

-$r$th~set only, \Inum{(iv)}~$\gamma_{r}$~is that sub-group of~$G$ which keeps all the

-symbols of the $r$th~set unchanged: then the groups $\dfrac{G_{r}}{g_{r}}$ and $\dfrac{\Gamma_{r}}{\gamma_{r}}$

-are simply isomorphic, and $n_{r}$,~$\nu_{r}$ being the orders of $g_{r}$,~$\gamma_{r}$, an

-$n_{r}$~to~$\nu_{r}$ correspondence is thus established between the substitutions

-of the groups $G_{r}$ and~$\Gamma_{r}$. Moreover, the substitutions of~$G$

-are given, each once and once only, by multiplying each

-substitution of~$G_{r}$ by the $\nu_{r}$~substitutions of~$\Gamma_{r}$ that correspond to

-it\footnotemark.

-\end{Theorem}

-\index{Bolza@\Topic{Bolza}, quoted}%

-\index{Klein and Fricke@\Topic{Klein} and \Topic{Fricke}, quoted}%

-\footnotetext{On intransitive groups, reference may be made to Bolza, ``On the

-  construction of intransitive groups,'' \textit{Amer.\ Journal}, Vol.~\VolNo{XI}, (1889), pp.~195--214.

-  The general isomorphism which underlies the construction of these

-  groups is considered by Klein and Fricke, \textit{Vorlesungen über die Theorie der

-  elliptischen Modulfunctionen}, Vol.~\VolNo{I}, (1890), pp.~402--406.}

-

-It is not necessary to give an independent proof of this

-theorem, since if, in the discussion of the two preceding paragraphs,

-$G_{\alpha}$,~$G_{\beta}$, $g_{a}$,~$g_{b}$,~$g$ be replaced by $G_{r}$,~$\Gamma_{r}$, $g_{r}$,~$\gamma_{r}$, $\{g_{r}, \gamma_{r}\}$, it

-will be found each step of the process there carried out may

-be repeated without alteration.

-

-If we regard $G_{\alpha}$ and $G_{\beta}$ as two given transitive groups

-in distinct sets of symbols, the determination of all the

-intransitive groups in the combined symbols, which reduce

-to $G_{\alpha}$ or $G_{\beta}$ when the symbols of the second or first set

-are neglected, involves a knowledge of the composition of

-the two groups. To each distinct $m$~to~$n$ isomorphism, that

-can be established between the two groups, there will correspond

-a distinct intransitive group. If $G_{\alpha}$ is a simple group,

-containing therefore only itself and identity self-conjugately,

-then to each substitution of~$G_{\beta}$ there must correspond either

-one or all of the substitutions of~$G_{\alpha}$; and the former can be

-the case only when $G_{\beta}$~contains a self-conjugate sub-group~$H$,

-such that $\dfrac{G_{\beta}}{H}$~is simply isomorphic with~$G_{\alpha}$. Hence, if the

-order of~$G_{\beta}$ is less than twice the order of~$G_{\alpha}$, the only possible

-intransitive group is the direct product of $G_{\alpha}$ and~$G_{\beta}$, unless $G_{\beta}$~is

-simply isomorphic with~$G_{\alpha}$.

-\index{Intransitive groups@\Topic{Intransitive groups}!properties of|)}%

-%% -----File: 179.png---Folio 163-------

-

-\begin{Remark}

-\Par{117.} In illustration of the preceding paragraphs, we will

-\index{Intransitive groups@\Topic{Intransitive groups}!of degree~$7$ with transitive sets of $3$ and~$4$}%

-determine the number of distinct intransitive groups of degree~$7$,

-when the symbols are interchanged in transitive sets of $4$ and $3$

-respectively. The four symbols will be referred to as the~$\alpha$'s, and

-the three symbols as the~$\beta$'s.

-

-If a transitive group of degree~$4$ contains operations of order~$3$,

-it must be either the symmetric or the alternating group. If it

-contains no operation of order~$3$, its order must be either $8$ or~$4$.

-By Sylow's theorem, the symmetric group of degree~$4$, being of

-order~$24$, contains a single set of conjugate sub-groups of order~$8$, so

-that there is only one type of transitive group of degree~$4$ and order~$8$.

-This is given by

-\[

-S = (1234),\quad T=(13),

-\]

-so that

-\[

-S^{4} = 1,\quad T^{2} = 1,\quad TST = S^{-1}.

-\]

-

-This group contains three self-conjugate sub-groups of order~$4$,

-namely,

-\[

-\begin{array}{l*{4}{c}}

-\Inum{(i)}   & 1, & (1234),   & (13)(24), & (1432), \\

-\Inum{(ii)}  & 1, & (12)(34), & (13)(24), & (14)(23), \\

-\Inum{(iii)} & 1, & (13),     & (13)(24), & (24).

-\end{array}

-\]

-

-The two latter are simply isomorphic; but as substitution groups,

-they are of distinct form. Hence for $G_{\alpha}$ in the construction of the

-intransitive group, we may take either $G_{24}$, the symmetric group of

-the~$\alpha$'s, or $G_{12}$, the alternating group of the~$\alpha$'s, or $G_{8}$, the above

-group of order~$8$, or finally $G_{4}$,~$G_{4}'$,~$G_{4}''$, the above three groups of

-order~$4$. The only transitive groups of $3$~symbols are $G_{6}$, the symmetric

-group, and $G_{3}$, the alternating group: one of these must be

-taken for~$G_{\beta}$.

-

-\Class{\Inum{I.}}{$G_{\alpha} = G_{24}$.}

-

-The only self-conjugate sub-groups of~$G_{24}$ are $G_{12}$ and~$G_{4}'$.

-Also $\dfrac{G_{24}}{G_{12}}$ is a group of order~$2$; and it may easily be verified

-that $\dfrac{G_{24}}{G_{4}'}$ is simply isomorphic with the symmetric group of degree~$3$.

-

-Hence \Inum{(i)}~if $G_{\beta}$ is~$G_{6}$, we may take

-\begin{alignat*}{4}

-g_{\alpha} &= G_{24},\quad & g_{\beta} &= G_{6}, &&\text{ so that } & n_{G} &= 144, \\

-g_{\alpha} &= G_{12},      & g_{\beta} &= G_{3}, &&\Ditto{ so that } & n_{G} &= 72, \\

-g_{\alpha} &= G_{4}',      & g_{\beta} &= 1,     &&\Ditto{ so that } & n_{G} &= 24; \\

-\intertext{and \Inum{(ii)}~if $G_{\beta}$ is~$G_{3}$, we must take}

-g_{\alpha} &= G_{24},      & g_{\beta} &= G_{3}, &&\text{ giving } & n_{G} &= 72.

-\end{alignat*}

-%% -----File: 180.png---Folio 164-------

-\index{Intransitive groups@\Topic{Intransitive groups}!of degree~$7$ with transitive sets of $3$ and~$4$}%

-

-\Class{\Inum{II.}}{$G_{\alpha} = G_{12}$.}

-

-The only self-conjugate sub-group of~$G_{12}$ is~$G_{4}'$; and $\dfrac{G_{12}}{G_{4}'}$~is a

-cyclical group of order~$3$.

-

-Hence \Inum{(i)}~if $G_{\beta}$ is~$G_{6}$, we must take

-\begin{alignat*}{4}

-g_{\alpha} &= G_{12},\quad & g_{\beta} &= G_{6}, &&\text{ giving } & n_{G} &= 72; \\

-\intertext{and \Inum{(ii)}~if $G_{\beta}$ is~$G_{3}$, we may take}

-g_{\alpha} &= G_{12},      & g_{\beta} &= G_{3}, &&\text{ giving } & n_{G} &= 36, \\

-\lintertext{or}

-g_{\alpha} &= G_{4}',      & g_{\beta} &= 1,  &&\Ditto{ giving }  & n_{G} &= 12.

-\end{alignat*}

-

-\Class{\Inum{III.}}{$G_{\alpha} = G_{8}$.}

-

-The self-conjugate sub-groups of~$G_{8}$, of order~$4$, are determined

-above.

-

-If \Inum{(i)}~$G_{\beta}$ is~$G_{6}$, we may take

-\begin{alignat*}{3}

-g_{\alpha} &= G_{8},   & g_{\beta} &= G_{6}, & \text{ giving }  & n_{G} = 48, \\

-g_{\alpha} &= G_{4},   & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24, \\

-g_{\alpha} &= G_{4}',  & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24, \\

-g_{\alpha} &= G_{4}'', \quad & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24.

-\end{alignat*}

-

-The two latter groups are simply isomorphic; but regarded as

-substitution groups, they are of distinct forms.

-

-If \Inum{(ii)}~$G_{\beta}$ is~$G_{3}$, we must take

-\[

-g_{\alpha} = G_{8},\quad g_{\beta} = G_{3}, \text{ giving } n_{G} = 24.

-\]

-

-\Class{\Inum{IV.}}{$G_{\alpha} = G_{4}$.}

-

-If \Inum{(i)}~$G_{\beta}$ is~$G_{6}$, we may take

-\[

-g_{\alpha} = G_{4},\quad g_{\beta} = G_{6}, \text{ giving } n_{G} = 24,

-\]

-or

-\[

-g_{\alpha} = G_{2},\quad g_{\beta} = G_{3}, \Ditto{ giving } n_{G} = 12;

-\]

-$G_{2}$ representing the group $\{1, (13)(24)\}$.

-

-If \Inum{(ii)}~$G_{\beta}$ is~$G_{3}$, we must take

-\[

-g_{\alpha} = G_{4},\quad g_{\beta} = G_{3}, \text{ giving } n_{G} = 12.

-\]

-

-\Class{\Inum{V.}}{$G_{\alpha} = G_{4}'$.}

-

-There are, exactly as in the last case, three possibilities: $G_{4}'$~taking

-the place of~$G_{4}$. These groups are not however simply isomorphic

-with the preceding three.

-

-\Class{\Inum{VI.}}{$G_{\alpha} = G_{4}''$.}

-

-There are in this case four possibilities. Of these three correspond

-to those of~\Inum{IV}, $G_{4}''$~taking the place of~$G_{4}$. The remaining

-one is given by $G_{\beta} = G_{6}$, $g_{\beta} = G_{3}$, $g_{\alpha} = G_{2}'$, where $G_{2}'$~represents the

-group $\{1, (13)\}$. Regarded as substitution groups all these are of

-distinct form from the groups of~\Inum{V}\@.

-%% -----File: 181.png---Folio 165-------

-

-There are thus $22$~distinct intransitive substitution groups of

-degree~$7$, in which the symbols are interchanged transitively in two

-sets of $4$~and~$3$ respectively.

-\end{Remark}

-

-\Par{118.} Let $\nu_{r}$ ($r = 0, 1, 2, \Dots, n$) be the number of substitutions

-\index{Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group}%

-of a group of degree~$n$ and order~$N$ which leave exactly

-$r$ symbols unchanged, so that

-\[

-N = \sum_{r = 0}^{r = n} \nu_{r}.

-\]

-

-Suppose first that the group is transitive; and in a sub-group,

-which keeps one symbol unchanged, let $\nu_{r}'$ ($r = 1,

-2, \Dots, n$) be the number of substitutions that leave exactly $r$

-symbols unchanged, so that

-\[

-\frac{N}{n} = \sum_{r = 1}^{r = n} \nu_{r}'.

-\]

-

-Each of the $n$~sub-groups, which leave a single symbol

-unchanged, have $\nu_{r}'$~substitutions which leave exactly $r$ symbols

-unchanged; and each of these substitutions belong to $r$~sub-groups

-which leave one symbol unchanged. Hence

-\[

-n\nu_{r}' = r\nu_{r},

-\]

-and therefore

-\[

-N = \sum_{r = 1}^{r = n} r\nu_{r};

-\]

-or the number of unchanged symbols in all the substitutions

-of a transitive group is equal to the order of the group.

-

-Suppose, next, that the group is intransitive; and consider a

-set of $s$~symbols among the~$n$, which are permuted transitively

-among themselves by the operations of the group. Let $N_{1}$ be

-the order of the self-conjugate sub-group~$H_{1}$, which leaves

-unchanged each of this set of $s$ symbols. Then if we consider

-the effect of the substitutions on this set of $s$ symbols only, the

-group reduces to a transitive group of order~$\dfrac{N}{N_{1}}$ with which the

-original group is multiply isomorphic. If $S'$ is any substitution

-of this group of order~$\dfrac{N}{N_{1}}$, and if $SH_{1}$~denote the corresponding $N_{1}$

-operations of the original group, then every substitution of the

-set~$SH_{1}$ produces the same effect on the $s$~symbols that $S'$

-%% -----File: 182.png---Folio 166-------

-\index{Intransitive groups@\Topic{Intransitive groups}!transitive sets of symbols in}%

-\index{Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group}%

-produces. Now the number of unchanged symbols in all the

-substitutions of the transitive group of degree~$s$ and order~$\dfrac{N}{N_{1}}$

-is~$\dfrac{N}{N_{1}}$; therefore, in all the substitutions of the original group,

-the number of symbols of the set of~$s$ that remain unchanged is~$N$.

-The same reasoning applies to each separate set of the $n$~symbols,

-which are permuted transitively among themselves by

-the operations of the group. Hence if there are $t_{1}$ such transitive

-sets, the total number of symbols which remain unchanged

-in all the substitutions of the group is~$Nt_{1}$; or

-\[

-Nt_{1} = \sum_{r = 1}^{r = n} r\nu_{r}.

-\]

-

-\Par{119.} The formula just obtained is the first of a series of

-similar \DPchg{formulae}{formulæ}, due to Herr Frobenius,\footnote

-  {``Ueber die Congruenz nach einem aus zwei endlichen Gruppen gebildeten

-  Doppelmodul,'' \textit{Crelle}, t.~\VolNo{CI}, (1887), p.~288.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-which are capable of

-many useful applications.

-

-{\Loosen Consider the symbol $(a_{1}, a_{2}, \Dots, a_{p})$, where the letters are

-$p$~distinct letters chosen from the $n$ which are operated on by

-the group, the sequence in which the $p$~letters are arranged

-being regarded as essential. The number of such symbols

-that can be formed from the $n$~letters is $n(n - 1)\Dots (n - p + 1)$.

-Every substitution of the group interchanges this set of

-symbols among themselves; and no substitution can leave one

-of the symbols unchanged unless it leaves each of the letters

-forming it unchanged. Moreover no substitution of the group,

-other than identity, can leave each of the set of symbols

-unchanged. Hence the group can be expressed as a substitution

-group in this set of $n(n - 1)\Dots (n - p + 1)$ symbols.

-Every substitution of the group which leaves exactly $r$ letters

-unchanged will, if $r < p$, leave none of the set of symbols

-unchanged; while if $r \geq p$, it will leave exactly}

-\[

-r(r - 1)\Dots (r - p + 1)

-\]

-unchanged. Hence, if the set of symbols are interchanged by

-the substitutions of the group in $t_{p}$~transitive sets, then

-\[

-Nt_{p} = \sum_{r = p}^{r = n} \frac{r!}{r - p!} \nu_{r}.

-\]

-%% -----File: 183.png---Folio 167-------

-

-From the symbol $(a_{1}, a_{2}, \Dots, a_{p})$, containing $p$~letters,

-may be formed $n - p$ symbols containing $p + 1$ letters, by

-adding any one of the remaining $n - p$ letters in the last place.

-If the symbol $(a_{1}, a_{2}, \Dots, a_{p})$ is one of a transitive set of

-$s$~symbols, to these there will correspond $s(n - p)$ symbols of

-$p + 1$ letters. No symbol of $p + 1$ letters, which is not included

-among these $s(n - p)$ symbols, can enter into a transitive set

-with any one of the $s(n - p)$; since if it did, the $s$~symbols of $p$~letters

-would not form a transitive set. Hence the $s(n - p)$

-symbols must form, by themselves, a number of transitive sets

-of the symbols of $p + 1$ letters; and this number clearly cannot

-be less than~$1$ nor greater than $n - p$. Accordingly, to every

-transitive set of the symbols of $p$~letters, there correspond

-$x$ ($1 \leq x \leq n - p$) transitive sets of the symbols of $p + 1$ letters;

-and therefore

-\[

-t_{p} \leq t_{p + 1} \leq (n - p)t_{p}.

-\]

-

-\begin{Ex}[1.] If $t_{p}$ and $t_{p + 1}$ are equal and if $p < n - 1$, shew that $t_{p + 1}$~is

-unity and that the group is $(p + 1)$-ply transitive.

-\end{Ex}

-

-\begin{Ex}[2.] Apply the method of §~119 to shew that no substitution,

-except identity, of a $k$-ply transitive group, which does not contain

-the alternating group, can displace less than $2k - 2$ letters.

-\end{Ex}

-

-

-\Note{\textit{Note to} §~108.}

-

-\begin{Remark}

-The results of M.~Jordan, stated on \PageRef{p.}{149}, may be established

-as follows. Let $G$~be a quadruply transitive group of degree~$n$ and

-order $n(n - 1)(n - 2)(n - 3)$, so that $n - 2$ is the power of a prime.

-There is a single operation of~$G$ which changes any four symbols

-into any other four symbols. Let $a$,~$b$, $c$,~$d$ be four of the symbols

-operated on by~$G$. The operations of~$G$ which permute these

-symbols among themselves form a sub-group~$H$, simply isomorphic

-with the symmetric group of degree~$4$. Suppose that the remaining

-$n - 4$ symbols are permuted by~$H$ in transitive sets of $n_{1}$, $n_{2}$,~\dots\

-symbols each.

-

-The only groups with which $H$~is multiply isomorphic are

-\Inum{(i)}~the symmetric group of degree~$3$, \Inum{(ii)}~a group of order~$2$. If

-then, when we consider the effect of~$H$ on a set of $n_{1}$~symbols which

-are permuted transitively by it, the group of degree~$n_{1}$ so obtained

-is one with which $H$~is multiply isomorphic, this group must be

-%% -----File: 184.png---Folio 168-------

-either the symmetric group of degree~$3$ or a group of order~$2$.

-Hence $n_{1}$ must be either $6$,~$3$ or~$2$. Since in this case $H$ will

-contain operations leaving all the $n_{1}$~symbols unchanged, the value~$6$

-for~$n_{1}$ is inadmissible. If $n_{1}$ is~$3$, the operations of~$H$, which

-give the substitutions

-\[

-1,\quad (ab)\,(cd),\quad (ad)\,(bc),\quad (ac)(bd),

-\]

-of $a$,~$b$, $c$,~$d$, leave the $3$~symbols unchanged; and if $n_{1}$ is~$2$, the

-operations of~$H$ which give all the even substitutions of $a$,~$b$, $c$,~$d$,

-leave the $2$~symbols unchanged.

-

-If, on the other hand, when we consider the effect of~$H$ on the

-set of $n_{1}$ symbols only, the transitive group of degree~$n_{1}$ so obtained

-is simply isomorphic with~$H$, then $n_{1}$~must be either $4$,~$6$, $8$,~$12$ or~$24$.

-In this case, the value~$4$ for~$n_{1}$ is inadmissible. In fact, if

-$n_{1}$ were~$4$ the operation of~$H$ which leaves $c$~and~$d$ unchanged

-would also leave two of the $n_{1}$ symbols unchanged; and this is

-impossible since no operation of~$H$, except identity, can leave more

-than $3$ symbols unchanged. For a similar reason, the value~$12$ for~$n_{1}$

-is inadmissible; while, if $n_{1}$ is~$6$, the sub-group that keeps one of

-the $n_{1}$~symbols unchanged must be cyclical.

-

-The only other possible value for $n_{1}$ is unity.

-

-Suppose, first, that $n_{1}$ is~$2$. If any of the remaining numbers

-$n_{2}$, $n_{3}$,~\dots\ differ from~$24$, it may be shewn immediately that

-$H$~contains operations which keep more than $3$ symbols fixed.

-Hence in this case $n$~is congruent to $6\ (\mod 24)$; and since $n - 2$

-cannot then be a power of a prime unless $n = 6$, this case gives the

-alternating group of degree~$6$.

-

-Next, suppose that $n_{1}$ is~$3$. The only admissible values for

-$n_{2}$, $n_{3}$,~\dots\ are $8$~and~$24$. In this case, the sub-group of~$H$ which

-keeps all the $n_{1}$ ($= 3$) symbols unchanged is a non-cyclical sub-group

-of order~$4$. But we have seen in §~105 that, if $n - 3$ is even, a

-sub-group of order~$n - 3$ which keeps $3$~symbols unchanged can only

-have a single operation of order~$2$. Hence this case cannot occur.

-

-Next, suppose that $n_{1}$ is~$8$. The only admissible values for

-$n_{2}$, $n_{3}$,~\dots\ are again~$24$. This case cannot occur since no number

-congruent to $10, \pmod{24}$, can be a power of a prime.

-

-The only remaining possibilities are:

-%[** TN: Reformatted; ad hoc dot spacing]

-\begin{alignat*}{3}

-&\Inum{(i)}   & n_{1} &= n_{2} = \ \dots\dots\dots\dots\dots &&= 24, \\

-&\Inum{(ii)}  & n_{1} &= 1,\quad n_{2} = n_{3} = \ \dots\dots &&= 24, \displaybreak[0]\\

-&\Inum{(iii)}\quad &  n_{1} &= 6,\quad n_{2} = n_{3} = \ \dots\dots &&= 24, \displaybreak[0]\\

-&\Inum{(iv)}  & n_{1} &= 1,\quad n_{2} = 6,\quad n_{3} = \dots &&= 24, \displaybreak[0]\\

-&\Inum{(v)}   & n_{1} &= 1,\quad n_{2} = 6,\quad n_{3} = \dots &&= 0, \\

-&\Inum{(vi)}  & n_{1} &= 1,\quad n_{2} = n_{3} = \ \dots\dots &&= 0.

-\end{alignat*}

-

-The first of them cannot occur, since then $n - 2$ is not the power

-%% -----File: 185.png---Folio 169-------

-of a prime. In the second case, an operation of~$H$ which leaves $c$

-and $d$ unchanged would be of the form

-\[

-(ab) (\alpha\beta) (\gamma\delta) \Dots,

-\]

-where there are $\frac{1}{2}(n - 3)$ independent transpositions; while an

-operation of~$H$, of order~$2$, which leaves none of the symbols $a$,~$b$,

-$c$,~$d$ unchanged, will consist of the product of $\frac{1}{2}(n - 1)$ independent

-transpositions. Now the operation

-\[

-(ab) (\alpha\beta) (\gamma\delta) \Dots

-\]

-occurs in the group, conjugate to~$H$, which permutes $a$,~$b$, $\alpha$,~$\beta$,

-among themselves; and as an operation of this group, it must be

-the product of $\frac{1}{2}(n - 1)$ independent transpositions. This is a

-contradiction: hence this case cannot occur.

-

-In case~\Inum{(iii)}, let $1$,~$2$, $3$, $4$, $5$,~$6$ be the six symbols that are

-permuted transitively. The self-conjugate sub-group of~$H$ of order~$4$

-will consist of identity and the three substitutions

-\begin{gather*}

-(ab) (cd) (34) (56) (\alpha\beta) (\gamma\delta) \Dots, \\

-(ac) (bd) (56) (12) (\alpha\gamma) (\beta\delta) \Dots, \\

-(ad) (bc) (12) (34) (\alpha\delta) (\beta\gamma) \Dots.

-\end{gather*}

-

-This sub-group will also occur as the self-conjugate sub-group of

-order~$4$ of the group, conjugate to~$H$, which permutes $\alpha$,~$\beta$, $\gamma$,~$\delta$

-among themselves. If $S$,~$S'$ are two operations, of order~$3$, of these

-sub-groups which give the same permutation of $1$,~$2$, $3$, $4$, $5$,~$6$, then

-$SS'^{-1}$~is an operation of~$G$, distinct from identity, which keeps the

-six symbols unchanged. Hence this case cannot occur. Precisely

-the same reasoning applies to case~\Inum{(iv)}.

-

-Finally then, cases \Inum{(v)} and~\Inum{(vi)}, in which $n$~is respectively $11$ and

-$5$ are the only remaining possibilities.

-

-When $n$ is~$5$, $G$~is the symmetric group of degree~$5$.

-

-When $n$ is $11$, the group~$G$, if it exists, is of degree~$11$ and

-order $11 · 10 · 9 · 8$. A sub-group of order~$8$ which keeps three

-symbols fixed must contain a single operation of order~$2$; hence

-it must either be cyclical or of the type given in Theorem~V, §~63.

-When expressed in $8$~symbols, this is generated by

-\[

-(1254) (3867)\quad \text{and}\quad (1758) (2643),

-\]

-while we may take a cyclical group of order~$8$ to be generated by

-\[

-(12835476).

-\]

-

-It is easy to verify that each of these substitutions transforms the

-group of order~$9$, generated by

-\[

-(123) (456) (789)\quad \text{and}\quad (147) (258) (369),

-\]

-into itself.

-

-If we now apply the method of §~109, we find that each of

-these doubly transitive groups of degree~$9$ is contained in a triply

-%% -----File: 186.png---Folio 170-------

-transitive group of degree~$10$; a repetition of the same process of

-trial shews that, while the group

-\[

-\{(123) (456) (789),\ (12835476)\}

-\]

-is not contained in a quadruply transitive group of degree~$11$ and

-order~$11·10·9·8$, the group

-\[

-\{(123) (456) (789),\ (1254) (3867),\ (1758) (2643)\}

-\]

-and the two substitutions

-\[

-(a2) (58) (64) (79),\ (ab) (57) (68) (49)

-\]

-actually generate such a group.

-

-A further trial will shew that this group and the substitution

-\[

-(bc) (47) (58) (69)

-\]

-generate a group of degree~$12$ and order $12·11·10·9·8$; but that

-there is no group of degree~$13$ which contains the last group as the

-sub-group that keeps one symbol fixed.

-

-The three substitutions, of order two, just given generate that

-sub-group of order~$24$, of the transitive group of degree~$11$ and

-order $11·10·9·8$ in the symbols

-\[

-a,\ b,\ c,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,

-\]

-which permutes $a$,~$b$, $c$,~$2$ among themselves.

-

-{\Loosen If $k \nless 4$, a $k$-ply transitive group, of degree~$n$ and order

-$n(n - 1)\Dots(n - k + 1)$, must contain a quadruply transitive group of

-degree $n - k + 4$ and order $(n - k + 4) (n - k + 3) (n - k + 2) (n - k + 1)$.

-Hence, if the group is neither the alternating group nor the symmetric

-group of degree~$n$, it must be either the group of degree~$11$

-or the group of degree~$12$ that have been determined above.}

-

-

-\Note{\textit{Note to} §~110.}

-

-It may be shewn that, with a single exception when $n = 12$, the

-inequality (\PageRef{p.}{152})

-\[

-k \ngtr \tfrac{1}{3} n + 1

-\]

-may be replaced by

-\[

-k < \tfrac{1}{3} n + 1.

-\]

-

-Since $k$ is an integer, it is only necessary to consider the case in

-\index{Substitution group@\Topic{Substitution group}!quintuply transitive, of degree~$12$}%

-which $n$~is a multiple of~$3$, so that we may write $3m$ for~$n$. If, in

-this case, $k = m + 1$, then $2k - 2 = 2m$; and a $(m + 1)$-ply transitive

-group of degree~$3m$, which does not contain the alternating group,

-can therefore have no substitution which displaces fewer than $2m$

-symbols. Its order must therefore be $3m (3m - 1)\Dots (2m + 1) 2m$.

-From M.~Jordan's results, which have just been proved, such a

-group exists only when $n = 12$; and therefore when $n$~is not~$12$, a

-$k$-ply transitive group of degree~$n$, which does not contain the

-alternating group, can only exist if $\DPtypo{k\frac{1}{3} < n + 1}{k < \frac{1}{3}n + 1}$.

-\end{Remark}

-%% -----File: 187.png---Folio 171-------

-

-

-\Chapter{IX.}{On Substitution Groups: Primitive and Imprimitive Groups.}

-

-\Par{120.} \First{We} have seen that the symbols permuted by the

-operations of an intransitive substitution group may be divided

-into sets, such that every substitution of the group permutes

-the symbols of each set among themselves. For a transitive

-group the symbols must, from this point of view, be regarded

-as forming a single set. It may however in particular cases

-be possible to divide the symbols permuted by a transitive

-group into sets in such a way, that every substitution of the

-group either interchanges the symbols of any set among themselves

-or else changes them all into the symbols of some other

-set. That this may be possible, it is clearly necessary that each

-set shall contain the same number of symbols.

-

-In the present Chapter we shall discuss those properties

-of transitive groups which depend on their possessing or not

-possessing the property indicated.

-

-\Par{121.} \begin{Definition}

-When the symbols operated on by a

-\index{Imprimitive groups@\Topic{Imprimitive groups}!definition of}%

-\index{Primitive groups@\Topic{Primitive groups}!definition of}%

-\index{Substitution group@\Topic{Substitution group}!primitive and imprimitive, definition of}%

-transitive substitution group can be divided into sets, each set

-containing the same number of distinct symbols and no symbol

-occurring in two different sets, and when the sets are such

-that all the symbols of any set are either interchanged among

-themselves or changed into the symbols of another set by every

-substitution of the group, the group is called \emph{imprimitive}.

-When no such division into sets is possible, the group is called

-%% -----File: 188.png---Folio 172-------

-\index{Imprimitive systems@\Topic{Imprimitive systems}!definition of}%

-\emph{primitive}. The sets of symbols which are interchanged by

-an imprimitive group are called \emph{imprimitive systems}.

-\end{Definition}

-

-A simple example of an imprimitive group is given by

-group~\Inum{VII} of §~17. An examination of the substitutions of this

-group will shew that they all either transform the systems of

-symbols $xyz$ and $abc$ into themselves or else interchange them,

-and that the same is true of the systems $xa$,~$yb$,~$zc$; so that, in

-this case, the symbols may be divided into two distinct sets of

-imprimitive systems.

-

-It follows at once, from the definition, that an imprimitive

-group cannot be more than simply transitive. For if it were

-doubly transitive, it would contain substitutions changing any

-two symbols into any other two, and of these the first pair

-might be chosen from the same imprimitive system and the

-second pair from distinct systems.

-

-The question as to whether a given group can be expressed

-as a transitive group of given degree, and the further question

-as to whether such a representation of the group, when possible,

-is imprimitive or primitive, finds its complete solution in the

-following investigation due to Herr Dyck\footnotemark.

-\index{Dyck@\Topic{Dyck}, quoted}%

-\footnotetext{Dyck, ``Gruppentheoretische Studien,~II,'' \textit{Math.\ Ann.}\DPnote{** No "Vol."} \VolNo{XXII}, (1883), pp.~86--95.}

-

-\Par{122.} In §~20 it was shewn how any group~$G$ of order~$N$

-could be represented as a substitution group of $N$~symbols.

-This form of the group, defined as the regular form, is simply

-transitive; for all its substitutions except identity displace all

-the symbols, and therefore there must be just one substitution

-changing a given symbol into any other. Let us now suppose

-that $N = \mu\nu$, and that $G$~has a sub-group~$H$ of order~$\mu$, consisting

-of the operations

-\[

-S_{1} (= 1),\ S_{2},\ S_{3},\ \Dots,\ S_{\mu},

-\]

-so that every operation of the group can be represented uniquely

-in the form

-\[

-S_{m}T_{n}\quad (m = 1, 2, 3, \Dots, \mu;\ n = 1, 2, 3, \Dots, \nu);\ T_{1} = 1.

-\]

-

-The tableau representing the group as a substitution group

-of $N$~symbols will, in terms of these symbols, take the form

-%% -----File: 189.png---Folio 173-------

-\index{Imprimitive systems@\Topic{Imprimitive systems}!of a regular group}%

-\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group|(}%

-given on the following page. Every symbol in this tableau is

-of the form

-\[

-S_{m} T_{n} S_{m'} T_{n'};

-\]

-and such a symbol will belong to the column headed by $S_{m}T_{n}$

-and to the line beginning with~$S_{m'}T_{n'}$. The symbols in any

-line (or column) differ in arrangement only from those in the

-leading line (or column); hence $T_{k}$~must occur in the line

-beginning with~$S_{m'} T_{n'}$. We may therefore suppose that

-\[

-S_{m} T_{n} S_{m'} T_{n'} = T_{k};

-\]

-and then

-\begin{align*}

-  S_{2} S_{m} T_{n} S_{m'} T_{n'} &= S_{2} T_{k}; \\

-  S_{3} S_{m} T_{n} S_{m'} T_{n'} &= S_{3} T_{k}, \\

-  \multispan{3}{\dotfill} \\

-  S_{\mu} S_{m} T_{n} S_{m'} T_{n'} &= S_{\mu} T_{k}.

-\end{align*}

-Since

-\[

-1,\ S_{2},\ S_{3},\ \Dots,\ S{\mu}

-\]

-form a group, these symbols differ from

-\[

-S_{m},\ S_{2}S_{m},\ S_{3}S_{m},\ \Dots,\ S_{\mu}S_{m}

-\]

-only in the sequence in which they are written; and therefore

-the set of symbols

-\[

-S_{p} T_{n} S_{m'} T_{n'}\quad (p = 1, 2, \Dots, \mu)

-\]

-is identical, except as regards arrangement, with the set

-\[

-S_{p}T_{k}\quad (p = 1, 2, \Dots, \mu).

-\]

-

-Hence the substitution of~$G$, represented by the line beginning

-with~$S_{m'} T_{n'}$, changes the set of symbols $S_{p} T_{n}$ into the

-set~$S_{p} T_{k}$ in some sequence or other.

-

-Every substitution of the group therefore changes the

-symbols of each of the $\nu$~sets, into which the first line of the

-tableau is divided, either into themselves or into the symbols of

-some other of the $\nu$ sets. Hence:---

-

-\begin{Theorem}[I.]

-If a group of order~$\mu\nu$ contains a sub-group

-of order~$\mu$, the regular form of the group will be imprimitive in

-such a way that the $\mu\nu$~symbols may be divided into $\nu$~imprimitive

-systems of $\mu$~symbols each.

-\end{Theorem}

-

-The converse of this theorem is also true. For if the $\mu\nu$~symbols

-\[

-1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu\nu},

-\]

-%% -----File: 190.png---Folio 174-------

-\begin{sidewaystable}

-\setlength{\arraycolsep}{2pt}

-\PageLabel{174}%

-\ifthenelse{\boolean{ForPrinting}}{}{\Squeeze[0.75]}{$

-\begin{array}{@{}*{5}{c}|c|*{5}{c}|c|*{5}{c}@{}}

-1  & S_{2} & S_{3} & \dots & S_{\mu} & \dots &

-T_{a} & S_{2}T_{a} & S_{3}T_{a} & \dots & S_{\mu}T_{a} & \dots &

-T_{\nu} & S_{2}T_{\nu} & S_{3}T_{\nu} & \dots & S_{\mu}T_{\nu} \\

-%

-S_{2} & S_{2}S_{2} & S_{3}S_{2} & \dots & S_{\mu}S_{2} & \dots &

-T_{a}S_{2} & S_{2}T_{a}S_{2} & S_{3}T_{a}S_{2} & \dots & S_{\mu}T_{a}S_{2} & \dots &

-T_{\nu}S_{2} & S_{2}T_{\nu}S_{2} & S_{3}T_{\nu}S_{2} & \dots & S_{\mu}T_{\nu}S_{2} \\

-%

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c}{\dotfill} \\

-%

-S_{\mu} & S_{2}S_{\mu}   & S_{3}S_{\mu}  & \dots & S_{\mu}S_{\mu}  & \dots & T_{a}S_{\mu}  & S_{2}T_{a}S_{\mu}  & S_{3}T_{a}S_{\mu}  & \dots & S_{\mu}T_{a}S_{\mu}  & \dots & T_{\nu}S_{\mu}  & S_{2}T_{\nu}S_{\mu}  & S_{3}T_{\nu}S_{\mu}  & \dots & S_{\mu}T_{\nu}S_{\mu} \\

-  \hline

-\multicolumn{5}{c|}{\dotfill} & \Strut\dots &

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c}{\dotfill} \\

-%

-\hline

-\Strut

-T_{b} & S_{2}T_{b} & S_{3}T_{b} & \dots & S_{\mu}T_{b} & \dots &

-T_{a}T_{b} & S_{2}T_{a}T_{b} & S_{3}T_{a}T_{b} & \dots & S_{\mu}T_{a}T_{b} & \dots &

-T_{\nu}T_{b} & S_{2}T_{\nu}T_{b} & S_{3}T_{\nu}T_{b} & \dots & S_{\mu}T_{\nu}T_{b} \\

-%

-S_{2}T_{b} & S_{2}S_{2}T_{b} & S_{3}S_{2}T_{b} & \dots & S_{\mu}S_{2}T_{b} & \dots &

-T_{a}S_{2}T_{b} & S_{2}T_{a}S_{2}T_{b} & S_{3}T_{a}S_{2}T_{b} & \dots & S_{\mu}T_{a}S_{2}T_{b} & \dots &

-T_{\nu}S_{2}T_{b} & S_{2}T_{\nu}S_{2}T_{b} & S_{3}T_{\nu}S_{2}T_{b} & \dots & S_{\mu}T_{\nu}S_{2}T_{b} \\

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c}{\dotfill} \\

-%

-S_{\mu}T_{b} & S_{2}S_{\mu}T_{b} & S_{3}S_{\mu}T_{b} & \dots & S_{\mu}S_{\mu}T_{b} & \dots &

-T_{a}S_{\mu}T_{b} & S_{2}T_{a}S_{\mu}T_{b} & S_{3}T_{a}S_{\mu}T_{b} & \dots & S_{\mu}T_{a}S_{\mu}T_{b} & \dots &

-T_{\nu}S_{\mu}T_{b} & S_{2}T_{\nu}S_{\mu}T_{b} & S_{3}T_{\nu}S_{\mu}T_{b} & \dots & S_{\mu}T_{\nu}S_{\mu}T_{b} \\

-\hline

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c}{\dotfill} \\

-%

-\hline

-\Strut

-T_{\nu} & S_{2}T_{\nu} & S_{3}T_{\nu} & \dots & S_{\mu}T_{\nu} & \dots &

-T_{a}T_{\nu} & S_{2}T_{a}T_{\nu} & S_{3}T_{a}T_{\nu} & \dots & S_{\mu}T_{a}T_{\nu} & \dots &

-T_{\nu}T_{\nu} & S_{2}T_{\nu}T_{\nu} & S_{3}T_{\nu}T_{\nu} & \dots & S_{\mu}T_{\nu}T_{\nu} \\

-%

-S_{2}T_{\nu} & S_{2}S_{2}T_{\nu} & S_{3}S_{2}T_{\nu} & \dots & S_{\mu}S_{2}T_{\nu} & \dots &

-T_{a}S_{2}T_{\nu} & S_{2}T_{a}S_{2}T_{\nu} & S_{3}T_{a}S_{2}T_{\nu} & \dots & S_{\mu}T_{a}S_{2}T_{\nu} & \dots &

-T_{\nu}S_{2}T_{\nu} & S_{2}T_{\nu}S_{2}T_{\nu} & S_{3}T_{\nu}S_{2}T_{\nu} & \dots & S_{\mu}T_{\nu}S_{2}T_{\nu} \\

-%

-\multicolumn{5}{c|}{\dotfill} & \Strut\dots &

-\multicolumn{5}{c|}{\dotfill} & \dots &

-\multicolumn{5}{c}{\dotfill} \\

-%

-S_{\mu}T_{\nu} & S_{2}S_{\mu}T_{\nu} & S_{3}S_{\mu}T_{\nu} & \dots & S_{\mu}S_{\mu}T_{\nu} & \dots &

-T_{a}S_{\mu}T_{\nu} & S_{2}T_{a}S_{\mu}T_{\nu} & S_{3}T_{a}S_{\mu}T_{\nu} & \dots & S_{\mu}T_{a}S_{\mu}T_{\nu} & \dots &

-T_{\nu}S_{\mu}T_{\nu} & S_{2}T_{\nu}S_{\mu}T_{\nu} & S_{3}T_{\nu}S_{\mu}T_{\nu} & \dots & S_{\mu}T_{\nu}S_{\mu}T_{\nu}

-\end{array}

-$}

-\end{sidewaystable}

-%% -----File: 191.png---Folio 175-------

-by whose permutations the group can be expressed in regular

-form, are divisible into $\nu$ imprimitive systems of $\mu$ each, let

-\[

-1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu}

-\]

-be that system which contains identity. Then this system

-and the systems

-\[

-S_{m},\ S_{2}S_{m},\ S_{3}S_{m},\ \Dots,\ S_{\mu}S_{m},\quad m = 2,\ 3,\ \Dots,\ \mu,

-\]

-having the symbol~$S_{m}$ in common, must have all their symbols

-in common; therefore the product of any two of the operations

-of this set of $\mu$ operations is another operation of the set. The

-set therefore forms a group.

-

-\Par{123.} Let us now represent the imprimitive system

-\[

-T_{m},\ S_{2}T_{m},\ S_{3}T_{m},\ \Dots,\ S_{\mu}T_{m}

-\]

-by the single symbol~$U_{m}$, for all values of~$m$. If we then, in

-the preceding tableau representing the group, pay attention only

-to the way in which the systems are permuted among themselves,

-without regarding the permutations of the symbols within the individual

-systems, we obtain a substitution group of the $\nu$ symbols

-\[

-U_{1},\ U_{2},\ U_{3},\ \Dots,\ U_{\nu}.

-\]

-

-This group is isomorphic with the original group~$G$; and if

-no substitution of the original group interchanges among

-themselves the symbols of each imprimitive system, the isomorphism

-must be simple. Now a substitution of~$G$, which

-does not change any imprimitive system into another, must, if

-it exists, be a substitution~$S'$ of the sub-group~$H$, which is

-constituted by

-\[

-1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu},

-\]

-such that $T_{m}S'$ belongs to the set~$U_{m}$ for each suffix~$m$; and

-therefore, for each suffix~$m$, we must have an equation

-\[

-T_{m}S' = S''T_{m},

-\]

-where $S''$ is another substitution of~$H$. The sub-group~$H$

-must therefore contain every substitution of~$G$ which is conjugate

-to~$S'$: in other words, it must contain a self-conjugate

-sub-group of~$G$. Hence:---

-

-\begin{Theorem}[II.]

-If $H$, of order~$\mu$, is a sub-group of~$G$ of

-\index{Transitive group@\Topic{Transitive group}!representation of any group as a|(}%

-order~$\mu\nu$, and if no self-conjugate sub-group of~$G$ is contained in~$H$,

-then $G$~can be expressed as a transitive group of degree~$\nu$.

-\end{Theorem}

-%% -----File: 192.png---Folio 176-------

-

-That the converse of this theorem is true is immediately

-obvious.

-

-\Par{124.} If the sub-group~$H$ of~$G$ is contained in a greater

-sub-group~$K$ of order~$\mu\nu'$, where $\nu = \nu'\nu''$, the operations of~$K$

-consist of the sets

-\[

-H,\ HT_{2},\ HT_{3},\ \Dots,\ HT_{\nu'},

-\]

-and the operations of~$G$ of the sets

-\[

-K,\ KR_{2},\ KR_{3},\ \Dots,\ KR_{\nu''};

-\]

-while the set~$KR_{m}$ is made up of the sets

-\[

-HR_{m},\ HT_{2} R_{m},\ HT_{3} R_{m},\ \Dots,\ HT_{\nu'} R_{m}.

-\]

-

-The regular form of the group has $\nu'\nu''$~imprimitive systems

-\index{Imprimitive systems@\Topic{Imprimitive systems}!of any transitive group}%

-corresponding to the sub-group~$H$, and $\nu''$~imprimitive systems

-corresponding to the sub-group~$K$; the above method of representing~$K$

-shews that each of the latter systems contain $\nu'$~complete

-systems of the former set.

-

-Now it has just been proved that, if $H$~contains no self-conjugate

-sub-group of~$G$, the group can be represented as a

-transitive substitution group of the symbols

-\[

-HT_{n} R_{m}\quad (n = 1, 2, \Dots, \nu',\ m = 1, 2, \Dots, \nu'').

-\]

-

-But from the division of the symbols in the regular form of

-the group into $\nu''$~imprimitive systems, it follows that the set of

-symbols

-\[

-HT_{n} R_{m}\quad (n = 1, 2, \Dots, \nu')

-\]

-must either be permuted among themselves or be changed into

-another set

-\[

-HT_{n} R_{m'}\quad (n = 1, 2, \Dots, \nu')

-\]

-by every substitution of the group. The representation of the

-group as a transitive group of $\nu\nu''$~symbols is therefore imprimitive.

-Hence:---

-

-\begin{Theorem}[III.]

-If a group~$G$ of order~$\mu\nu'\nu''$ has a sub-group~$H$

-of order~$\mu$, which contains no self-conjugate sub-group of~$G$;

-and if $H$~is contained in a sub-group~$K$ of~$G$ of order~$\mu\nu'$;

-then the representation of~$G$ as a transitive group of degree~$\nu'\nu''$

-\(in respect of~$H$\) is imprimitive, the $\nu'\nu''$~symbols being

-divisible into $\nu''$~systems of $\nu'$~symbols each.

-\end{Theorem}

-%% -----File: 193.png---Folio 177-------

-

-\begin{Corollary}[I.]

-A transitive group of order~$\mu\nu$ and degree~$\nu$

-\index{Representation of a group@\Topic{Representation of a group}!in primitive form}%

-will be primitive if, and only if, a sub-group of order~$\mu$ that

-keeps one symbol fixed is a maximum sub-group.

-\end{Corollary}

-

-\begin{Corollary}[II.]

-A group, which contains other self-conjugate

-operations besides identity, cannot be represented in primitive

-form.

-\end{Corollary}

-

-For if a sub-group~$H$ of order~$\mu$ contains a self-conjugate

-operation, the group (of order~$\mu\nu$) cannot be represented as

-a transitive group of degree~$\nu$ in respect of~$H$; and if $H$

-contains none of the self-conjugate operations, and is not a

-self-conjugate sub-group, it cannot be a maximum sub-group.

-

-In particular, a group whose order is the power of a prime

-cannot be represented as a primitive group.

-

-\begin{Corollary}[III.]

-An Abelian group when represented as a

-transitive substitution group, must be in regular form.

-\end{Corollary}

-

-\begin{Corollary}[IV.]

-A simple group can always be represented

-in primitive form.

-\end{Corollary}

-

-\Par{125.} Every possible representation of a group as a

-transitive substitution group is given by the method of the

-preceding paragraphs. There is another method of dealing

-with the same problem which we may shortly consider here in

-view of its utility in many special cases, though it does not in

-general lead to all possible modes of representation. Let

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{\nu}

-\]

-be a conjugate set of sub-groups (or operations) of a given

-group~$G$, and let

-\[

-I_{1},\ I_{2},\ \Dots,\ I_{\nu}

-\]

-be a set of sub-groups of~$G$, such that $I_{r}$~is the greatest sub-group

-containing $H_{r}$ self-conjugately. The latter set of groups

-are not necessarily all distinct; in fact, we have seen in §~55

-that, when the order of~$G$ is the power of a prime, they cannot

-be all distinct.

-

-If $S$ is any operation of~$G$, then

-\[

-\Sub{\PadTo{S^{-1}H_{1}S}{H_{1}},

-     \PadTo{S^{-1}H_{2}S}{H_{2}}, \Dots,

-     \PadTo{S^{-1}H_{\nu}S}{H_{\nu}}}

-    {S^{-1}H_{1}S, S^{-1}H_{2}S, \Dots, S^{-1}H_{\nu}S}

-\]

-%% -----File: 194.png---Folio 178-------

-is a substitution performed on the set of symbols $H_{1}$, $H_{2}$,~\dots, $H_{\nu}$;

-and if for $S$ each substitution of the group is written in turn,

-these substitutions form a transitive substitution group of

-degree~$\nu$. The substitution corresponding to~$S$, followed by

-the substitution

-\[

-\Sub{\PadTo{T^{-1}H_{1}T}{H_{1}},

-     \PadTo{T^{-1}H_{2}T}{H_{2}}, \Dots,

-     \PadTo{T^{-1}H_{\nu}T}{H_{\nu}}}

-    {T^{-1}H_{1}T, T^{-1}H_{2}T, \Dots, T^{-1}H_{\nu}T}

-\]

-corresponding to~$T$, gives the substitution

-\[

-\Sub{\PadTo{T^{-1}S^{-1}H_{1}ST}{H_{1}},

-     \PadTo{T^{-1}S^{-1}H_{2}ST}{H_{2}}, \Dots,

-     \PadTo{T^{-1}S^{-1}H_{\nu}ST}{H_{\nu}}}

-    {T^{-1}S^{-1}H_{1}ST, T^{-1}S^{-1}H_{2}ST, \Dots, T^{-1}S^{-1}H_{\nu}ST};

-\]

-and therefore the substitutions form a group isomorphic with~$G$,

-since the product of the substitutions corresponding to $S$ and

-$T$ is the substitution corresponding to~$ST$.

-

-Moreover, since there are operations of~$G$ which transform

-$H_{1}$ into each of the other sub-groups (or operations) of the

-conjugate set, the substitution group is transitive in the $\nu$~symbols.

-The substitution group will be simply isomorphic

-with~$G$ if, and only if, there is no operation of~$G$ which

-transforms each of the $\nu$~sub-groups into itself. Now the only

-operations of~$G$ which transform $H_{1}$ into itself are the operations

-of~$I_{1}$; and hence the substitution group will be simply

-isomorphic with~$G$ if, and only if, the conjugate set of sub-groups

-\[

-I_{1},\ I_{2},\ \Dots,\ I_{\nu},

-\]

-has no common sub-group except identity. This will be the

-case only when $I_{1}$~contains no self-conjugate sub-group of~$G$.

-

-It has been seen (§~123) that, when this condition is satisfied,

-$G$~can be represented as a transitive substitution group whose

-degree is~$\nu$, the ratio of the orders of $G$~and~$I_{1}$. That the

-form there obtained is identical with the form obtained in the

-present paragraph may be easily verified. Thus in the earlier

-form, the substitution corresponding to~$S$ is

-\[

-\Sub{\PadTo{I_{1}T_{1}S}{I_{1}T_{1}},

-     \PadTo{I_{1}T_{2}S}{I_{1}T_{2}}, \Dots,

-     \PadTo{I_{1}T_{\nu}S}{I_{1}T_{\nu}}}

-    {I_{1}T_{1}S, I_{1}T_{2}S, \Dots, I_{1}T_{\nu}S}, \quad (T_{1} = 1),

-\]

-%% -----File: 195.png---Folio 179-------

-or in abbreviated form

-\[

-\Sub{I_{1}T_{x}}{I_{1}T_{x'}},

-\]

-if

-\[

-T_{x}S = iT_{x'},

-\]

-$i$ being some operation of~$I_{1}$.

-

-In the present mode of representation, the substitution

-corresponding to~$S$ can be written in the form

-\[

-\Sub{\PadTo{S^{-1} T_{1}^{-1} H_{1} T_{1}S}{T_{1}^{-1} H_{1} T_{1}},

-     \PadTo{S^{-1} T_{2}^{-1} H_{1} T_{2}S}{T_{2}^{-1} H_{1} T_{2}}, \Dots,

-     \PadTo{S^{-1} T_{\nu}^{-1} H_{1} T_{\nu}S}{T_{\nu}^{-1} H_{1} T_{\nu}}}

-    {S^{-1} T_{1}^{-1} H_{1} T_{1}S,

-     S^{-1} T_{2}^{-1} H_{1} T_{2}S, \Dots,

-     S^{-1} T_{\nu}^{-1} H_{1} T_{\nu}S},

-\]

-since each operation of the set~$I_{1} T_{n}$ will transform $H_{1}$ into the

-same sub-group. Now, if a corresponding abbreviated form be

-used, this substitution may be written

-\[

-\Sub{T_{x}^{-1} H_{1} T_{x}}{T_{x'}^{-1} H_{1} T_{x'}},

-\]

-and therefore the symbols~$I_{1} T_{x}$ in the one form are permuted

-by the substitutions in identically the same manner as the

-corresponding symbols $T_{x}^{-1} H_{1} T_{x}$ in the other form.

-

-It should be noticed that, if $G$~contains self-conjugate operations

-other than identity, these operations necessarily occur

-in~$I_{1}$; and therefore in such a case the present method cannot

-lead to a substitution group which is simply isomorphic with~$G$.

-In any case, if $K$~is the greatest self-conjugate sub-group of~$G$

-contained in~$I_{1}$, the substitution group is simply isomorphic

-with~$\dfrac{G}{K}$.

-\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group|)}%

-\index{Transitive group@\Topic{Transitive group}!representation of any group as a|)}%

-

-\begin{Remark}

-\Par{126.} As an illustration of the preceding paragraphs, we will

-\index{Transitive group@\Topic{Transitive group}!number of distinct modes of representing the alternating group of degree~$5$ as a}%

-determine the different modes in which the alternating group of

-degree~$5$ can be represented as a transitive group.

-

-The only cyclical sub-groups contained in~$G_{60}$, the alternating

-group of degree~$5$, are groups of orders $2$,~$3$ and~$5$; and of each of

-these cyclical sub-groups there is a single conjugate set.

-

-The non-cyclical sub-groups may be determined as follows. The

-lowest possible order for such a sub-group is~$4$; since this is the

-highest power of~$2$ that divides~$60$, there is a single conjugate set of

-sub-groups of order~$4$. The next lowest possible order is~$6$. Now

-no operation of order~$3$ is permutable with an operation of order~$2$,

-as the group contains no operations of order~$6$; on the other hand,

-every sub-group of order~$3$ is permutable with an operation of order~$2$;

-thus

-\[

-(12) (45) (123) (12) (45) = (132).

-\]

-%% -----File: 196.png---Folio 180-------

-\index{Transitive group@\Topic{Transitive group}!number of distinct modes of representing the alternating group of degree~$5$ as a}%

-There is therefore a single set of conjugate sub-groups of order~$6$.

-The next lowest possible order is~$10$. The group contains no operation

-of order~$10$; but every sub-group of order~$5$ is permutable with

-an operation of order~$2$; thus

-\[

-(14) (23) (12345) (14) (23) = (15432).

-\]

-There is therefore a single conjugate set of sub-groups of order~$10$.

-The next lowest possible order is~$12$. If the group contains a

-sub-group of this order, it must be transitive in $4$~symbols. Now the

-alternating group of $4$~symbols is of order~$12$. Hence $G_{60}$~must

-contain a single conjugate set of sub-groups of order~$12$. The only

-other possible orders are $15$,~$20$ and~$30$. The reader will readily

-verify directly that there are no sub-groups of these orders. This

-can also be seen indirectly, since $G_{60}$~is a simple group and therefore,

-if there were a sub-group of order~$15$, the group could be expressed

-transitively in $4$~symbols. Since the group contains operations of

-order~$5$, this is clearly impossible.

-

-Hence finally, since each of the sub-groups leads to a transitive

-representation of the group, $G_{60}$~can be represented as a transitive

-substitution group in $30$,~$20$, $15$, $12$, $10$, $6$ and $5$~symbols, and in

-one distinct form in each case. The second method, as given in

-§~125, does not lead to all these modes of representation. The

-group will be found to contain $15$~conjugate operations (or sub-groups)

-of order~$2$: $10$~conjugate sub-groups and $20$~conjugate

-operations of order~$3$: $5$~conjugate sub-groups of order~$4$: $6$~conjugate

-sub-groups and two sets of $12$~conjugate operations, each of

-order~$5$: $10$~conjugate sub-groups of order~$6$: $6$~conjugate sub-groups

-of order~$10$: and $5$~conjugate sub-groups of order~$12$. Hence, by

-using the second method, the representation of the group as transitive

-in $30$~symbols would be missed.

-

-Since a sub-group of order~$2$ is contained in sub-groups of orders

-$4$,~$6$,~$10$ and~$12$, the $30$~symbols permuted by~$G_{60}$, when it is expressed

-as a transitive group of degree~$30$, can be divided into sets

-of imprimitive systems, containing respectively $2$,~$3$,~$5$ and $6$~symbols

-each. Similarly, when $G_{60}$~is represented as a transitive group of

-degree $20$,~$15$ or~$12$, it is imprimitive. When expressed as a group

-of order $10$,~$6$ or~$5$, it is primitive.

-

-\Par{127.} As a further illustration, we shall determine all the

-distinct forms of imprimitive groups of degree~$6$. Let $G$ be such

-a group, and $H$ that sub-group of~$G$ which interchanges among

-themselves the symbols of each imprimitive system.

-

-We will first suppose that there are two systems of three symbols

-each, viz.\

-\[

-1,\ 2,\ 3 \quad\text{and}\quad 4,\ 5,\ 6.

-\]

-%% -----File: 197.png---Folio 181-------

-\index{Imprimitive groups@\Topic{Imprimitive groups}!of degree~$6$}%

-

-In this case, $H$~cannot consist of the identical operation only;

-for there must be a substitution changing $1$ into~$2$, and this must

-permute $1$,~$2$ and~$3$ among themselves, and therefore also $4$,~$5$ and~$6$

-among themselves.

-

-Let $H$~contain substitutions which leave $4$,~$5$ and~$6$ unchanged.

-These must (§~114) form a self-conjugate sub-group of~$H$, which will

-be either

-\[

-\{(123),\ (12)\}\quad \text{or}\quad \{(123)\}.

-\]

-

-Since $G$~has substitutions interchanging the systems, $H$~must

-similarly contain

-\[

-\{(456),\ (45)\}\quad \text{or}\quad \{(456)\}.

-\]

-

-In the first alternative, $H$~is the group

-\[

-\lintertext{\Inum{(i)}} \{(123),\ (12),\ (456),\ (45)\};

-\]

-for $H$~contains this group, and on the other hand, this is the most

-general group that interchanges the six symbols in two intransitive

-systems of three each. The order of this group is~$2^{2} · 3^{2}$.

-

-In the second alternative, $H$~contains the self-conjugate sub-group

-\[

-\{(123),\ (456)\}.

-\]

-

-Now if $H$~is of order~$2^{2} · 3^{2}$, it is necessarily of the form~\Inum{(i)}. If

-it is of order~$2 · 3^{2}$, it must contain a substitution of order~$2$ which

-transforms the sub-group just given into itself. This may be taken,

-without loss of generality, to be~$(12) (45)$; and then $H$~is the

-group

-\[

-\lintertext{\Inum{(ii)}} \{(123),\ (456),\ (12) (45)\}.

-\]

-

-If $H$ is of order~$3^{2}$, it is the group

-\[

-\lintertext{\Inum{(iii)}} \{(123),\ (456)\}.

-\]

-

-Next, let $H$~contain no substitutions which leave $4$,~$5$ and~$6$

-unchanged. Then $H$~is simply isomorphic with a group of degree~$3$,

-and therefore it must be of the form

-\[

-\lintertext{\Inum{(iv)}} \{(123) (456),\ (12) (45)\},

-\]

-or

-\[

-\lintertext{\Inum{(v)}} \{(123) (456)\}.

-\]

-

-Now $\dfrac{G}{H}$~is of order~$2$; therefore $G$~must have a substitution of

-order $2$ or~$4$, which interchanges the systems. If the order of~$H$ is

-odd, this substitution must be of order~$2$. When the substitution

-is of order~$2$, we may, without loss of generality, take it to be

-$(14) (25) (36)$ or $(14) (26) (35)$. If $H$ is of the form \Inum{(i)},~\Inum{(ii)}, or~\Inum{(iv)},

-we get for~$G$, in each case, the same group whichever of these substitutions

-we take. When $H$ is of the form~\Inum{(iii)}, we get two groups which

-are easily seen to be conjugate in the symmetric group. These we

-do not regard as distinct. When $H$~is of the form~\Inum{(v)}, we get two

-%% -----File: 198.png---Folio 182-------

-\index{Imprimitive groups@\Topic{Imprimitive groups}!of degree~$6$}%

-distinct groups, one of which is simply isomorphic with the symmetric

-group of three symbols, while the other is a cyclical group.

-In these two latter cases, the symbols can be divided into three imprimitive

-systems of two each.

-

-If the substitution which interchanges the systems is of order~$4$,

-its square must occur in~$H$; we may therefore take it to be

-$(1425) (36)$. When $H$~is of form~\Inum{(i)}, this gives the same group as

-before; but when $H$~is of either of the forms \Inum{(ii)} or~\Inum{(iv)}, we get new

-forms for~$G$. There are therefore eight distinct forms of groups of

-degree~$6$, in which the symbols form two imprimitive systems of

-three symbols each.

-

-Secondly, suppose that there are three systems of symbols,

-containing two each, viz.\

-\[

-1,\ 2;\quad 3,\ 4;\quad \text{and}\quad 5,\ 6.

-\]

-

-The self-conjugate sub-group~$H$ is of order $2^{3}$,~$2^{2}$, $2$ or~$1$.

-Corresponding to the first three cases, the forms of~$H$ are easily

-seen to be

-\begin{alignat*}{2}

-&\Inum{(i)}       && \{(12),\ (34),\ (56)\}, \\

-&\Inum{(ii)}      && \{(12) (34),\ (34) (56),\ (12) (56)\}, \\

-\lintertext{and}

-&\Inum{(iii)}\quad&& \{(12) (34) (56)\}.

-\end{alignat*}

-

-Again, since $\dfrac{G}{H}$~interchanges the three systems, it must be

-simply isomorphic with a group of degree~$3$, and its order is therefore

-either $3$ or~$6$. First, let its order be~$3$. It must then contain a

-substitution of order~$3$ which, without loss of generality, may be

-taken to be $(135)(246)$; this gives, with the three above forms of~$H$,

-three distinct forms for~$G$. The form of~$G$ corresponding to the form~\Inum{(iii)}

-of~$H$ is, however, the same as one of those already determined.

-

-If $\dfrac{G}{H}$~is of order~$6$, $G$~must contain as a self-conjugate sub-group

-one of the three groups just obtained. Also if $\dfrac{G}{H}$~were cyclical,

-there would be a substitution in~$G$, not belonging to~$H$ and

-permutable with $(135) (246)$. This is clearly impossible, and therefore

-$G$~must contain a substitution which transforms $(135) (246)$

-into its inverse. We may take this to be $(13) (24)$ or $(14) (23) (56)$.

-With the form~\Inum{(i)} for~$H$, these two substitutions lead to the same

-group. When $H$~is of the form~\Inum{(ii)}, they give two distinct forms

-for~$G$. When $H$~is of the form~\Inum{(iii)}, $G$~admits the imprimitive

-systems $1$,~$3$,~$5$, and $2$,~$4$,~$6$.

-

-Lastly, if $H$~is the identical operation, $G$~is necessarily of order~$6$;

-no new forms can arise.

-%% -----File: 199.png---Folio 183-------

-

-There are therefore five distinct forms of groups of degree~$6$, in

-which the symbols form three imprimitive systems of two symbols

-each but do not at the same time form two imprimitive systems of

-three symbols each.

-\end{Remark}

-

-\Par{128.} An actual test to determine whether any transitive

-group is primitive or imprimitive may be applied as follows.

-Consider the effect of the substitutions of the group~$G$ on $r$ of

-the symbols which are permuted transitively by it. Those

-substitutions, which permute the $r$~symbols, say

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{r}

-\]

-among themselves, form a sub-group~$H$. Now suppose that

-every substitution, which changes $a_{1}$ into one of the $r$~symbols,

-belongs to~$H$. Then if $S$~is a substitution, which does not

-permute the $r$~symbols among themselves, it must change

-them into a new set

-\[

-b_{1},\ b_{2},\ \Dots,\ b_{r},

-\]

-which has no symbol in common with the previous set; and

-every operation of the set~$HS$ changes all the $a$'s into~$b$'s.

-Moreover, since $G$~is transitive, $H$~must permute the $a$'s

-transitively; and therefore the set~$HS$ must contain substitutions

-changing $a_{1}$ into each one of the~$b$'s.

-

-Suppose now, if possible, that the group contains a substitution~$S'$,

-which changes some of the $a$'s into~$b$'s, and the

-remainder into new symbols. We may assume that $S'$~changes

-$a_{1}$ into~$b_{1}$, and $a_{2}$~into a new symbol~$c_{2}$. Among the set~$HS$

-there is at least one substitution,~$T$, which changes $a_{1}$ into~$b_{1}$.

-Hence $TS'^{-1}$ changes $a_{1}$ into itself and some new symbol into~$a_{2}$.

-This however contradicts the supposition that every

-substitution, which changes $a_{1}$ into one of the set of~$a$'s,

-belongs to~$H$. Hence no substitution such as~$S'$ can belong to~$G$;

-and every substitution, which changes one of the $a$'s into

-one of the~$b$'s, must change all the $a$'s into~$b$'s.

-

-If the substitutions of the group are not thus exhausted,

-there must be another set of $r$~symbols

-\[

-c_{1},\ c_{2},\ \dots,\ c_{r},

-\]

-which are all distinct from the previous sets, such that some

-substitution changes all the $a$'s into~$c$'s. We may now repeat

-%% -----File: 200.png---Folio 184-------

-the previous reasoning to shew that every substitution, which

-changes an $a$ into a~$c$, must change all the $a$'s into~$c$'s. By

-continuing this process, we finally divide the symbols into a

-number of distinct sets of $r$~each, such that every substitution of

-the group must change the~$a$'s either into themselves or into

-some other set: and therefore also must change every set either

-into itself or into some other set. The group must therefore be

-imprimitive. Hence:---

-

-\begin{Theorem}[IV.]

-If, among the symbols permuted by a transitive

-\index{Primitivity@\Topic{Primitivity}, test of}%

-group, it is possible to choose a set such that every

-substitution of the group, which changes a chosen symbol of the

-set either into itself or into another of the set, permutes all the

-symbols of the set among themselves; then the group is imprimitive,

-and the set of symbols forms an imprimitive system.

-\end{Theorem}

-

-\begin{Corollary}[I.]

-If $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ are a part of the symbols

-permuted by a primitive group, there must be substitutions of

-the group, which replace some of this set of symbols by others

-of the set, and the remainder by symbols not belonging to the

-set.

-\end{Corollary}

-

-\begin{Corollary}[II.]

-A sub-group of a primitive group, which

-keeps one symbol unchanged, must contain substitutions which

-displace any other symbol.

-\end{Corollary}

-

-If the sub-group~$H$, that leaves $a_{1}$ unchanged, leaves every

-symbol of the set $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ unchanged, then $H$~must be

-transformed into itself by every substitution which changes any

-one of these symbols into any other. Every substitution, which

-changes one of the set into another, must therefore permute

-the set among themselves; and the group, contrary to supposition,

-is imprimitive.

-

-\Par{129.} It has already been seen that, in particular cases, it

-may be possible to distribute the symbols, which are permuted by

-an imprimitive group, into imprimitive systems in more than

-one way. When this is possible, suppose that two systems

-which contain~$a_{1}$ are

-\begin{align*}

-&a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1},\ \Dots,\ a_{m}, \\

-\lintertext{and}

-&a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1}',\ \Dots\Dots\Dots,\ a_{n}';

-\end{align*}

-%% -----File: 201.png---Folio 185-------

-and that the symbols common to the two systems are

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{r}.

-\]

-

-A substitution of the group, which changes $a_{1}$ into~$a_{r+1}'$,

-must change $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ into $r$~symbols of that system of

-the first set which contains~$a_{r+1}'$, while it changes the system of

-the second set that contains~$a_{1}$ into itself. Hence the latter

-system contains at least $r$~symbols of that system of the first

-set in which $a_{r+1}'$ occurs. By considering the effect of the

-inverse substitution, it is clear that the system

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1}',\ \Dots,\ a_{n}'

-\]

-cannot have more than $r$~symbols in common with the system

-of the first set that contains~$a_{r+1}'$. Hence the $n$~symbols of this

-system can be divided into sets of~$r$, such that each set is

-contained in some system of the first set. It follows that~$n$, and

-therefore also~$m$, must be divisible by~$r$.

-

-Suppose now that $b_{1}$ is any symbol which is not contained

-in either of the above systems. A substitution that changes

-$a_{1}$ into $b_{1}$ must change the two systems into two others, which

-have $r$~symbols

-\[

-b_{1},\ b_{2},\ \Dots,\ b_{r}

-\]

-in common; and since no two systems of either set have a

-common symbol, these $r$~symbols must be distinct from

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{r}.

-\]

-

-Further, from the mode in which the set $b_{1}$, $b_{2}$,\ \Dots, $b_{r}$ has

-been obtained, any operation, which changes one of the symbols

-$a_{1}$, $a_{2}$,~\Dots, $a_{r}$ into one of the symbols $b_{1}$, $b_{2}$,~\Dots, $b_{r}$, must

-change all the symbols of the first set into those of the second.

-Hence the symbols operated on by the group can be divided

-into systems of $r$ each, by taking together the sets of $r$~symbols

-which are common to the various pairs of the two

-given sets of imprimitive systems; and the group is imprimitive

-in regard to this new set of systems of $r$~symbols each.

-Hence:---

-

-\begin{Theorem}[V.]

-If the symbols permuted by a transitive group

-\index{Imprimitive systems@\Topic{Imprimitive systems}!properties of}%

-can be divided into imprimitive systems in two distinct ways, $m$~being

-the number of symbols in each system of one set and $n$~in

-%% -----File: 202.png---Folio 186-------

-each system of the other; and if some system of the first set has

-$r$~symbols in common with some system of the second set; then

-\Inum{(i)}~$r$~is a factor of both $m$ and~$n$, and \Inum{(ii)}~the symbols can be

-divided into a set of systems of $r$~each, in respect of which the

-group is imprimitive.

-\end{Theorem}

-

-This result may also be regarded as an immediate consequence

-of Theorem~III, §~124. For if $H_{1}$~and~$H_{2}$ are two

-sub-groups of~$G$ each of which contains the sub-group~$R$, and

-if $K$~is the greatest common sub-group of $H_{1}$ and~$H_{2}$, then $K$

-contains~$R$. Now if $R$~contains no self-conjugate sub-group

-of~$G$, then $G$~can be represented as a transitive group whose

-degree is the order of~$G$ divided by the order of~$R$. If the

-respective orders of $H_{1}$ and $H_{2}$ are $m$~times and $n$~times the

-order of~$R$, the symbols can be divided in two distinct ways

-into sets of imprimitive systems, the systems containing $m$

-and $n$~symbols respectively. Also, if the order of~$K$ is $r$~times

-the order of~$R$, then $r$~is a factor of~$m$ and of~$n$; by considering

-the sub-group~$K$, the symbols may be divided into a set of

-systems which contain $r$~symbols each.

-

-\begin{Remark}

-It might be expected that, just as we can form a new set of

-imprimitive systems by taking together the symbols which are

-common to pairs of systems of two given sets, so we might form

-another new set of systems by combining all the systems of one set

-which have any symbols in common with a single system of the

-other set. A very cursory consideration will shew however that

-this is not in general the case. In fact, it is sufficient to point out

-that, with the notation already used, the number of symbols in such

-a new system would be~$\dfrac{mn}{r}$; and this number is not necessarily a

-factor of the degree of the group. Also, even if this number is a

-factor of the degree of the group, it will not in general be the case

-\index{Imprimitive systems@\Topic{Imprimitive systems}!properties of}%

-that the symbols so grouped together form an imprimitive system.

-\end{Remark}

-

-\Par{130.} We may now discuss, more fully than was possible in

-§~106, the form of a self-conjugate sub-group of a given transitive

-group. Such a sub-group must clearly contain one or more

-operations displacing every symbol operated on by the group.

-For if every operation of the sub-group keeps the symbol $a_{1}$

-unchanged, then since it is self-conjugate, every operation will

-keep $a_{2}$, $a_{3}$,~\Dots, unchanged: and the sub-group must reduce

-to the identical operation only.

-%% -----File: 203.png---Folio 187-------

-

-Suppose now, if possible, that $H$~is an intransitive self-conjugate

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of a primitive group must be transitive}%

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of an imprimitive group}%

-sub-group of a transitive group~$G$; and that $H$

-permutes the $n$~symbols of~$G$ in the separate transitive sets

-$a_{1}$, $a_{2}$,~\Dots, $a_{n_{1}}$; $b_{1}$, $b_{2}$,~\Dots, $b_{n_{2}}$;~\Dots. If $S$~is any operation

-of~$G$ which changes $a_{1}$ into~$b_{1}$, then, since

-\[

-S^{-1}HS = H,

-\]

-it must change all the $a$'s into~$b$'s; and since

-\[

-SHS^{-1} = H,

-\]

-$S^{-1}$ must change all the $b$'s into~$a$'s. Hence the number of

-symbols in the two sets, and therefore the number of symbols

-in each of the sets, must be the same.

-

-Moreover every operation of~$G$, since it transforms $H$ into

-itself, must either permute the symbols of any set among

-themselves, or it must change them all into the symbols of

-some other set. Hence $G$~must be imprimitive, and $H$~must

-consist of those operations of~$G$ which permute the symbols of

-each imprimitive system among themselves.

-

-Conversely, when $G$~is imprimitive, it is immediately obvious

-that those operations of~$G$, if any such exist, which permute the

-symbols of each of a set of imprimitive systems among themselves,

-form a self-conjugate sub-group. Hence:---

-

-\begin{Theorem}[VI.]

-A self-conjugate sub-group of a primitive

-group must be transitive; and if an imprimitive group has an

-intransitive self-conjugate sub-group, it must consist of the

-operations which permute among themselves the symbols of each

-of a set of imprimitive systems.

-\end{Theorem}

-

-If $G$~is an imprimitive group of degree~$mn$, and if there are

-$n$ imprimitive systems of $m$~symbols each, then we have seen

-in §~123 that $G$~is isomorphic with a group~$G'$ of degree~$n$. In

-particular instances, it may at once be evident, from the order of~$G$,

-that this isomorphism cannot be simple. For example, if

-the order of~$G$ has a factor which does not divide~$n!$, this is

-certainly the case: and more generally, if it is known independently

-that $G$~cannot be expressed as a transitive group of

-degree~$n$, then $G$~must certainly be multiply isomorphic with~$G'$.

-In such instances the self-conjugate sub-group of~$G$, which

-%% -----File: 204.png---Folio 188-------

-corresponds to the identical operation of~$G'$, is that intransitive

-self-conjugate sub-group, which interchanges among themselves

-the symbols of each imprimitive system.

-

-If $G$~is soluble, a minimum self-conjugate sub-group of~$G$

-must have for its order a power of a prime. Also, if $G$~has an

-intransitive self-conjugate sub-group, it must have an intransitive

-minimum self-conjugate sub-group. Hence if $G$~is soluble

-and has intransitive self-conjugate sub-groups, the symbols

-permuted by~$G$ must be capable of division into imprimitive

-systems, such that the number in each system is the power of

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of an imprimitive group}%

-a prime.

-

-\Par{131.} Let $G$ be a $k$-ply transitive group of degree~$n$ ($k > 2$),

-and let $G_{r}$~be that sub-group of~$G$ which keeps $r$ ($< k$) given

-symbols unchanged, so that $G_{r}$~is $(k - r)$-ply transitive in the

-remaining $n - r$ symbols. Also, let $H$ be a self-conjugate

-sub-group of~$G$, and let $H_{r}$ be that sub-group of~$H$ which keeps

-the same $r$~symbols unchanged; so that $H_{r}$ is the common

-sub-group of $H$ and~$G_{r}$. Since every operation of~$G_{r}$ transforms

-both $H$ and $G_{r}$ into themselves, every operation of~$G_{r}$ must be

-permutable with~$H_{r}$; \ie\ $H_{r}$~is a self-conjugate sub-group of~$G_{r}$.

-Now, if $r = k - 2$, $G_{k - 2}$~is doubly transitive in the $n - k + 2$

-symbols on which it operates; it is therefore primitive. Hence,

-unless $H_{k - 2}$ consists of the identical operation only, it must be

-transitive in the $n - k + 2$ symbols.

-

-If $H_{k - 2}$ is the identical operation, $H$~contains no operation,

-except identity, which displaces less than $n - k + 3$ symbols.

-Suppose, first, that $H$~contains operations, other than identity,

-which leave one or more symbols unchanged. Then, since $H$~is

-a self-conjugate sub-group and $G$~is $k$-ply transitive, it may

-be shewn, exactly as in §~110, that $H$~must contain operations

-displacing not more than $2k - 2$ symbols. Hence $H_{k - 2}$ can

-consist of the identical operation alone, only if

-\[

-n - k + 3 \ngtr 2k - 2,

-\]

-or

-\[

-k \nless \tfrac{1}{3}n + \tfrac{5}{3}.

-\]

-

-When this inequality holds, we have seen (\PageRef{p.}{152}) that $G$~contains

-the alternating group. Hence in this case, if $G$~does

-not contain the alternating group, it follows that $H_{k - 2}$~is transitive

-in the $n - k + 2$ symbols on which it operates.

-%% -----File: 205.png---Folio 189-------

-

-Since $H$~is self-conjugate and $G$~is $k$-ply transitive, $H$~must

-\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of a $k$-ply transitive group is, in general, $(k - 1)$-ply transitive}%

-contain a sub-group conjugate to~$H_{k - 2}$ which keeps any other

-$k - 2$ symbols unchanged. Hence $H_{k - 3}$~must be doubly transitive

-in the $n - k + 3$ symbols on which it operates; and so on.

-Finally, if $G$~is not the symmetric group (the alternating group,

-being simple, contains no self-conjugate sub-group) $H$~must be

-$(k - 1)$-ply transitive.

-

-Suppose, next, that $H$~contains no operation, except identity,

-which leaves any symbol unchanged. Then if, with the notation

-of §~110, $j = k - 1$ for every operation of~$H$, the argument there

-used does not apply. For it is impossible to choose the operation~$T$

-so that $c_{k}$~is a symbol which does not occur in~$S$.

-

-The self-conjugate sub-group~$H$ contains a single operation

-changing a given symbol~$a_{1}$ into any other symbol~$a_{r}$. Also $G$~contains

-operations which leave $a_{1}$ unchanged and change $a_{r}$

-into any other symbol~$a_{s}$. Hence the operations of~$H$, other

-than identity, form a single conjugate set in~$G$; and therefore

-$H$~must be an Abelian group of order~$p^{m}$ and type $(1, 1, \Dots,

-\text{to~$m$ units})$; $p$~being a prime. Further, since $G$~is by supposition

-at least triply transitive, it must contain operations which

-transform any two operations of~$H$, other than identity, into

-any other two. If $p$~were an odd prime, and $P_{1}$~and~$P_{2}$ were

-two of the generating operations of~$H$, it follows that $G$~would

-have an operation~$S$ such that

-\[

-S^{-1} P_{1} S = P_{1}, \quad S^{-1} P_{2} S = P_{1}^{\alpha};

-\]

-and this is impossible. Hence $p$~must be~$2$. Further if $G$~were

-more than triply transitive, and if $A$,~$B$,~$C$ were three independent

-generating operations of~$H$, then $G$~would have an operation~$\Sigma$

-such that

-\[

-\Sigma^{-1} A\Sigma = A, \quad

-\Sigma^{-1} B\Sigma = B, \quad

-\Sigma^{-1} C\Sigma = AB.

-\]

-This again is impossible, and therefore $k$~must be~$3$. Hence:---

-

-\begin{Theorem}[VII.]

-A self-conjugate sub-group of a $k$-ply

-transitive group of degree~$n$ \($2 < k < n$\), is in general at least

-$(k - 1)$-ply transitive\footnotemark.

-\index{Jordan@\Topic{Jordan}, quoted}%

-\footnotetext{Jordan, \textit{Traité des Substitutions}, p.~65; where, however, the exceptional

-  case is overlooked.}%

-The only exception is that a triply

-transitive group of degree~$2^{m}$ may have a self-conjugate sub-group

-of order~$2^{m}$.

-\end{Theorem}

-%% -----File: 206.png---Folio 190-------

-

-\Par{132.} For the further discussion of the self-conjugate

-\index{Direct product@\Topic{Direct product} of two groups!represented as a transitive group}%

-sub-groups of a primitive group, it is necessary to consider in

-what forms the direct product of two groups can be represented

-as a transitive group.

-

-Let $G$~be the direct product of two groups $H_{1}$ and~$H_{2}$, and

-suppose that $G$~can be represented as a transitive group

-of degree~$n$. When $G$ is thus represented, we will suppose

-that $H_{1}$~is transitive in the $n$~symbols that $G$~permutes. We

-have seen in §~107 that every substitution of $n$~symbols, which

-is permutable with each of the substitutions of a group

-transitive in the $n$~symbols, must displace all the $n$~symbols.

-It follows that every substitution of~$H_{2}$ must displace all the $n$~symbols

-on which $G$~operates; and that the order of~$H_{2}$ is

-equal to or is a factor of~$n$.

-

-If the order of~$H_{2}$ is equal to~$n$, then $H_{2}$~is transitive in the

-$n$~symbols, so that the order of~$H_{1}$ cannot be greater than~$n$.

-In this case, $H_{1}$~and~$H_{2}$ must (§~107) be two simply isomorphic

-groups of order~$n$, which have no self-conjugate operations

-except identity. Further, if $H_{1}$~and $H_{2}$ in this case are not

-simple groups, let $K$ be a self-conjugate sub-group of~$H_{1}$.

-Since every operation of~$H_{2}$ is permutable with every operation

-of~$H_{1}$, $K$~is a self-conjugate sub-group of~$G$. Now the order of~$K$

-is less than~$n$, the degree of~$G$; therefore $K$~is intransitive

-and $G$~is imprimitive. On the other hand, we have seen

-(\textit{loc.~cit.})\ that, if $H_{1}$~and $H_{2}$ are simple, the sub-group of~$G$

-that keeps one symbol fixed is a maximum sub-group: and

-therefore $G$~is primitive. Hence:---

-

-\begin{Theorem}[VIII.]

-If the direct product~$\Gamma$ of $H_{1}$ and $H_{2}$ can

-be represented as a transitive group of degree~$n$, in such a way

-that $H_{1}$~and $H_{2}$ are transitive sub-groups of~$\Gamma$, then $H_{1}$~and $H_{2}$

-must be simply isomorphic groups of order~$n$, which have no

-self-conjugate operations except identity. When this condition

-is satisfied, $\Gamma$~will be primitive if, and only if, $H_{1}$~and $H_{2}$ are

-simple.

-\end{Theorem}

-

-\Par{133.} Suppose now that a primitive group~$G$, of degree~$n$,

-has two distinct minimum self-conjugate sub-groups $H_{1}$ and~$H_{2}$.

-Then every operation of~$H_{1}$ (or~$H_{2}$) is permutable with~$H_{2}$

-%% -----File: 207.png---Folio 191-------

-(or~$H_{1}$), and $H_{1}$,~$H_{2}$ have no common operation except

-identity. Hence (§~34) the group $\{H_{1}, H_{2}\}$, which we will call~$\Gamma$,

-is the direct product of $H_{1}$ and~$H_{2}$. Now $\Gamma$~is a self-conjugate

-sub-group of~$G$: it is therefore transitive in

-the $n$~symbols which $G$~permutes. Also $H_{1}$ and~$H_{2}$, being

-self-conjugate sub-groups of~$G$, are transitive. Hence, by

-Theorem~VIII (§~132), $H_{1}$~and $H_{2}$ are simply isomorphic, and $n$~is

-equal to the order of~$H_{1}$. Moreover, since $H_{1}$~is a minimum

-self-conjugate sub-group of~$G$ which contains no self-conjugate

-operations except identity, it must (Theorem~IV, §~94) be either

-a simple group of composite order, or the direct product of

-several simply isomorphic simple groups of composite order. It

-follows that $G$~cannot have two distinct minimum self-conjugate

-sub-groups unless the degree of~$G$ is equal to or is a power

-of the order of some simple group of composite order.

-

-\Par{134.} Let now $\Gamma$ be a minimum self-conjugate sub-group

-of a doubly transitive group~$G$, and suppose that $\Gamma$~is the direct

-product of the $\alpha$~simply isomorphic simple groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{\alpha}.

-\]

-

-Since $G$~is primitive, $\Gamma$~is transitive. If $H_{1}$~is a cyclical

-group of prime order~$p$, the order of~$\Gamma$ is~$p^{\alpha}$; therefore the

-degree of~$\Gamma$, or what is the same thing, the degree of~$G$, is~$p^{\alpha}$.

-

-If $H_{1}$~is a simple group of composite order, and if $\alpha > 2$,

-then (§~132) $H_{1}$~cannot be transitive. The intransitive systems

-of~$H_{1}$, since they form a set of imprimitive systems for~$\Gamma$, must

-each contain the same number $m$ of symbols. If $m$~is less than

-the order of~$H_{1}$, a sub-group of~$H_{1}$ which leaves unchanged one

-symbol of one intransitive system will leave unchanged one

-symbol of each intransitive system. Now we shall see, in §~136,

-that the operations of an imprimitive self-conjugate sub-group

-of a doubly transitive group must displace all or all but one of

-the symbols. Hence $m$~cannot be less than the order of~$H_{1}$.

-We may similarly shew that, if $m$~is equal to the order of~$H_{1}$,

-some of the operations of~$\Gamma$ must keep more than one symbol

-fixed; and therefore, if $\alpha > 2$, the group assumed cannot exist.

-If $\alpha = 2$, $H_{1}$~may be transitive. But in this case $\{H_{1}, H_{2}\}$

-certainly contains operations which leave more than one symbol

-unchanged; and again the group assumed cannot exist. Hence

-%% -----File: 208.png---Folio 192-------

-finally no doubly transitive group can contain a minimum self-conjugate

-sub-group of the type assumed.

-

-No general law can be stated regarding self-conjugate

-sub-groups of simply transitive primitive groups; but for

-groups which are at least doubly transitive the preceding

-results may be summed up as follows:---

-

-\begin{Theorem}[IX.]

-A group~$G$ which is at least doubly transitive

-\index{Doubly transitive groups@\Topic{Doubly transitive groups}!generally contain simple self-conjugate sub-groups}%

-\index{Primitive groups@\Topic{Primitive groups}!when soluble, have a power of a prime for degree}%

-either must be simple or must contain a simple group~$H$ as a

-self-conjugate sub-group. In the latter case no operation of~$G$,

-except identity, is permutable with every operation of~$H$. The

-only exceptions to this statement are that a triply transitive group

-of degree~$2^{m}$ may have a self-conjugate sub-group of order~$2^{m}$;

-and that a doubly transitive group of degree~$p^{m}$, where $p$~is a

-prime, may have a self-conjugate sub-group of order~$p^{m}$.

-\end{Theorem}

-

-\begin{Corollary}

-If a primitive group is soluble, its degree

-must be the power of a prime\footnotemark.

-\end{Corollary}

-\index{Galois@\Topic{Galois}, quoted}%

-\footnotetext{This result, stated in a somewhat different form, is given, among many

-\PageLabel{192}%

-  others, in the letter written by Galois to his friend Chevalier on the evening

-  of May~29th, 1832, the day before the duel in which he was killed. The letter

-  was first printed in the \textit{Revue Encyclopédique} (1832), p.~568; it was reprinted

-  in the collection of Galois's mathematical writings in \textit{Liouville's Journal}, t.~\VolNo{XI}

-  (1846), pp.~381--444}

-%[** TN: See http://www.gutenberg.org/ebooks/40213]

-

-In fact, if a group is soluble, so also is its minimum

-self-conjugate sub-group. The latter must be therefore an

-Abelian group of order~$p^{\alpha}$: and since this group must be

-transitive, its order is equal to the degree of the primitive

-group.

-

-\begin{Remark}

-\Par{135.} As illustrating the occurrence of an imprimitive self-conjugate

-sub-group in a primitive group, we will construct a

-primitive group of degree~$36$ which has an imprimitive self-conjugate

-sub-group. For this purpose, let\footnote

-  {The commas in the symbols for the substitutions are here used to prevent

-  confusion among the one-digit and two-digit numbers.}

-\begin{multline*}

-S = (1, 2, 3)(4, 5, 6)(7, 8, 9)(10, 11, 12)(13, 14, 15) \\

-(16, 17, 18)(19, 20, 21)(22, 23, 24)(25, 26, 27) \\

-(28, 29, 30)(31, 32, 33)(34, 35, 36),

-\end{multline*}

-and

-\begin{multline*}

-A = (3, 4)(5, 6)(9, 10)(11, 12)(15, 16)(17, 18) \\

-(21, 22)(23, 24)(27, 28)(29, 30)(33, 34)(35, 36);

-\end{multline*}

-so that $\{S, A\}$ is an intransitive group of degree~$36$, the symbols

-being interchanged in $6$~transitive systems containing $6$~symbols

-each. This group is simply isomorphic with

-\[

-\{(123) (456),\ (34) (56)\};

-\]

-%% -----File: 209.png---Folio 193-------

-and it may be easily verified that this group is simply isomorphic with

-the alternating group of $5$~symbols, the order of which is~$60$.

-

-Also let

-\begin{multline*}

-J = (2, 7) (3, 13) (4, 19) (5, 25) (6, 31) (9, 14) (10, 20) (11, 26) \\

-(12, 32) (16, 21) (17, 27) (18, 33) (23, 28) (24, 34) (30, 35).

-\end{multline*}

-

-Then

-\begin{multline*}

-JSJ = (1, 7, 13) (19, 25, 31) (2, 8, 14) (20, 26, 32) \\

-  (3,  9, 15) (21, 27, 33) (4, 10, 16) (22, 28, 34) \\

-  (5, 11, 17) (23, 29, 35) (6, 12, 18) (24, 30, 36),

-\end{multline*}

-and

-\begin{multline*}

-JAJ = (13, 19) (25, 31) (14, 20) (26, 32) (15, 21) (27, 33) \\

-  (16, 22) (28, 34) (17, 23) (29, 35) (18, 24) (30, 36);

-\end{multline*}

-and $\{JSJ, JAJ\}$ is similar to and simply isomorphic with~$\{S, A\}$.

-

-Now it may be directly verified that $S$~and~$A$ are, each of them,

-permutable with $JSJ$ and~$JAJ$; and therefore every operation of

-the group $\{S, A\}$ is permutable with every operation of the group

-$\{JSJ, JAJ\}$. Also these two groups can have no common operation,

-since the symbols into which $\{S, A\}$ changes any given symbol are

-all distinct from those into which $\{JSJ, JAJ\}$ change it. Hence

-$\{S, A, JSJ, JAJ\}$ is the direct product of $\{S, A\}$ and~$\{JSJ, JAJ\}$;

-it is therefore a group of order~$3600$. It is also, from its mode of

-formation, a transitive group of degree~$36$; and it interchanges the

-symbols in two and only two distinct sets of imprimitive systems, of

-which

-\[

-1,\ 2,\ 3,\ 4,\ 5,\ 6 \quad\text{and}\quad 1,\ 7,\ 13,\ 19,\ 25,\ 31

-\]

-may be taken as representatives.

-

-Now $J$~does not interchange the $36$~symbols in either of these

-systems, and therefore it cannot occur in $\{S, A, JSJ, JAJ\}$.

-Further

-\[

-J\{S, A, JSJ, JAJ\}J = \{S, A, JSJ, JAJ\};

-\]

-and therefore $\{J, S, A\}$ is a transitive group of degree~$36$ and order~$7200$.

-Also, since $J$~does not interchange the symbols in either of

-the two sets of imprimitive systems of $\{S, A, JSJ, JAJ\}$, it follows

-that $\{J, S, A\}$ is primitive.

-

-\Par{136.} We have seen that, with a single exception, a $k$-ply

-\index{Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group}%

-transitive group ($k > 2$) cannot have an imprimitive self-conjugate

-sub-group; while the example in the preceding section illustrates

-the occurrence of an imprimitive self-conjugate sub-group in a simply

-transitive primitive group. We will now consider, from a rather

-different point of view, the possibility of an imprimitive self-conjugate

-sub-group in a doubly transitive group. Let $G$~be a

-doubly transitive group of degree~$mn$, and let $H$~be an imprimitive

-self-conjugate sub-group of~$G$. Suppose that $m$~is the smallest

-number, other than unity, of symbols which occur in an imprimitive

-system; and let

-\[

-a_{1},\ a_{2},\ \Dots,\ a_{m}

-\]

-%% -----File: 210.png---Folio 194-------

-form an imprimitive system. Since $G$~is doubly transitive, it must

-contain a substitution~$S$, which leaves $a_{1}$ unchanged and changes $a_{2}$

-into~$a_{m + 1}$, a symbol not contained in the given set. If $S$~changes

-the given set into

-\[

-a_{1},\ a_{m + 1},\ \Dots,\ a_{2m - 1},

-\]

-then, since

-\[

-S^{-1}HS = H,

-\]

-this new set must form an imprimitive system for~$H$. Also, since

-$m$~is the smallest number of symbols that can occur in an imprimitive

-system, the two sets have no symbol in common except~$a_{1}$.

-

-Now $a_{m + 1}$ may be any symbol not contained in the original

-system. Hence it must be possible to distribute the $mn$~symbols

-into sets of imprimitive systems of $m$~each, such that every pair of

-symbols occurs in one system and no pair in more than one system.

-This implies that $mn - 1$ is divisible by $m - 1$, or that $n - 1$ is

-divisible by~$m - 1$.

-

-Consider now a substitution of~$H$ which leaves $a_{1}$ unchanged.

-It must permute among themselves the remaining $m - 1$ symbols

-of each of the $\dfrac{mn - 1}{m - 1}$ systems in which $a_{1}$~occurs. If $a_{r}$~is any

-other symbol, a similar statement applies to it. Now no two

-systems have more than one symbol in common. Hence every

-substitution, which leaves both $a_{1}$ and $a_{r}$ unchanged, must leave all

-the symbols unchanged. The sub-group~$H$ is therefore such that

-each of its substitutions displaces all or all but one of the symbols.

-Moreover the substitutions, which leave $a_{1}$ unchanged, permute

-among themselves the remaining symbols of each system in which

-$a_{1}$~occurs; therefore the order of~$H$ must be~$mn\mu$, where $\mu$~is a

-factor of~$m - 1$.

-

-It has been seen in §~112 that, for certain values of~$mn$, groups

-satisfying these conditions actually exist. The doubly transitive

-groups, of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$ there obtained, have

-\index{Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group}%

-imprimitive self-conjugate sub-groups of orders $p^{m}(p - 1)$ and~$p^{m}$.

-\end{Remark}

-

-\Par{137.} \begin{Ex}[1.] Shew that a transitive group of order~$N$ and

-degree~$2n$, which is imprimitive in respect of two systems of $n$~symbols

-each, must have a self-conjugate sub-group of order~$\frac{1}{2}N$.

-\end{Ex}

-

-\begin{Ex}[2.] Shew that an imprimitive group of order~$Np^{2}$ and

-degree~$np$, where $p$~is an odd prime, which has $p$~imprimitive

-systems of $n$~symbols each, must contain a self-conjugate sub-group

-whose order is a multiple of~$p$.

-\end{Ex}

-

-\begin{Ex}[3.] Shew that, if $n$~is the smallest number of symbols in

-which a group~$G$ can be represented as a transitive group, then $n^{\alpha}$~is

-the smallest number of symbols in which the direct product of $\alpha$

-groups, simply isomorphic with~$G$, can be represented as a transitive

-group.

-\end{Ex}

-%% -----File: 211.png---Folio 195-------

-

-\begin{Ex}[4.] Prove that if a group~$G$, of order~$N$, represented as a

-transitive substitution group, is imprimitive only when its degree is~$N$;

-then either $G$~is Abelian, or $N$~is the product of two distinct

-primes.

-\index{Dyck@\Topic{Dyck}, quoted}%

-\Attrib{Dyck.}

-\end{Ex}

-

-\begin{Ex}[5.] Shew that if, in the tableau of \PageRef{p.}{174}, the sub-group

-\[

-1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu}

-\]

-is a self-conjugate sub-group~$H$ of the given group~$G$, then, in each

-square compartment of the tableau, every horizontal line contains

-the same $\mu$~symbols.

-

-Shew also that, if each square compartment of the tableau is

-regarded as a single symbol, the permutations of these symbols

-given by the tableau represents the group~$\dfrac{G}{H}$ in regular form.

-\Attrib{Dyck.}

-\end{Ex}

-%% -----File: 212.png---Folio 196-------

-

-

-\Chapter{X.}{On Substitution Groups: Transitivity and

-Primitivity: (Concluding Properties.)}

-

-\Par{138.} \First{The} determination of all distinct transitive sub-groups

-of the symmetric group of $n$ symbols has been carried

-out for values of~$n$ up to~$12$\footnotemark.

-\index{Cole@\Topic{Cole}, quoted}%

-\index{Jordan@\Topic{Jordan}, quoted}%

-\index{Miller@\Topic{Miller}, quoted}%

-\footnotetext{Jordan, \textit{Comptes Rendus}, t.~\VolNo{LXXIII} (1871), pp.~853--857; \textit{ib.}\ t.~\VolNo{LXXV}

-  (1872), pp.~1754--1757.

-

-  And see the \textit{Bulletin of the New York Mathematical Society}, Cole, 1st~series,

-  Vol.~\VolNo{II}, pp.~184--190; 250--258: Miller, 1st~series, Vol.~\VolNo{III}, pp.~168,~169; 242--245:

-  2nd series, Vol.~\VolNo{I}, pp.~67--72; 255--258: Vol.~\VolNo{II}, pp.~138--145. \textit{Quarterly

-  Journal of Mathematics}, Cole, Vol.~\VolNo{XXVI}, pp.~372--388: Vol.~\VolNo{XXVII}, pp.~35--50:

-  Miller, Vol.~\VolNo{XXVII}, pp.~99--118; Vol.~\VolNo{XXVIII}, pp.~193--231. These memoirs

-  were all published between 1892 and~1896; in them further references will be

-  found.}

-This investigation, if carried out

-for sufficiently great values of~$n$, would involve the expression

-of all types of group in all possible transitive forms.

-

-From the point of view of one of the chief problems of pure

-group-theory, namely, the determination of all distinct types

-of group of a given order, this analysis of the symmetric group

-of $n$~symbols is not a succinct process, as it continually involves

-the redetermination of groups which have been already obtained.

-Thus a simple group of degree~$mn$, in which the symbols

-are permuted in $m$~imprimitive systems of $n$~each, would in

-this analysis have been already obtained as a group of degree~$m$.

-With reference then to the more restricted problem of determining

-all types of simple groups, it would certainly be

-sufficient to find all primitive sub-groups of the symmetric

-group.

-

-\Par{139.} We shall proceed to determine a superior limit to the

-order of a primitive group of degree~$n$, other than the alternating

-or the symmetric group.

-%% -----File: 213.png---Folio 197-------

-

-Let $G$~be a primitive group of degree~$n$, and suppose that $G$~contains

-\index{Primitive groups@\Topic{Primitive groups}!with a transitive sub-group of smaller degree}%

-a sub-group~$H$ which leaves $n - m$ symbols unchanged

-and is transitive in the remaining~$m$. Since $G$~is primitive, $H$

-and the sub-groups conjugate to it must generate a transitive

-self-conjugate sub-group of~$G$; and therefore there must be

-some sub-group~$H'$, conjugate to~$H$, such that the $m$~symbols

-operated on by~$H$ and the $m$ operated on by~$H'$ are not all

-distinct. Suppose $H'$ is chosen so that these two sets of $m$~symbols

-have as great a number in common as possible, say~$s$;

-and represent by

-\begin{alignat*}{2}

-&\alpha_{1},\ \alpha_{2},\ \Dots,\ \alpha_{r},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s}, \\

-\lintertext{and}

-&\beta_{1},\ \beta_{2},\ \Dots,\ \beta_{r},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s},

-\end{alignat*}

-where $r + s = m$, the symbols operated on by $H$~and~$H'$ respectively.

-Then $\{H, H'\}$ is a transitive group in the $2r + s$ symbols

-$\alpha$,~$\beta$, and~$\gamma$, which leaves unaltered all the remaining symbols

-of~$G$.

-

-If $S$~is an operation of~$H$ which changes $\alpha_{2}$ into~$\alpha_{1}$, $S^{-1}H'S$

-does not affect~$\alpha_{1}$. Hence, unless $S$ interchanges the $\alpha$'s among

-themselves, the $m$~symbols operated on by~$H'$ and the $m$

-operated on by~$S^{-1}H'S$ will have more than~$s$ in common.

-Every operation of~$H$ which changes one~$\alpha$ into another must

-therefore interchange all the $\alpha$'s among themselves; hence $H$~must

-be imprimitive.

-

-If then $H$~is primitive, $s$~must be equal to~$m - 1$. In any

-case, if $s = m - 1$, $\{H, H'\}$~is a doubly transitive and therefore

-primitive group of degree~$m + 1$, which leaves the remaining

-$n - m - 1$ symbols of~$G$ unchanged. We may reason about this

-sub-group as we have done about~$H$. Among the sub-groups

-conjugate to~$\{H, H'\}$, there must be one at least which operates

-on $m$ of the symbols displaced by~$\{H, H'\}$. This, with $\{H, H'\}$,

-generates a triply transitive group of degree~$m + 2$, which

-leaves $n - m - 2$ symbols unchanged. Proceeding thus, we find

-finally that $G$~itself must be $(n - m + 1)$-ply transitive.

-

-\Par{140.} If $s$ is less than $m - 1$, we may again deal with the

-sub-group $\{H, H'\}$, or~$H_{1}$, exactly as we have dealt with~$H$. It

-is a transitive group of degree~$m_{1}$ ($> m$), which leaves $n - m_{1}$

-symbols unchanged. If, among the sub-groups conjugate to~$H_{1}$,

-%% -----File: 214.png---Folio 198-------

-none operates on more than $s_{1}$ of the symbols affected by~$H_{1}$,

-and if $H_{1}'$~is a suitably chosen conjugate sub-group, then

-$\{H_{1}, H_{1}'\}$ is a transitive group of degree $2m_{1} - s_{1}$, which leaves

-$n - 2m_{1} + s_{1}$ symbols unchanged. Continuing this process, we

-must, before arriving at a group of degree~$n$, reach a stage at

-which the number~$s_{r}$ is equal to~$m_{r} - 1$.

-

-For suppose, if possible, that among the groups conjugate

-to~$K$, of degree~$\rho + \sigma$, none displaces more than $\sigma$ of the

-symbols acted on by~$K$, while at the same time $2\rho + \sigma = n$. If

-\begin{alignat*}{2}

-&\alpha_{1},\ \alpha_{2},\ \Dots,\ \alpha_{\rho},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{\sigma}, \\

-\lintertext{and}

-&\beta_{1},\ \beta_{2},\ \Dots,\ \beta_{\rho},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{\sigma}

-\end{alignat*}

-are the symbols affected by $K$ and $K'$ respectively, then since $G$~is

-primitive, it must contain an operation~$S$ which changes $\alpha_{1}$ into~$\alpha_{2}$

-without at the same time changing all the $\alpha$'s into~$\alpha$'s. If

-then we transform~$K'$ by~$S$, the two groups $K'$ and $S^{-1}K'S$ must

-operate on more than $\sigma$ common symbols, contrary to supposition.

-

-Hence $G$ must in this case certainly contain a transitive

-sub-group of degree~$n - 1$, and therefore is itself at least

-doubly transitive\footnotemark.

-\index{Jordan@\Topic{Jordan}, quoted}%

-\index{Netto@\Topic{Netto}, quoted}%

-\index{Marggraff@\Topic{Marggraff}, quoted}%

-\footnotetext{The results contained in §§~139,~140 are due to Jordan (\textit{Liouville's Journal},

-  Vol.~\VolNo{XVI}, 1871) and Netto (\textit{Crelle's Journal}, Vol.~\VolNo{CIII}, 1889). They have been

-  extended by Marggraff: ``Ueber primitiven Gruppen mit transitiven Untergruppen

-  geringeren Grades,'' (\textit{Inaugural Dissertation}, Giessen, 1892). With the

-  notation used in the text, Marggraff shews that, unless the symbols affected by~$H$

-  can be divided into imprimitive systems of $r$~symbols each, in at least $r + 1$

-  distinct ways, $G$~will be $(n - m + 1)$-ply transitive. In particular, if $H$~is a

-  cyclical group of degree~$m$, $G$~is $(n - m + 1)$-ply transitive. He also shews that

-  in any case $m \geq \frac{1}{2}n$.}%

-

-\Par{141.} Returning to the case in which $H$~is primitive and $G$

-\index{Primitive groups@\Topic{Primitive groups}!with a transitive sub-group of smaller degree}%

-therefore $(n - m + 1)$-ply transitive, we at once obtain an

-inferior limit for~$m$. We have seen, in fact, in Theorem~IV,

-§~110, that a group of degree~$n$, other than the alternating or

-the symmetric group, cannot be more than $(\frac{1}{3}n + 1)$-ply transitive.

-Hence

-\[

-n - m + 1 \ngtr \tfrac{1}{3}n + 1,

-\]

-or

-\[

-m \nless \tfrac{2}{3}n.

-\]

-We may sum up these results as follows:---

-%% -----File: 215.png---Folio 199-------

-

-\begin{Theorem}[I.]

-A primitive group~$G$ of degree~$n$, which has a

-\index{Primitive groups@\Topic{Primitive groups}!limit to the order of, for a given degree}%

-sub-group~$H$ that keeps $n - m$ symbols unchanged and is transitive

-in the remaining $m$~symbols, is at least doubly transitive. If

-$H$~is primitive and $G$~does not contain the alternating group, $m$~cannot

-be less than~$\frac{2}{3}n$, and $G$~is $(n - m + 1)$-ply transitive.

-\end{Theorem}

-

-\begin{Corollary}

-The order of a primitive group of degree~$n$

-cannot exceed $\dfrac{n!}{2 · 3 \Dots p}$, where $2$, $3$,~\Dots\ $p$ are the distinct

-primes which are less than~$\frac{2}{3}n$.

-\end{Corollary}

-

-If $q^{\alpha}$~is the highest power of a prime~$q$ that divides~$n!$, the

-sub-groups of order~$q^{\alpha}$ of the symmetric group form a single

-conjugate set, and each of them must contain circular substitutions

-of order~$q$. Hence if $q < \frac{2}{3}n$, it follows by the theorem

-that no primitive group of degree~$n$, other than the alternating

-or the symmetric group, can contain a sub-group of order~$q^{\alpha}$;

-and therefore $q^{\alpha - 1}$ is the highest power of~$q$ that can divide

-the order of the group.

-

-\Par{142.} The ratio of $2 · 3 \Dots p$ to~$n$ increases rapidly as $n$~increases,

-and it is at once obvious that, when $n > 7$, this ratio is

-greater than unity; hence for values of~$n$ greater than~$7$, the

-symmetric group can have no primitive sub-group of order~$(n - 1)!$.

-

-The order of the greatest imprimitive sub-group of the

-symmetric group is $a!\left(\dfrac{n}{a}!\right)^{\alpha}$, where $a$~is the smallest factor of~$n$.

-When $n > 4$, this is less than~$(n - 1)!$.

-

-The order of the greatest intransitive sub-group of the

-symmetric group, other than the sub-groups that keep one

-symbol fixed, is $2!(n - 2)!$. This is always less than $(n - 1)!$.

-

-Hence when $n > 7$, the only sub-groups of order~$(n - 1)!$ of

-the symmetric group are the sub-groups which each keep one

-symbol fixed; and these form a conjugate set of $n$~sub-groups.

-

-When $n = 7$, a sub-group of order~$(n - 1)!$ must be intransitive,

-and therefore the same result holds in this case; this also

-is true when $n$~is $3$,~$4$, or~$5$.

-%% -----File: 216.png---Folio 200-------

-

-Lastly, when $n = 6$, there may, by the foregoing theorem,

-\index{Symmetric group@\Topic{Symmetric group}!of degree~$n$ has a single set of conjugate sub-groups of order~$\Factorial{(n - 1)}$ except when $n$ is~$6$}%

-\index{Symmetric group@\Topic{Symmetric group}!of degree~$6$ has $12$~simply isomorphic sub-groups of order~$\Factorial{5}$ which form two distinct conjugate sets}%

-be primitive sub-groups of order~$5!$. That such sub-groups

-actually exist may be verified at once by considering the

-symmetric group of $5$~symbols. This group contains $6$~cyclical

-sub-groups of order~$5$, and each of them is self-conjugate in a

-sub-group of order~$20$. Hence, since the only self-conjugate

-sub-group contained in the symmetric group of degree~$5$ is the

-corresponding alternating group, the symmetric group of degree~$5$

-can be expressed as a doubly transitive group of degree~$6$.

-The symmetric group of degree~$6$ therefore contains a set of $6$

-conjugate doubly transitive sub-groups of order~$5!$, which are

-simply isomorphic with the intransitive sub-groups that each

-keep one symbol fixed. Finally, if the $12$~sub-groups of order~$5!$,

-which are thus accounted for, do not exhaust all the sub-groups

-of this order, any other would have in common with

-each of the~$12$ a sub-group of order~$20$; and therefore the

-operations of order~$3$ contained in it would be distinct from

-those in the previous~$12$. But these clearly contain all the

-operations of order~$3$ of the symmetric group, and therefore

-there can be no other sub-groups of order~$5!$. Hence:---

-

-\begin{Theorem}[II.]

-The symmetric group of degree~$n$ \($n \neq 6$\)

-contains $n$~and only~$n$ sub-groups of order~$(n - 1)!$, which form a

-single conjugate set. The symmetric group of degree~$6$ contains

-$12$~sub-groups of order~$5!$, which are simply isomorphic with one

-another and form two conjugate sets of $6$ each.

-\end{Theorem}

-

-\Par{143.} We shall now discuss certain further limitations on

-the order of a primitive group of given degree. Though it will

-be seen that these do not lead to general results, similar to that

-given by Theorem~I, §~141, yet in many special cases they are

-of considerable assistance in determining the possible existence

-of groups of given orders and degrees.

-

-We consider first a group~$G$ of order~$N$ and of prime degree~$p$.

-If $G$~is not cyclical, it must contain substitutions which keep

-only one symbol unchanged. For let $P$ be a substitution of~$G$ of

-order~$p$. The only substitutions permutable with~$P$ are its own

-powers (§~107); and the only substitutions permutable with~$\{P\}$

-are substitutions which keep one symbol unchanged and are regular

-%% -----File: 217.png---Folio 201-------

-in the remaining $p - 1$ symbols (§~112)\footnotemark.

-\footnotetext{It is shewn in §~112 that $\{P\}$~is permutable with a circular substitution of

-\PageLabel{201}%

-  $p - 1$ symbols, which leaves one symbol~$a_{1}$ unchanged. If there are other substitutions

-  which leave $a_{1}$ unchanged and are permutable with~$\{P\}$, some such

-  substitution will leave two symbols unchanged. This is clearly impossible.

-  Hence the group of order~$p(p - 1)$ is the greatest group of the $p$~symbols in which

-  $\{P\}$~is self-conjugate.}%

-Now if the only substitutions

-permutable with~$\{P\}$ are its own, then $\{P\}$~is one of $\dfrac{N}{p}$

-conjugate sub-groups; and these contain $N\left(1 - \dfrac{1}{p}\right)$ substitutions

-of order~$p$. In this case, $G$~would contain exactly exactly $\dfrac{N}{p}$

-substitutions whose orders are not divisible by~$p$. But this is

-clearly impossible, since $\dfrac{N}{p}$~is the order of a sub-group which

-keeps one symbol unchanged, and there are $p$~such sub-groups.

-Hence there must be substitutions in~$G$, other than those of~$\{P\}$,

-which are permutable with~$\{P\}$; and each of these substitutions

-keeps one symbol unchanged.

-

-It follows from §~134 that $G$, if soluble, must contain a

-\index{Primitive groups@\Topic{Primitive groups}!of prime degree}%

-self-conjugate sub-group of order~$p$: therefore no group of

-prime degree~$p$, which contains more than one sub-group of

-order~$p$, can be soluble.

-

-If $1 + kp$ is the number of sub-groups of order~$p$ contained

-in~$G$, then

-\[

-N = p \frac{p - 1}{d} (1 + kp),

-\]

-where $d$ is a factor of $p - 1$; and a sub-group of order~$p$ is

-transformed into itself by every substitution of a cyclical

-sub-group of order~$\dfrac{p - 1}{d}$. When $d$~is odd, a substitution which

-generates this cyclical sub-group is an odd substitution; and

-$G$~then contains a self-conjugate sub-group of order~$\frac{1}{2}N$.

-

-If both $p$ and $\frac{1}{2}(p - 1)$ are primes, the order of a group of

-degree~$p$, which contains more than one sub-group of order~$p$,

-must be divisible by~$\frac{1}{2}(p - 1)$. For if the order is not divisible

-by~$\frac{1}{2}(p - 1)$, the order of the sub-group, within which a sub-group

-of order~$p$ is self-conjugate, must be~$2p$. Now the

-%% -----File: 218.png---Folio 202-------

-substitutions of order~$2$ in this sub-group consist of $\frac{1}{2}(p - 1)$

-transpositions, so that they are odd substitutions. The group

-must therefore contain a self-conjugate sub-group in which

-these operations of order~$2$ do not occur. In such a sub-group,

-the only operations permutable with those of a sub-group of

-order~$p$ are its own; and we have seen that no such group can

-exist. The order of the group must therefore, as stated above,

-be divisible by~$\frac{1}{2}(p - 1)$.

-

-\Par{144.} Let $G$ be a primitive group of degree~$n$ and order~$N$;

-and let $p$ be a prime, which is a factor of~$N$ but not of either $n$

-or $n - 1$. Moreover, suppose that $n$~is congruent to $\nu, \pmod{p}$;

-$\nu$~being less than~$p$. If $n < p^{2}$, and if $p^{\alpha}$ is the highest power of~$p$

-which divides~$N$, the sub-groups of order~$p^{\alpha}$ must be Abelian

-groups of type $(1, 1, \dots \text{ to $\alpha$ units})$. In fact, such a sub-group

-must be intransitive, and, since $n < p^{2}$, the number of symbols

-in each transitive system of the sub-group must be~$p$. In

-any case the number of symbols left unchanged by a sub-group

-of order~$p^{\alpha}$ is of the form $kp + \nu$.

-

-Suppose now that, in a sub-group of order~$\dfrac{N}{n}$ which leaves

-one symbol unchanged, a sub-group~$H$ of order~$p^{\alpha}$ is one of

-$\dfrac{N}{p^{\alpha}mn}$ conjugate sub-groups. Then each of the $n$~sub-groups

-that keep one symbol unchanged contains $\dfrac{N}{p^{\alpha}mn}$ sub-groups of

-order~$p^{\alpha}$; and each sub-group of order~$p^{\alpha}$ belongs to $kp + \nu$

-sub-groups that keep one symbol unchanged. Hence $G$~contains

-$\dfrac{N}{p^{\alpha}m (kp + \nu)}$ sub-groups of order~$p^{\alpha}$, and any one of

-them, say~$H$, is contained self-conjugately in a sub-group~$I$ of

-order~$p^{\alpha}m (kp + \nu)$. This sub-group~$I$ must interchange transitively

-among themselves the $kp + \nu$ symbols left unchanged by~$H$.

-For let $a$~and~$b$ be any two of these symbols; and let $S$ be

-an operation which changes $a$ into $b$ and transforms $H$ into~$H'$.

-There must be an operation~$T$ which keeps $b$ unchanged and

-transforms $H'$ into~$H$, since in the sub-group that keeps $b$

-unchanged there is only one conjugate set of sub-groups of

-order~$p^{\alpha}$. Then $ST$~changes $a$ into $b$ and transforms~$H$ into

-%% -----File: 219.png---Folio 203-------

-itself; and therefore $I$~contains substitutions which change $a$

-into~$b$. Now it may happen that the existence of a sub-group

-such as~$I$ requires that $G$ is either the alternating or the symmetric

-group. When this is the case, we infer that the order

-of a primitive group of degree~$n$, other than the alternating or

-the symmetric group, cannot be divisible by~$p$.

-

-\begin{Remark}

-\Par{145.} As a simple example, we will shew that the order of

-a group of degree~$19$ cannot be divisible by~$7$, unless it contains

-the alternating group. It follows from Theorem~I, Corollary,

-§~141, that the order of a group of degree~$19$, which does not

-contain the alternating group, cannot be divisible by a power of~$7$

-higher than the first, and that if the group contains a substitution

-of order~$7$, the substitution must consist of two cycles of $7$~symbols

-each. The sub-group of order~$7$ must therefore leave $5$~symbols

-unchanged; hence, by §~144, it must be contained self-conjugately in

-a sub-group whose order is divisible by~$5$. Now (§~83) a group of

-order~$35$ is necessarily Abelian; so that the group of degree~$19$ must

-contain a substitution of order~$5$ which is permutable with a substitution

-of order~$7$. Such a substitution of order~$5$ must clearly

-consist of a single cycle, and its presence in a group of degree~$19$

-requires that the latter should contain the alternating group. It

-follows that, if a group of degree~$19$ does not contain the

-alternating group, its order is not divisible by~$7$.

-

-As a second example, we will determine the possible forms for

-the order of a group of degree~$13$, which does not contain the alternating

-group. It follows, from Theorem~I, Corollary, §~141, that

-the order of such a group must be of the form $2^{\alpha} · 3^{\beta} · 5^{\gamma} · 11^ {\delta} · 13$;

-where $\alpha$,~$\beta$, $\gamma$,~$\delta$ do not exceed $9$,~$4$, $1$,~$1$ respectively.

-

-Suppose, first, that $\gamma$~is unity, if possible. A substitution of

-order~$5$ must consist of two cycles of $5$~symbols each; and a sub-group

-of order~$5$ must therefore be self-conjugate in a sub-group of

-order~$15$. There is then a substitution of order~$3$ which is permutable

-with a substitution of order~$5$. Such a substitution must,

-as in the last example, consist of a single cycle; and its existence

-would imply that the group contains the alternating group. It

-follows that, for the group as specified, $\gamma$~must be zero.

-

-Suppose, next, that $\delta$~is unity, if possible. The group is then

-(Theorem~I, §~141) triply transitive; and the order of the sub-group,

-that keeps two symbols fixed and is transitive in the remaining~$11$,

-is $2^{\alpha} · 3^{\beta} · 11$; and this sub-group must contain more than one sub-group

-of order~$11$. We have seen in §~143 that no such group can

-exist. Therefore $\delta$~must be zero.

-

-The two smallest numbers of the form $2^{m} 3^{n}$ which are congruent

-to unity, $\pmod{13}$, are $3^{3}$ and~$2^{4} 3^{2}$; and every number of this form,

-%% -----File: 220.png---Folio 204-------

-which is congruent to unity, $\pmod{13}$, can be written $(2^{4}3^{2})^{x}3^{3y}$.

-Hence the order of every group of order~$13$, which contains no odd

-substitution, must be of the form $(2^{4}3^{2})^{x} · 3^{3y} · z · 13$, where $z$~is $2$,~$3$

-or~$6$. Since $3^{4}$~is the highest power of~$3$ that can divide the order

-of the group, the only admissible values of $x$~and~$y$ are \Inum{(i)}~$x = 0$,

-$y = 1$: \Inum{(ii)}~$x = 2$, $y = 0$: \Inum{(iii)}~$x = 1$, $y = 0$.

-

-Suppose, first, that $x = 0$, $y = 1$. The order of the group is

-$2 · 3^{3} · 13$, $3^{4} · 13$, or $2 · 3^{4} · 13$. There must be $13$~sub-groups of order~$3^{4}$

-(or~$3^{3}$), and since $13$~is not congruent to unity, $\pmod{9}$, there must

-be sub-groups of order~$3^{3}$ (or~$3^{2}$) common to some two sub-groups of

-order~$3^{4}$ (or~$3^{3}$). Such a sub-group must be self-conjugate (Theorem~III,

-§~80). This case therefore cannot occur.

-

-Next, suppose that $x = 2$, $y = 0$. Then $z$~must be~$2$, and the

-order of the group is $2^{9} · 3^{4} · 13$. Now it is easy to verify that $2^{9} · 3^{4}$

-is not a possible order either for an intransitive or for an imprimitive

-group of degree~$12$. The order of a sub-group of the group of degree~$12$

-which keeps one symbol fixed is $2^{7} · 3^{3}$. This sub-group can have

-no substitution consisting of a single cycle of $3$~symbols, since no

-such substitution can occur in the original group. Hence it must

-permute the $11$ symbols in two transitive sets of $9$~and $2$~symbols

-respectively. It must therefore contain a self-conjugate sub-group

-of order $2^{6} · 3^{3}$ which keeps $3$~of the $12$~symbols unchanged; and this

-sub-group must occur self-conjugately in $3$~of the $12$~sub-groups which

-keep one symbol unchanged. This however makes the group of

-degree~$12$ imprimitive, contrary to supposition. Hence this case

-cannot occur.

-

-Finally, then, the only possible values of $x$~and~$y$ are $x = 1$,

-$y = 0$. The order of a group of degree~$13$, which has more than

-one sub-group of order~$13$ and no odd substitutions, is $2^{5} · 3^{2} · 13$,

-$2^{4} · 3^{3} · 13$, or $2^{5} · 3^{3} · 13$. The order of a group of order~$13$ with odd

-substitutions will be twice one of the preceding three numbers.

-

-A further and much more detailed examination would be necessary

-to determine whether groups of degree~$13$ correspond to any or

-all of these orders. We shall see in Chapter~XIV that there is a

-group of degree~$13$ and order~$2^{4} · 3^{3} · 13$; and M.~Jordan\footnote

-  {\textit{Comptes Rendus}, t.~\VolNo{LXXV} (1872), p.~1757.}

-\index{Jordan@\Topic{Jordan}, quoted}%

-states

-that this is the only group of degree~$13$ which contains more than

-one sub-group of order~$13$.

-\end{Remark}

-

-\begin{Ex} If $n$ ($> 3$) and $2n + 1$ are primes, shew that there is no

-triply transitive group of degree~$2n + 3$ which does not contain the

-alternating group.

-\end{Ex}

-

-\Par{146.} As a further illustration, and for the actual value of

-the results themselves, we proceed to determine all types of

-primitive groups for degrees not exceeding~$8$.

-%% -----File: 221.png---Folio 205-------

-

-\Class{\Inum{(i)}}{$n = 3$.}

-\index{Primitive groups@\Topic{Primitive groups}!of degrees $3$,~$4$ and~$5$}%

-

-The symmetric group of $3$~symbols has a single sub-group,

-viz.\ the alternating group. Both these groups are necessarily

-primitive.

-

-\Class{\Inum{(ii)}}{$n = 4$.}

-

-Since a group whose order is the power of a prime cannot

-(§~124) be represented in primitive form, a primitive group of

-degree~$4$ must contain $3$ as a factor of its order. Hence the

-only primitive groups of degree~$4$ are the symmetric and the

-alternating groups.

-

-\Class{\Inum{(iii)}}{$n = 5$.}

-

-Since $5$~is a prime, every group of degree~$5$ is a primitive

-group. The symmetric group of degree~$5$ contains $6$~cyclical

-sub-groups of order~$5$; therefore, by Sylow's theorem, every

-group of degree~$5$ must contain either $1$~or $6$ sub-groups of

-order~$5$. Since the alternating group is simple, every sub-group

-that contains $6$~sub-groups of order~$5$ must contain the

-alternating group. Hence, besides the alternating and the

-symmetric groups, we have only sub-groups which contain a

-sub-group of order~$5$ self-conjugately. In such a group, an

-operation of order~$5$ can be permutable with its own powers

-only. Hence (§~112) the only sub-groups of the type in question

-other than cyclical sub-groups, are groups of orders $20$ and~$10$.

-These are defined by

-\begin{align*}

-&\{(12345),\ (2354)\}, \\

-\lintertext{and}

-&\{(12345),\ (25) (34)\}.

-\end{align*}

-

-\Class{\Inum{(iv)}}{$n = 6$.}

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$6$}%

-

-If the order of a primitive group of degree~$6$ is not divisible

-by~$5$, the order must (§~141) be equal to or be a factor of~$2^{3} · 3$.

-The order of a sub-group that keeps one symbol fixed is equal to

-or is a factor of~$2^{2}$. Hence the sub-group must keep two symbols

-fixed, and therefore (§~128) the group cannot be primitive. Hence

-the order of every primitive group of degree~$6$ is divisible by~$5$,

-and every such group is at least doubly transitive. The

-symmetric group contains $36$~sub-groups of order~$5$; and hence,

-since no transitive group of degree~$6$ can contain a self-conjugate

-sub-group of order~$5$, every primitive group of degree~$6$,

-%% -----File: 222.png---Folio 206-------

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$6$}%

-which does not contain the alternating group, must have $6$~sub-groups

-of order~$5$.

-

-If $G$ is such a group, the sub-group of $G$ that keeps one

-symbol fixed is a transitive group of degree~$5$ which has a

-self-conjugate sub-group of order~$5$. If this transitive group

-of degree~$5$ were cyclical, every operation of the doubly transitive

-group~$G$ of order~$30$ would displace all or all but one of the

-symbols. Since $6$~is not the power of a prime, this is impossible

-(§~105). Hence the sub-group of~$G$ which keeps one symbol

-fixed must be of one of the two types given above; and the order

-of~$G$ must be $120$ or~$60$. Now we have seen, in §~142, that the

-symmetric group of degree~$6$ has a single conjugate set of primitive

-sub-groups of order~$120$ and a single set of order~$60$. Hence

-there is a single type of primitive group of degree~$6$, corresponding

-to each of the orders $120$ and~$60$. These are defined by

-\begin{align*}

-&\{(126)(354),\ (12345),\ (2354)\}, \\

-\lintertext{and}

-&\{(126)(354),\ (12345),\ (25)(34)\}:

-\end{align*}

-where the last two substitutions in each case generate a sub-group

-that keeps one symbol unchanged.

-

-\Class{\Inum{(v)}}{$n = 7$.}

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$7$|(}%

-

-Every transitive group of degree~$7$ is primitive; and if it

-does not contain the alternating group, its order must (§~141) be

-equal to or be a factor of $7 · 6 · 5 · 4$. A cyclical sub-group of order~$7$

-must (footnote, \PageRef{p.}{201}), in a group of degree~$7$ that contains

-more than one such sub-group, be self-conjugate in a group of

-order $21$ or~$42$. Now neither $20$ nor $40$ is congruent to unity,

-$\pmod{7}$; and therefore $5$~cannot be a factor of the order of such

-a group. Hence the order of a transitive group of degree~$7$,

-that does not contain the alternating group, is equal to or is a

-factor of $7 · 6 · 4$. But $8$~is the only factor of $7 · 6 · 4$ which is

-congruent to unity, $\pmod{7}$; and therefore, if the group contains

-more than one sub-group of order~$7$, its order must be equal to

-$7 · 6 · 4$ and it must contain $8$~sub-groups of order~$7$.

-

-Such a group must be doubly transitive; for if a sub-group

-of order~$24$ interchanges the symbols in two intransitive

-systems, it is easily shewn that the group would contain

-substitutions displacing three symbols only, and therefore that

-it would contain the alternating group. A sub-group of degree~$24$,

-%% -----File: 223.png---Folio 207-------

-transitive in $6$~symbols, must contain $4$~sub-groups of order~$3$.

-For if it contained only one, it would necessarily have

-circular substitutions of $6$~symbols, and the group of order

-$7 · 6 · 4$ would have a self-conjugate sub-group of order $7 · 6 · 2$;

-which is not the case. Hence the sub-groups of order~$24$ must

-be simply isomorphic with the symmetric group of $4$~symbols.

-

-The actual construction of the group is now reduced to

-a limited number of trials. A group of degree~$6$, simply

-isomorphic with the symmetric group of $4$~symbols, and containing

-no odd substitutions, may always be represented in the

-form

-\[

-\{(234)(567),\ (2763)(45)\};

-\]

-and we have to find a circular substitution of the seven symbols

-$1$,~$2$, $3$, $4$, $5$, $6$,~$7$ such that the group generated by it shall be

-permutable with this group. Moreover since, in the required

-group, every operation of order~$3$ transforms some operation of

-order~$7$ into its square, we may assume without loss of generality

-that the circular substitution of order~$7$ contains the sequence

-\dots$12$\dots\ and is transformed into its own square by~$(234)(567)$.

-There are only three circular substitutions satisfying these

-conditions, viz.\

-\begin{align*}

-&(1235476), \\

-&(1236457), \\

-\lintertext{and}

-&(1237465).

-\end{align*}

-

-It appears on trial that the group generated by the first of

-these is not permutable with the sub-group of order~$24$, while

-the groups generated by the other two are. There are therefore

-just two groups of order $7 · 6 · 4$ which contain the given group

-of order~$24$. Now in the symmetric group of $7$~symbols, a

-sub-group of order $7 · 6 · 4$ must, from the foregoing discussion,

-be one of a set of $30$ conjugate sub-groups. These all enter in

-the alternating group; and therefore, in that group, they must

-form two sets of $15$ conjugate sub-groups each. Each of these

-contains $7$~sub-groups of the type

-\[

-\{(234)(567),\ (2763)(45)\};

-\]

-and the alternating group contains a conjugate set of $105$ such

-such sub-groups. Hence each sub-group of this set will enter

-in two, and only in two, sub-groups of the alternating group of

-%% -----File: 224.png---Folio 208-------

-order $7 · 6 · 4$; and in the symmetric group these two sub-groups

-are conjugate. Finally then, the sub-groups of order $7 · 6 · 4$

-form a single conjugate set in the symmetric group. They are

-defined by

-\[

-\{(1236457),\ (234)(567),\ (2763)(45)\},

-\]

-the two latter substitutions giving a sub-group that keeps

-one symbol fixed.

-

-These groups are simple; for since they are expressed as

-transitive groups of degree~$7$, there can be no self-conjugate

-sub-group whose order divides~$24$, while it is evident that a

-self-conjugate sub-group that contains an operation of order~$7$

-must coincide with the group itself. Also since there are $8$~sub-groups

-of order~$7$, these groups can be expressed as doubly

-transitive groups of eight symbols.

-

-A group of degree~$7$, which has only one sub-group of order~$7$,

-must either be cyclical or be contained in the group of order

-$7 · 6$ given by §~112. Such groups are defined by

-\begin{align*}

-&\{(1234567),\ (243756)\}, \\

-\lintertext{or}

-&\{(1234567),\ (235)(476)\}, \\

-\lintertext{or}

-&\{(1234567),\ (27)(45)(36)\}.

-\end{align*}

-

-The simple group of order~$168$, which here occurs as a

-\PageLabel{208}%

-transitive group of degree~$7$, is the only simple group of that

-order. For, if possible, let there be a simple group~$G$ of order~$168$

-and of a distinct type from the above. It certainly cannot

-be expressed as a group of degree~$7$; and therefore it must

-have $21$~sub-groups of order~$8$. If two of these sub-groups

-have a common sub-group of order~$4$, it must be contained

-self-conjugately (§~80) in a sub-group of order $24$ or~$56$; and

-this is inconsistent with the suppositions made. If on the

-other hand, $2$~is the order of the greatest sub-group common

-to two sub-groups of order~$8$, such a common sub-group of order~$2$

-must, on the suppositions made, be self-conjugate in a sub-group

-of order~$12$. But a group of order~$12$, which has a self-conjugate

-operation of order~$2$, must have a self-conjugate

-sub-group of order~$3$; and therefore $G$ would only contain $7$~sub-groups

-of order~$3$, and could be expressed as a group of

-degree~$7$; contrary to supposition. No other supposition is

-possible with regard to the sub-groups of order~$8$, since $21$~is

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$7$|)}%

-%% -----File: 225.png---Folio 209-------

-not congruent to unity, $\pmod{8}$. Hence, finally, there is no

-simple group of order~$168$ distinct from the group of degree~$7$.

-

-\Class{\Inum{(vi)}}{$n = 8$.}

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|(}%

-

-The order of a primitive group of degree~$8$, which does not

-contain the alternating group, cannot (§~141) be divisible by~$5$.

-Suppose, if possible, that the order of such a group is $2^{\alpha + 3} · 3$

-($\alpha = 0, 1, 2, 3$). A substitution of order~$3$ must consist of two

-cycles; and therefore the sub-group of order~$\DPchg{2^{\alpha}3}{2^{\alpha} · 3}$, which keeps

-one symbol fixed, must interchange the others in two intransitive

-systems of $3$ and $4$ respectively. In this sub-group, a

-sub-group of order~$3$ must be one of four conjugate sub-groups,

-and therefore $\alpha$~is either $2$ or~$3$. Now a group of order~$2^{5} · 3$

-or $2^{6} · 3$ is soluble, as is seen at once by considering the sub-groups

-of order $2^{5}$ or~$2^{6}$. Hence a primitive group of order~$2^{5} · 3$

-or $2^{6} · 3$ must contain a transitive self-conjugate sub-group of

-order~$8$, whose operations are all of order~$2$.

-

-If $7$ is a factor of the order of the group, the group must be

-doubly transitive; and from the case of $n = 7$, it follows that

-the possible orders are $8 · 7$, $8 · 7 · 2$, $8 · 7 · 3$, $8 · 7 · 6$, and

-$8 · 7 · 6 · 4$. Moreover, for the orders $8 · 7 · 2$ and $8 · 7 · 6$, the

-group contains odd substitutions and therefore it contains self-conjugate

-sub-groups of order $8 · 7$ and $8 · 7 · 3$ respectively.

-

-A simple group of order~$8 · 7 · 3$ is necessarily identical in

-type with the group of this order determined on \PageRef{p.}{208};

-and a group of order~$8 · 7 · 3$, which is not simple, is certainly

-soluble. Hence a composite group of order~$8 · 7 · 3$, and

-a group of order $8 · 7 · 6$ which does not contain a simple

-sub-group of order~$8 · 7 · 3$, must both, if expressible as primitive

-groups of degree~$8$, contain transitive self-conjugate

-sub-groups of order~$8$ whose operations are all of order~$2$.

-With the possible exception then of groups of order $8 · 7 · 6 · 4$,

-the only primitive groups of degree~$8$, which do not contain a

-self-conjugate sub-group of order~$8$, are the simple group of

-order~$8 · 7 · 3$ and any group of $8 · 7 · 6$ which contains this self-conjugately.

-We have seen that the simple group of order

-$8 · 7 · 3$ contains a single set of $8$ conjugate sub-groups of order~$21$,

-and therefore it can be expressed in one form only as a

-group of degree~$8$. A group of degree~$8$ and order~$8 · 7 · 6$,

-%% -----File: 226.png---Folio 210-------

-which contains this self-conjugately, can occur only in one

-form, if at all; for, if it exists, it must be triply transitive, and

-it must be given by combining the simple group with any

-operation of order~$2$ which transforms one of its operations of

-order~$7$ into its own inverse. That such a group does exist has

-been shewn in §~113. These two groups are actually given by

-\begin{align*}

-&\{(15642378),\ (1234567),\ (243756)\}, \\

-\lintertext{and}

-&\{(1627) (5438),\ (1234567),\ (235) (476)\};

-\end{align*}

-where in each case the last two substitutions give a sub-group

-that keeps one symbol fixed.

-

-To avoid unnecessary prolixity we shall, in dealing with the

-primitive groups of degree~$8$, which contain a transitive self-conjugate

-sub-group of order~$8$ whose operations are all of

-order~$2$, anticipate some of the results of the next Chapter. It

-will there be shewn that, if $G$~is an Abelian group of order~$N$,

-and $L$~a group of isomorphisms (§~156) of~$G$, a group of degree~$N$

-may be formed which has a transitive self-conjugate sub-group

-simply isomorphic with $G$, while at the same time the sub-group

-that keeps one symbol fixed is isomorphic with~$L$ (§~158).

-The group of isomorphisms of a group of order~$8$, whose

-operations are all of order~$2$, will be shewn in Chapter~XIV to

-be identical with the simple group of order~$168$. This group

-has a single set of conjugate sub-groups of each of the orders $7$

-and $21$, but no sub-group of order $14$ or~$42$. When expressed

-as a group of degree~$7$, it has a single set of conjugate sub-groups

-of order~$12$ (or~$24$) which leave no symbols unchanged.

-There are therefore primitive groups of degree~$8$ containing

-transitive self-conjugate sub-groups of order~$8$ corresponding

-to each of the orders $8 · 7$, $8 · 7 · 3$, $2^{5} · 3$, $2^{6} · 3$, and $8 · 7 · 6 · 4$;

-and in each case there is a single type of such group.

-

-It remains to determine whether there can be any type

-of group, of degree~$8$ and order $8 · 7 · 6 · 4$, other than that just

-obtained. Such a group must be one of $15$~conjugate sub-groups

-in the alternating group of degree~$8$, and can therefore

-itself be expressed as a group of degree~$14$. Since it

-certainly cannot be expressed as a group of degree~$7$, the

-group of degree~$14$ must be transitive. The order of the sub-group,

-in this form, that keeps one symbol fixed is~$2^{5} · 3$. If

-%% -----File: 227.png---Folio 211-------

-this keeps only one symbol unchanged, it must interchange the

-remaining symbols in four intransitive systems of $3$,~$3$, $3$ and~$4$

-respectively, since a substitution of order~$3$ must clearly consist

-of $4$~cycles. A group of order~$2^{5} · 3$ cannot however be so

-expressed; and therefore the sub-group that keeps one symbol

-fixed must keep two fixed. The group of degree~$14$ is therefore

-imprimitive, and the group must contain a sub-group of order~$2^{6} · 3$.

-Moreover, since the group cannot be expressed as a

-transitive group of degree~$7$, this sub-group of order~$2^{6} · 3$ must

-contain (§~123) a sub-group which is self-conjugate in the group

-itself. The order of this sub-group must be a power of~$2$;

-since the group is primitive, it cannot be less than~$2^{3}$. On the

-other hand, the order cannot be greater than~$2^{3}$ since the group

-contains a simple sub-group of order~$7 · 6 · 4$. Hence finally,

-there is no type of primitive group of degree~$8$ and order

-$8 · 7 · 6 · 4$ other than that already obtained.

-

-There is no difficulty now in actually constructing the

-primitive groups of degree~$8$ which have a self-conjugate sub-group

-of order~$8$. They are all contained in the group of order

-$8 · 7 · 6 · 4$; and this may be constructed from a sub-group

-keeping one symbol fixed, which is given on \PageRef{p.}{208}, by the

-method of §~109. It will thus be found that the group in

-question is given by

-\[

-\{(81)(26)(37)(45),\ (1236457),\ (234)(567),\ (2763)(45)\};

-\]

-while the groups of orders $8 · 7 · 3$ and $8 · 7$ are given by omitting

-respectively the last and the two last of the four generating

-operations.

-

-The construction of the two remaining groups, of order~$2^{5} · 3$

-and $2^{6} · 3$, is left as an exercise for the reader.

-

-It may be noticed that it has been shewn incidentally, in

-discussing above the possibility of a second type of group of

-degree~$8$ and order $8 · 7 · 6 · 4$, that the alternating group of

-degree~$8$ can be expressed as a doubly transitive group of

-degree~$15$.

-

-It may similarly be shewn that the alternating group of

-degree~$7$ can be expressed as a doubly transitive group of

-degree~$15$, and the alternating group of degree~$6$ as a simply

-transitive and primitive group of degree~$15$.

-\index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|)}%

-%% -----File: 228.png---Folio 212-------

-

-\Par{147.} We have seen in §~105 that a doubly transitive group,

-\index{Doubly transitive groups@\Topic{Doubly transitive groups}!the sub-groups of, which keep two symbols fixed}%

-of degree~$n$ and order~$n (n - 1)$, can exist only when $n$~is the

-power of a prime. For such a group, the identical operation is

-the only one which keeps more than one symbol unchanged.

-We shall now go on to consider the sub-groups of a doubly

-transitive group, of degree~$n$ and order~$n(n - 1)m$, which keep

-two symbols fixed. The order of any such sub-group is~$m$;

-since the group contains operations changing any two symbols

-into any other two, the sub-groups which keep two symbols

-fixed must form a single conjugate set.

-

-Suppose first that the sub-group, which keeps two symbols

-unchanged, displaces all the other symbols. The sub-group that

-keeps $a$ and $b$ unchanged cannot then be identical with that

-which keeps $c$ and $d$ unchanged, unless the symbols $c$ and $d$ are

-the same pair as $a$ and~$b$. Since there are $\frac{1}{2}n(n - 1)$ pairs

-of $n$~symbols, the conjugate set contains $\frac{1}{2}n(n - 1)$ sub-groups;

-and each sub-group of order~$m$ keeping two symbols fixed

-must be self-conjugate in a sub-group of order~$2m$, which

-consists of the operations of the sub-group of order~$m$ and of

-those operations interchanging the two symbols that the sub-group

-of order~$m$ keeps fixed.

-

-Suppose next that all the operations of a sub-group~$H$,

-which keeps two symbols fixed, keep $x$~symbols fixed, while

-none of the remaining $n - x$ symbols are unchanged by all the

-operations of~$H$. From $x$~symbols $\frac{1}{2}x(x - 1)$ pairs can be

-formed, and therefore the sub-group that keeps one pair unchanged

-must keep $\frac{1}{2}x(x - 1)$ pairs unchanged. In this case,

-the conjugate set contains $\dfrac{n(n - 1)}{x(x - 1)}$ distinct sub-groups of order~$m$,

-and $H$~is therefore self-conjugate in a group~$K$ of order

-$x(x - 1)m$. The operations of this sub-group which do not

-belong to~$H$ interchange among themselves the $x$~symbols that

-are left unchanged by~$H$. Now since the group itself is doubly

-transitive, there must be operations which change any two of

-these $x$~symbols into any other two; and any such operation

-being permutable with~$H$ must belong to~$K$. Hence if we

-consider the effect of~$K$ on the $x$~symbols only which are left

-unchanged by~$H$, $K$~reduces to a doubly transitive group of

-%% -----File: 229.png---Folio 213-------

-degree~$x$ and order~$x(x - 1)$. It follows that $x$ must be a prime

-or the power of a prime.

-

-\begin{Remark}

-\Par{148.} The preceding paragraph suggests the combinatorial

-problem of forming from $n$~distinct symbols $\dfrac{n(n - 1)}{x(x - 1)}$ sets of $x$~symbols,

-such that every pair of symbols occurs in one set of~$x$ and

-no pair occurs in more than one.

-

-There is one class of cases in which a solution of this problem is

-given immediately by the theory of Abelian groups. Let $G$ be an

-Abelian group of order~$p^{m}$, where $p$~is a prime, and type $(1, 1, \Dots\DPchg{\,}{,}

-\text{to $m$ units})$. We have seen, in §~49, that $G$ has $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of

-order~$p$, and $\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sub-groups of order~$p^{2}$. Now any pair

-of sub-groups of order~$p$ generates a sub-group of order~$p^{2}$, and therefore

-every pair of sub-groups of order~$p$ occurs in one and only one

-sub-group of order~$p^{2}$. Moreover, every sub-group of order~$p^{2}$ contains

-$p + 1$ sub-groups of order~$p$. When $p$~is a prime and $m$~any integer,

-it is therefore always possible to form from $\dfrac{p^{m} - 1}{p - 1}$ symbols

-$\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sets of $p + 1$ symbols each, such that every pair of

-the symbols occurs in one set of $p + 1$ and no pair occurs in more

-than one set.

-

-Supposing that, for given values of $n$ and~$x$, such a distribution

-is possible, it is still of course an open question as to whether there

-is a doubly transitive substitution group of the $n$~symbols, such that

-every substitution which keeps any two symbols unchanged keeps

-also unchanged the whole set of~$x$ in which they occur. When $x$~is

-greater than~$3$, the question as to the existence of such groups is

-one which still remains to be investigated. There is however an

-important class of groups, to be considered later (Chapter~XIV),

-that possess a closely analogous property. These groups are doubly

-transitive; and from the $n$~symbols upon which they operate, we can

-form $\dfrac{n(n - 1)}{x(x - 1)}$ sets of~$x$, that are interchanged transitively by the

-substitutions of the group: the sets being such that every pair

-occurs in one set and no pair in more than one set.

-

-If $n(n - 1)m$ is the order of such a group, and if $H$~is a

-sub-group of order~$m$ which keeps a given pair fixed, then $H$~must

-interchange among themselves the remaining $x - 2$ symbols of that

-set of~$x$ which contains the pair kept unchanged by~$H$. $H$~contains,

-as a self-conjugate sub-group, the group~$h$ which leaves every symbol

-of the set of~$x$ unchanged; and if $m''$~is the order of this sub-group,

-while $m = m'm''$, then $m'$~is the order of the group to which $H$

-%% -----File: 230.png---Folio 214-------

-reduces when we consider its effect only on the $x - 2$ symbols. Now

-$h$~is self-conjugate in the group~$K$ that interchanges all the symbols

-of the set of~$x$ among themselves. But since the original group is

-doubly transitive, it must contain substitutions which change any

-two of the set of~$x$ into any other two, and every such substitution

-must belong to~$K$. Hence $K$~must be doubly transitive in the $x$~symbols,

-and therefore finally the order of the group, to which $K$~reduces

-when we consider its effect on the $x$~symbols only, is

-$x(x - 1)m'$. Since the order of~$h$, which keeps unchanged each of

-the $x$~symbols is~$m''$, the order of~$K$ is~$x(x - 1)m$.

-

-\Par{149.} When $x = 3$, $n$~must be of the form $6k + 1$ or $6k + 3$, since

-\index{Doubly transitive groups@\Topic{Doubly transitive groups}!with a complete set of triplets}%

-otherwise $\dfrac{n(n - 1)}{x(x - 1)}$ would not be an integer. The substitutions of

-a doubly transitive group of degree~$n$, which possesses a complete

-set of $\frac{1}{6}n(n - 1)$ triplets, must be such that every substitution which

-leaves two given symbols unchanged also leaves a third definite

-symbol unchanged.

-

-The smallest possible value of~$n$ is~$7$; and the group of Ex.~2,

-§~35, which is one of the groups obtained in §~146, satisfies all the

-conditions. In fact, the group is clearly a doubly transitive group

-of degree~$7$; and since its order is $7 · 6 · 4$, the order of the sub-group

-which keeps two symbols fixed is~$4$. Now in the sub-group

-\[

-\{(267) (345),\ (23) (47)\},

-\]

-which keeps one symbol fixed, the only operations that keep $1$~and~$2$

-unchanged are

-\[

-(35) (67),\ (36) (57),\ (37) (56).

-\]

-

-These with identity form a sub-group of order~$4$, which must

-therefore be the sub-group that keeps three symbols fixed. The

-complete set of triplets in this case is

-\[

-124,\ 137,\ 156,\ 235,\ 267,\ 346,\ 457.

-\]

-

-The next smallest value of~$n$ is~$9$, and in this case again, a group

-with the required properties exists.

-\end{Remark}

-

-\begin{Ex} Shew that the group

-\[

-\{(26973854),\ (456) (798)\}

-\]

-is an imprimitive group of order~$48$, each imprimitive system

-containing two symbols; and that the sub-group, which keeps the

-symbols of one imprimitive system unchanged, is isomorphic with

-the symmetric group of three symbols. Prove that this group is

-permutable with

-\[

-\{(123) (456) (789),\ (147) (258) (369)\},

-\]

-and thence that

-\[

-\{(123) (456) (789),\ (26973854),\ (456) (798)\},

-\]

-is a doubly transitive group of degree~$9$, which possesses a complete

-set of $12$~triplets.

-\end{Ex}

-%% -----File: 231.png---Folio 215-------

-

-\begin{Remark}

-The reader is not to infer from the examples given that, when

-$n$~is of the form $6k + 1$ or $6k + 3$, there is always a doubly transitive

-group of degree~$n$ which possesses a complete set of triplets. It is

-a good exercise to verify that there is no such group when $n$~is~$13$.

-

-The case $n = 13$, $x = 4$ is the simplest case that can occur of the

-division of $n$~symbols into sets of~$x$ in the manner of §~148 when $x$~is

-greater than~$3$. We shall see in Chapter~XIV that there is a

-doubly transitive group of degree~$13$ such that from the $13$~symbols

-permuted by the group a complete set of $13$ quartets can be

-formed, which are themselves permuted by the operations of the

-group. Of the operations forming a sub-group that keeps two given

-symbols fixed, half will keep fixed the two other symbols, which form

-a quartet with the two given symbols, and half will permute them.

-

-On the question of the independent formation of a complete set

-of triplets of $n$~symbols, and in certain cases of the group of

-degree~$n$ which interchanges the triplets among themselves, reference

-may be made to the memoirs mentioned in the subjoined footnote\footnotemark.

-\end{Remark}

-\index{Heffter@\Topic{Heffter}, quoted}%

-\index{Moore@\Topic{Moore}, quoted}%

-\index{Netto@\Topic{Netto}, quoted}%

-\footnotetext{Netto: ``Substitutionentheorie,'' pp.~220--235; ``Zur Theorie der Tripelsysteme,''

-  \textit{Math.\ Ann.}\ Vol.~\VolNo{XLII}, (1892), pp.~143--152. Moore: ``Concerning

-  triple systems,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XLIII}, (1893), pp.~271--285. Heffter: ``Ueber

-  Tripelaysteme,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLIX}, (1897), pp.~101--112.}

-

-\Par{150.} We shall conclude the present Chapter with some

-applications of substitution groups, which enable us to complete

-and extend certain earlier results.

-

-We have seen in §~107 that the substitutions of $n$~symbols,

-\index{Substitutions@\Topic{Substitutions} which are permutable!with a given substitution}%

-which are permutable with each of the substitutions of a

-regular substitution group~$G$ of order~$n$ of the same $n$ symbols,

-form another regular substitution group of order~$n$; and that,

-if $G$~is Abelian, the latter group coincides with~$G$. Hence the

-only substitutions of $n$~symbols, which are permutable with a

-circular substitution of the $n$~symbols, are the powers of the

-circular substitution.

-

-Let now $S$ be a regular substitution of order~$m$, in $mn$~symbols.

-It must permute the symbols in $n$~cycles of $m$~symbols

-each; and so we may take

-\[

-S = (a_{11} a_{12} \Dots a_{1m})

-    (a_{21} a_{22} \Dots a_{2m}) \Dots

-    (a_{n1} a_{n2} \Dots a_{nm}).

-\]

-

-If $T$ is permutable with~$S$, and if it changes $a_{rp}$ into~$a_{rq}$, it

-clearly must permute the $m$~symbols

-\[

-a_{r1},\ a_{r2},\ \Dots,\ a_{rm}

-\]

-%% -----File: 232.png---Folio 216-------

-among themselves; and therefore, so far as regards its effect on

-these $m$~symbols, $T$~must be a power of

-\[

-(a_{r1} a_{r2}\Dots a_{rm}).

-\]

-

-Again, if $T$~changes $a_{rp}$ into~$a_{sq}$, it must change the set

-\[

-a_{r1},\ a_{r2},\ \Dots,\ a_{rm},

-\]

-into the set

-\[

-a_{s1},\ a_{s2},\ \Dots,\ a_{sm};

-\]

-as otherwise it would not be permutable with~$S$.

-

-Now the totality of the substitutions of the $mn$~symbols,

-\index{Substitutions@\Topic{Substitutions} which are permutable!with a given substitution}%

-which are permutable with~$S$, form a group~$G_{S}$. This group

-must, from the properties of~$T$ just stated, be imprimitive,

-interchanging the symbols in $n$ imprimitive systems of $m$

-symbols each; and the symbols in any cycle of~$S$ will form an

-imprimitive system. Moreover, the self-conjugate sub-group~$H_{S}$

-of this group, which permutes the symbols of each system

-among themselves, is the group of order~$m^{n}$ generated by

-\[

-(a_{11} a_{12}\Dots a_{1m}),\

-(a_{21} a_{22}\Dots a_{2m}),\ \Dots,\

-(a_{n1} a_{n2}\Dots a_{nm}).

-\]

-

-In fact, every substitution of this group is clearly permutable

-with~$S$; and conversely, every substitution of the $mn$~symbols,

-which does not permute the systems, must belong to

-this group.

-

-Now $\dfrac{G_{S}}{H_{S}}$ is capable of representation as a group of degree~$n$,

-for none of its operations changes every one of the $n$~systems

-into itself. Hence $n!$~is the greatest possible order of~$\dfrac{G_{S}}{H_{S}}$. On

-the other hand, every operation of the group, generated by

-\[

-(a_{11} a_{21}\Dots a_{n1})

-(a_{12} a_{22}\Dots a_{n2})\Dots

-(a_{1m} a_{2m}\dots a_{nm})

-\]

-and

-\[

-(a_{11} a_{21}) (a_{12} a_{22}) \Dots (a_{1m} a_{2m}),

-\]

-is clearly permutable with~$S$; and this group, being simply

-isomorphic with the group

-\[

-\{(a_{1} a_{2}\Dots a_{n}),\ (a_{1} a_{2})\},

-\]

-\ie\ with the symmetric group of $n$~symbols, is of order~$n!$.

-

-Hence, finally, the order of~$G_{S}$ is~$m^{n} · n!$; and $G_{S}$~is generated

-by

-\begin{gather*}

-(a_{11} a_{12}\Dots a_{n1})

-(a_{12} a_{22}\Dots a_{n2}) \Dots

-(a_{1m} a_{2m}\Dots a_{nm}), \\

-(a_{11} a_{21})(a_{12} a_{22})\Dots (a_{1m} a_{2m}),

-\end{gather*}

-and

-\[

-(a_{11} a_{12}\Dots a_{1m}).

-\]

-%% -----File: 233.png---Folio 217-------

-

-\Par{151.} Let $h_{r}$ be a regular substitution group of order~$m$ in

-the $m$~symbols

-\[

-a_{r1},\ a_{r2},\ \Dots,\ a_{rm},

-\]

-and let $S_{rt}$ be one of its substitutions. Then if for $r$ we write

-in turn $1$, $2$,~\Dots, $n$, and if for each value of~$t$ from $1$ to~$m$ we

-form the substitution

-\[

-S_{1t} S_{2t} \Dots S_{nt},

-\]

-the set of $m$~substitutions so formed constitute an intransitive

-group~$H$ in the $mn$~symbols, simply isomorphic with~$h_{r}$.

-

-The method of §~150 can be applied directly to determine

-\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%

-the group~$G_{H}$ of degree~$mn$, each of whose substitutions are\DPnote{[** [sic]]}

-permutable with every substitution of~$H$. The order of this

-group is~$m^{n} · n!$; and it can be generated by

-\begin{gather*}

-(a_{11} a_{21} \Dots a_{n1})

-(a_{12} a_{22} \Dots a_{n2}) \Dots

-(a_{1m} a_{2m} \Dots a_{nm}), \\

-(a_{11} a_{21}) (a_{12} a_{22}) \Dots (a_{1m} a_{2m}),

-\end{gather*}

-and

-\[

-h_{1}';

-\]

-where $h_{1}'$~is the regular group in the symbols

-\[

-a_{11},\ a_{12},\ \Dots,\ a_{1m},

-\]

-each of whose substitutions is permutable with every substitution

-of~$h_{1}$.

-

-This group will contain~$H$ if, and only if, $H$~is an Abelian

-group. Moreover, the only self-conjugate substitutions of~$G_{H}$

-are the substitutions of~$H$ contained in it. For if $G_{H}$

-contained other self-conjugate substitutions $S_{1}$, $S_{2}$,~\dots, every

-operation of~$G_{H}$ would be permutable with every operation of

-the group $\{H,\ S_{1},\ S_{2},\ \dots\}$. Now $G_{H}$~is transitive, so that

-$S_{1}$, $S_{2}$,~\dots\ must displace all the symbols; and therefore

-$\{H,\ S_{1},\ S_{2},\ \dots\}$ has all its substitutions regular in the $mn$~symbols.

-If its order is~$mn_{1}$ where $n = n_{1}n_{2}$, the order of the

-group formed of all the substitutions of $mn$~symbols, which are

-permutable with each of its operations, is $(mn_{1})^{n_{2}} · n_{2}!$; and this

-number is less than~$m^{n} · n!$. Thus the supposition, that $G_{H}$~has

-self-conjugate operations other than the operations of~$H$ which

-it contains, leads to an impossibility.

-

-By means of this and the preceding section, the reader will

-have no difficulty in forming the group of $n$~symbols, which is

-permutable with every operation of any given group in the $n$~symbols.

-%% -----File: 234.png---Folio 218-------

-

-\Par{152.} If a group, whose order is a power of a prime~$p$, be

-\index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%

-expressed as a transitive substitution group, its degree must

-also be a power of~$p$ (§~123). Moreover such a group, since it

-has self-conjugate operations, must necessarily be imprimitive.

-

-The greatest value of~$m$, for which a group of order~$p^{m}$ can

-be expressed as a transitive group of degree~$p^{n}$, where $n$~is

-regarded as given, is determined at once by considering the

-symmetric group of degree~$p^{n}$. The highest power of~$p$ that

-divides~$p^{n}!$ is~$p^{\nu}$, where

-\[

-\nu = p^{n - 1} + p^{n - 2} + \Dots + p + 1.

-\]

-

-Hence the symmetric group of degree~$p^{n}$ contains a set of

-conjugate sub-groups of order~$p^{\nu}$ and it contains no groups whose

-order is a higher power of~$p$. Also, these groups are transitive in

-the $p^{n}$~symbols; for any one of them must contain a circular

-substitution of order~$p^{n}$. There are therefore groups of order~$p^{\nu}$

-which can be expressed as transitive groups of degree~$p^{n}$; but

-no group of order~$p^{\nu'}$ ($\nu' > \nu$) can be so expressed. Moreover, in

-order that a group of order~$p^{m}$ may be capable of representation

-as a transitive group of degree~$p^{n}$, it must be simply isomorphic

-with a transitive sub-group of the above substitution group of

-order~$p^{\nu}$.

-

-This group may be constructed synthetically as follows.

-Since the sub-group of order~$p^{\nu - n}$, that leaves one symbol

-unchanged, is contained in a sub-group of order~$p^{\nu - n + 1}$ (§~55),

-there must be imprimitive systems containing $p$~symbols each.

-If then we distribute the $p^{n}$~symbols into $p^{n - 1}$~sets of $p$~each,

-and with each set of~$p$ form a circular substitution, the $p^{n - 1}$

-permutable and independent circular substitutions will generate

-an intransitive group of order~$p^{p^{n - 1}}$. It will be the self-conjugate

-sub-group of the group of order~$p^{\nu}$, which permutes the

-symbols of each system among themselves.

-

-Again, the systems of $p$~symbols each may be taken in sets

-of~$p$ to form systems of $p^{2}$~symbols each, since the previously

-considered sub-group of order~$p^{n - \nu + 1}$ must be contained in a sub-group

-of order~$p^{n - \nu + 2}$. Hence we may form systems of $p^{2}$~symbols

-by combining the previous systems in sets of~$p$; and then with

-each $p^{2}$~symbols we can form a circular substitution, whose \DPchg{$p^{th}$}{$p$th}~power

-is the product of the $p$~circular substitutions of order~$p$,

-%% -----File: 235.png---Folio 219-------

-which have been previously formed from the $p^{2}$~symbols. The

-symbols of any set of~$p^{2}$ will then be interchanged by a transitive

-group of order~$p^{p + 1}$; and since there are $p^{n - 2}$ such sets, we obtain

-in this way an intransitive group of $p^{p^{n - 1} + p^{n - 2}}$. The group thus

-formed is that self-conjugate sub-group of the original group,

-which interchanges among themselves the symbols of each

-system of~$p^{2}$. This process may be continued, taking greater

-and greater systems, till at the last step we combine the $p$~systems

-of $p^{n - 1}$~symbols each into a single system by means of

-a circular substitution of order~$p^{n}$. The order of the resulting

-group is clearly~$p^{\nu}$, as it should be.

-

-The self-conjugate operations of this group form a sub-group

-of order~$p$.

-

-For suppose, if possible, they form a sub-group of order~$p^{r}$.

-Every operation of this sub-group displaces all the symbols;

-and therefore, when expressed as a substitution group in the $p^{n}$~symbols,

-it must interchange them transitively in $p^{n - r}$~sets of

-$p^{r}$ each.

-

-Now (§~151) those substitutions of the $p^{n}$~symbols, which

-are permutable with every operation of this sub-group, form a

-group of order~$p^{rp^{n - r}} · p^{n - r}!$; this number is only divisible by~$p^{\nu}$,

-\index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%

-as it must be, when $r = 1$.

-

-\begin{Ex} Shew that, for the group of degree~$p^{2}$ and order~$p^{p + 1}$, the

-factor groups $\dfrac{H_{r + 1}}{H_{r}}$ (of §~53) are all of type~$(1)$ except the last,

-which is of type~$(1, 1)$.

-\end{Ex}

-

-The fact that $\nu$~is a function of~$p$ when $n$~is given, explains

-why, in classifying all groups of order~$p^{n}$, some of the lower

-primes may behave in an exceptional manner. Thus we saw, in

-§~73, that for certain groups of order~$p^{4}$ it was necessary to

-consider separately the case $p = 3$. The present article makes

-it clear that, while there may be more than one type of group

-of order~$p^{4}$ ($p > 3$), which can be expressed as a transitive group

-of degree~$p^{2}$, there is only a single type of group of order~$3^{4}$

-which can be expressed transitively in $9$~symbols.

-

-\begin{Remark}

-\Par{153.} In the memoirs referred to in the footnote on \PageRef{p.}{155},

-M.~Mathieu has demonstrated the existence of a remarkable group,

-%% -----File: 236.png---Folio 220-------

-of degree~$12$ and order~$12 · 11 · 10 · 9 · 8$\DPtypo{.}{}, which is quintuply transitive.

-The generating operations of this group have been given in

-the note to §~108, at the end of Chapter~VIII\@. The verification of

-some of the more important properties of this group, as stated in

-the succeeding example, forms a good exercise on the results of this

-and the two preceding Chapters.

-\end{Remark}

-

-\begin{Ex} Shew that the substitutions

-\begin{gather*}

-(1254) (3867),\quad       (1758) (2643), \\

-(12) (48) (57) (69),\quad (a2) (58) (46) (79), \\

-(ab) (57) (68) (49),\quad (bc) (47) (58) (69),

-\end{gather*}

-generate a quintuply transitive group of degree~$12$ and order

-\[

-12 · 11 · 10 · 9 · 8.

-\]

-

-Prove that this group is simple; that a sub-group of degree~$11$

-and order $11 · 10 · 9 · 8$, which leaves one symbol unchanged, is a

-simple group; and that a sub-group of degree~$10$ and order~$10 · 9 · 8$,

-which leaves two symbols unchanged, contains a self-conjugate sub-group

-simply isomorphic with the alternating group of degree~$6$.

-

-Shew also that the group of degree~$12$ contains \Inum{(i)}~$1728$

-sub-groups of order~$11$ each of which is self-conjugate in a group

-of order~$55$: \Inum{(ii)}~$2376$ sub-groups of order~$5$, each of which is self-conjugate

-in a group of order~$40$: \Inum{(iii)}~$880$ sub-groups of order~$27$,

-each of which is self-conjugate in a group of order~$108$: \Inum{(iv)}~$1485$

-sub-groups of order~$64$.

-

-Prove further that the group is a maximum sub-group of the

-alternating group of degree~$12$.

-\end{Ex}

-%% -----File: 237.png---Folio 221-------

-

-

-\Chapter{XI.}{On the Isomorphism of a Group with Itself.}

-

-\Par{154.} \First{It} is shewn in §~24 that, if all the operations of a

-group are transformed by one of themselves, which is not self-conjugate,

-a correspondence is thereby established among the

-operations of the group which exhibits the group as simply

-isomorphic with itself.

-

-In an Abelian group every operation is self-conjugate, and

-the only correspondence established in the manner indicated is

-that in which every operation corresponds to itself. If however

-in an Abelian group we take, as the operation which corresponds

-to any given operation~$S$, its inverse~$S^{-1}$, then to $ST$ will

-correspond $S^{-1}T^{-1}$ or~$(ST)^{-1}$; and the correspondence exhibits

-the group as simply isomorphic with itself. For this particular

-correspondence, a group of order~$2$ is the only one in which

-each operation corresponds to itself.

-

-It is therefore possible for every group, except a group of

-order~$2$, to establish a correspondence between the operations of

-the group, which shall exhibit the group as simply isomorphic

-with itself. Moreover, we shall see that in general there are

-such correspondences which cannot be established by either of

-the processes above given. We devote the present Chapter to

-a discussion of the isomorphism of a group with itself. It will

-be seen that, for many problems of group-theory, and in particular

-for the determination of the various types of group

-which are possible when the factor-groups of the composition-series

-are given, this discussion is most important.

-%% -----File: 238.png---Folio 222-------

-

-\Par{155.}

-\begin{Definition}

-A correspondence between the operations

-\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!definition of}%

-of a group, such that to every operation~$S$ there

-corresponds a single operation~$S'$, while to the product~$ST$ of

-two operations there corresponds the product~$S'T'$ of the

-corresponding operations, is said to define an \emph{isomorphism of

-the group with itself}.

-\end{Definition}

-

-That isomorphism in which each operation corresponds to

-itself is called the \emph{identical isomorphism}. In every isomorphism

-of a group with itself, the identical operation corresponds to

-itself; and the orders of two corresponding operations are

-the same. For if $1$~and~$S$ were corresponding operations, so

-also would be $1 · 1$ and~$S^{2}$; and therefore more than one operation

-would correspond to~$1$. Again, if $S$~and~$S'$, of orders $n$~and~$n'$,

-are corresponding operations, so also are $S^{n}$ and~$S'^{n}$; and therefore

-$n$~must be a multiple of~$n'$. Similarly $n'$~must be a multiple

-of~$n$; and therefore $n$~and~$n'$ are equal.

-

-If the operations of a group of order~$N$ are represented by

-\[

-1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1},

-\]

-and if, for a given isomorphism of the group with itself, $S'_{r}$~is

-the operation that corresponds to~$S_{r}$ ($r = 1, 2, \Dots, N - 1$), the

-isomorphism will be completely represented by the symbol

-\[

-\Isom{1, S_{1}, S_{2}, \Dots, S_{N - 1}}

-     {1, S_{1}', S_{2}', \Dots, S_{N - 1}'}.

-\]

-

-In this symbol, two operations in the same vertical line are

-corresponding operations. When no risk of confusion is

-thereby introduced, the simpler symbol

-\[

-\Isom{S}{S'}

-\]

-is used.

-

-\Par{156.} An isomorphism of a group with itself, thus defined,

-is not an operation. The symbol of an isomorphism however

-defines an operation. It may, in fact, be regarded as a substitution

-performed upon the $N$~letters which represent the

-operations of the group. Corresponding to every isomorphism

-there is thus a definite operation; and it is obvious that the

-operations, which correspond to two distinct isomorphisms, are

-%% -----File: 239.png---Folio 223-------

-themselves distinct. The totality of these operations form a

-group. For let

-\[

-\Isom{S}{S'}

-\quad\text{and}\quad

-\Isom{S'}{S''}

-\]

-be any two isomorphisms of the group with itself. Then if, as

-hitherto, we use curved brackets to denote a substitution, we

-have

-\[

-\Sub{S}{S'} \Sub{S'}{S''} = \Sub{S}{S''}.

-\]

-

-But since $\Isom{S}{S'}$ is an isomorphism, the relation

-\[

-S_{p} S_{q} = S_{r}

-\]

-requires that

-\[

-S_{p}' S_{q}' = S_{r}'.

-\]

-

-And since $\Isom{S'}{S''}$ is an isomorphism, the relation

-\[

-S_{p}' S_{q}' = S_{r}'

-\]

-requires that

-\[

-S_{p}'' S_{q}'' = S_{r}''.

-\]

-

-Hence if

-\[

-S_{p} S_{q} = S_{r},

-\]

-then

-\[

-S_{p}'' S_{q}'' = S_{r}'';

-\]

-and therefore $\Isom{S}{S''}$ is an isomorphism.

-

-The product of the substitutions which correspond to two

-isomorphisms is therefore the substitution which corresponds to

-some other isomorphism.

-

-The set of substitutions which correspond to the isomorphisms

-of a given group with itself, therefore form a group.

-

-\begin{Definition}

-A group, which is simply isomorphic with

-\index{Group@\Topic{Group}!group of isomorphisms of a}%

-the group thus derived from a given group, is called \emph{the

-group of isomorphisms} of the given group.

-\end{Definition}

-

-It is not, of course, necessary always to regard this group as

-a group of substitutions performed on the symbols of the

-operations of the given group. But however the group of

-isomorphisms may be represented, each one of its operations

-corresponds to a definite isomorphism of the given group. To

-avoid an unnecessarily cumbrous phrase, we may briefly apply

-%% -----File: 240.png---Folio 224-------

-the term ``isomorphism'' to the operations of the group of

-isomorphisms. So long, at all events, as we are dealing with

-the properties of a group of isomorphisms, no risk of confusion

-is thereby introduced. Thus we shall use the phrase ``the

-isomorphism $\Sub{S}{S'}$'' as equivalent to ``the operation of the

-group of isomorphisms which corresponds to the isomorphism

-$\Isom{S}{S'}$.''

-

-\Par{157.} If $\Sigma$~is some operation of a group~$G$, while for $S$ each

-\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!cogredient and contragredient, definition of}%

-\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!class of, definition of}%

-operation of the group is put in turn, the symbol

-\[

-\Sub{S}{\Sigma^{-1}S \Sigma}

-\]

-defines an isomorphism of the group. For if

-\[

-S_{p} S_{q} = S_{r},

-\]

-then

-\[

-\Sigma^{-1} S_{p} \Sigma · \Sigma^{-1} S_{q}\Sigma

-  = \Sigma^{-1} S_{p} S_{q}\Sigma

-  = \Sigma^{-1} S_{r} \Sigma;

-\]

-and $\Sigma_{-1} S_{r}\Sigma$ is an operation of the group. An isomorphism of a

-group, which is thus formed on transforming the operations of the

-group by one of themselves, is called a \emph{cogredient} isomorphism.

-All others are called \emph{contragredient} isomorphisms\footnotemark.

-\index{Holder@\Topic{Hölder}, quoted}%

-\index{Klein@\Topic{Klein}, quoted}%

-\footnotetext{Klein, ``Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen

-  vom fünften Grade'' (1884), p.~232. Hölder, \textit{Math.\ Ann.}, Vol.~\VolNo{XLIII}

-  (1893), p.~314.}%

-If $\Sub{S}{S'}$ is

-any contragredient isomorphism, the isomorphisms

-\[

-\Sub{S}{S'} \Sub{S}{\Sigma^{-1}S\Sigma},

-\]

-when for $\Sigma$ each operation of the group is taken successively,

-are said to form a \emph{class} of contragredient isomorphisms.

-

-\begin{Theorem}[I.]

-The totality of the cogredient isomorphisms of

-a group~$G$ form a group isomorphic with~$G$; this group is

-a self-conjugate sub-group of the group of isomorphisms of~$G$\footnotemark.

-\end{Theorem}

-\footnotetext{Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

-  (1895), p.~326.}

-

-The product of the isomorphisms

-\[

-\Sub{S}{\Sigma^{-1}S\Sigma} \quad\text{and}\quad

-\Sub{S}{\Sigma'^{-1}S\Sigma'}

-\]

-%% -----File: 241.png---Folio 225-------

-is given by

-\begin{align*}

-\Sub{S}{\Sigma^{-1}S\Sigma}

-\Sub{S}{\Sigma'^{-1}S\Sigma'}

-& =

-\Sub{S}{\Sigma^{-1}S\Sigma}

-\Sub{\Sigma^{-1}S\Sigma}{\Sigma'^{-1}\Sigma^{-1}S\Sigma\Sigma'} \\

-& =

-\Sub{S}{\Sigma'^{-1}\Sigma^{-1}S\Sigma\Sigma'} \\

-& =

-\Sub{S}{\Sigma''^{-1}S\Sigma''},

-\end{align*}

-where

-\[

-\Sigma \Sigma' = \Sigma''.

-\]

-

-The product of two cogredient isomorphisms is therefore another

-cogredient isomorphism; hence the cogredient isomorphisms

-form a group. Moreover, if we take the isomorphism

-\[

-\Sub{S}{\Sigma^{-1}S\Sigma}

-\]

-as corresponding to the operation~$\Sigma$ of the group~$G$, then to

-every operation of~$G$ there will correspond a definite cogredient

-isomorphism, so that to the product of any two operations

-of~$G$ there corresponds the product of the two corresponding

-isomorphisms. The group~$G$ and its group of cogredient

-isomorphisms are therefore isomorphic. If $G$~contains no

-self-conjugate operation, identity excepted, no two isomorphisms

-corresponding to different operations of~$G$ can be identical;

-and therefore, in this case, $G$~is simply isomorphic with its

-group of cogredient isomorphisms. If however $G$~contains self-conjugate

-operations, forming a self-conjugate sub-group~$H$,

-then to every operation of~$H$ there corresponds the identical

-isomorphism; and the group of cogredient isomorphisms is

-simply isomorphic with~$\dfrac{G}{H}$.

-

-Let now

-\[

-\Sub{S}{S'}

-\]

-be any isomorphism. Then

-\begin{align*}

-\Sub{S}{S'}^{-1}

-\Sub{S}{\Sigma^{-1}S\Sigma}

-\Sub{S}{S'}

-& =

-\Sub{S'}{S}

-\Sub{S}{\Sigma^{-1}S\Sigma}

-\Sub{S}{S'} \\

-& =

-\Sub{S'}{\Sigma^{-1}S\Sigma}

-\Sub{\Sigma^{-1}S\Sigma}{\Sigma'^{-1}S'\Sigma'} \\

-& =

-\Sub{S'}{\Sigma'^{-1}S'\Sigma'}.

-\end{align*}

-%% -----File: 242.png---Folio 226-------

-

-The isomorphism $\Sub{S}{S'}$ therefore transforms every cogredient

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!contains the group of cogredient isomorphisms self-conjugately}%

-isomorphism into another cogredient isomorphism. It follows

-that the group of cogredient isomorphisms is self-conjugate

-within the group of isomorphisms.

-

-\Par{158.} Let $G$ be a group of order~$N$, whose operations are

-\[

-1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};

-\]

-and let $L$ be the group of isomorphisms of~$G$. We have seen

-in §~20 that $G$~may be represented as a transitive group of

-substitutions performed on the $N$~symbols

-\[

-1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};

-\]

-and that, when it is so represented, the substitution which

-corresponds to the operation~$S_{x}$ is

-\[

-\Sub{\PadTo{S_{x}}{1},

-     \PadTo{S_{1}S_{x}}{S_{1}},

-     \PadTo{S_{2}S_{x}}{S_{2}}, \Dots,

-     \PadTo{S_{N-1}S_{x}}{S_{N-1}}}

-    {S_{x}, S_{1}S_{x}, S_{2}S_{x}, \Dots, S_{N-1}S_{x}},

-\]

-or more shortly

-\[

-\Sub{S}{SS_{x}}.

-\]

-

-When $G$ is thus represented, we will denote it by~$G'$. We

-have already seen that $L$~can be represented as an intransitive

-substitution group of the same $N$~symbols; a typical substitution

-of~$L$, when it is so represented, is

-\[

-\Sub{1, S_{1}, S_{2}, \Dots, S_{N-1}}

-    {1, S'_{1}, S'_{2}, \Dots, S'_{N-1}},

-\]

-or more shortly

-\[

-\Sub{S}{S'}.

-\]

-

-When $L$ is thus represented, we will denote it by~$L'$. It is

-clear that the two substitution groups $G'$ and $L'$ have no

-substitution in common except identity. For every substitution

-of~$L'$ leaves the symbol~$1$ unchanged; and no substitution of~$G'$,

-except identity, leaves $1$ unchanged.

-

-Now

-\begin{align*}

-\Sub{S}{S'}^{-1}

-\Sub{S}{SS_{x}}

-\Sub{S}{S'}

-& =

-\Sub{S'}{SS_{x}}

-\Sub{SS_{x}}{S'S'_{x}} \\

-& =

-\Sub{S'}{S'S'_{x}} \\

-& =

-\Sub{S}{SS'_{x}}.

-\end{align*}

-%% -----File: 243.png---Folio 227-------

-

-Every operation of~$L'$ is therefore permutable with~$G'$.

-Hence if $M$~is the order of~$L$, the group~$\{G', L'\}$, which we will

-call~$K'$, is a transitive group of degree~$N$ and order~$NM$,

-containing $G'$ self-conjugately. Further $\Sub{S}{S'}$ transforms $\Sub{S}{SS_{x}}$

-into $\Sub{S}{SS'_{x}}$; and these two substitutions of~$G'$ correspond to

-the operations $S_{x}$ and~$S_{x}'$ of~$G$. Hence the isomorphism,

-established on transforming the substitutions of~$G'$ by any

-substitution $\Sub{S}{S'}$ of~$L'$, is the isomorphism denoted by the

-symbol~$\Sub{S}{S'}$.

-

-{\Loosen Since $\Sub{S}{S_{x}SS_{x}^{-1}}$ is a substitution of~$L'$, the substitution

-$\Sub{S}{S_{x}SS_{x}^{-1}} \Sub{S}{SS_{x}}$, or $\Sub{S}{S_{x}S}$, belongs to~$K'$. Hence $K'$~contains

-the set of substitutions}

-\[

-\Sub{S}{S_{x}S},\quad (x = 0, 1, 2, \Dots, N - 1).

-\]

-

-These form (§~107) a transitive group~$G''$, simply isomorphic

-with~$G'$ and such that every substitution of~$G''$ is

-permutable with every substitution of~$G'$. Moreover (\lc),

-the substitutions of~$G''$ are the only substitutions of the

-$N$~symbols which are permutable with each of the substitutions

-of~$G'$.

-

-Suppose now that $\Sigma$~is any substitution of the $N$~symbols

-which is permutable with~$G'$. When the substitutions of~$G'$

-are transformed by~$\Sigma$, the resulting isomorphism is identical

-with that given by some substitution, say $\Sub{S}{S'}$, of $L'$.

-Hence $\Sigma \Sub{S}{S'}^{-1}$ is a substitution of the $N$~symbols which is

-permutable with every substitution of~$G'$. It therefore belongs

-to~$G''$; and hence $\Sigma$~belongs to~$K'$. It follows that $K'$~contains

-every substitution of the $N$~symbols which is permutable

-with~$G'$.

-%% -----File: 244.png---Folio 228-------

-

-The only substitutions common to $G'$ and $G''$ are the self-conjugate

-substitutions of either. The factor group $\dfrac{K'}{\{G', G''\}}$

-is simply isomorphic with~$\dfrac{L}{g}$, where $g$~is the group of cogredient

-isomorphisms of $G$ contained in~$L$. The groups $G'$ and $G''$ are

-identical only when $G'$~is Abelian; in this case, $g$~consists of the

-identical operation alone.

-

-\begin{Definition}

-A group~$K$, simply isomorphic with the

-\index{Group@\Topic{Group}!holomorph of a}%

-\index{Holomorph@\Topic{Holomorph}!definition of}%

-substitution group~$K'$ which has just been constructed, we

-shall call the \emph{holomorph} of~$G$.

-\end{Definition}

-

-\Par{159.} An isomorphism must change any set of operations,

-which are conjugate to each other, into another set which are

-conjugate. For if

-\[

-\Sub{S}{S'}

-\]

-be the isomorphism, and if

-\[

-S_{z}^{-1} S_{x} S_{z} = S_{y},

-\]

-then

-\[

-{S'_{z}}^{-1} S'_{x} S'_{z} = S'_{y},

-\]

-so that $S'_{x}$ and $S'_{y}$ are conjugate operations when $S_{x}$~and~$S_{y}$ are

-conjugate. A~cogredient isomorphism changes every set of

-conjugate operations into itself; and all the members of a class

-of contragredient isomorphisms permute the conjugate sets in

-the same way. If

-\[

-\Sub{S}{S'}

-\]

-is an isomorphism which changes every conjugate set of operations

-of~$G$ into itself, and if

-\[

-\Sub{S}{S''}

-\]

-is any isomorphism of~$G$, then the isomorphism

-\[

-\Sub{S}{S''}^{-1} \Sub{S}{S'}\Sub{S}{S''}

-\]

-changes every conjugate set into itself. It follows that those

-isomorphisms, which change every conjugate set of operations

-into itself, form a self-conjugate sub-group of the complete group

-of isomorphisms. This sub-group clearly contains the group of

-cogredient isomorphisms and may be identical with it.

-%% -----File: 245.png---Folio 229-------

-

-If now $\Sub{S}{S'}$ is any isomorphism of~$G$ of order~$n$, the

-substitutions

-\[

-\Sub{S}{S'}\quad \text{and}\quad \Sub{S}{SS_{x}},\quad(x = 0, 1, \Dots, N - 1),

-\]

-generate a group of order~$N_{n}$. When $J$~is used to represent

-the isomorphism, this group may be denoted by $\{J, G\}$; as shewn

-above, it contains $G$ self-conjugately. Suppose that $n$~is prime

-and not a factor of~$N$. The operation~$J$ is not permutable

-with every operation of~$G$; and therefore there must be operations~$S$

-of~$G$ which are permutable with no operation of the

-conjugate set to which $J$~belongs. The number of operations

-which in $\{J, G\}$ are conjugate to such an operation~$S$ must be a

-multiple of~$n$; and since $n$~is not a factor of~$N$, this conjugate

-set of operations must be made up of~$n$ distinct conjugate sets

-of operations in~$G$. The isomorphism~$J$ must therefore interchange

-some of the conjugate sets of~$G$.

-

-The same result is clearly true if the order~$n$ of~$J$ has any

-prime factor not contained in~$N$. Hence:---

-

-\begin{Theorem}[II.]

-An isomorphism of a group~$G$, whose order

-contains a prime factor which does not occur in the order of~$G$,

-must interchange some of the conjugate sets of~$G$.

-\end{Theorem}

-

-\Par{160.} If the isomorphism $\Sub{S}{S'}$ or~$J$ leaves no operation

-except identity unchanged, it must in $\{J, G\}$ be one of $N$

-conjugate operations. For if

-\[

-S_{x}^{-1} J S_{x} = S_{y}^{-1} J S_{y},

-\]

-$J$~would be permutable with $S_{y}S_{x}^{-1}$, which is not the case.

-

-These $N$~conjugate operations are

-\[

-J,\ JS_{1},\ JS_{2},\ \Dots,\ JS_{N-1},

-\]

-and since the first transforms every operation of~$G$, except

-identity, into a different one, the same must be true of all the

-set. If now $J$~transformed any operation~$S$ into a conjugate

-operation $\Sigma^{-1} S \Sigma, J \Sigma^{-1}$ would transform $S$ into itself; hence

-$J$~must transform every conjugate set of~$G$ into a different

-conjugate set.

-%% -----File: 246.png---Folio 230-------

-

-\begin{Remark}

-The special case in which the order of~$J$ is two may here be

-considered. Representing the $N$~operations conjugate to~$J$ by

-\[

-J,\ J_{1},\ J_{2},\ \Dots,\ J_{N-1},

-\]

-the $N$~operations of~$G$ are

-\[

-J^{2},\ JJ_{1},\ JJ_{2},\ \Dots,\ JJ_{N-1}.

-\]

-

-Now

-\[

-J^{-1} · JJ_{x} · J = J_{x}J = (JJ_{x})^{-1},

-\]

-so that $J$~transforms every operation of~$G$ into its own inverse.

-But if

-\[

-S' = S^{-1},

-\]

-and

-\[

-T' = T^{-1},

-\]

-then

-\[

-S'T' = S^{-1}T^{-1} = (TS)^{-1}.

-\]

-

-Now as $S'T'$ is the operation into which the isomorphism

-transforms~$ST$, it must be~$(ST)^{-1}$, and therefore

-\[

-ST = TS.

-\]

-

-The group~$G$ is therefore an Abelian group of odd order.

-\end{Remark}

-

-\Par{161.} Any sub-group~$H$ of~$G$ is transformed by an isomorphism

-into a simply isomorphic sub-group~$H'$; but $H$~and~$H'$

-are not necessarily conjugate within~$G$. If however the set of

-conjugate sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{m},

-\]

-are the only sub-groups of~$G$ of a given type, every isomorphism

-must interchange them among themselves; and if no

-isomorphism transforms each one of the set into itself, the

-group of isomorphisms can be represented as a transitive group

-of degree~$m$.

-

-Suppose now that no operation of~$G$ is permutable with

-each of the conjugate sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{m},

-\]

-so that $G$, or what in this case is the same thing (since $G$~can

-have no self-conjugate operation except identity) the group of

-cogredient isomorphisms of~$G$, can be expressed as the transitive

-group of $m$~symbols that arises on transforming the set of $m$~sub-groups

-by each operation of~$G$. Let $J$~be any operation, of

-order~$\mu$, that transforms~$G$ and each of the set of $m$ conjugate

-sub-groups, into itself. Then $J^{\mu}$~is the lowest power of~$J$ that

-can occur in~$G$, since no operation of~$G$ transforms each of the $m$~sub-groups

-%% -----File: 247.png---Folio 231-------

-into itself. Now in $\{J, G\}$, the greatest sub-group

-that contains $H_{r}$ self-conjugately is~$\{J, I_{r}\}$, $I_{r}$~being the greatest

-sub-group of~$G$ that contains~$H_{r}$ self-conjugately. Also, in

-$\{J, G\}$ the set of sub-groups $\{J, I_{r}\}$, ($r = 1, 2, \dots, m$), is a

-complete conjugate set. Now the set of groups

-\[

-I_{1},\ I_{2},\ \Dots,\ I_{m}

-\]

-have by supposition no common operation except identity; and

-therefore the greatest common sub-group of

-\[

-\{J, I_{1}\},\ \{J, I_{2}\},\ \Dots,\ \{J, I_{m}\}

-\]

-is~$\{J\}$. Hence $\{J\}$~is a self-conjugate sub-group of~$\{J, G\}$; and

-since $G$~is also a self-conjugate sub-group of~$\{J, G\}$, while $\{J\}$

-and~$G$ have no common operation except identity, $J$~must be

-permutable with every operation of~$G$. Every operation

-therefore which is permutable with~$G$, and with each of the

-sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{m},

-\]

-is permutable with every operation of~$G$. Thus finally, no

-contragredient isomorphism can transform each of the sub-groups~$H_{r}$

-($r = 1, 2, \Dots, m$) into itself. Hence:---

-

-\begin{Theorem}[III.]

-If the conjugate set of $m$ sub-groups

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{m}

-\]

-contains all the sub-groups of~$G$ of a given type, and if no

-operation of~$G$ is permutable with each sub-group of the set, the

-group of isomorphisms of~$G$ can be represented as a transitive

-group of degree~$m$.

-\end{Theorem}

-

-\begin{Corollary}[I.]

-If $G$ contains $kp + 1$ sub-groups of order~$p^{\alpha}$,

-where $p^{\alpha}$~is the highest power of a prime~$p$ that divides the

-order of~$G$; and if the greatest sub-group~$I$, that contains a

-sub-group of order~$p^{\alpha}$ self-conjugately, contains no self-conjugate

-sub-group of~$G$; then the group of isomorphisms of~$G$ can

-be represented as a transitive group of degree~$kp + 1$.

-\end{Corollary}

-

-For it has been seen that the groups of order~$p^{\alpha}$ contained

-in~$G$ form a single conjugate set.

-

-\begin{Corollary}[II.]

-If the conjugate set of $m$ sub-groups of~$G$

-\[

-H_{1},\ H_{2},\ \Dots,\ H_{m}

-\]

-is changed into itself by every isomorphism of~$G$, and if no

-%% -----File: 248.png---Folio 232-------

-operation of~$G$ is permutable with every one of these sub-groups;

-then the group of isomorphisms of~$G$ can be expressed

-as a transitive group of degree~$m$.

-\end{Corollary}

-

-In fact, under the conditions stated, the reasoning applied

-to prove the theorem may be used to shew that no isomorphism

-of~$G$ can transform each of the $m$ sub-groups into itself.

-

-\Par{162.} \begin{Definition}

-Any sub-group of a group~$G$ which is

-\index{Characteristic sub-group@\Topic{Characteristic sub-group}!definition of}%

-\index{Characteristic sub-group@\Topic{Characteristic sub-group}!groups with no, are either simple or the direct product of simply isomorphic simple groups}%

-transformed into itself by every isomorphism of~$G$, is called\footnote

-  {Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte}, 1895,

-  p.~183.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-a \emph{characteristic sub-group} of~$G$.

-\end{Definition}

-

-A characteristic sub-group of a group~$G$ is necessarily a

-self-conjugate sub-group of~$G$; but a self-conjugate sub-group

-is not necessarily characteristic. A simple group, having no

-self-conjugate sub-groups, can have no characteristic sub-groups.

-Let now $G$ be any group, and let $K$~be the holomorph

-of~$G$. A characteristic sub-group of~$G$ is then a self-conjugate

-sub-group of~$K$; and conversely, every self-conjugate sub-group

-of~$K$ which is contained in~$G$ is a characteristic sub-group of~$G$.

-

-Suppose now a chief-series of~$K$ formed which contains~$G$.

-If $G$~has no characteristic sub-group, it must be the

-last term but one of this series, the last term being identity.

-It follows by §~94 that $G$~must be the direct product of a

-number of simply isomorphic simple groups. Hence:---

-

-\begin{Theorem}[IV.]

-A group, which has no characteristic sub-group,

-must be either a simple group or the direct product of

-simply isomorphic simple groups.

-\end{Theorem}

-

-The converse of this theorem is clearly true.

-

-\Par{163.} Suppose now that $G$~is a group which has characteristic

-\index{Characteristic series@\Topic{Characteristic series}!definition of}%

-sub-groups; and let

-\[

-G,\ G_{1},\ \Dots,\ G_{r},\ G_{r+1},\ \Dots,\ 1

-\]

-be a series of such sub-groups, each containing all that follow it

-and chosen so that, for each consecutive pair $G_{r}$ and $G_{r+1}$, there

-is no characteristic sub-group of~$G$ contained in~$G_{r}$ and containing~$G_{r+1}$,

-except $G_{r+1}$~itself. Such a series is called\footnote

-  {Frobenius, ``Ueber auflösbare Gruppen, II,'' \textit{Berliner Sitzungsberichte}, 1895,

-  p.~1027.}

-a \emph{characteristic series} of~$G$.

-%% -----File: 249.png---Folio 233-------

-\index{Characteristic series@\Topic{Characteristic series}!invariance of}%

-

-It may clearly be possible to choose such a series in more

-than one way. If

-\[

-G,\ G_{1}',\ \Dots,\ G_{r}',\ G_{r+1}',\ \Dots,\ 1

-\]

-be a second characteristic series of~$G$, then

-\begin{align*}

-&K,\ J,\ \Dots,\ H,\ G,\ G_{1},\ \Dots,\ G_{r},\ G_{r+1},\ \Dots,\ 1 \\

-\lintertext{and}

-&K,\ J,\ \Dots,\ H,\ G,\ G_{1}',\ \Dots,\ G_{r}',\ G_{r+1}',\ \Dots,\ 1

-\end{align*}

-are two chief-series of~$K$. In fact, if $K$~had a self-conjugate

-sub-group contained in~$G_{r}$ and containing $G_{r+1}$, then $G$~would

-have a characteristic sub-group contained in~$G_{r}$ and containing

-$G_{r+1}$. The two chief-series of~$K$ coincide in the terms from $K$

-to~$G$ inclusive. Hence the two sets of factor-groups

-\[

-\frac{G}{G_{1}},\ \frac{G_{1}}{G_{2}},\ \Dots,\ \frac{G_{r}}{G_{r+1}},\ \Dots

-\]

-and

-\[

-\frac{G}{G_{1}'},\ \frac{G_{1}'}{G_{2}'},\ \Dots,\ \frac{G_{r}'}{G_{r+1}'},\ \Dots

-\]

-must be equal in number and, except possibly as regards the

-sequence in which they occur, identical in type. Moreover, each

-factor-group must be either a simple group or the direct product

-of simply isomorphic simple groups.

-

-\Par{164.} We will now shew how to determine a characteristic

-\index{Characteristic series@\Topic{Characteristic series}!of a group whose order is a power of a prime|(}%

-series for a group whose order is the power of a prime\footnotemark.

-\footnotetext{Frobenius, \lc, pp.~1028, 1029.}%

-

-{\Loosen First, let the group~$G$ be Abelian; and suppose that it is

-generated by a set of independent operations, of which $n_{s}$ are of

-the order~$p^{m}$, ($s = 1, 2, \dots, r$), while}

-\[

-m_{1} > m_{2} > \Dots > m_{r}.

-\]

-

-The sub-group~$G_{\mu}$ (§~42), formed of the operations of~$G$

-which satisfy the relation

-\[

-S^{p^{\mu}} = 1,

-\]

-is clearly a characteristic sub-group. As a first step towards

-forming the characteristic series, we may take the set of

-groups

-\[

-G_{m_{1}} (= G),\ G_{m_{1} - 1},\ G_{m_{1} - 2},\ \Dots,\ G_{2},\ G_{1},\ 1;

-\]

-for this is a set of characteristic sub-groups such that each

-contains the one that follows it.

-%% -----File: 250.png---Folio 234-------

-

-Now the sub-group~$H_{\nu}$ (§~45), formed of the distinct

-operations that remain when every operation of~$G$ is raised

-to the power~$p^{\nu}$, is also a characteristic sub-group; and since

-the operations common to two characteristic sub-groups also

-form a characteristic sub-group, the sub-group~$K_{\mu,\nu}$ (common to

-$G_{\mu}$ and~$H_{\nu}$) is characteristic. It follows from this that $G_{1}$~will be

-a characteristic sub-group only when $r = 1$. If $r > 1$, $G_{1}$~is not

-contained in~$H_{m_{r - 1} - 1}$, and the common sub-group~$K_{1, m_{r - 1} - 1}$ of

-these two is characteristic. If $r > 2$, this sub-group again is

-not contained in~$H_{m_{r - 2} - 1}$; and the common sub-group~$K_{1, m_{r - 2} - 1}$

-of $G_{1}$ and~$H_{m_{r - 2} - 1}$ is a characteristic sub-group contained

-in~$K_{1, m_{r - 1} - 1}$. Continuing thus, we form between $G_{1}$ and~$1$ the

-series

-\[

-G_{1},\ K_{1, m_{r - 1} - 1},\ K_{1, m_{r - 2} - 1},\ \Dots,\ K_{1, m_{1} - 1},\ 1.

-\]

-

-In a similar way, between $G_{\alpha}$ and~$G_{\alpha-1}$ we introduce such of

-the series

-\[

-\{G_{\alpha - 1}, K_{\alpha, m_{r - 1} - \alpha} \},\

-\{G_{\alpha - 1}, K_{\alpha, m_{r - 2} - \alpha} \},\ \Dots,\

-\{G_{\alpha - 1}, K_{\alpha, m_{1} - \alpha} \}

-\]

-as are distinct, the symbol $m_{s} - \alpha$ being replaced by zero where

-it is negative.

-

-From the original series we thus form a new one, in which

-again each group is characteristic and contains the following.

-This series may be shewn to be a characteristic series.

-

-Let

-\[

-P_{m_{s}, 1},\ P_{m_{s}, 2},\ \Dots,\ P_{m_{s}, n_{s}}

-\]

-be the $n_{s}$~generating operations of~$G$, whose orders are~$p^{m_{s}}$.

-Then if $\{G_{\alpha - 1}, K_{\alpha, m_{s} - \alpha}\}$ and $\{G_{\alpha - 1}, K_{\alpha, m_{s - 1} - \alpha}\}$ are distinct, the

-generating operations of the latter differ only from those of the

-former in containing the set

-\[

-P_{m_{s}, x}^{p^{m_{s} - \alpha + 1}},\quad (x = 1, 2, \Dots, n_{s}),

-\]

-in the place of

-\[

-P_{m_{s}, x}^{p^{m_{s} - \alpha} },\quad (x = 1, 2, \Dots, n_{s}).

-\]

-

-Now any permutation of the $n_{s}$~generating operations

-\[

-P_{m_{s}, x},\quad (x = 1, 2, \Dots, n_{s}),

-\]

-among themselves, the remaining generating operations being

-unaltered, must clearly give an isomorphism of~$G$ with itself;

-and therefore no sub-group of~$G$, contained in $\{G_{\alpha - 1}, K_{\alpha, m_{s} - \alpha} \}$

-%% -----File: 251.png---Folio 235-------

-and containing $\{G_{\alpha - 1}, K_{\alpha, m_{s - 1} - \alpha}\}$, can be a characteristic sub-group.

-This result being true for every pair of distinct groups

-which succeed each other in the series that has been formed,

-it follows that the series is a characteristic series. It may be

-noticed that, if $\Gamma$~and~$\Gamma'$ are any two consecutive sub-groups in

-a characteristic series of~$G$, the order of~$\dfrac{\Gamma}{\Gamma'}$ must be~$p^{\nu}$, where $\nu$~is

-one of the $r$~numbers~$n_{s}$.

-

-Secondly, suppose that $G$~is not Abelian. We may first

-consider the series of sub-groups

-\[

-G,\ H_{n},\ H_{n - 1},\ \Dots,\ H_{1},\ 1

-\]

-of §~53. Each of these is clearly a characteristic sub-group, and

-each contains the succeeding. Moreover, $\dfrac{H_{r + 1}}{H_{r}}$ is an Abelian

-group; and, by the process that we have just investigated, a

-characteristic series may be formed for it. To each group in

-this series will correspond a characteristic sub-group of~$H_{r + 1}$

-containing~$H_{r}$; and the set of groups so obtained forms part of

-a complete characteristic series of~$G$. When between each

-consecutive pair of groups in the above series the groups thus

-formed are interpolated, the resulting series of groups is a

-characteristic series for~$G$.

-

-\Par{165.} The isomorphisms of a given group with itself are

-closely connected with the composition of every composite group

-in which the given group enters as a self-conjugate sub-group.

-Let $G$~be any composite group and $H$~a self-conjugate sub-group

-of~$G$. Then since every operation of~$G$ transforms~$H$

-into itself, to every such operation will correspond an isomorphism

-of~$H$ with itself. If $S$~is an operation of~$G$ not

-contained in~$H$, and if the isomorphism of~$H$ arising on

-transforming its operations by~$S$ is contragredient, so also is

-the isomorphism arising from each of the set of operations~$SH$.

-In this case, no one of this set of operations is permutable with

-every operation of~$H$. If however the isomorphism arising from~$S$

-is cogredient, there must be some operation~$h$ of~$H$ which

-gives the same isomorphism as~$S$; and then $Sh^{-1}$~is permutable

-with every operation of~$H$. In this case, the set of operations~$SH$

-will give all the cogredient isomorphisms of~$H$.

-\index{Characteristic series@\Topic{Characteristic series}!of a group whose order is a power of a prime|)}%

-%% -----File: 252.png---Folio 236-------

-

-Suppose now that $H_{1}$ is that sub-group of~$G$ which is formed

-of all the operations of~$G$ that are permutable with every

-operation of~$H$. Then to every operation of~$G$, not contained

-in~$\{H, H_{1}\}$, must correspond a contragredient isomorphism of~$H$;

-and to every operation of the factor group $\dfrac{G}{\{H, H_{1}\}}$ corresponds

-a class of contragredient isomorphisms. If then $L$~is the

-group of isomorphisms of~$H$, and if $L_{1}$~is that self-conjugate

-sub-group of~$L$ which gives the cogredient isomorphisms of~$H$,

-$\dfrac{G}{\{H, H_{1}\}}$ must be simply isomorphic with a sub-group of~$\dfrac{L}{L_{1}}$.

-

-If now $H$~contains no self-conjugate operation except

-identity, $H$~and~$H_{1}$ can contain no common operation except

-identity; and since each of them is a self-conjugate sub-group

-of~$G$, every operation of~$H$ is permutable with every operation

-of~$H_{1}$. In this case, $\{H, H_{1}\}$ is the direct product of $H$ and~$H_{1}$.

-

-If, further, $L$~coincides with~$H$, so that $H$~admits of no

-contragredient isomorphisms, $\dfrac{G}{\{H, H_{1}\}}$ must reduce to identity.

-In this case, $G$~is the direct product of $H$ and~$H_{1}$.

-

-\begin{Definition}

-A group, which contains no self-conjugate

-\index{Complete group@\Topic{Complete group}!definition of}%

-\index{Complete group@\Topic{Complete group}!groups which contain a, self-conjugately are direct products}%

-\index{Group@\Topic{Group}!complete}%

-operation except identity and which admits of no contragredient

-isomorphism, is called\footnote

-  {Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}

-  (1895), p.~325.}

-\index{Holder@\Topic{Hölder}, quoted}%

-a \emph{complete} group.

-\end{Definition}

-

-The result of the present paragraph may be expressed in

-the form:---

-

-\begin{Theorem}[V.]

-A group, which contains a complete group as

-a self-conjugate sub-group, must be the direct product of the

-complete group and some other group\footnotemark.

-\end{Theorem}

-\footnotetext{\textit{Ibid.}\ p.~325.}%

-

-\Par{166.} \begin{Theorem}[VI.]

-If $G$ is a group with no self-conjugate

-operations except identity; and if the group of cogredient isomorphisms

-of~$G$ is a characteristic sub-group of~$L$, the group of

-isomorphisms of~$G$; then $L$~is a complete group\footnotemark.

-\end{Theorem}

-\footnotetext{Hölder (\textit{loc.~cit.}\ p.~331) gives a theorem which is similar but not quite

-  equivalent to Theorem~VI.}%

-%% -----File: 253.png---Folio 237-------

-

-With the notation of §~158, the operations of~$L$ may be

-represented by the substitutions

-\[

-\Sub{S}{S'}.

-\]

-The group of cogredient isomorphisms, which we will call~$G'$, is

-given by the substitutions

-\[

-\Sub{S}{S_{x}^{-1} SS_{x}}\DPtypo{;}{,}\quad (x = 0, 1, \Dots, N - 1)\DPtypo{,}{;}

-\]

-it is simply isomorphic with~$G$.

-

-Now

-\begin{align*}

-\Sub{S}{S'}^{-1} \Sub{S}{S_{x}^{-1} SS_{x}} \Sub{S}{S'}

-  &= \Sub{S'}{S_{x}'^{-1} S'S_{x}'}, \\

-  &= \Sub{S}{S_{x}'^{-1} SS_{x}'}:

-\end{align*}

-and therefore no operation of~$L$ is permutable with every

-operation of~$G'$. Hence every isomorphism of~$G'$ is given on

-transforming its operations by those of~$L$. Suppose now

-that $J$~is an operation which transforms $L$ into itself. Since

-$G'$~is by supposition a characteristic sub-group of~$L$, the

-operation~$J$ transforms $G'$ into itself. If $J$~does not belong to~$L$,

-we may assume that $J$~is permutable with every operation of~$G'$.

-For if it is not, it must give the same isomorphism of~$G'$

-as some operation~$S$ of~$L$; and then $JS^{-1}$~is permutable with

-every operation of~$G'$, and is not contained in~$L$. Now $J$~being

-permutable with every operation of~$G'$, we have

-\[

-J^{-1} s^{-1} gsJ = s^{-1}gs,

-\]

-where $s$~is any operation of~$L$, and $g$~any operation of~$G'$.

-

-Moreover

-\[

-JgJ^{-1} = g,

-\]

-and therefore

-\[

-J^{-1} s^{-1} JgJ^{-1} sJ = s^{-1}gs.

-\]

-

-Hence $s$ and $J^{-1}sJ$ give the same isomorphism of~$G'$. Now

-no two distinct operations of~$L$ give the same isomorphism of~$G'$,

-so that $s$ and $J^{-1}sJ$ must be identical; in other words, $J$~is

-permutable with every operation of~$L$. Hence $L$~admits of

-no contragredient isomorphisms. Moreover, $G'$~has no self-conjugate

-operations, and no operation of~$L$ is permutable with

-%% -----File: 254.png---Folio 238-------

-every operation of~$G'$; hence $L$~has no self-conjugate operations.

-It is therefore a complete group.

-

-\begin{Corollary}

-If $G$~is a simple group of composite order, or

-\index{Complete group@\Topic{Complete group}!group of isomorphisms of a simple group of composite order is a}%

-if it is the direct product of a number of isomorphic simple

-groups of composite order, the group of isomorphisms~$L$ of~$G$ is

-a complete group.

-\end{Corollary}

-

-For suppose, if possible, in this case that $G'$~is not a

-characteristic sub-group of~$L$; and that, by a contragredient

-isomorphism of~$L$, $G$~is transformed into~$G''$. Then $G''$~is a

-self-conjugate sub-group of~$L$, and each of the groups $G'$ and~$G''$

-transforms the other into itself. Hence (§~34) either every

-operation of~$G'$ is permutable with every operation of~$G''$, or $G'$

-and~$G''$ must have a common sub-group. The former supposition

-is impossible since no operation of~$L$ is permutable with

-every operation of~$G$. On the other hand, if $G'$~and~$G''$ have a

-common sub-group, it is a self-conjugate sub-group of~$L$ and it

-therefore is a characteristic sub-group of $G'$. Now (§~162) $G$~has

-no characteristic sub-groups, and therefore the second supposition

-is also impossible. It follows that, in this case, $G'$~is a characteristic

-sub-group of~$L$, and that $L$~is a complete group.

-

-\Par{167.} \begin{Theorem}[VII.]

-If $G$~is an Abelian group of odd order,

-and if $K$~is the holomorph of~$G$; then when $G$~is a characteristic

-sub-group of~$K$, the latter group is a complete group.

-\end{Theorem}

-

-If $N$~is the order of~$G$, then $K$~can be expressed (§~158) as a

-transitive group of degree~$N$. When $K$~is so expressed, those

-operations of~$K$ which leave one symbol unchanged form a

-sub-group~$H$, which is simply isomorphic with the group of

-isomorphisms of~$G$. Now (§~160) an Abelian group of odd order

-admits of a single isomorphism of order two, which changes every

-operation into its own inverse. The corresponding substitution

-of~$H$ is a self-conjugate substitution in~$H$, and is one of $N$~conjugate

-substitutions in~$K$. These are the only substitutions

-of~$K$ which transform every substitution of~$G$ into its inverse.

-If $G$~is a characteristic sub-group of~$K$, every isomorphism of~$K$

-must transform~$G$, and therefore also the set of $N$ conjugate

-substitutions of order two, into itself. Also, no substitution of~$K$

-can be permutable with each one of these $N$~substitutions,

-since each of them keeps just one symbol unchanged. Hence

-%% -----File: 255.png---Folio 239-------

-(Theorem~III, Cor.~II, §~161) the group of isomorphisms of~$K$

-can be expressed as a transitive group of degree~$N$, which

-contains $G$ as a transitive self-conjugate sub-group. But

-when the group of isomorphisms of~$K$ is so expressed, $K$~itself

-consists of all the substitutions of the $N$~symbols which are

-permutable with~$G$; and at the same time, every isomorphism

-of~$K$ transforms $G$ into itself. Hence the group of isomorphisms

-of~$K$ must coincide with $K$~itself; \ie\ $K$~admits of no

-contragredient isomorphisms. Also $K$~obviously contains no

-self-conjugate operation except identity; hence it is a complete

-group.

-

-\begin{Corollary}

-The holomorph of an Abelian group of order~$p^{m}$,

-\index{Complete group@\Topic{Complete group}!holomorph of an Abelian group of order~$p^{m}$ and type $(1, 1, \dots, 1)$ is a}%

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of a cyclical group|(}%

-where $p$~is an odd prime, and type $(1, 1, \dots\DPchg{\,}{,} \text{to $m$ units})$, is a

-complete group.

-\end{Corollary}

-% [** TN: Logically (but not typographycally), corollary ends here.]

-

-For if, in this case, $G$~is not a characteristic

-sub-group of~$K$, let $G'$~be a sub-group of~$K$ which, in the group

-of isomorphisms of~$K$, is conjugate to~$G$. Then $G$~and~$G'$ being

-both self-conjugate in~$K$ must have a common sub-group, since

-$K$~cannot contain their direct product. But the common sub-group

-of $G$ and~$G'$, being self-conjugate in~$K$, is a characteristic

-sub-group of~$G$. This is impossible (§~162); hence $G$~is a

-characteristic sub-group of~$K$.

-

-\begin{Ex} Shew that the holomorph of an Abelian group of degree~$2^{m}$

-and type $(1, 1, \dots\DPchg{\,}{,} \text{to $m$ units})$ is a complete group.

-\end{Ex}

-

-\begin{Remark}

-\Par{168.} We shall now discuss the groups of isomorphisms of

-certain special groups; and we shall begin with the case of a

-cyclical group~$G$, of prime order~$p$, generated by an operation~$P$.

-Every isomorphism of such a group must interchange among themselves

-the $p - 1$ operations

-\[

-P,\ P^{2},\ \Dots,\ P^{p - 1};

-\]

-and if any isomorphism replaces $P$ by~$P^{\alpha}$, it must replace $P^{2}$ by~$P^{2\alpha}$,

-and so on. Moreover, the symbol

-\[

-\Sub{\PadTo{P^{\alpha}}{P}\Add{,}

-     \PadTo{P^{2\alpha}}{P^{2}}\Add{,} \Dots\Add{,}

-     \PadTo{P^{(p - 1)\alpha}}{P^{p - 1}}}

-    {P^{\alpha}\Add{,} P^{2\alpha}\Add{,} \Dots\Add{,} P^{(p - 1)\alpha}}

-\]

-does actually represent an isomorphism whatever number $\alpha$~may be

-from $1$ to $p - 1$; for each operation occurs once in the second line,

-and the change indicated leaves the multiplication-table of the group

-unaltered. If $\alpha = p$, the symbol does not represent an isomorphism:

-if $\alpha = p + \alpha'$, the symbol represents the same isomorphism as

-\[

-\Sub{\PadTo{P^{\alpha'}}{P}\Add{,}

-     \PadTo{P^{2\alpha'}}{P^{2}}\Add{,} \Dots\Add{,}

-     \PadTo{P^{(p - 1)\alpha'}}{P^{p - 1}}}

-    {P^{\alpha'}\Add{,} P^{2\alpha'}\Add{,} \Dots\Add{,} P^{(p - 1)\alpha'}}.

-\]

-%% -----File: 256.png---Folio 240-------

-

-The group of isomorphisms of a group of prime order~$p$ is

-therefore a group of order~$p - 1$. Now the $n$th~power of the

-isomorphism

-\[

-\Sub{P}{P^{\alpha}}

-\]

-is

-\[

-\Sub{P}{P^{\alpha^{n}}}.

-\]

-

-Hence if $\alpha$~is a primitive root of the congruence

-\[

-\alpha^{p - 1} - 1 \equiv 0,\ (\mod p),

-\]

-the group of isomorphisms is a cyclical group generated by the

-isomorphism

-\[

-\Sub{P}{P^{\alpha}}.

-\]

-

-Finally, if $S$~is an operation satisfying the relations

-\[

-S^{p - 1} = 1,\quad S^{-1}PS = P^{\alpha},

-\]

-where $\alpha$~is a primitive root of~$p$, $\{S, P\}$~is the holomorph of~$G$.

-

-The reader will at once observe that this group of order~$p(p - 1)$

-is identical with the doubly transitive group of §~112. It is a

-complete group.

-

-\Par{169.} We shall consider next the case of any cyclical group.

-\index{Defining relations@\Topic{Defining relations} of a group!for the holomorph of a cyclical group}%

-\index{Holomorph@\Topic{Holomorph}!of a cyclical group|(}%

-

-Suppose, first, that $G$~is a cyclical group of order~$p^{n}$, where $p$~is

-an odd prime; and let it be generated by an operation~$S$. The group

-contains $p^{n - 1}(p - 1)$ operations of order~$p^{n}$; if $S'$~is any one of these,

-\[

-\Sub{S}{S'}

-\]

-defines a distinct isomorphism. The group of isomorphisms is therefore

-a group of order~$p^{n - 1}(p - 1)$. Moreover, since the congruence

-\[

-\alpha^{p^{n - 1}(p - 1)} - 1 \equiv 0,\ (\mod p^{n}),

-\]

-has primitive roots, the group of isomorphisms is a cyclical group.

-The holomorph of~$G$ is defined by

-\[

-S^{p^{n}} = 1,\quad J^{p^{n - 1}(p - 1)} =  1,\quad J^{-1}SJ = S^{\alpha},

-\]

-where $\alpha$~is a primitive root of the congruence

-\[

-\alpha^{p^{n - 1}(p - 1)} - 1 \equiv 0,\ (\mod p^{n}).

-\]

-

-If $G$~is a cyclical group of order~$2^{n}$, it follows, in the same way,

-that the group of isomorphisms is an Abelian group of order~$2^{n - 1}$.

-In this case, however, the congruence

-\[

-\alpha^{2^{n - 1}} - 1 \equiv 0,\ (\mod 2^{n}), \quad n > 2,

-\]

-has no primitive root, and therefore the group of isomorphisms is not

-cyclical. The congruence

-\[

-\alpha^{2^{n - 2}} - 1 \equiv 0\Add{,}\ (\mod 2^{n})

-\]

-%% -----File: 257.png---Folio 241-------

-however has primitive roots, and a primitive root~$\alpha$ of this congruence

-can always be found to satisfy the condition

-\[

-\alpha^{2^{n - 3}}  \equiv  1 + 2^{n - 1},\ (\mod 2^{n}).

-\]

-

-The powers of the isomorphism

-\[

-\Sub{S}{S^{\alpha}}

-\]

-then form a cyclical group of order~$2^{n - 2}$; and the only isomorphism

-of order~$2$ contained in it is

-\[

-\Sub{S}{S^{1 + 2^{n - 1}}}.

-\]

-

-Hence

-\[

-\Sub{S}{S^{\alpha}} \quad\text{and}\quad \Sub{S}{S^{-1}},

-\]

-the latter not being contained in the sub-group generated by the

-former, are two permutable and independent isomorphisms of orders

-$2^{n - 2}$ and~$2$. They generate an Abelian group of order~$2^{n - 1}$ which is

-the group of isomorphisms of~$G$. The corresponding holomorph is

-given by

-\begin{gather*}

-S^{2^{n}} = 1,\quad J_{1}^{2^{n - 2}} = 1,\quad J_{2}^{2} = 1,\quad J_{1}J_{2} = J_{2}J_{1}, \\

-J_{1}^{-1}SJ_{1} = S^{\alpha},\quad J_{2}SJ_{2} = S^{-1},

-\end{gather*}

-where $\alpha$~satisfies the conditions given above.

-

-If $G$~is a cyclical group of order~$4$, its group of isomorphisms is

-clearly a group of order~$2$.

-

-\Par{170.} It is now easy to construct the group of isomorphisms of

-\index{Complete group@\Topic{Complete group}!holomorph of a cyclical group of odd order is a}%

-\index{Defining relations@\Topic{Defining relations} of a group!for the holomorph of a cyclical group}%

-any cyclical group~$G$, and the corresponding holomorph. If the order

-of~$G$ is $2^{n}p_{1}^{m_{1}} p_{2}^{m_{2}} \Dots$, where $p_{1}$, $p_{2}$,~\Dots\ are odd primes, $G$~is the

-direct product of cyclical groups of orders $2^{n}$, $p_{1}^{m_{1}}$, $p_{2}^{m_{2}}$,~\Dots; and

-every isomorphism of~$G$ transforms each of these groups into itself.

-Hence if the groups of isomorphisms of these cyclical groups be

-formed, and their direct product be then constructed, every operation

-of the group so formed will give a distinct isomorphism of the group~$G$.

-Moreover, the order

-\[

-2^{n - 1}p_{1}^{m_{1} - 1}(p_{1} - 1)p_{2}^{m_{2} - 1}(p_{2} - 1) \Dots

-\]

-of this group is equal to the number of operations of~$G$ whose order

-is $2^{n}p_{1}^{m_{1}}p_{2}^{m_{2}} \Dots$, or in other words to the number of isomorphisms

-of which $G$~is capable. The group thus formed is therefore the group

-of isomorphisms of~$G$. The corresponding holomorph is clearly the

-direct product of the holomorphs of the cyclical groups of orders $2^{n}$,

-$p_{1}^{m_{1}}$,~etc.

-

-When the order of~$G$ is odd, the holomorph~$K$ is easily shewn

-to be a complete group. Suppose it to be expressed transitively,

-as in §~158, in $N$~symbols, where $N$~is the order of~$G$; if $G$~is

-not a characteristic sub-group of~$K$, let $G'$ be a group into which

-$G$~is transformed by a contragredient isomorphism of~$K$. Then $G'$~is

-%% -----File: 258.png---Folio 242-------

-a self-conjugate sub-group of~$K$; and since $G'$~is cyclical, every

-sub-group of~$G$' is a self-conjugate sub-group of~$K$. Hence a

-generating operation of~$G'$ must be a circular substitution of

-$N$~symbols. The $N$~operations of order~$2$ which transform each

-operation of~$G'$ into its own inverse therefore each keep one symbol

-fixed; hence each of them must transform every operation of~$G$

-into its inverse. But there is only one such set of operations

-of order~$2$, and therefore $G'$~cannot differ from~$G$. It follows by

-Theorem~VII, §~167 that, as $G$~is a characteristic sub-group of~$K$, the

-group~$K$ itself is complete.

-

-If the order of~$G$ is even, $K$~must contain a self-conjugate

-operation other than identity, namely the operation of order~$2$

-contained in~$G$. Moreover, $K$~admits of a contragredient isomorphism

-whose square is cogredient. From the mode of formation of~$K$,

-it is clearly sufficient to verify this when the order of~$G$ is a

-power of~$2$. The holomorph~$K$ is given by

-\begin{gather*}

-S^{2^{n}} = 1,\quad J_{1}^{2^{n - 2}} = 1,\quad J_{2}^{2} = 1,\quad J_{1}J_{2} = J_{2}J_{1}, \\

-J_{1}^{-1}SJ_{1} = S^{\alpha},\quad J_{2}SJ_{2} = S^{-1};

-\end{gather*}

-where $\alpha$~is a primitive root of

-\[

-\alpha^{2^{n - 2}} - 1 \equiv 0\ (\mod 2^{n}),

-\]

-such that

-\[

-\alpha^{2^{n - 3}} + 1 \nequiv 0\ (\mod 2^{n}).

-\]

-

-In this group, $J_{2}$~is one of $2^{n - 1}$ conjugate operations

-\[

-J_{2},\ J_{2}S^{2},\ J_{2}S^{4},\ \Dots.

-\]

-

-Now it may be directly verified that $K$~admits the isomorphism

-represented by

-\[

-\Sub{S, \PadTo{J_{1}S^{\efrac{1}{2} (1 - \alpha)}}{J_{1}}, \PadTo{J_{2}S}{J_{2}}}

-    {S, J_{1}S^{\efrac{1}{2} (1 - \alpha)}, J_{2}S},

-\]

-Moreover, since this isomorphism changes $J_{2}$ into~$J_{2}S$, it cannot be a

-cogredient isomorphism. Finally, the square of this isomorphism is

-\[

-\Sub{S, \PadTo{J_{1}S^{1 - \alpha}}{J_{1}}, \PadTo{J_{2}S^{2}}{J_{2}}}

-    {S, J_{1}S^{1 - \alpha}, J_{2}S^{2}},

-\]

-or

-\[

-\Sub{S, \PadTo{S^{-1}J_{1}S}{J_{1}}, \PadTo{S^{-1}J_{2}S}{J_{2}}}

-    {S, S^{-1}J_{1}S, S^{-1}J_{2}S},

-\]

-which is a cogredient isomorphism.

-

-\Par{171.} We shall next consider the group of isomorphisms of an

-Abelian group of order~$p^{n}$ and type $(1, 1, \dots\DPchg{\,}{,} \text{to $n$ units})$. Such a

-group is generated by $n$~independent permutable operations of

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of a cyclical group|)}%

-\index{Holomorph@\Topic{Holomorph}!of a cyclical group|)}%

-order~$p$, say

-\[

-P_{1},\ P_{2},\ \Dots,\ P_{n}.

-\]

-%% -----File: 259.png---Folio 243-------

-

-Since every operation of the group is self-conjugate and of order~$p$,

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!group of isomorphisms of}%

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$}%

-while the group contains no characteristic sub-group, there must

-be isomorphisms transforming any one operation of the group into

-any other. We may therefore begin by determining under what

-conditions the symbol

-\[

-\Sub{P_{r}}

-    {P_{1}^{a_{1r}} P_2^{a_{2r}} \Dots P_{n}^{a_{nr}}} \quad (r = 1, 2, \Dots, n),

-\]

-defines an isomorphism. This symbol replaces the operation

-$P_{1}^{x_{1}} P_{2}^{x_{2}} \Dots P_{n}^{x_{n}}$ by $P_{1}^{y_{1}} P_{2}^{y_{2}} \Dots P_{n}^{y_{n}}$, where

-\[

-\begin{aligned}

-y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\

-y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\

-\multispan{3}{\dotfill} \\

-y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},

-\end{aligned}

-\quad

-(\mod p).

-\]

-

-Unless the $p^{n}$ operations $P_{1}^{y_{1}} P_{2}^{y_{2}} \Dots P_{n}^{y_{n}}$ thus formed are all

-distinct, when for $P_{1}^{x_{1}} P_{2}^{x_{2}} \Dots P_{n}^{x_{n}}$ is put successively each of the $p^{n}$~operations

-of the group, the symbol does not represent an isomorphism.

-On the other hand, when this condition is satisfied, the symbol

-represents a permutation of the operations among themselves which

-leaves the multiplication table of the group unchanged; it is therefore

-an isomorphism.

-

-If this condition is satisfied, $x_{1}$, $x_{2}$,~\Dots, $x_{n}$ must be definite

-numbers $(\mod p)$, when $y_{1}$, $y_{2}$,~\Dots, $y_{n}$ are given; and therefore

-the above set of $n$ simultaneous congruences must be capable

-of definite solution with respect to the~$x$'s. The necessary and

-sufficient condition for this is that the determinant

-\[

-\begin{vmatrix}

-a_{11}, & a_{12}, & \Dots, & a_{1n} \\

-a_{21}, & a_{22}, & \Dots, & a_{2n} \\

-\hdotsfor{4} \\

-a_{n1}, & a_{n2}, & \Dots, & a_{nn}

-\end{vmatrix}

-\]

-should not be congruent to zero $(\mod p)$.

-

-Every distinct set of congruences of the above form, for

-which this condition is satisfied, represents a distinct isomorphism

-of the group, two sets being regarded as distinct if the

-congruence

-\[

-a_{rs} \equiv a'_{rs}\ (\mod p)

-\]

-does not hold for each corresponding pair of coefficients. Moreover,

-to the product of two isomorphisms will correspond the set of congruences

-which results from carrying out successively the operations

-indicated by the two sets that correspond to the two isomorphisms.

-%% -----File: 260.png---Folio 244-------

-

-The group of isomorphisms is therefore simply isomorphic with

-the group of operations defined by all sets of congruences

-\[

-\begin{aligned}

-  y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\

-  y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\

-  \multispan{3}{\dotfill} \\

-  y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},

-\end{aligned}

-\quad (\mod p)

-\]

-for which the relation

-\[

-\begin{vmatrix}

-  a_{11}, & a_{12}, & \Dots, & a_{1n} \\

-  a_{21}, & a_{22}, & \Dots, & a_{2n} \\

-  \hdotsfor{4} \\

-  a_{n1}, & a_{n2}, & \Dots, & a_{nn} \\

-\end{vmatrix}

-\nequiv 0\ (\mod p)

-\]

-is satisfied.

-

-\Par{172.} The group thus defined is of great importance in many

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!definition of}%

-branches of analysis. It is known as the \emph{linear homogeneous} group.

-In a subsequent Chapter we shall consider some of its more

-important properties; we may here conveniently determine its

-order. Let this be represented by~$N_{n}$, so that $N_{r}$~represents the

-number of distinct solutions of the congruences

-\[

-\begin{vmatrix}

-  a_{11}, & a_{12}, & \Dots, & a_{1r} \\

-  a_{21}, & a_{22}, & \Dots, & a_{2r} \\

-  \hdotsfor{4} \\

-  a_{r1}, & a_{r2}, & \Dots, & a_{rr} \\

-\end{vmatrix}

-\nequiv 0\ (\mod p)

-\]

-

-Since the group of isomorphisms of the Abelian group of order~$p^{n}$

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!group of isomorphisms of}%

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$}%

-transforms every one of its operations (identity excepted) into

-every other, it can be represented as a transitive substitution group

-of $p^{n} - 1$ symbols, and therefore, if $M$~is the order of the sub-group

-that keeps $P_{1}$ unchanged,

-\[

-N_{n} = (p^{n} - 1) M.

-\]

-

-Now, in the congruences corresponding to an operation that keeps

-$P_{1}$ unchanged, we have

-\[

-a_{11} \equiv 1,\ a_{12} \equiv a_{13} \equiv \Dots \equiv a_{1n} \equiv 0,\ (\mod p).

-\]

-

-Hence $M$~is equal to the number of solutions of the congruences

-\[

-\begin{vmatrix}

-  1, & 0, & \Dots, & 0 \\

-  a_{21}, & a_{22}, & \Dots, & a_{2n} \\

-  \hdotsfor{4} \\

-  a_{n1}, & a_{n2}, & \Dots, & a_{nn} \\

-\end{vmatrix}

-\nequiv 0\ (\mod p)

-\]

-

-The value of the determinant does not depend on the values of

-$a_{21}$, $a_{31}$,~\Dots, $a_{n1}$, and therefore

-\[

-M = p^{n - 1} N_{n - 1}.

-\]

-%% -----File: 261.png---Folio 245-------

-

-Hence

-\[

-N_{n} = (p^{n} - 1) p^{n - 1} N_{n - 1},

-\]

-and therefore immediately

-\[

-N_{n} = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n-1}).

-\]

-

-The reader will notice that an independent proof of this result

-\index{Holomorph@\Topic{Holomorph}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$}%

-has already been obtained in §~48. The discussion there given of

-the number of distinct ways, in which independent generating

-operations of an Abelian group of type $(1, 1, \Dots, 1)$ may be

-\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!holomorph of}%

-chosen, is clearly equivalent to a determination of the order of the

-corresponding group of isomorphisms.

-

-{\Loosen The holomorph of an Abelian group, of order~$p^{n}$ and type

-$(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$, can similarly be represented as a group of

-linear transformations to the prime modulus~$p$. Consider, in fact,

-the set of transformations}

-\[

-\begin{aligned}

-  y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n} + b_{1}, \\

-  y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n} + b_{2}, \\

-  \multispan{3}{\dotfill} \\

-  y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n} + b_{n},

-\end{aligned}

-\quad (\mod p);

-\]

-where the coefficients take all integral values $(\mod p)$ consistent

-with

-\[

-\begin{vmatrix}

-a_{11}, & a_{12}, &\Dots, &a_{1n} \\

-a_{21}, & a_{22}, &\Dots, &a_{2n} \\

-\hdotsfor{4} \\

-a_{n1}, & a_{n2}, &\Dots, &a_{nn}

-\end{vmatrix}

-\nequiv 0.

-\]

-

-The set of transformations clearly forms a group whose order is~$N_{n} p^{n}$.

-The sub-group formed by all the transformations

-\[

-y_{1} \equiv x_{1} + b_{1},\

-y_{2} \equiv x_{2} + b_{2},\ \Dots,\

-y_{n} \equiv x_{n} + b_{n},\ (\mod p),

-\]

-is an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$, and it

-is a self-conjugate sub-group. Moreover, the only operations of the

-group, which are permutable with every operation of this self-conjugate

-sub-group, are the operations of the sub-group itself; and,

-since the order of the group is equal to the order of the holomorph

-of the Abelian group, it follows that the group of transformations

-must be simply isomorphic with the holomorph of the Abelian

-group. In the simplest instance, where $p^{n}$~is~$2^{2}$, the holomorph is

-simply isomorphic with the alternating group of four symbols. In

-any case the holomorph, when expressed as in §~158, is a doubly

-transitive group of degree~$p^{n}$.

-\end{Remark}

-

-\Par{173.} It has been seen in §~142 that, except when $n = 6$, the

-symmetric group of degree~$n$ has $n$~and only~$n$ sub-groups

-of order~$n - 1!$,\DPnote{** No () in original} which form a conjugate set. Hence by

-Theorem~III, §~161, the group of isomorphisms of the symmetric

-group of degree~$n$ can be expressed, except when $n = 6$,

-%% -----File: 262.png---Folio 246-------

-as a transitive group of degree~$n$. The symmetric group of $n$~symbols

-however consists of all possible substitutions that can

-be performed on the $n$~symbols, and therefore it must coincide

-with its group of isomorphisms. Hence\footnotemark:---

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}, (1895), p.~345.}%

-

-\begin{Theorem}[VIII.]

-The symmetric group of $n$~symbols is a

-\index{Complete group@\Topic{Complete group}!symmetric group is a, except for degree~$6$}%

-\index{Symmetric group@\Topic{Symmetric group}!is a complete group, except for degree~$6$}%

-complete group, except when $n = 6$.

-\end{Theorem}

-

-\begin{Corollary}

-Except when $n = 6$, the alternating group of $n$~symbols

-\index{Alternating group@\Topic{Alternating group}!group of isomorphisms of}%

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of the alternating group}%

-admits of one and only one class of contragredient

-isomorphisms.

-\end{Corollary}

-

-For with this exception, the alternating group of degree~$n$

-has just $n$ sub-groups of order~$\frac{1}{2} (n - 1)!$.

-

-\begin{Remark}

-\Par{174.} The alternating group of degree~$6$ occurs as a special

-case of another class of groups of which we will determine the

-groups of isomorphisms. These are the doubly and the triply transitive

-groups that have been defined in §§~112,~113.

-

-The doubly transitive group of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$|(}%

-there considered has a single set of $p^{m}$ conjugate cyclical sub-groups

-of order~$p^{m} - 1$. Its group of isomorphisms can therefore be expressed

-as a transitive group of degree~$p^{m}$. Let $p^{m}(p^{m} - 1) \mu$~be the

-order of the group of isomorphisms. The order of a sub-group that

-keeps two symbols fixed is~$\mu$; and every operation of this sub-group

-must transform a cyclical sub-group of order~$p^{m} - 1$ into itself. With

-the notation of §~112, we will consider the sub-group which keeps $0$

-and $i$ fixed. Every operation of this sub-group must transform the

-cyclical sub-group generated by the congruence\footnote

-  {This and all subsequent congruences in the present section are to be taken,

-  $\pmod{p}$.}

-\[

-x' \equiv ix

-\]

-into itself; and no operation of the sub-group can be permutable

-with the given operation. If $S$~is an operation of the sub-group

-which transforms

-\[

-x' \equiv ix

-\]

-into

-\[

-x' \equiv i^{\alpha}x,

-\]

-then $S$, when represented as a substitution, is given by

-\[

-\Sub{\PadTo{i^{\alpha + 1}}{i^{2}},

-     \PadTo{i^{2\alpha + 1}}{i^{3}},

-     \PadTo{i^{3\alpha + 1}}{i^{4}}, \Dots}

-    {i^{\alpha + 1}, i^{2\alpha + 1}, i^{3\alpha + 1}, \Dots}.

-\]

-

-Now $S$~must transform the sub-group of order~$p^{m}$ into itself; it

-must therefore be permutable with that operation of this sub-group

-which changes $0$ into~$i$.

-%% -----File: 263.png---Folio 247-------

-

-This operation is given by

-\[

-x' \equiv x + i,

-\]

-and if we denote it by~$T$, then $S^{-1}TS$ changes $i^{y + 1}$ into $i^{\alpha z + 1}$, where

-\[

-i^{\efrac{y}{\alpha} + 1} + i \equiv i^{z + 1}.

-\]

-

-Now $T$ changes $i^{y + 1}$ into $i^{y + 1} + i$. Hence, since $S$~and $T$ are

-permutable, we must for all values of~$y$ have the simultaneous

-congruences

-\[

-i^{\efrac{y}{\alpha} + 1} + i \equiv i^{z + 1},

-\]

-and

-\[

-i^{y + 1} + i \equiv i^{\alpha z + 1}.

-\]

-

-Eliminating $z$, the congruence

-\[

-1 + i^{y} \equiv (1 + i^{\efrac{y}{\alpha}})^{\alpha}

-\]

-must hold for all values of~$y$ from $0$ to~$p^{m} - 1$. If $\alpha$~is not a power

-of~$p$, this congruence involves an identity of the form

-\[

-a_{1}i^{\alpha_{1}} + a_{2}i^{\alpha_{2}} + \Dots \equiv 0,

-\]

-where all the indices are less than $p^{m} - 1$; and this is impossible.

-Hence the only possible values of~$\alpha$ are $p$, $p^{2}$, $p^{3}$,~\Dots; and the

-greatest possible value of~$\mu$ is~$m$.

-

-Now the congruence

-\[

-x' \equiv x^{p}

-\]

-defines a substitution performed on the $p^{m}$~symbols permuted by the

-group, and this substitution is permutable with the group. For if

-we denote this operation by~$J$, and any operation

-\[

-x' \equiv \alpha x + \beta

-\]

-of the group by $\Sigma$, then $J^{-1} \Sigma J$ is

-\[

-x' \equiv \alpha^{p}x + \beta^{p},

-\]

-another operation of the group. Moreover, $J$~clearly transforms

-\[

-x' \equiv ix

-\]

-into its $p$th~power.

-

-The group of isomorphisms of the doubly transitive group of

-order~$p^{m}$ is therefore the group defined by

-\[

-x' \equiv ix,\quad x' \equiv x + i,\quad x' \equiv x^{p};

-\]

-where $i$~is a primitive root of the congruence

-\[

-i^{p^{m - 1}} \equiv 1.

-\]

-

-From this it immediately follows that the group of isomorphisms

-of the triply transitive group, of degree~$p^{m} + 1$ and order

-\[

-(p^{m} + 1) p^{m} (p^{m} - 1),

-\]

-%% -----File: 264.png---Folio 248-------

-defined by the congruences

-\[

-x' \equiv \frac{\alpha x + \beta}{\gamma x + \delta},

-\]

-where $\alpha$,~$\beta$, $\gamma$,~$\delta$ satisfy the conditions of §~113, is the group of order

-$(p^{m} + 1) p^{m} (p^{m} - 1) m$ obtained by combining the previous congruences

-with

-\[

-x' \equiv x^{p}.

-\]

-

-In fact, it may be immediately verified that the operation given

-by this congruence is permutable with the group, and does not give

-a cogredient isomorphism of the group. Moreover, by Theorem~III,

-§~161, the group of isomorphisms of the given group can be expressed

-as a transitive group of degree~$p^{m} + 1$; and therefore, among a class of

-contragredient isomorphisms, there must be some transforming into

-itself a sub-group which keeps one symbol fixed. Hence the order

-of the group of isomorphisms cannot exceed

-\[

-(p^{m} + 1) p^{m} (p^{m} - 1) m.

-\]

-

-When $p$ is an odd prime, the triply transitive group of degree

-$p^{m} + 1$, which may be defined by

-\[

-x' \equiv -\frac{1}{x},\quad x' \equiv ix,\quad x' \equiv x + i,

-\]

-contains as a self-conjugate sub-group a doubly transitive group

-defined by

-\[

-x' \equiv -\frac{1}{x},\quad x' \equiv i^{2}x,\quad x' \equiv x + i.

-\]

-

-It will be shewn in Chapter~XIV that this is a simple group.

-When $p^{m}$~is equal to~$3^{2}$, it is easy to verify that this group is simply

-isomorphic with the alternating group of degree~$6$.

-

-The group of isomorphisms of this simple group can be expressed

-as a transitive group of degree~$p^{m} + 1$, and must clearly contain the

-triply transitive group of order $(p^{m} + 1)p^{m}(p^{m} - 1)$ self-conjugately.

-It therefore coincides with the group of order $(p^{m} + 1)p^{m}(p^{m} - 1)m$,

-which has just been determined. The latter group is therefore

-(Theorem~VI, Cor.\ §~166) a complete group. Further, if the simple

-group be denoted by~$G$ and the group of isomorphisms by~$L$, the

-factor group~$\dfrac{L}{G}$ will be determined by

-\[

-x' \equiv ix,\quad x' \equiv x^{p},

-\]

-when all operations of~$G$ are treated as the identical operation.

-Denoting these operations by $I$~and~$J$, then $J^{-1}IJ$ is~$I^{p}$, which is

-the same as $I$ multiplied by an operation of~$G$. Hence $\dfrac{L}{G}$~is an

-Abelian group generated by two independent operations of orders $2$

-and~$m$.

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$|)}%

-%% -----File: 265.png---Folio 249-------

-

-The group of isomorphisms of the alternating group of degree~$6$

-is therefore a group of order~$1440$; and the symmetric group of

-degree~$6$ admits a single class of contragredient isomorphisms\footnotemark.

-\end{Remark}

-\footnotetext{Hölder, \textit{loc.~cit.}\ p.~343.}%

-\index{Holder@\Topic{Hölder}, quoted}%

-

-\Par{175.} Let $P$~be any group whose order is the power of a

-prime, and let

-\[

-P,\ P_{1},\ P_{2},\ \Dots,\ P_{n},\ 1

-\]

-of orders

-\[

-p^{\alpha},\ p^{\alpha_{1}},\ p^{\alpha_{2}},\ \Dots,\ P^{\alpha_{n}},\ 1

-\]

-be a characteristic series (§~163) of~$P$. Every isomorphism

-of~$P$ must transform $P_{r}$ and $P_{r + 1}$ into themselves, and therefore

-also $\dfrac{P_{r}}{P_{r + 1}}$ into itself. Suppose now that an isomorphism~$I$

-of~$P$ transforms every operation of $P_{r + 1}$ into itself and

-every operation of $\dfrac{P_{r}}{P_{r + 1}}$ into itself. If $S$~is any operation of~$P_{r}$,

-not contained in~$P_{r + 1}$, $I$~must transform $S$ into~$SA$, where

-$A$~is some operation of~$P_{r + 1}$; so that, if $p^{\mu}$~is the order of~$A$,

-$I$~transforms the operations of the set~$SP_{r + 1}$ cyclically

-among themselves in sets of~$p^{\mu}$. Similarly, if $S'$~is an operation

-of~$P_{r}$ not contained in the set~$SP_{r + 1}$, $I$~will transform the

-operations of the set $S'P_{r + 1}$ cyclically among themselves in

-sets of~$p^{\mu'}$. Hence the order of the isomorphism~$I$ is a multiple

-of~$p$; and any isomorphism, that transforms every operation

-of~$P_{r + 1}$ into itself and every operation of~$\dfrac{P_{r}}{P_{r + 1}}$ into itself and is

-of order prime to~$p$, will transform every operation of~$P_{r}$ into

-itself. Therefore, the only isomorphism of~$P$, that transforms

-every operation of each of the groups

-\[

-\frac{P}{P_{1}},\ \frac{P_{1}}{P_{2}},\ \Dots,\ \frac{P_{n - 1}}{P_{n}},\ P_{n}

-\]

-into itself and is of order prime to~$p$, is the identical isomorphism.

-Now each of these groups is an Abelian group, whose

-operations are all of order~$p$; and it has been shewn that,

-if $p^{m}$~is the order of such a group, the order of its group of

-isomorphisms is

-\[

-(p^{m} - 1)(p^{m} - p)\Dots (p^{m} - p^{m - 1}).

-\]

-%% -----File: 266.png---Folio 250-------

-

-Every isomorphism of~$P$, whose order is relatively prime to~$p$,

-must therefore be such that its order is a factor of one of the

-expressions of the above form, obtained by writing

-\[

-\alpha - \alpha_{1},\ \alpha_{1} - \alpha_{2},\ \Dots,\ \alpha_{n},

-\]

-in succession for~$m$. If then $k$~is the greatest of these numbers,

-the order of any isomorphism of~$P$, whose order is relatively

-prime to~$p$, is a factor of

-\[

-(p - 1) (p^{2} - 1) \Dots (p^{k} - 1).

-\]

-

-Herr Frobenius\footnote

-  {``Ueber auflösbare Gruppen,~\VolNo{II},'' \textit{Berliner Sitzungsberichte}, 1895, p.~1028.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\index{Symbol@\Topic{Symbol}!theta@$\vartheta(P)$ and $\theta(P)$, definition of}%

-has introduced the symbol~$\vartheta(P)$ to denote

-this product. If $G$~is a group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}}$, where

-$p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are distinct primes, and if $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ are

-groups of orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n}^{\alpha_{n}}$ contained in~$G$, we shall use

-the symbol~$\vartheta(G)$ to denote the least common multiple of

-\[

-\vartheta(P_{1}),\ \vartheta(P_{2}),\ \Dots,\ \vartheta(P_{n}).

-\]

-

-\Par{176.} If $P'$~is a sub-group of a group~$P$ of order~$p^{\alpha}$, $\vartheta(P')$~is

-not necessarily a factor of~$\vartheta(P)$. For instance, the group of

-order~$p^{4}$, generated by the four operations $A$,~$B$, $C$,~$D$ of order~$p$,

-of which $A$~and $B$ are self-conjugate while

-\[

-D^{-1}CD = CA,

-\]

-has a characteristic sub-group $\{A, B\}$ of order~$p^{2}$, and it

-has no characteristic sub-group of order~$p^{3}$. Hence

-\[

-\vartheta(P) = (p - 1) (p^{2} - 1).

-\]

-

-The sub-group~$P'$ however, which is generated by $A$,~$B$ and~$C$,

-is an Abelian group whose operations are all of order~$p$; and

-therefore

-\[

-\vartheta(P') = (p - 1) (p^{2} - 1) (p^{3} - 1).

-\]

-

-So again, for the Abelian group~$P$ of order~$p^{3}$, generated by

-$A$~and~$B$, where

-\[

-A^{p^{2}} = 1,\quad B^{p} = 1,\quad AB = BA,

-\]

-we have

-\[

-\vartheta(P) = p - 1;

-\]

-while its sub-group~$P'$, generated by $A^{p}$~and~$B$, is an Abelian

-group whose operations are all of order~$p$, so that

-\[

-\vartheta(P') = (p - 1) (p^{2} - 1).

-\]

-%% -----File: 267.png---Folio 251-------

-

-Suppose now that $P$~is an Abelian group of order~$p^{\alpha}$,

-generated by $\rho$~independent and permutable operations. For

-such a group, we define\footnote

-  {Frobenius, \textit{loc.~cit.}\ p.~1030.}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\index{Symbol@\Topic{Symbol}!theta@$\vartheta(P)$ and $\theta(P)$, definition of}%

-a new symbol~$\theta(P)$ by the equation

-\[

-\theta(P) = (p - 1) (p^{2} - 1) \Dots (p^{\rho} - 1).

-\]

-

-In forming the characteristic series of~$P$ in §~164, we

-commenced with the series of groups~$G_{r}$ ($r = 1, 2, \Dots$), such

-that $G_{r}$~consists of all the operations of~$P$ satisfying the

-relation

-\[

-S^{p^{r}} = 1.

-\]

-

-Since $P$~is generated by $\rho$~independent operations, the order

-of~$G_{1}$ is~$p^{\rho}$, and the order of~$\dfrac{G_{r + 1}}{G_{r}}$ cannot be greater than~$p^{\rho}$. If

-the generating operations of~$P$ are all of the same order~$\beta$, the

-series

-\[

-P,\ G_{\beta - 1},\ G_{\beta - 2},\ \Dots,\ G_{1},\ 1

-\]

-is a complete characteristic series of~$P$, and each factor-group

-$\dfrac{G_{r + 1}}{G_{r}}$ is of order~$p^{\rho}$. In this case, therefore,

-\[

-\vartheta(P) = \theta(P).

-\]

-

-If however the generating operations are not all of the same

-order, $G_{1}$~will not be the last term of the complete characteristic

-series; nor will $G_{r + 1}$~and $G_{r}$ be consecutive terms in the series, if

-the order of~$\dfrac{G_{r + 1}}{G_{r}}$ is~$p^{\rho}$. Hence, in this case, $\vartheta(P)$~will be a

-factor of~$\theta(P)$.

-

-If now $P'$~is any sub-group of~$P$, then it has been seen

-(§~46) that the number of independent generating operations of~$P'$

-is equal to or less than~$\rho$. Hence $\theta(P')$~is equal to or is a

-factor of~$\theta(P)$.

-

-\Par{177.} \begin{Theorem}[IX.]

-If a group~$G$, of order~$N$, is transformed

-into itself by an operation~$S$, whose order is relatively

-prime to~$N\vartheta(G)$, every operation of~$G$ is permutable with~$S$\footnotemark.

-\end{Theorem}

-\footnotetext{Frobenius, \textit{loc.~cit.}\ p.~1030.}

-

-Let $p_{1}^{\alpha_{1}}$~be the highest power of a prime~$p_{1}$ which divides~$N$.

-%% -----File: 268.png---Folio 252-------

-\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!limitation on the order of}%

-The number of sub-groups of~$G$, and therefore also of $\{S, G\}$,

-whose order is~$p_{1}^{\alpha_{1}}$ is a factor of~$N$; and since the order of~$S$ is

-relatively prime to~$N$, one at least of them, say~$P_{1}$, must be

-transformed into itself by~$S$. Now the order of~$S$ is relatively

-prime both to~$p_{1}$ and to~$\vartheta(P_{1})$; and therefore the isomorphism of~$P$

-given on transforming its operations by~$S$ must be the

-identical isomorphism. In other words, $S$~is permutable with

-every operation of~$P_{1}$. In the same way it follows that, if

-$p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ are the highest powers of $p_{2}$, $p_{3}$,~\Dots\ that divide~$N$,

-there must be sub-groups of orders $p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ with every

-operation of each of which $S$~is permutable. But these groups

-of orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ generate~$G$; therefore $S$~is permutable

-with every operation of~$G$.

-

-\begin{Corollary}

-If the order of an isomorphism of~$G$ is relatively

-prime to~$N$, it must be a factor of~$\vartheta(G)$.

-\end{Corollary}

-

-It follows immediately, from the theorem, that there is no

-isomorphism of~$G$ whose order contains a prime factor not

-occurring in~$N\vartheta(G)$. Suppose, if possible, that the group of

-isomorphisms of~$G$ contains an operation~$S$ of order~$q^{m}$, where

-$q$~is a prime which does not divide~$N$, and that the highest

-power of~$q$ that divides~$\vartheta(G)$ is~$q^{m'}$, where $m' < m$. If $p_{r}^{\alpha_{r}}$~is

-the highest power of any prime~$p_{r}$ which divides~$N$, $S$~must

-transform some sub-group~$P_{r}$ of order~$p_{r}^{\alpha_{r}}$ into itself. Since

-$q^{m}$~is not a factor of~$\vartheta(P_{r})$, some power of~$S$ must be permutable

-with every operation of~$P_{r}$. Hence, as in the theorem,

-it follows that some power of~$S$, certainly~$S^{q^{m - 1}}$, must be permutable

-with every operation of~$G$. But no operation of the

-group of isomorphisms of~$G$, except identity, is permutable

-with every operation of~$G$. Hence the group of isomorphisms

-cannot contain an operation of order~$q^{m}$, if $m > m'$; and therefore

-there is no isomorphism of~$G$ whose order contains a

-higher power of~$q$ than~$q^{m'}$. If then $q^{r} q'^{r'} \Dots$, a number

-relatively prime to~$N$, is the order of an isomorphism of~$G$, all

-the numbers $q^{r}$, $q'^{r'}$,~\Dots\ are factors of~$\vartheta(G)$, and so also is

-their product.

-

-\Par{178.} The method, by which it has been shewn in §~175 that

-the order of any isomorphism of~$P_{r}$ which transforms every

-operation of each of the groups $\dfrac{P_{r}}{P_{r + 1}}$ and $P_{r + 1}$ into itself is a

-%% -----File: 269.png---Folio 253-------

-power of~$p$, may be used to obtain the following more general

-result.

-

-\begin{Theorem}[X.]

-If $H$~is a self-conjugate sub-group of~$G$, the

-order of an isomorphism of~$G$, which transforms every operation

-of each of the groups $\dfrac{G}{H}$ and $H$ into itself, is a factor of the order

-of~$H$.

-\end{Theorem}

-

-If $S$~is any operation of~$G$ not contained in~$H$, the

-isomorphism will change $S$ into~$Sh$, where $h$~is some operation

-of~$H$. If then $m$~is the order of~$h$, the isomorphism transforms

-\[

-S,\ Sh,\ Sh^{2},\ \Dots,\ Sh^{m - 1}

-\]

-cyclically; and therefore it transforms all the operations of the

-set~$SH$ in cycles of $m$~each. If $S'$~is any operation of~$G$ not contained

-in~$SH$, the isomorphism will interchange the operations

-of the set~$S'H$ among themselves in cycles of $m'$~each, where $m'$

-again is the order of some operation of~$H$. The isomorphism,

-when expressed as a substitution performed on the operations

-of~$G$, will consist of a number of cycles of $m$, $m'$,~\Dots\ symbols;

-and its order is therefore the least common multiple of

-$m$, $m'$,~\Dots. Now if $q$~is any prime that divides the order

-of~$H$, and $q^{n}$~the highest power of~$q$ that occurs as the order of

-a cyclical operation of~$H$, no power of~$q$ higher than~$q^{n}$ can

-occur in any of the numbers $m$, $m'$,~\Dots; and $q^{n}$~is therefore

-the highest power of~$q$ that can occur in their least common

-multiple. This least common multiple, which is the order of

-the isomorphism, must therefore divide the order of~$H$.

-

-\Par{179.} \begin{Ex}[1.] Shew that, for the group of order~$p^{3}$ defined by

-\begin{gather*}

-P^{p} = 1, \quad Q^{p} = 1, \quad R^{p} = 1, \quad Q^{-1}PQ = PR, \\

-RP = PR,\quad RQ = QR,

-\end{gather*}

-the symbol

-\[

-\Sub{\PadTo{P^{x} Q^{y} R^{z}}{P},

-     \PadTo{P^{\alpha} Q^{\beta} R^{\gamma}}{Q},

-     \PadTo{R^{n}}{R}}

-    {P^{x} Q^{y} R^{z}, P^{\alpha} Q^{\beta} R^{\gamma}, R^{n}}

-\]

-gives an isomorphism if

-\[

-\beta x - \alpha y \equiv n,\ (\mod p).

-\]

-Hence determine the order of the group of isomorphisms.

-\end{Ex}

-%% -----File: 270.png---Folio 254-------

-

-\begin{Ex}[2.] Shew that, for the group of order~$p^{3}$ defined by

-\[

-P^{p^{2}} = 1, \quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p},

-\]

-the symbol

-\[

-\Sub{\PadTo{P^{x} Q^{y}}{P}\Add{,} \PadTo{P^{\alpha p}Q}{Q}}

-    {P^{x} Q^{y}, P^{\alpha p}Q}

-\]

-gives an isomorphism if $x$ is not a multiple of~$p$; and determine the

-order of the group of isomorphisms.

-\end{Ex}

-

-\begin{Ex}[3.] Shew that the group of isomorphisms of the group of

-order~$2^{n}$, defined by (§~63)

-\[

-P^{2^{n - 1}} = 1,\quad Q^{2} = P^{2^{n - 2}}, \quad Q^{-1}PQ = P^{-1},

-\]

-is of order~$2^{2n - 3}$, when $n > 3$. If $n = 3$, its order is~$24$ and it is

-simply isomorphic with the last type but one of §~84.

-\end{Ex}

-

-\begin{Ex}[4.] If $G$~is a complete group of order~$N$, shew that the order

-of~$K$, the holomorph of~$G$, is~$N_{2}$; and that the order of the holomorph

-of~$K$ is~$2N^{4}$.

-\end{Ex}

-%% -----File: 271.png---Folio 255-------

-

-

-\Chapter{XII.}{On the Graphical Representation of Group\protect\footnotemark.}

-\index{Dyck@\Topic{Dyck}, quoted}%

-\footnotetext{The investigations of this Chapter are due to Dyck, ``Gruppentheoretische

-  Studien,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XX}, (1882), pp.~1--44. We have followed Dyck's

-  memoir closely except in two respects. Firstly, we have used a rather more

-  definite geometrical operation than that of the memoir; and secondly, we have

-  not specially considered a regular and symmetric division of a closed surface,

-  apart from a merely regular division.}

-

-\Par{180.} \First{Our} discussions hitherto have been confined exclusively

-to groups of finite order. When however, as we now

-propose to do, we consider a group in relation to the operations

-that generate it, it becomes almost necessary to deal, incidentally

-at least, with groups whose order is not finite; for it is not

-possible to say a~priori what must be the number and the nature

-of the relations between the given generating operations, which

-will ensure that the order of the resulting group is finite.

-

-Many of the definitions given in respect of finite groups

-may obviously be extended at once to groups containing an

-infinite number of operations. Among these may be specially

-mentioned the definitions of a sub-group, of conjugate operations

-and sub-groups, of self-conjugate sub-groups, of the

-relation of isomorphism between two groups and of the

-factor-group given by this relation. In regard to the last of

-them, the isomorphism between two groups, one at least of

-which is not of finite order, may be such that to one operation

-of the one group there correspond an infinitely great number of

-operations of the other. On the other hand, all the results

-obtained for finite groups, which depend directly or indirectly on

-the order of the group, necessarily become meaningless when

-the group is not a group of finite order.

-%% -----File: 272.png---Folio 256-------

-

-\Par{181.} Suppose that

-\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations}%

-\index{Group@\Topic{Group}!general}%

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}

-\]

-represent any $n$~distinct operations which can be performed,

-directly or inversely, on a common object, and that between

-these operations no relations exist. Then the totality of the

-operations represented by

-\[

-\Dots S_{p}^{\alpha} S_{q}^{\beta} S_{r}^{\gamma} \Dots,

-\]

-where the number of factors is any whatever and the indices

-are any positive or negative integers, form a group~$G$ of infinite

-order, which is generated by the $n$~operations. If, moreover,

-whenever such a succession of factors as $S_{p}^{\alpha} S_{p}^{\beta}$ occurs in the

-above expression, it is replaced by $S_{p}^{\alpha + \beta}$, each operation of

-the group can be expressed in one way and in one way only by

-an expression of the above form, which is then called \emph{reduced}.

-

-It will sometimes be convenient to avoid the use of negative

-indices in the expression of any operation of the group. To this

-end we may write

-\[

-S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1,

-\]

-so that $S_{n + 1}$ is a definite operation of the group; then

-\[

-S_{r}^{-1} = S_{r + 1}S_{r + 2} \Dots S_{n}S_{n + 1}S_{1} \Dots S_{r - 1}, \quad (r = 1, 2, \Dots, n).

-\]

-By using these relations to replace all negative powers of

-operations wherever they occur, we may represent every

-operation of the group in a single definite way by means of

-the $n + 1$ operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},\ S_{n + 1},

-\]

-with positive indices only.

-

-The group, thus defined and represented, is the most

-general group conceivable that is generated by $n$ distinct

-operations. Any two such groups, for which $n$~is the same, are

-simply isomorphic with each other.

-

-Suppose now that

-\[

-\Bar{S}_{1},\ \Bar{S}_{2},\ \Dots,\ \Bar{S}_{n}

-\]

-represent $n$ distinct operations, but that, instead of being entirely

-independent, they are connected by a relation of the form

-\[

-\Bar{S}_{p}^{a} \Bar{S}_{q}^{b} \Dots \Bar{S}_{r}^{c} = 1,

-\]

-which will be represented by

-\[

-f(\Bar{S}_{i}) = 1.

-\]

-%% -----File: 273.png---Folio 257-------

-\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations!relation of special groups to}%

-\index{Group@\Topic{Group}!special}%

-\index{Isomorphism@\Topic{Isomorphism} of general and special groups}%

-

-If $\Bar{G}$ is the group generated by these operations, an isomorphism

-may be established between $G$ and $\Bar{G}$ by taking $\Bar{S}_{i}$

-($i = 1, 2, \Dots, n$) as the operation of~$\Bar{G}$ that corresponds to

-the operation~$S_{i}$ of~$G$.

-

-Then to every operation of $G$

-\[

-\Dots S_{p}^{\alpha} S_{q}^{\beta} S_{r}^{\gamma} \Dots

-\]

-will correspond a single definite operation

-\[

-\Dots \Bar{S}_{p}^{\alpha} \Bar{S}_{q}^{\beta} \Bar{S}_{r}^{\gamma} \Dots

-\]

-of~$\Bar{G}$; for the supposition that two distinct operations of~$\Bar{G}$

-correspond to the same operation of~$G$ leads to the result that

-between the generating operations of~$G$ there is a relation,

-which is not the case. On the other hand, to the identical

-operation of~$\Bar{G}$ there will correspond an infinite number of

-distinct operations of~$G$, namely those which are formed by

-combining together in every possible way all operations of~$G$ of

-the form

-\[

-R^{-1} f(S_{i})R,

-\]

-where $R$~is any operation of~$G$. These operations of~$G$ form a

-self-conjugate sub-group~$H$; the corresponding factor-group

-$\dfrac{G}{H}$ is simply isomorphic with~$\Bar{G}$.

-

-If between the generating operations of~$\Bar{G}$ there are several

-independent relations

-\[

-f_{1}(\Bar{S}_{i}) = 1,\quad

-f_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad

-f_{m}(\Bar{S}_{i}) = 1,

-\]

-it may be shewn exactly as before that the groups $G$ and $\Bar{G}$ are

-isomorphic in such a way that to the identical operation of~$\Bar{G}$

-there corresponds that self-conjugate sub-group of~$G$, which is

-formed by combining in every possible way all the operations

-of~$G$ of the form

-\[

-R^{-1} f_{j}(S_{i}) R,\quad (j = 1, 2, \Dots, m).

-\]

-

-\Par{182.} We may at once extend the result of the preceding

-paragraph in the following way:---

-

-\begin{Theorem}[I.]

-If $G$~is the group generated by the $n$~operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},

-\]

-between which the $m$~relations

-\[

-f_{1}(S_{i}) = 1,\quad

-f_{2}(S_{i}) = 1,\quad \Dots,\quad

-f_{m}(S_{i}) = 1,

-\]

-%% -----File: 274.png---Folio 258-------

-exist; and if $\Bar{G}$~is the group generated by the $n$~operations

-\[

-\Bar{S}_{1},\ \Bar{S}_{2},\ \Dots,\ \Bar{S}_{n},

-\]

-which are connected by the same $m$~relations

-\[

-f_{1}(\Bar{S}_{i}) = 1,\quad

-f_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad

-f_{m}(\Bar{S}_{i}) = 1,

-\]

-as hold between the generating operations of~$G$, and by the

-further $m'$~relations

-\[

-g_{1}(\Bar{S}_{i}) = 1,\quad

-g_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad

-g_{m'}(\Bar{S}_{i}) = 1;

-\]

-then $\Bar{G}$~is simply isomorphic with the factor-group~$\dfrac{G}{H}$; where $H$~is

-that self-conjugate sub-group of~$G$, which results from combining

-in every possible way all operations of the form

-\[

-R^{-1} g_{j}(S_{i}) R,\quad (j= 1, 2, \Dots, m'),

-\]

-\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations!relation of special groups to}%

-\index{Isomorphism@\Topic{Isomorphism} of general and special groups}%

-$R$ being any operation of~$G$.

-\end{Theorem}

-

-{\Loosen In proving this theorem, it is sufficient to notice that, if we

-take $\Bar{S}_{i}$ ($i = 1, 2, \Dots, n$) as the operation of~$\Bar{G}$ which corresponds

-to the operation~$S_{i}$ of~$G$, then to each operation of~$G$ a single

-definite operation of~$\Bar{G}$ will correspond, while to the identical

-operation of~$\Bar{G}$ there corresponds the self-conjugate sub-group~$H$

-of~$G$.}

-

-\begin{Remark}

-The theorem just stated is of such a general nature that it is

-perhaps desirable to illustrate it by considering shortly some simple

-examples.

-

-Let us take first the case of a group~$G$, generated by two independent

-operations $S_{1}$ and~$S_{2}$, subject to no relations; and let us

-suppose that the single relation

-\[

-\Bar{S}_{1} \Bar{S}_{2} = \Bar{S}_{2} \Bar{S}_{1}

-\]

-holds between the generating operations of~$\Bar{G}$. The self-conjugate

-sub-group~$H$ of~$G$ then consists of all the operations

-\[

-\Dots S_{1}^{\alpha_{n}} S_{2}^{\beta_{n}} \Dots

-\]

-of~$G$ which reduce to identity if we regard $S_{1}$~and $S_{2}$ as permutable;

-or, in other words, of those operations of~$G$ for which the relations

-\[

-\sum \alpha_{n} = 0,\quad \sum \beta_{n} = 0

-\]

-simultaneously hold.

-

-In respect of this sub-group, the operations of~$G$ can be divided

-into an infinite number of classes of the form

-\[

-S_{1}^{p} S_{2}^{q} H.

-\]

-%% -----File: 275.png---Folio 259-------

-

-For the operations of the class $S_{1}^{p} S_{2}^{q}H$, multiplied by those of

-the class $S_{1}^{p'} S_{2}^{q'}H$, give always operations of the class

-\[

-S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}H,

-\]

-since $H$~is a self-conjugate sub-group; and, because

-\[

-S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}

-  = S_{1}^{p + p'} · S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q} · S_{2}^{q + q'},

-\]

-while $S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q}$ belongs to~$H$, the class $S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'} H$ is the

-same as $S_{1}^{p + p'} S_{2}^{q + q'} H$. Hence the operations of any two given classes,

-multiplied in either order, give the same third class; and therefore

-the group~$\dfrac{G}{H}$ is an Abelian group generated by two permutable,

-but otherwise unrestricted, operations.

-

-As a second illustration, we will choose a case in which $\Bar{G}$~is of

-finite order. Let $G$ be generated by the operations $S$~and~$T$, which

-satisfy the relations

-\[

-S^{3} = 1,\quad T^{3} = 1,\quad (ST)^{3} = 1;

-\]

-and for~$\Bar{G}$, suppose that the generating operations satisfy the \DPtypo{addtional}{additional}

-relation

-\[

-(ST^{2})^{3} = 1.

-\]

-

-Then $H$~is formed by combining in all possible ways the

-operations

-\[

-R^{-1} (ST^{2})^{3} R.

-\]

-

-Now it may be easily verified that in~$G$, the operation~$ST^{2}$

-belongs to a set of three conjugate operations

-\[

-ST^{2},\quad TST,\quad T^{2}S;

-\]

-and that these three operations are permutable among themselves,

-while their product is identity. Hence $H$~consists of the Abelian

-group

-\[

-(ST^{2})^{3\alpha} (TST)^{3\beta};

-\]

-and in respect of~$H$, $G$~may be divided into $27$~classes of the form

-\[

-S^{x} (ST^{2})^{y} (TST)^{z} H,\quad (x, y, z = 0, 1, 2).

-\]

-

-The group~$\Bar{G}$ will be defined by the laws according to which

-these $27$~classes combine among themselves; and the reader will

-have no difficulty in verifying that it is isomorphic with the non-Abelian

-group of order~$27$, whose operations are all of order~$3$ (§~73).

-\end{Remark}

-

-\begin{Ex} If $S_{1} (= 1)$, $S_{2}$,~\Dots, $S_{N}$ are the operations of a group~$G$

-of finite order, prove that the totality of the operations of the form

-$S_{p}^{-1} S_{q}^{-1} S_{p} S_{q}$ generate a self-conjugate sub-group~$H$; and that the

-group~$\dfrac{G}{H}$ is an Abelian group. Shew also that, if $H'$~is a self-conjugate

-sub-group of~$G$ and if $\dfrac{G}{H'}$~is Abelian, then $H'$~contains~$H$.

-\index{Miller@\Topic{Miller}, quoted}%

-\Attrib{Miller, \textit{Quarterly Journ.\ of Math.}, Vol.~\VolNo{XXVIII}, (1896), p.~266.}

-\end{Ex}

-%% -----File: 276.png---Folio 260-------

-

-\Par{183.} For the further discussion of a group, as defined by its

-\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%

-generating operations and the relations between them, a suitable

-graphical mode of representation becomes of the greatest assistance.

-To this we shall now proceed.

-

-In the simple case in which the group is generated by a

-single unrestricted operation, such a representation may be

-constructed as follows. Let $C_{1}$~and $C_{-1}$ be two circles which

-touch each other; $C_{2}$~and $C_{-2}$ the inverses of~$C_{-1}$ in~$C_{1}$ and $C_{1}$

-in~$C_{-1}$; $C_{3}$~and $C_{-3}$ the inverses of~$C_{-2}$ in~$C_{1}$ and $C_{2}$ in~$C_{-1}$, and

-so on. These circles (fig.~1) divide the plane in which they are

-% [Illustration: Fig. 1.]

-\Figure{1}{fig01}

-drawn into an infinite number of crescent-shaped spaces. Suppose

-now that the space between $C_{1}$~and~$C_{-1}$ is left white, and

-the spaces between $C_{1}$~and~$C_{2}$ and between $C_{-1}$~and~$C_{-2}$ (on either

-side of this white space) are coloured black; the next pair on

-either side left white, the next coloured black, and so on. Then

-any white space may be transformed into any other (and any

-black into any other) by an even number of inversions at the

-circles $C_{-1}$ and~$C_{1}$; and if $S$~denote\DPnote{** [sic]} the operation consisting

-of an inversion at~$C_{-1}$ followed by an inversion at~$C_{1}$,

-the space between $C_{-1}$~and~$C_{1}$ will be transformed into

-another perfectly definite white space by the operation~$S^{n}$,

-while conversely the operation necessary to transform the space

-between $C_{-1}$~and~$C_{1}$ into any other given white space will be a

-definite power of~$S$. Hence if one of the white spaces, say

-%% -----File: 277.png---Folio 261-------

-that between $C_{-1}$~and~$C_{1}$, is taken to correspond to the identical

-operation, there is then a unique correspondence between the

-white spaces and the operations of the group generated by

-the unrestricted operation~$S$; and the figure that has been

-constructed gives a graphical representation of the group. It

-should be noticed that the actual geometrical process of inversion,

-which has been here used to construct the spaces

-corresponding to the operations of the group, is in no way

-essential to the graphical representation. It is however convenient

-as giving definiteness to the construction; and later,

-when we deal with the case of a general group, such definiteness

-becomes almost a necessity.

-

-In a precisely similar manner, the group generated by a

-single relation~$S$, satisfying the relation

-\[

-S^{n} = 1,

-\]

-may be treated. In this case, we take two circles $C_{-1}$ and~$C_{1}$

-intersecting at an angle~$\dfrac{\pi}{n}$, and from these form, as before,

-the circles obtained by successive inversions. This gives a

-\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%

-% [Illustration: Fig. 2.]

-\Figure{2}{fig02}

-finite series of $n$~circles, each of which intersects the two next to

-it on either side at angles~$\dfrac{\pi}{n}$, while the $n$~circles divide the

-%% -----File: 278.png---Folio 262-------

-plane into $2n$~spaces. If these are left white and coloured

-black in alternate succession, and if one of the white is taken

-to correspond to the identical operation, there is a unique

-correspondence between the white spaces and the operations of

-the group generated by~$S$, where $S$~represents the result of

-successive inversions first at~$C_{-1}$ and then at~$C_{1}$.

-

-This operation obviously satisfies the relation

-\[

-S^{n} = 1

-\]

-and no simpler relation; so that the figure gives a graphical

-representation of a cyclical group of order~$n$.

-

-The systems of circles in figures 1~and 2 have a common

-\index{Graphical representation@\Topic{Graphical representation}!of a general group|(}%

-geometrical property which may be noticed here as it will be of

-use in the sequel. Successive inversions at any one of the pairs

-$C_{-1}$~and~$C_{1}$, $C_{1}$~and~$C_{2}$, $C_{2}$~and~$C_{3}$ are equivalent to the operation~$S$;

-and therefore successive inversions at $C_{-1}$~and~$C_{r}$ are equivalent

-to the operation~$S^{r}$. Hence the result of an even

-number of inversions at any of the circles in either figure is

-equivalent to some operation of the group that the figure

-represents.

-

-\Par{184.} We may now proceed to construct a graphical

-representation of the group which is generated by $n$~operations

-subject to no relations. To this end, suppose $n + 1$ circles

-drawn, each of which is external to all the others while each

-touches two and only two of the rest. Such a system can be

-drawn in an infinite variety of ways: we will suppose, to give

-definiteness and simplicity to the resulting figure, that the

-$n + 1$ points of contact lie on a circle, which cuts the $n + 1$

-circles orthogonally. If these $n + 1$ points taken in order are

-$A_{1}$, $A_{2}$,~\Dots, $A_{n + 1}$, the successive circles are $A_{n + 1}A_{1}$, $A_{1}A_{2}$,~\Dots,

-$A_{n}A_{n + 1}$. We will suppose that only so much of these

-circles is drawn as lies within the common orthogonal circle

-$A_{1}A_{2}\Dots A_{n + 1}$. The $n + 1$ circular arcs $A_{n + 1}A_{1}$, $A_{1}A_{2}$,~\Dots\

-then bound a finite simply connected plane figure which we

-will denote provisionally by~$P$. Suppose now that $P$~is inverted

-in each of its sides, that the resulting figures are inverted in

-each of their new sides, and so on continually. Then from

-their mode of formation no two of the figures thus arising can

-%% -----File: 279.png---Folio 263-------

-overlap either wholly or in part; and when the process is

-continued without limit, every point in the interior of the

-orthogonal circle $A_{1}A_{2} \Dots A_{n + 1}$ will lie in one and only one of

-the figures thus formed from~$P$ by successive inversions.

-

-If $P'$~is any one of the figures or polygons so formed, the set

-of inversions by which it is derived from~$P$ is perfectly definite.

-For suppose, if possible, that $P'$~is derived from~$P$ by two

-distinct sets of inversions represented by $\Sigma$~and~$\Sigma'$. Then

-$\Sigma\Sigma'^{-1}$~is a set of inversions in the sides of~$P$ which transforms~$P$

-into itself. But every set of inversions necessarily transforms~$P$

-into some polygon lying outside it, and therefore

-\[

-\Sigma\Sigma'^{-1} = 1;

-\]

-or the set of inversions composing~$\Sigma$ is identical term for term

-%[Illustration: Fig. 3.]

-\Figure{3}{fig03}

-with the set composing~$\Sigma'$. It immediately follows that the

-polygons can be divided into two sets, according as they are

-derived from~$P$ by an even or an odd number of inversions. The

-%% -----File: 280.png---Folio 264-------

-latter we shall suppose coloured black, and the former (including~$P$)

-left white. Every white polygon will be surrounded

-by black polygons and vice~versâ. Since there is only one

-definite set of inversions that will transform~$P$ into any other

-white polygon~$P'$, the $n + 1$ corners of~$P'$ will correspond one

-by one to the $n + 1$ corners of~$P$; and when the perimeters of

-the two polygons are described in the same direction of rotation

-with regard to their interiors, the angular points that correspond

-will occur in the same cyclical order. On the other hand, in

-order that the corresponding angular points of a white and a

-black polygon may occur in the same cyclical order, their perimeters

-must be described in opposite directions. In consequence

-of these results, we may complete our figure (fig.~3) by lettering

-every angular point of every polygon with the same letter that

-occurs at the corresponding angular point of the polygon~$P$.

-

-\Par{185.} If now $T_{1}$, $T_{2}$,~\Dots, $T_{n + 1}$ represent inversions at

-$A_{1}A_{2}$, $A_{2}A_{3}$,~\Dots, $A_{n + 1}A_{1}$, the operation $T_{r - 1}T_{r}$ leaves the

-corner~$A_{r}$ of~$P$ unchanged and it transforms~$P$ into the next

-white polygon which has the corner~$A_{r}$ in common with~$P$, the

-direction of turning round~$A_{r}$ coinciding with the direction

-$A_{n + 1}A_{n} \Dots A_{1}$ of describing the perimeter of~$P$. For brevity,

-we shall describe this transformation of~$P$ as a positive rotation

-round~$A_{r}$. If then we denote the operation $T_{r - 1}T_{r}$ by the

-single symbol~$S_{r}$, we may say that $S_{r}$~produces a positive

-rotation of~$P$ round~$A_{r}$. Let $P_{1}$~be the new polygon so

-obtained; and let $P_{1}'$~be the polygon into which any other

-white polygon~$P'$ is changed by a positive rotation round the

-corner of~$P'$ that corresponds to~$A_{r}$. Then if $\Sigma$~is the set of

-inversions that changes $P$ into~$P'$, it also changes $P_{1}$ into~$P_{1}'$: so

-that $\Sigma^{-1}S_{r}\Sigma$ changes $P'$ into~$P_{1}'$, \ie\ produces a positive rotation

-round the corner~$A_{r}$ of~$P'$; and $S_{r}\Sigma$~changes $P$ into~$P_{1}'$.

-

-Let us now represent the operations

-\[

-T_{n + 1}T_{1},\ T_{1}T_{2},\ \Dots,\ T_{n}T_{n + 1},

-\]

-by

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n + 1},

-\]

-so that

-\[

-S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1.

-\]

-

-Then every operation, consisting of a pair of inversions in

-the sides of~$P$, can be represented in terms of

-\[

-S_{1},\ S_{2},\ S_{3},\ \Dots,\ S_{n}.

-\]

-%% -----File: 281.png---Folio 265-------

-For an inversion at~$A_{r}A_{r + 1}$, followed by an inversion at~$A_{s}A_{s + 1}$,

-is given by~$T_{r}T_{s}$; and

-\begin{align*}

-T_{r}T_{s}

-  &= T_{r}T_{r + 1} · T_{r + 1}T_{r + 2} · \Dots · T_{s - 1}T_{s} \\

-  &= S_{r + 1}S_{r + 2}\Dots S_{s}.

-\end{align*}

-

-If $s > r$, this is of the form required. If $s < r$, the term~$S_{n + 1}$

-that then occurs may be replaced by

-\[

-S_{n}^{-1} S_{n - 1}^{-1} \Dots S_{2}^{-1} S_{1}^{-1}.

-\]

-Hence finally, every operation consisting of an even number of

-inversions in the sides of~$P$ can be expressed in terms of

-\[

-S_{1},\ S_{2},\ S_{3},\ \Dots,\ S_{n};

-\]

-and with a restriction to positive indices, every such operation

-can be expressed in terms of

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},\ S_{n + 1}.

-\]

-

-Now it has been seen that no two operations, each consisting

-of a set of inversions in the sides of~$P$, can be identical unless

-the component inversions are identical term for term. Hence

-no two reduced operations of the form

-\[

-f(S_{i})

-\]

-are identical; in other words, the $n$ generating operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}

-\]

-are subject to no relations.

-

-If then we take the polygon~$P$ to correspond to the identical

-operation of the group~$G$ generated by

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},

-\]

-each white polygon may be taken as associated with the

-operation which will transform~$P$ into it. The foregoing discussion

-makes it clear that in this way a unique correspondence

-is established between the operations of~$G$ and the system of

-white polygons; or in other words, that the geometrical figure

-gives a complete graphical representation of the group.

-

-Moreover, since the operation $\Sigma^{-1}S_{r} \Sigma$ is a positive rotation

-round the corner~$A_{r}$ of the polygon~$\Sigma$ (calling now $P$ the

-polygon~$1$), a simple rule may be formulated for determining

-by a mere inspection of the figure what operation of the group

-any given white polygon corresponds to.

-%% -----File: 282.png---Folio 266-------

-

-This rule may be stated as follows. Let a continuous line

-be drawn inside the orthogonal circle from a point in the white

-polygon~$1$ to a point in any other white polygon, so that every

-consecutive pair of white polygons through which the line

-passes have a common corner, a positive rotation round which

-leads from the first to the second of the pair. This is always

-possible. Then if the common corners of each consecutive pair

-of white polygons through which the line passes, starting from~$1$,

-are $A_{p}$, $A_{q}$,~\Dots, $A_{r}$, $A_{s}$, the final white polygon corresponds

-to the operation

-\[

-S_{s}S_{r} \Dots S_{q}S_{p}\footnotemark.

-\]

-

-\footnotetext

-  {The reader who refers to Prof.~Dyck's memoir should notice that the

-  definition of the operation~$S_{r}$ above given is not exactly equivalent to that used

-  by Prof.~Dyck. With his notation, the white polygon here considered would

-  correspond to the operation $S_{p}S_{q} \Dots S_{r}S_{s}$.}%

-\begin{Remark}

-\Par{186.} The graphical representation of a general group we have

-\index{Graphical representation@\Topic{Graphical representation}!of a special group|(}%

-thus arrived at is only one of an infinite number that could be constructed;

-and we choose this in preference to others mainly because

-the form of the figure and the relative positions of the successive polygons

-are readily apprehended by the eye. As regards the mere establishment

-of such a representation we might, still using the process of

-inversion for the purpose of forming a definite figure, have started

-with $n + 1$ circles each exterior to and having no point in common

-with any of the others. Taking as the figure~$P$ the space external

-to all the circles and inverting it continually in the circles, we should

-form a series of black and white spaces of which the latter would

-again give a complete picture of the group. It is however only

-necessary to begin the construction of such a figure in order to

-convince ourselves that it would not appeal to the eye in the same

-way as the figure actually chosen.

-

-Moreover, as in the representation of a cyclical group, the process

-of inversion is in no way essential to the representation at which we

-arrive. Any arbitrary construction, which would give us the series

-of white and black polygons having, in the sense of the geometry of

-position, the same relative configuration as our actual figure, would

-serve our purpose equally well.

-\end{Remark}

-

-\Par{187.} If $\Sigma$~is the operation which transforms $P$ into~$P'$, and

-if $Q$~is the black polygon which has the side~$A_{r}A_{r + 1}$ in common

-with~$P$, then $\Sigma$~transforms $Q$ into the black polygon which has

-in common with~$P'$ the side corresponding to~$A_{r}A_{r + 1}$. If then

-we take $Q$ to correspond to the identical operation, any black

-polygon will correspond to the same operation as that white

-\index{Graphical representation@\Topic{Graphical representation}!of a general group|)}%

-%% -----File: 283.png---Folio 267-------

-polygon with which it has the side~$A_{r}A_{r + 1}$ in common. In this

-way we may regard our figure as divided in a definite way into

-double-polygons, each of which represents a single operation of

-the group.

-

-\Par{188.} We have next to consider how, from the representation

-of a general group whose $n$~generating operations are subject

-to no relations, we may obtain the representation of a special

-group generated by $n$ operations connected by a series of

-relations

-\[

-F_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).

-\]

-

-It has been seen (§~181) that to the identical operation of the

-special group there corresponds a self-conjugate sub-group~$H$ of

-the general group; or in other words, that the set of operations

-$\Sigma H$ of the general group give one and only one operation in the

-special group.

-

-Hence, to obtain from our figure for the general group one

-that will apply to the special group, we must regard all the

-double-polygons of the set~$\Sigma H$ as equivalent to each other;

-and if from each such set of double-polygons we choose one as

-a representative of the set, the totality of these representative

-polygons will have a unique correspondence with the operations

-of the special group.

-

-We shall first shew that a set of representative double-polygons

-can always be chosen so as to form a single simply

-connected figure. Starting with the double-polygon,~$P_{1}$, that

-corresponds to the identical operation of the general group as

-the one which shall correspond to the identical operation

-of the special group, we take as a representative of some

-other operation of the special group a double-polygon,~$P_{2}$,

-which has a side in common with~$P_{1}$. Next we take as

-a representative of some third operation of the special group

-a double-polygon which has a side in common with either

-$P_{1}$ or~$P_{2}$; and we continue the choice of double-polygons

-in this way until it can be carried no further. The set of

-double-polygons thus arrived at of necessity forms a single

-simply connected figure~$C$, bounded by circular arcs; and no

-two of the double-polygons belonging to it correspond to the

-same operation of the special group. Moreover, in~$C$ there is

-%% -----File: 284.png---Folio 268-------

-one double-polygon corresponding to each operation of the

-special group. To shew this, let $C'$ be the figure formed by

-combining with~$C$ every double-polygon which has a side in

-common with~$C$; and form $C''$ from~$C'$, $C'''$~from~$C''$, and so on,

-as $C'$~has been formed from~$C$. From the construction of~$C$ it

-follows that every polygon in~$C'$ is equivalent, in respect of the

-special group, to some polygon in~$C$. Similarly, every polygon

-in~$C''$ is equivalent to some polygon in~$C'$ and therefore to some

-polygon in~$C$; and so on. Hence finally, every polygon in the

-complete figure of the general group is equivalent to some

-polygon in~$C$, in respect of the special group; and therefore,

-since no two polygons of~$C$ are equivalent in respect of the

-special group, the figure~$C$ is formed of a complete set of representative

-double-polygons for the special group.

-

-Suppose now that $S$~is a double-polygon outside~$C$, with a

-side $A_{r}' A_{r + 1}'$ belonging to the boundary of~$C$. Within~$C$ there

-must be just one polygon, say~$ST$, of the set~$SH$. If this

-polygon lay entirely inside~$C$, so as to have no side on the

-boundary of~$C$, every polygon having a side in common with it

-would belong to~$C$. Now since $S$~and $ST$ are equivalent, every

-polygon having a side in common with~$S$ is equivalent to some

-polygon having a side in common with~$ST$. Hence since $C$~contains

-no two equivalent polygons, $ST$~must have a side on

-the boundary of~$C$; and if this side is~$A_{r}'' A_{r + 1}''$, the operation~$T$

-of~$H$ transforms $A_{r}' A_{r + 1}'$ into $A_{r}'' A_{r + 1}''$. Moreover, no operation

-of~$H$ can transform $A_{r}' A_{r + 1}'$ into another side of~$C$; for if this

-were possible, $C$~would contain two polygons equivalent to~$S$.

-It is also clear that, regarded as sides of polygons within~$C$,

-$A_{r}' A_{r + 1}'$~and $A_{r}'' A_{r + 1}''$ belong to polygons of different colours.

-Hence a correspondence in pairs of the sides of~$C$ is established:

-to each portion $A_{r}' A_{r + 1}'$ of the boundary of~$C$, which forms

-a side of a white (or black) polygon of~$C$, there corresponds

-another definite portion $A_{r}'' A_{r + 1}''$, forming a side of a black (or

-white) polygon of~$C$, such that a certain operation of~$H$ and its

-inverse will change one into the other, while no other operation

-of~$H$ will change either into any other portion of the boundary

-of~$C$.

-

-The system of double-polygons forming the figure~$C$, and

-the correspondence of the sides of~$C$ in pairs, will now give a

-%% -----File: 285.png---Folio 269-------

-complete graphical representation of the group. For the figure

-has been formed so that there is a unique correspondence

-between the white polygons of~$C$ and the operations of the

-group, such that until we arrive at the boundary the previously

-obtained rule will apply; and when we arrive at a polygon on

-the boundary, the correspondence of the sides in pairs enables

-the process to be continued.

-

-\Par{189.} From the mode in which the figure~$C$ has been formed,

-no two of the figures~$CH$ can have a polygon in common, when

-for $H$~is taken in turn each operation of the self-conjugate

-sub-group~$H$ of the general group~$G$; also the complete

-set contains every double-polygon of our original figure. This

-set of figures, or rather the division of the original figure into

-this set, will then represent in a graphical form the self-conjugate

-sub-group~$H$ of~$G$. Moreover, the operations which

-transform corresponding pairs of sides of~$C$ into each other will,

-when combined and repeated, clearly suffice to transform~$C$ into

-any one of the figures~$CH$ and will therefore form a set of

-generating operations of~$H$.

-

-\begin{Remark}

-\Par{190.} A simple example, in which the process described in the

-\index{Graphical representation@\Topic{Graphical representation}!examples of|(}%

-preceding paragraphs is actually carried out, will help to familiarize

-the reader with the nature of the process and will also serve to

-introduce a further modification of our figure. The example we

-propose to consider is the special group with two generating operations

-which are connected by the relations

-\[

-S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad S_{1}S_{2} = S_{2}S_{1}.

-\]

-

-As a first step, we will take account only of the relation

-\[

-S_{1}S_{2} = S_{2}S_{1},

-\]

-and form for this special group the figure~$C$. All operations

-\[

-\Dots S_{1}^{\alpha_{n}} S_{2}^{\beta_{n}} \Dots,

-\]

-for which $\Sigma \alpha_{n}$ and $\Sigma \beta_{n}$ have given values, are in the special group

-identical. We may thus select from the figure for the general group

-the set of polygons

-\[

-S_{2}^{\alpha} S_{1}^{\beta}\quad (\alpha, \beta = -\infty\ \text{to}\ +\infty)

-\]

-as a set of representative polygons; and a reference to the diagram\footnote

-  {In fig.~4 the orthogonal circle, which is not shewn, is taken to be a straight

-  line.}

-(fig.~4) makes it clear that this set of polygons forms a figure with

-a single bounding curve. The black polygon which corresponds to

-the operation~$\Sigma$ has here been chosen as that which has the side

-$A_{1}A_{2}$ in common with the white polygon~$\Sigma$.

-\index{Graphical representation@\Topic{Graphical representation}!of a special group|)}%

-%% -----File: 286.png---Folio 270-------

-

-Each double-polygon, except those of the set~$S_{1}^{m}$, contributes two

-sides to the boundary of~$C$, one belonging to a white polygon and

-% [Illustration: Fig. 4.]

-\Figure[\textwidth]{4}{fig04}

-one to a black. The polygons, which border~$C$ and have sides in

-common with $S_{2}^{\alpha} S_{1}^{\beta}$, are $S_{1} S_{2}^{\alpha} S_{1}^{\beta}$ and $S_{1}^{-1} S_{2}^{\alpha} S_{1}^{\beta}$; and these, regarded

-as operations of the special group, are equivalent to $S_{2}^{\alpha} S_{1}^{\beta + 1}$ and

-$S_{2}^{\alpha} S_{1}^{\beta - 1}$. Hence the correspondence between the sides of~$C$ is such

-that

-

-\Inum{(i)}\quad to the side~$A_{1}A_{3}$ of the white polygon $S_{2}^{\alpha} S_{1}^{\beta}$ corresponds the

-side~$A_{1}A_{3}$ of the black polygon $S_{2}^{\alpha} S_{1}^{\beta - 1}$;

-

-\Inum{(ii)}\quad to the side~$A_{1}A_{3}$ of the black polygon $S_{2}^{\alpha} S_{1}^{\beta}$ corresponds the

-side~$A_{1}A_{3}$ of the white polygon $S_{2}^{\alpha}S_{1}^{\beta + 1}$.

-

-When we now take account of the additional relations

-\[

-S_{1}^{3} = 1,\quad S_{2}^{3} = 1,

-\]

-the figure~$C$ is found to reduce to a set of nine double-polygons,

-which is completely represented by fig.~5.

-

-In addition to the correspondences between the sides of~$C$ to

-which those just written simplify when the indices of $S_{1}$~and~$S_{2}$ are

-reduced $(\mod 3)$, we have now also the correspondences, indicated

-in the figure by curved lines with \DPchg{arrowheads}{arrow-heads}, which result from the

-new relations. Our figure may be further modified in such a way that

-its form takes direct account of these four new correspondences.

-Thus without in any way altering the configuration of the double-polygons,

-from the point of view of geometry of position, we may

-continuously deform the figure so that the pairs of corresponding

-sides indicated by the curved \DPchg{arrowheads}{arrow-heads} are brought to actual

-coincidence. When this is done, the resulting figure will have the

-form shewn in fig.~6. The correspondence in pairs of the sides of

-the boundary is indicated in the figure by full and dotted lines.

-%% -----File: 287.png---Folio 271-------

-% Illustration Fig. 5.

-\begin{sidewaysfigure}[hp!]

-\centering

-\Graphic{\textheight}{fig05}

-\end{sidewaysfigure}

-%\Figure[\textwidth]{5}{fig05}

-%% -----File: 288.png---Folio 272-------

-The two unmarked portions $A_{1}A_{3}A_{1}$ correspond, as also do the two

-similar portions marked with a full line, and the two marked with a

-dotted line.

-%[Illustration: Fig. 6.]

-\Figure{6}{fig06}

-

-It will be noticed that, in this final form of the figure for the

-special group, direct account is taken of the finite order of the

-generating operations $S_{1}$~and~$S_{2}$ and also of the operation~$S_{1}S_{2}$. The

-simplification of the figure that results by thus taking account

-directly of the finite order of the generating operations, and the

-greater ease with which the eye follows this simplified representation,

-are immediately obvious on a comparison of figs.\ 5~and~6.

-\index{Graphical representation@\Topic{Graphical representation}!examples of|)}%

-\end{Remark}

-

-\Par{191.} In the applications of this graphical representation

-of a group that we have specially in view, namely to groups

-of finite order, the generating operations themselves are

-necessarily of finite order. The generating operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},

-\]

-of such a group may be taken as of orders

-\[

-m_{1},\ m_{2},\ \Dots,\ m_{n};

-\]

-%% -----File: 289.png---Folio 273-------

-and if

-\[

-S_{1} S_{2} \Dots S_{n} S_{n + 1} = 1,

-\]

-then $S_{n + 1}$~will be of finite order~$m_{n + 1}$. We shall therefore next

-consider a group generated by $n$~operations which satisfy the

-relations

-\[

-S_{1}^{m_{1}} = 1,\quad S_{2}^{m_{2}} = 1,\ \Dots,\quad S_{n}^{m_{n}} = 1,\quad (S_{1} S_{2} \Dots S_{n})^{m_{n + 1}} = 1.

-\]

-

-The simple example we have given makes it clear that,

-\index{Graphical representation@\Topic{Graphical representation}!of a group of finite order|(}%

-at least in some particular cases, relations of this form may

-be directly taken into account in constructing our figure; in

-such a way that in the complete figure, consisting of a finite or

-an infinite number of double-polygons, the correspondence in

-pairs of the sides of the boundary, if any, will depend upon

-further relations between the generating operations.

-

-We may, in fact, always take account of relations of the

-form in question in the construction of our figure as follows.

-

-Let us take as before $n + 1$ arcs of circles

-\[

-A_{n + 1}A_{1},\ A_{1}A_{2},\ \Dots,\ A_{n - 1}A_{n},\ A_{n}A_{n + 1},

-\]

-bounding a polygonal figure~$P$ of $n + 1$ corners; but now,

-instead of supposing the circles $A_{r - 1}A_{r}$ and $A_{r}A_{r + 1}$ to touch at~$A_{r}$,

-let them cut at an angle (measured inside~$P$) of

-\[

-\frac{\pi}{m_{r}}, \quad (r = 1, 2, \Dots, n),

-\]

-while $A_{n}A_{n + 1}$ and $A_{n + 1}A_{1}$ cut at an angle~$\dfrac{\pi}{m_{n + 1}}$. Such a figure

-can again be chosen in an infinite variety of ways: we will

-suppose that it is drawn so that the $n + 1$ circles have a common

-orthogonal circle. This clearly is always possible; but it is not

-now necessarily the case that this orthogonal circle is real. Let

-the figure~$P$ be now inverted in each of its sides; let the new

-figures so formed be inverted in each of their new sides; and so

-on continually. Then since the angles of~$P$ are sub-multiples of

-two right angles, no two of the figures thus formed can overlap

-in part without coinciding entirely. Moreover, when the process

-is completely carried out, every point within the orthogonal

-circle when it is real, and every point in the plane of the figure

-when the orthogonal circle is evanescent or imaginary, will lie

-in one and in only one of the polygons thus formed from~$P$ by

-successive inversions.

-%% -----File: 290.png---Folio 274-------

-

-\Par{192.} Exactly as with the general group, these polygons

-are coloured white or black according as they are derivable

-from~$P$ by an even or an odd number of inversions. The

-corners of any white polygon correspond one by one to the

-corners of $P$; so that, when the perimeters of the polygons are

-described in the same direction, corresponding corners occur in

-the same cyclical order.

-

-If now the operation of successive inversions at $A_{n + 1}A_{1}$ and

-$A_{1}A_{2}$ is represented by~$S_{1}$, and that of successive inversions at

-$A_{r - 1}A_{r}$ and $A_{r}A_{r + 1}$ by~$S_{r}$, ($r = 2, 3, \Dots, n$); all operations,

-consisting of an even number of inversions in the sides of~$P$,

-can be represented in terms of

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}.

-\]

-

-Moreover, from the construction of the polygon~$P$, these

-operations satisfy the relations

-\[

-S_{1}^{m_{1}} = 1,\quad S_{2}^{m_{2}} = 1,\quad \Dots, \quad S_{n}^{m_{n}} = 1,\quad S_{n + 1}^{m_{n + 1}} = 1,

-\]

-where

-\[

-S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1.

-\]

-

-Again, if $P'$~is any white polygon of the figure, which can be

-derived from~$P$ by the operation~$\Sigma$, a positive rotation (§~185) of~$P'$

-round its corner~$A_{r}'$ is effected by the operation $\Sigma^{-1}S_{r}\Sigma$;

-and, if $P''$~is the polygon so obtained, $P''$~is derived from~$P$ by

-the operation~$S_{r}\Sigma$. It is to be observed that a positive rotation

-of a polygon round its $A_{r}$~corner is now an operation of finite

-order~$m_{r}$.

-

-Suppose now that two operations $\Sigma$ and $\Sigma'$ transform~$P$ into

-the same polygon~$P'$, so that $\Sigma\Sigma'^{-1}$ leaves $P$ unchanged. If

-this operation, written at length, is

-\[

-S_{p}^{\alpha} \Dots S_{q}^{\beta} S_{r}^{\gamma} S_{s}^{\delta},

-\]

-and if $P$~is transformed into~$P_{1}$ by a positive rotation round~$A_{s}$

-repeated $\delta$~times, $P_{1}$~into~$P_{2}$ by a rotation round its

-corner~$A_{r}$ repeated $\gamma$~times, and so on; then the operation

-may be indicated by a broken line drawn from $P$~to~$P_{1}$, from

-$P_{1}$~to~$P_{2}$, and so on, the line returning at last to~$P$. But the

-operation indicated by such a line is clearly equivalent to complete

-rotations, (\ie\ rotations each of which lead to identity),

-%% -----File: 291.png---Folio 275-------

-round each of the corners which the broken line includes. In

-other words, $\Sigma\Sigma'^{-1}$~reduces to identity when account is taken of

-the relations which the generating operations satisfy. Hence

-finally, to every white polygon~$P'$ will correspond one and only

-one of the operations of the group, namely that operation

-which transforms $P$ into~$P'$. The same is clearly true of the

-black polygons; and by taking $P$ and a chosen black polygon

-which has a side in common with~$P$ as corresponding to the

-identical operation, the required unique correspondence is

-established between the complete set of double-polygons in the

-figure and the operations of the group, the relations which the

-generating operations satisfy being directly indicated by the

-configuration of the figure. Moreover, as with the general

-group (§~185), a simple rule may be stated for determining,

-from an inspection of the figure, the polygon that corresponds

-to any given operation of the group.

-

-\Par{193.} The number of polygons in the figure and therefore

-the order of the group will still, in general, be infinite. We

-may now proceed, just as in the previous case of a quite general

-group, to derive from the figure representing the group~$G$,

-generated by $n$ operations satisfying the relations

-\[

-S_{1}^{m_{1}} = 1,\quad

-S_{2}^{m_{2}} = 1,\quad \Dots,\quad

-S_{n}^{m_{n}} = 1,\quad (S_{1} S_{2} \Dots S_{n})^{m_{n + 1}} = 1,

-\]

-a suitable representation of the more special group~$\Bar{G}$, generated

-by $n$~operations which satisfy the above relations and in addition

-the further $m$~relations

-\[

-f_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).

-\]

-

-As has been seen in §~182, if $H$~is the self-conjugate sub-group

-of~$G$ which is formed by combining all possible operations

-of the form

-\[

-R^{-1} f_{j}(S_{i})R,

-\]

-and if $\Sigma$~is any operation of~$G$, then the set of operations~$\Sigma H$,

-regarded as operations of~$\Bar{G}$, are all equivalent to each other.

-From each set of polygons $\Sigma H$ in the figure of~$G$, we may

-therefore choose one to represent the corresponding operation

-of~$\Bar{G}$; and, as was shewn with the general group, a complete

-set of such representative polygons may be selected to form a

-%% -----File: 292.png---Folio 276-------

-connected figure, \ie\ a figure which does not consist of two or

-more portions which are either isolated or connected only by

-corners. Moreover, as in the former case, the sides of this

-figure~$C$ will be connected in pairs $A_{r}' A_{r + 1}'$ and $A_{r}'' A_{r + 1}''$, which

-are transformed into each other by some operation~$T$ of~$H$ and

-its inverse, while no other operation of~$H$ will transform either

-$A_{r}' A_{r + 1}'$ or $A_{r}'' A_{r + 1}''$ into any other side of~$C$.

-

-It is not now however necessarily the case that the figure~$C$,

-as thus constructed, is simply connected. Let us suppose then

-that $C$~has one or more inner boundaries as well as an outer

-boundary, and denote one of these inner boundaries by~$L$. If

-the sides of~$L$ do not all correspond in pairs, and if $A_{r}' A_{r + 1}'$~is a

-side of~$L$ such that the other side $A_{r}'' A_{r + 1}''$ corresponding to it

-does not belong to~$L$, we may replace the double-polygon~$P''$ in~$C$

-of which $A_{r}'' A_{r + 1}''$~is a side by the double-polygon, not

-previously belonging to~$C$, of which $A_{r}' A_{r + 1}'$~is a side. If $P''$~has

-a side on the boundary~$L$, the new figure~$C'$ thus obtained

-will have one inner boundary less than~$C$; and if $P''$~has no side

-on the boundary~$L$, the new inner boundary~$L'$ that is thus

-formed from~$L$ will contain one double-polygon less than~$L$,

-while the number of inner boundaries is not increased. This

-process may be continued till the new inner boundary~$L_{1}$ which

-replaces~$L$ is such that all of its sides correspond in pairs.

-

-Let now $A_{s}A_{s + 1}$ and $A_{s}' A_{s + 1}'$ be a pair of corresponding

-sides of~$L_{1}$, such that $A_{s}A_{s + 1}$~is transformed into $A_{s}' A_{s + 1}'$ by an

-operation~$h$ of the self-conjugate sub-group~$H$. A side~$A_{t}A_{t + 1}$

-of another boundary of~$C$ may be chosen such that $A_{s}A_{s + 1}$

-and $A_{t}A_{t + 1}$ are sides of a simply connected portion, say~$B$, of~$C$;

-while no side of~$L_{1}$ except $A_{s}A_{s + 1}$ forms part of the boundary of~$B$.

-The polygons of~$B$ are equivalent, in respect of the special

-group, to those of~$Bh$. Moreover, since the sides of~$L_{1}$ correspond

-in pairs, no side of~$Bh$, except $A_{s}' A_{s + 1}'$ can coincide with

-a side of~$L_{1}$. Hence when $B$~is replaced by~$Bh$, the inner

-boundary~$L_{1}$ will be got rid of and no new inner boundary will

-be formed. Finally then, $C$~may always be chosen so as to form

-a single simply connected figure.

-

-The simply connected plane figure~$C$, which has thus been

-constructed, with the correspondence of the sides of its boundary

-%% -----File: 293.png---Folio 277-------

-in pairs, will now give a complete graphical representation of

-the special group. The rule already formulated will determine

-the operation of the group to which each white polygon

-corresponds; and when, in carrying out this rule, we come to

-a polygon on the boundary, the correspondence of the sides of

-the boundary in pairs will enable the process to be continued.

-

-The correspondence of the sides of~$C$ in pairs involves a

-correspondence of the corners in sets of two or more. Thus if

-$A_{r}$~is a corner of~$C$ and if, of the $m_{r}$ white polygons which in

-the complete figure have a corner at~$A_{r}$, $n_{1}$~lie within~$C$, there

-must within~$C$ be $m_{r} - n_{1}$ white polygons equivalent to the

-remainder, and each of these must have an $A_{r}$~corner on the

-boundary. If $A_{r}'$~is a corner of~$C$ such that there are $n_{2}$~white

-polygons, lying within~$C$ and having a corner at~$A_{r}'$, and if one

-of the sides of the boundary with a corner at~$A_{r}'$ corresponds

-to one of the sides of the boundary with a corner at~$A_{r}$, these

-$n_{2}$~white polygons must be equivalent to $n_{2}$ of the white polygons,

-lying outside~$C$ and having a corner at~$A_{r}$. If

-\[

-n_{1} + n_{2} < m_{r},

-\]

-there must be a third corner~$A_{r}''$, contributing $n_{3}$~more white

-polygons towards the set. With this we proceed as before; and

-the process may be continued till the whole of the $m_{r}$~white

-polygons surrounding~$A_{r}$ are accounted for. The set of corners

-$A_{r}$, $A_{r}'$, $A_{r}''$,~\Dots\ will then form a set of corresponding corners,

-which are equivalent to each other in respect of the special

-group; and the whole of the corners of~$C$ may be divided into

-such sets. At each set of corresponding corners~$A_{r}$ of~$C$ there

-must clearly be also $m_{r}$~black polygons belonging to~$C$; and

-the sum of the angles of~$C$ at a set of corresponding corners

-must be equal to four right angles.

-

-\Par{194.} When the order of the group is finite, we may still

-further so modify our figure as to take account of the correspondence

-of the sides of the boundary in pairs. We may, in

-fact, by a suitable bending and stretching of the figure, bring

-corresponding sides of the boundary to actual coincidence.

-When this is done, the figure will no longer be a piece of a

-plane with a single boundary, but will form a continuous

-%% -----File: 294.png---Folio 278-------

-\index{Regular division@\Topic{Regular division} of a surface, representation of a group by means of}%

-surface, which is unbounded and in general is multiply connected.

-Every point~$A_{r}$ on the surface, which in the plane

-figure did not lie on the boundary, will be a corner common

-to $2m_{r}$~polygons alternately black and white; and, in consequence

-of what has just been seen in regard to the correspondence

-of corners of the boundary, the same is true

-for every point~$A_{r}$ on the surface which in the plane figure

-consisted of a set of corresponding corners of the boundary. If

-$N$~is the order of the group, the continuous unbounded surface

-will be divided into $2N$~polygons, black and white. The configuration

-of the set of white polygons with respect to any one

-of them will, from the point of view of geometry of position, be

-the same as that with respect to any other; and the like is

-true for the black polygons. Such a division of a continuous

-unbounded surface is described as a \emph{regular} division; and we

-have finally, as a graphical representation of any group of finite

-order~$N$, a division of a continuous surface into $2N$~polygons,

-half black and half white, which is regular with respect to each

-% [Illustration: Fig. 7.]

-\Figure{7}{fig07}

-\index{Graphical representation@\Topic{Graphical representation}!of a group of finite order|)}%

-%% -----File: 295.png---Folio 279-------

-set. The correspondence between the operations of the group

-and the white polygons on the surface is given by the rule that

-a single positive rotation of the white polygon~$\Sigma$ round its

-corner~$A_{r}$ leads to the white polygon~$S_{r}\Sigma$.

-

-\begin{Remark}

-\Par{195.} We may again here illustrate this final modification of the

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-graphical representation of a finite group by a simple example. For

-this purpose, we choose the group defined by

-\begin{gather*}

-S_{1}^{4} = 1,\quad S_{2}^{4} = 1,\quad S_{3}^{4} = 1, \\

-S_{1}S_{2}S_{3} = 1,\quad S_{2}^{-1}S_{1}S_{2}S_{1} = 1.

-\end{gather*}

-

-This group (§~74) is a non-Abelian group of order~$8$, containing

-a single operation of order~$2$. The reader will have no difficulty in

-verifying that the plane figure for this group is given by fig.~7; and

-that opposite sides of the octagonal boundary correspond. The

-single operation of order~$2$ is

-\[

-S_{1}^{2} (= S_{2}^{2} = S_{3}^{2});

-\]

-this corresponds to a displacement of the triangles among themselves

-in which all the six corners remain fixed. If now corresponding

-sides of the boundary are brought to coincidence, the continuous

-surface formed will be a double-holed anchor-ring, or sphere with

-% [Illustration: Fig. 8.]

-\Figure[0.6\textwidth]{8}{fig08}

-%% -----File: 296.png---Folio 280-------

-two holes through it. A view of one half of the surface divided

-into black and white triangles, is given in fig.~8. The half of the

-surface, not shewn, is divided up in a similar manner; and the

-operation of order~$2$ replaces each triangle of one half by the

-corresponding triangle of the other, an operation which clearly leaves

-the six corners of the polygons undisplaced.

-\end{Remark}

-

-\Par{196.} The form of the plane figure~$C$, which with the correspondence

-\index{Genus@\Topic{Genus} of a group, definition of}%

-\index{Group@\Topic{Group}!genus of a}%

-of its bounding sides in pairs represents the group,

-is capable of indefinite modification by replacing individual

-polygons on the boundary by equivalent polygons. If however

-we reckon a pair of corresponding sides of the boundary as a

-single side and a set of corresponding corners of the boundary

-as a single corner, it is clear that, however the figure may be

-modified, the numbers of its corners, sides and polygons remain

-each constant. This may be immediately verified on replacing

-any single boundary polygon by its equivalent.

-

-If now $A$~be the number of corners, and $E$~the number of

-sides in the figure~$C$ when reckoned as above, $2N$~being the

-number of polygons, then the connectivity\footnote

-  {Forsyth, \textit{Theory of Functions}, p.~325.}

-\index{Forsyth@\Topic{Forsyth}, quoted}%

-$2p + 1$ of the closed

-surface is given by the equation

-\[

-2p = 2 + E - 2N - A.

-\]

-

-When the group and its generating operations are given,

-the integer~$p$ is independent of the form of the plane figure~$C$,

-which as has been seen is capable of considerable modification.

-The plane figure~$C$ however depends directly on the set of

-generating operations that is chosen for the group. For a given

-group of finite order, such a set is not in general unique; and

-the number of generating operations as well as their order will

-in general vary from one set to another. It does not necessarily

-follow, and in fact it is not generally the case, that the connectivity

-of the surface by whose regular division the group is

-represented, is independent of the choice of generating operations.

-There must however obviously be a lower limit to the

-number~$p$ for any given group of finite order, whatever

-generating operations are chosen; this we shall call the \emph{genus}

-of the group\footnotemark.

-\index{Hurwitz@\Topic{Hurwitz}, quoted}%

-\footnotetext{Hurwitz, ``Algebraische Gebilde mit eindeutigen Transformationen in

-  sich,'' \textit{Math.\ Ann.}\DPnote{** No comma or "Vol."}\ \VolNo{XLI}, (1893), p.~426.}%

-%% -----File: 297.png---Folio 281-------

-

-\Par{197.} We shall now shew that there is a limit to the order

-of a group which can be represented by the regular division of

-a surface of given connectivity $2p + 1$. If $N$~is the order of

-such a group, generated by the $n$~operations

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n},

-\]

-which satisfy the relations

-\begin{gather*}

-S_{1}^{m_{1}} = 1,\quad

-S_{2}^{m_{2}} = 1,\quad \Dots, \quad

-S_{n}^{m_{n}} = 1, \\

-S_{1}S_{2} \Dots S_{n} = 1;

-\end{gather*}

-the surface will be divided in $2N$~polygons of $n$~sides each.

-Let $A_{1}$, $A_{2}$,~\Dots, $A_{n}$ be the angular points of one of these

-polygons; and suppose that on the surface there are $C_{1}$~corners

-in the set to which $A_{1}$ belongs, $C_{2}$~in the set to which $A_{2}$ belongs,

-and so on. Round each corner~$A_{r}$ there are $2m_{r}$~polygons; and

-each polygon has one and only one corner of the set to which

-$A_{r}$ belongs. Hence

-\[

-C_{r}m_{r} = N,

-\]

-and so

-\[

-C_{1} + C_{2} + \Dots + C_{n} = N \sum_{1}^{n} \frac{1}{m_{r}}.

-\]

-

-Again, each side belongs to two and only to two polygons, so

-that the number of sides is

-\[

-Nn.

-\]

-

-Using these values for $A$~and~$E$ in the formula of §~196,

-we obtain the equation

-\[

-2(p - 1) = N\left(n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}\right).

-\]

-

-A complete discussion of this equation for the cases $p = 0$

-and $p = 1$ will be given in the next chapter.

-

-When $p$~is a given integer greater than unity, we can

-determine the greatest value that is possible for~$N$ by finding

-the least possible positive value of the expression

-\[

-n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}.

-\]

-

-If $n > 4$, this quantity is not less than~$\frac{1}{2}$, since $m_{r}$~cannot be

-less than~$2$.

-

-If $n = 4$, the simultaneous values

-\[

-m_{1} = m_{2} = m_{3} = m_{4} = 2

-\]

-%% -----File: 298.png---Folio 282-------

-are not admissible, since they make the expression zero. Its

-least value in this case will therefore be given by

-\[

-m_{1} = m_{2} = m_{3} = 2,\quad m_{4} = 3;

-\]

-and the expression is then equal to~$\frac{1}{6}$.

-

-If $n = 3$, we require the least positive value of

-\[

-K = 1 - \frac{1}{m_{1}} - \frac{1}{m_{2}} - \frac{1}{m_{3}}.

-\]

-

-Now the three sets of values

-\begin{alignat*}{3}

-m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 3, \\

-m_{1} &= 2,\quad & m_{2} &= 4,\quad & m_{3} &= 4, \\

-\lintertext{and}

-m_{1} &= 2,\quad & m_{2} &= 3,\quad & m_{3} &= 6,

-\end{alignat*}

-each make $K$ zero; and therefore no positive value of~$K$ can be

-less than the least of those given by

-\begin{alignat*}{3}

-m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 4, \\

-m_{1} &= 2,\quad & m_{2} &= 4,\quad & m_{3} &= 5, \\

-\lintertext{and}

-m_{1} &= 2,\quad & m_{2} &= 3,\quad & m_{3} &= 7.

-\end{alignat*}

-

-These sets of values give for~$K$ the values $\frac{1}{12}$,~$\frac{1}{20}$ and~$\frac{1}{42}$.

-Hence finally, the absolutely least positive value of the expression

-is~$\frac{1}{42}$, and therefore the greatest admissible value of~$N$ is

-\[

-84(p - 1).

-\]

-

-Hence\footnotemark:---

-\footnotetext{Hurwitz, \textit{loc.~cit.}\ p.~424.}

-\index{Hurwitz@\Topic{Hurwitz}, quoted}%

-

-\begin{Theorem}[II.]

-The order of a group, that can be represented

-\index{Limitation@\Topic{Limitation}!on the order of a group of given genus}%

-by the regular division of a surface of connectivity $2p + 1$, cannot

-exceed $84(p - 1)$, $p$~being greater than unity.

-\end{Theorem}

-

-\Par{198.} If, when a group is represented by the regular division

-of an unbounded surface, we draw a line from any point inside

-the white polygon~$1$ (or any other polygon) returning after any

-path on the surface to the point from which it started, it will

-represent a relation between the generating operations of the

-group. For in following out along the line so drawn the rule

-that determines the operation of the group corresponding to

-each white polygon, some operation

-\[

-F(S_{i})

-\]

-%% -----File: 299.png---Folio 283-------

-will be found to correspond to the final polygon; and this

-being the white polygon~$1$, it follows that

-\[

-F(S_{i}) = 1.

-\]

-

-If the surface is simply connected, any such line can be

-\index{Defining relations@\Topic{Defining relations} of a group!limitation on the number of, when the genus is given}%

-\index{Limitation@\Topic{Limitation}!on the number of defining relations of a group of given genus}%

-continuously altered till it shrinks to a point; and therefore the

-$n + 1$ relations between the $n$~generating operations completely

-define the group, since all other relations can be deduced from

-them.

-

-If however the surface is of connectivity $2p + 1$, there are

-$2p$~independent closed paths that can be drawn on the surface,

-no one of which can by continuous displacement either be

-shrunk up to a point or brought to coincidence with another;

-and every closed path on the surface can by continuous displacement

-either be brought to a point or to coincidence with

-a path constructed by combination and repetition of the $2p$~independent

-paths\footnotemark.

-\footnotetext{Forsyth, \textit{Theory of Functions}, p.~330.}%

-\index{Forsyth@\Topic{Forsyth}, quoted}%

-Any one of these $2p$~independent paths will

-give a relation between the $n$~generating operations of the

-group, which cannot be deduced from the $n + 1$ relations on

-which the angles of the polygons depend. Moreover, every relation

-between the generating operations can be represented by a

-closed path on the surface; and therefore there can be no further

-relation independent of the original relations and those obtained

-from the $2p$~independent paths. There cannot therefore be

-more than $2p$~independent relations between the $n$~generating

-operations of a group, in addition to the $n + 1$ relations that give

-the order of the generating operations and of their product;

-$2p + 1$ being the connectivity of the surface by whose regular

-division into $n$-sided polygons the group is represented.

-

-The $2p$~relations given by $2p$~independent paths on the

-surface are not, however, necessarily independent. In fact we

-have already had an example to the contrary in §~195. On the

-closed surface, by the regular division of which the group there

-considered is represented, four independent closed paths can be

-drawn. Any three of the corresponding relations can be derived

-from the fourth by transformation.

-

-The only known cases in which the $2p$~relations are independent

-are those of a class of groups of genus one (§~205).

-%% -----File: 300.png---Folio 284-------

-

-\begin{Ex} Draw the figure of the group generated by $S_{1}$,~$S_{2}$,~$S_{3}$, where

-\[

-S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{8} = 1,\quad S_{1} S_{2} S_{3} = 1.

-\]

-

-Shew from the figure that the special group, given by the

-additional relation

-\[

-(S_{1} S_{3}^{4})^{2} = 1,

-\]

-is a finite group of order~$48$; and that it can be represented by the

-regular division of a surface of connectivity~$5$.

-\end{Ex}

-

-

-\Note{\textit{Note to} §~194.}

-

-\begin{Remark}

-If in the process of bending and stretching, described in §~194,

-by means of which the plane figure~$C$ is changed into an unbounded

-surface, the angles of the polygons all remain unaltered, the circles

-of the plane figure will become continuous curves on the surface.

-These curves on the surface, which we will still call circles, are

-necessarily re-entrant. It is not however necessarily the case that,

-on the surface, a circle will not cut itself.

-

-In the plane figure for the general group, an inversion at any

-circle of the figure leaves the figure unchanged geometrically but

-interchanges the black and white polygons. Each circle is, in fact, a

-line of symmetry for the figure such that, in respect of it, there

-is corresponding to every white polygon a symmetric black polygon

-and vice~versâ.

-

-Similarly on the surface a circle which does not cut itself may

-be a line of symmetry, such that a reflection at it is an operation of

-order two which leaves the surface and its division into polygons

-unchanged, but interchanges black and white polygons. When this

-is the case, every circle on the surface will be a line of symmetry and

-no circle will cut itself. On the other hand no such operation can

-ever be connected with a circle which cuts itself.

-

-When such lines of symmetry exist, Prof.~Dyck speaks of the

-division of the surface as regular and symmetric.

-\end{Remark}

-%% -----File: 301.png---Folio 285-------

-

-

-\Chapter{XIII.}{On the Graphical Representation of Groups:

-Groups of Genus Zero and Unity: Cayley's Colour Groups.}

-

-\Par{199.} \First{We} shall now proceed to a discussion in the cases

-$p = 0$ and $p = 1$ of the relation

-\[

-2(p - 1) = N \left(n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}\right),

-\]

-which connects the number and the orders of the generating

-operations of a group with the order of the group itself; and to

-the consideration of the corresponding groups.

-

-For any given value of~$p$, other than $p = 1$, we may regard

-this relation as an equation connecting the positive integers

-$N$,~$n$, $m_{1}$, $m_{2}$,~\Dots, $m_{n}$. It does not however follow from the

-investigations of the last Chapter that there is always a

-group or a set of groups corresponding to a given solution of

-the equation. In fact, for values of~$p$ greater than~$1$, this is not

-necessarily the case. We shall however find that, when $p = 0$,

-there is a single type of group corresponding to each solution

-of the equation; and that, when $p = 1$, there is an infinite

-number of types of group, all characterized by a common

-property, corresponding to each solution of the equation.

-When $p = 0$, the groups are (§~196) of genus zero; and all

-possible groups of genus zero are found by putting $p = 0$

-%% -----File: 302.png---Folio 286-------

-in the equation. The groups thus obtained are of special

-importance in many applications of group-theory; for this

-reason, they will be dealt with in considerable detail.

-

-\Par{200.} When $p = 0$, the equation may be written in the form

-\index{Groups@\Topic{Groups}!of genus zero|(}%

-\[

-2\left(1 - \frac{1}{N}\right) = \sum_{1}^{n} \left(1 - \frac{1}{m_{r}}\right);

-\]

-in this form, it is clear that the only admissible values of~$n$ are

-$2$ and~$3$.

-

-First, let $n = 2$. The only possible solution then is

-\[

-N = m_{1} = m_{2} = n,

-\]

-$n$~being any integer. The corresponding group is a cyclical

-group of order~$n$.

-

-Secondly, let $n = 3$. In this case, one at least of the three

-integers $m_{1}$,~$m_{2}$,~$m_{3}$ must be equal to~$2$, as otherwise the right-hand

-side of the equation would be not less than~$2$. We may

-therefore without loss of generality put $m_{1} = 2$. If now both

-$m_{2}$ and $m_{3}$ were greater than~$3$, the right-hand side would still

-be not less than~$2$; and therefore we may take $m_{2}$ to be either

-$2$ or~$3$. When $m_{1}$~and~$m_{2}$ are both~$2$, the equation becomes

-\[

-\frac{2}{N} = \frac{1}{m_{3}};

-\]

-giving

-\[

-m_{3} = n,\quad N = 2n,

-\]

-where $n$~is any integer.

-

-When $m_{1}$~is~$2$ and $m_{2}$~is~$3$, the equation is

-\[

-\frac{2}{N} + \frac{1}{6} = \frac{1}{m_{3}}.

-\]

-

-This has three solutions in positive integers; namely,

-\begin{alignat*}{2}

-m_{3} &= 3,\quad & N &= 12; \\

-m_{3} &= 4,\quad & N &= 24; \\

-\lintertext{and}

-m_{3} &= 5,\quad & N &= 60.

-\end{alignat*}

-%% -----File: 303.png---Folio 287-------

-

-The solutions of the equation for the case $p = 0$ may therefore

-be tabulated in the form:---

-\[

-\begin{array}{*{5}{c|}}

-\TEntry{}  & \TEntry{m_{1}} & \TEntry{m_{2}} & \TEntry{m_{3}} & \TEntry{N} \\

-\cline{2-5}

-\Strut

-\Inum{I}   & n & n &   & n \\

-\cline{2-5}

-\Strut

-\Inum{II}  & 2 & 2 & n & 2n \\

-\cline{2-5}

-\Strut

-\Inum{III} & 2 & 3 & 3 & 12 \\

-\cline{2-5}

-\Strut

-\Inum{IV}  & 2 & 3 & 4 & 24 \\

-\cline{2-5}

-\Strut

-\Inum{V}   & 2 & 3 & 5 & 60 \\

-\cline{2-5}

-\end{array}

-\]

-

-\Par{201.} That a single type of group actually exists, corresponding

-to each of these solutions, may be seen at once by

-returning to our plane figure. The sum of the internal angles

-of the triangle $A_{1} A_{2} A_{3}$ formed by circular arcs is, in each of

-these cases, greater than two right angles; and the common

-orthogonal circle is therefore imaginary. The complete figure

-will therefore divide the whole plane into black and white

-triangles, so that there are no boundaries to consider. Moreover,

-the number of white triangles in each case must be equal

-to the corresponding value of~$N$; for the preceding investigation

-shews that this is a possible value, and on the other hand the

-process, by which the figure is completed from a given original

-triangle, is a unique one. There is therefore a group corresponding

-to each solution; and the correspondence which has

-been established in any case between the operations of a group

-and the polygons of a figure, proves that there cannot be two

-distinct types of group corresponding to the same solution.

-

-\Par{202.} The plane figure for $p = 0$ does not, in fact, differ

-essentially from the figure drawn on a continuous simply connected

-surface in space. The former may be regarded as the

-stereographic projection of the latter. The five distinct types

-are represented graphically by the following figures.

-

-The first is a cyclical group, and the figure (fig.~9) does not

-differ essentially from fig.~2 in §~183.

-

-The group given by the second solution of the equation

-\index{Group@\Topic{Group}!dihedral}%

-is called the dihedral group. It is represented by fig.~10.

-%% -----File: 304.png---Folio 288-------

-\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%

-\index{Graphical representation@\Topic{Graphical representation}!examples of|(}%

-%[Illustration: Fig. 9.]

-\Figure{9}{fig09}

-%[Illustration: Fig. 10.]

-\Figure{10}{fig10}

-%% -----File: 305.png---Folio 289-------

-\index{Group@\Topic{Group}!tetrahedral}%

-\index{Group@\Topic{Group}!octohedral}%

-\index{Group@\Topic{Group}!icosahedral}%

-%[Illustration: Fig. 11.]

-\Figure{11}{fig11}

-

-The group given by the third solution of the equation is

-represented in fig.~11. It is known as the tetrahedral group.

-

-To the fourth solution of the equation corresponds the group

-represented by fig.~12. It is known as the octohedral group.

-

-To the fifth solution of the equation corresponds the group

-represented in fig.~13. It is known as the icosahedral group.

-

-The four last groups are identical with the groups of rotations

-which will bring respectively a double pyramid on an $n$-sided

-base, a tetrahedron, an octohedron, and an icosahedron to coincidence

-with itself in each case\footnotemark.

-\index{Klein@\Topic{Klein}, quoted}%

-\footnotetext{Klein, ``Vorlesungen über das Ikosaeder,'' Chap.~\DPchg{\textsc{i}}{I}.}%

-

-When the figures are drawn on a sphere, and the three circles

-of the original triangle and therefore also all the circles of the

-figure are taken to be great-circles of the sphere, the actual displacements

-of the triangles among themselves which correspond

-%% -----File: 306.png---Folio 290-------

-\PageLabel{290}%

-%[Illustration: Fig. 12.]

-\Figure{12}{fig12}

-%[Illustration: Fig. 13.]

-\Figure{13}{fig13}

-\index{Graphical representation@\Topic{Graphical representation}!examples of|)}%

-%% -----File: 307.png---Folio 291-------

-to the operations of the group can be effected by real rotations

-about diameters of the sphere; thus the statement of the preceding

-sentence may be directly verified.

-

-\Par{203.} In terms of their generating operations, the five types

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus zero}%

-of group of genus zero are given by the relations:---

-\begin{alignat*}{5}

-&\Inum{I.}   & S_{1}^{n} &= 1, & S_{2}^{n} &= 1, &           &     & S_{1}S_{2} &= 1; \\

-&\Inum{II.}  & S_{1}^{2} &= 1, & S_{2}^{2} &= 1, & S_{3}^{n} &= 1, & S_{1}S_{2}S_{3} &= 1; \\

-&\Inum{III.}\quad & S_{1}^{2} &= 1,\quad & S_{2}^{3} &= 1,\quad & S_{3}^{3} &= 1,\quad & S_{1}S_{2}S_{3} &= 1; \\

-&\Inum{IV.}  & S_{1}^{2} &= 1, & S_{2}^{3} &= 1, & S_{3}^{4} &= 1, & S_{1}S_{2}S_{3} &= 1; \\

-&\Inum{V.}   & S_{1}^{2} &= 1, & S_{2}^{3} &= 1, & S_{3}^{5} &= 1, & S_{1}S_{2}S_{3} &= 1.

-\end{alignat*}

-

-The first of these does not require special discussion.

-

-In the dihedral group, we have

-\[

-S_{1} S_{3} S_{1} = S_{1} S_{2} = S_{3}^{-1}.

-\]

-

-The dihedral group of order~$2n$ therefore contains a cyclical

-sub-group of order~$n$ self-conjugately; and every operation of

-the group which does not belong to this self-conjugate sub-group

-is of order~$2$. The operations of the group are given,

-each once and once only, by the form

-\[

-S_{1}^{\alpha} S_{3}^{\beta},\quad (\alpha = 0, 1;\ \beta = 0, 1, 2, \Dots, n - 1).

-\]

-

-When $n = 3$, this group is simply isomorphic with the

-symmetric group of three symbols.

-

-In the tetrahedral group, since

-\begin{gather*}

-(S_{1} S_{2})^{3} = 1, \\

-S_{2}^{-1} S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2}^{-1},

-\end{gather*}

-and therefore $S_{1}$, $S_{2}^{-1} S_{1}S_{2}$, $S_{2}S_{1}S_{2}^{-1}$ are permutable with each

-other. These operations of order~$2$ (with identity) form a self-conjugate

-sub-group of order~$4$; and the $12$~operations of the

-group are therefore given by the form

-\[

-S_{2}^{\alpha} (S_{2}^{-1} S_{1} S_{2})^{\beta} S_{1}^{\gamma},

-\]

-or

-\[

-S_{2}^{\alpha} S_{1}^{\beta} S_{2} S_{1}^{\gamma},\quad (\alpha = 0, 1, 2;\ \beta, \gamma = 0, 1).

-\]

-

-If

-\[

-S_{1} = (12) (34),\quad S_{2} = (123),

-\]

-then

-\[

-S_{1} S_{2} = (134);

-\]

-and therefore the tetrahedral group is simply isomorphic with

-the alternating group of four symbols.

-%% -----File: 308.png---Folio 292-------

-

-If, in the octohedral group, we write

-\[

-S_{3}^{2} = S',

-\]

-then

-\[

-S_{2}S' = S_{1}S_{3} = S_{1} S_{2}^{2} S_{1};

-\]

-and therefore

-\[

-(S_{2}S')^{3} = 1.

-\]

-

-Hence $S_{2}$ and $S'$ generate a tetrahedral group.

-

-Again

-\[

-S_{1}S_{2}S_{1} = (S_{2}S')^{2},

-\]

-and

-\[

-S_{1}S'S_{1} = S_{2}S' S_{2}^{-1},

-\]

-so that this is a self-conjugate sub-group. The operations of

-the group are given, each once and once only, in the form

-\[

-S_{1}^{\alpha} S_{2}^{\beta} S_{3}^{2\gamma} S_{2} S_{3}^{2\delta},\quad

-  (\alpha, \gamma, \delta = 0, 1;\ \beta = 0, 1, 2).

-\]

-

-If

-\[

-S_{1} = (12),\quad S_{2} = (234),

-\]

-then

-\[

-S_{1}S_{2} = (1342);

-\]

-and therefore the octohedral group is simply isomorphic with

-the symmetric group of four symbols.

-

-The icosahedral group is simple. It is, in fact, simply

-isomorphic with the alternating group of five symbols which

-has been shewn (§~111) to be a simple group. Thus if

-\[

-S_{1} = (12)(34),\quad S_{2} = (135),

-\]

-then

-\[

-S_{1}S_{2} = (12345);

-\]

-so that the substitutions $S_{1}$ and $S_{2}$ satisfy the relations

-\[

-S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad (S_{1}S_{2})^{5} = 1.

-\]

-

-They must therefore generate an icosahedral group or one

-of its sub-groups. On the other hand, from the substitutions

-$S_{1}$ and $S_{2}$ all the even substitutions of five symbols may be

-formed, and these are $60$~in number. The group therefore

-cannot be a sub-group of the icosahedral group; the only

-alternative is that the two are identical.

-

-As the icosahedral group has no self-conjugate group, we

-cannot in this case so easily construct a form which will

-represent each operation of the group just once in terms of

-the generating operations. It is however not difficult to verify

-that this is true of the set of forms\footnote

-  {Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}\ \VolNo{XX}, (1882), p.~35, and

-  Klein, \textit{loc.~cit.}\ p.~26.}

-\index{Dyck@\Topic{Dyck}, quoted}%

-\index{Klein@\Topic{Klein}, quoted}%

-\[

-\begin{gathered}

-  S_{3}^{\alpha},\quad S_{3}^{\alpha} S_{1} S_{3}^{\beta},\quad

-  S_{3}^{\alpha} S_{1} S_{3}^{2} S_{1} S_{3}^{\beta}, \\

-  S_{3}^{\alpha} S_{1} S_{3}^{2} S_{1} S_{3}^{3} S_{1},

-\end{gathered}

-\quad (\alpha, \beta = 0, 1, 2, 3, 4).

-\]

-\index{Groups@\Topic{Groups}!of genus zero|)}%

-%% -----File: 309.png---Folio 293-------

-

-\Par{204.} We shall next deal with the equation in the case

-$p = 1$. In this case alone, the order of the group disappears

-from the equation, which merely gives a relation between the

-number and order of the generating operations. This may be

-written in the form

-\[

-2 = \sum_{1}^{n} \left(1 - \frac{1}{m_{r}}\right);

-\]

-and $n$~must therefore be either $4$ or~$3$.

-

-When $n$~is~$4$, the equation becomes

-\[

-2 = \frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{4}},

-\]

-and the only solution is clearly

-\[

-m_{1} = m_{2} = m_{3} = m_{4} = 2.

-\]

-

-When $n$~is~$3$, the equation takes the form

-\[

-1 = \frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}},

-\]

-and is easily seen to have three solutions, viz.\

-\begin{alignat*}{3}

-m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 3; \\

-m_{1} &= 2, & m_{2} &= 4, & m_{3} &= 4; \\

-\lintertext{and}

-m_{1} &= 2, & m_{2} &= 3, & m_{3} &= 6.

-\end{alignat*}

-

-\Par{205.} Take first the solution

-\[

-n = 4,\quad m_{1} = m_{2} = m_{3} = m_{4} = 1.

-\]

-The corresponding general group is defined by the relations

-\begin{gather*}

-S_{1}^{2} = 1,\quad S_{2}^{2} = 1, \quad S_{3}^{2} = 1,\quad S_{4}^{2} = 1, \\

-S_{1}S_{2}S_{3}S_{4} = 1.

-\end{gather*}

-

-If we proceed to form the plane figure representing this

-group, the sum of the internal angles of the quadrilateral

-$A_{1}A_{2}A_{3}A_{4}$ is equal to four right angles, and the four circles

-that form it therefore pass through a point. If this point be

-taken at infinity, the four circles (and therefore all the circles of

-the figure) become straight lines. The plane figure will now

-take the form given in fig.~14, and the four generating operations

-are actual rotations through two right angles about lines

-through $A_{1}$,~$A_{2}$,~$A_{3}$ and~$A_{4}$, perpendicular to the plane of the

-figure.

-%% -----File: 310.png---Folio 294-------

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-

-%[Illustration: Fig. 14.]

-\Figure{14}{fig14}

-

-Every operation of the group is therefore, in this form of

-representation, either a rotation through two right angles

-about a corner of the figure or a translation; and it will clearly

-be the former or the latter according as it consists of an odd

-or an even number of factors, when expressed in terms of the

-generating operations. The operations which correspond to

-translations form a sub-group; for if two operations each

-consist of an even number of factors, so also does their

-product. Moreover, this sub-group is self-conjugate, since the

-number of factors in $\Sigma^{-1}S\Sigma$ is even if the number in~$S$ is

-even. This self-conjugate sub-group is generated by the two

-operations

-\[

-S_{1}S_{2}\quad \text{and}\quad S_{2}S_{3};

-\]

-for

-\begin{gather*}

-S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3}, \\

-S_{2}S_{1} = (S_{1}S_{2})^{-1},\quad

-S_{3}S_{2} = (S_{2}S_{3})^{-1},\quad

-S_{3}S_{1} = (S_{2}S_{3})^{-1} (S_{1}S_{2})^{-1};

-\end{gather*}

-and therefore every operation containing an even number of

-factors can be represented in terms of $S_{1}S_{2}$ and~$S_{2}S_{3}$. Lastly,

-these two operations are permutable with each other; for

-\[

-S_{2}S_{3} · S_{1}S_{2} = S_{2}S_{3} · S_{4}S_{3} = S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3};

-\]

-and therefore every operation of the group is contained, once

-and only once, in the form

-\[

-S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma},\quad (\alpha = 0, 1:\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).

-\]

-

-The results thus arrived at may also be verified very simply

-by purely kinematical considerations. If a group generated

-%% -----File: 311.png---Folio 295-------

-by $S_{1}$,~$S_{2}$,~$S_{3}$ and~$S_{4}$ is of finite order, there must, since it is of

-genus~$1$, be either one or two additional relations between the

-generating operations; and any such relation is expressible by

-equating the symbol of some operation of the general group to

-unity. Such a relation is therefore either of the form

-\[

-S_{1} (S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1,

-\]

-or

-\[

-(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1.

-\]

-

-The operation $S_{1} (S_{1}S_{2})^{b} (S_{2}S_{3})^{c}$ of the general group, consisting

-of an odd number of factors, must be a rotation round

-some corner of the figure, say a rotation round the corner~$A_{r}$ of

-the white quadrilateral~$\Sigma$; it is therefore identical with $\Sigma^{-1}S_{r}\Sigma$.

-

-Now the relation

-\[

-\Sigma^{-1}S_{r}\Sigma = 1,

-\]

-gives

-\[

-S_{r} = 1.

-\]

-

-A relation of the first of the two forms is therefore inconsistent

-with the supposition that the group is actually generated

-by $S_{1}$,~$S_{2}$ and~$S_{3}$. It, in fact, reduces the generating operations

-and the relations among them to

-\[

-S_{1}^{2} = 1,\quad S_{2}^{2} = 1,\quad S_{3}^{2} = 1,\quad S_{1}S_{2}S_{3} = 1,

-\]

-which define a group of genus zero.

-

-The only admissible relations for a group of genus~$1$ are

-\index{Groups@\Topic{Groups}!of genus one|(}%

-therefore those of the form

-\[

-(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1.

-\]

-

-A single relation of this form reduces the operations of the

-general group to those contained in

-\begin{gather*}

-S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma}, \\

-(\alpha = 0, 1;\ \beta = 0, 1, \dots, b - 1;\ \gamma = -\infty, \dots, 0, \dots, \infty);

-\end{gather*}

-and the group so defined is still of infinite order.

-

-Finally, two independent relations

-\begin{gather*}

-(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1, \\

-(S_{1}S_{2})^{b'} (S_{2}S_{3})^{c'} = 1,

-\end{gather*}

-where

-\[

-\frac{b}{b'} \neq \frac{c}{c'},

-\]

-must necessarily lead to a group of finite order. If $m$~is the

-greatest common factor of $b$~and~$b'$, so that

-\[

-b = b_{1}m,\quad b' = b_{1}'m,

-\]

-%% -----File: 312.png---Folio 296-------

-where $b_{1}$ and~$b_{1}'$ are relatively prime; and if

-\[

-b_{1}x - b_{1}'y = 1;

-\]

-the two relations give

-\[

-(S_{2}S_{3})^{cb_{1}' - c'b_{1}} = 1,

-\]

-and

-\[

-(S_{1}S_{2})^{m} = (S_{2}S_{3})^{c'y - cx}.

-\]

-

-Every operation of the group is now contained, once and

-only once, in the form

-\begin{gather*}

-S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma}, \\

-(\alpha = 0, 1;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, cb_{1}' - b_{1}c' - 1);

-\end{gather*}

-%[** TN: Minus sign in next line printed as a tilde in the original]

-and the order of the group is $2(bc' - b'c)$.

-

-\Par{206.} Corresponding to the solution

-\[

-n = 3,\quad m_{1} = m_{2} = m_{3} = 3,

-\]

-we have the general group generated by $S_{1}$,~$S_{2}$,~$S_{3}$, where

-\begin{gather*}

-S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{3} = 1, \\

-S_{1} S_{2} S_{3} = 1.

-\end{gather*}

-

-The sum of the three angles of the triangle $A_{1} A_{2} A_{3}$ is two

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-% [Illustration: Fig.~15.]

-\Figure{15}{fig15}

-%% -----File: 313.png---Folio 297-------

-right angles, and therefore again the circles in the plane figure

-(fig.~15) may be taken as straight lines. When the figure is

-thus chosen, the generating operations are rotations through~$\frac{2}{3}\pi$

-about the angles of an equilateral triangle; and every operation

-of the group is either a translation or a rotation.

-

-The three operations

-\[

-S_{1} S_{2}^{2},\quad S_{2} S_{1} S_{2},\quad S_{2}^{2} S_{1},

-\]

-when transformed by~$S_{2}$, are interchanged among themselves.

-When transformed by~$S_{1}$ they become

-\[

-S_{2}^{2} S_{1},\quad

-S_{1}^{2} S_{2} S_{1} S_{2} S_{1},\quad

-S_{1}^{2} S_{2}^{2} S_{1}^{2},

-\]

-and since

-\[

-(S_{1} S_{2})^{3} = 1,

-\]

-the two latter are $S_{1} S_{2}^{2}$ and $S_{2}S_{1} S_{2}$ respectively.

-

-Hence the three operations generate a self-conjugate sub-group;

-and since

-\[

-S_{1} S_{2}^{2} · S_{2} S_{1} S_{2} · S_{2}^{2} S_{1} = 1,

-\]

-this sub-group is generated by $S_{1} S_{2}^{2}$ and $S_{2} S_{1} S_{2}$.

-

-These two operations are permutable; for

-\[

-S_{1} S_{2}^{2} · S_{2} S_{1} S_{2}

-  = S_{1}^{2} S_{2}

-  = S_{2} S_{1} S_{2} S_{1} S_{2} · S_{2}

-  = S_{2} S_{1} S_{2} · S_{1} S_{2}^{2}.

-\]

-

-Hence finally, every operation of the group is represented,

-once and only once, by the form

-\[

-S_{1}^{\alpha} (S_{1}S_{2}^{2})^{\beta} (S_{2}S_{1}S_{2})^{\gamma},\quad

-  (\alpha = 0, 1, 2;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).

-\]

-

-This result might also be arrived at by purely kinematical

-considerations; for an inspection of the figure shews that the

-two simplest translations are

-\[

-S_{1} S_{2}^{2} \quad\text{and}\quad S_{2} S_{1} S_{2},

-\]

-and that every translation in the group can be obtained by the

-combination and repetition of these. Every operation in which

-the index~$\alpha$ is not zero must be a rotation through $\frac{2}{3}\pi$ or~$\frac{4}{3}\pi$

-about one of the angles of the figure; it is therefore necessarily

-identical with an operation of the form

-\[

-\Sigma^{-1} S_{r}^{\alpha} \Sigma.

-\]

-

-If now the group generated by $S_{1}$,~$S_{2}$,~$S_{3}$ is of finite order,

-there must be either one or two additional relations between

-them. A relation of the form

-\[

-S_{1}^{a} (S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,

-\]

-%% -----File: 314.png---Folio 298-------

-where $a$~is either $1$ or~$2$, is equivalent to

-\[

-\Sigma^{-1} S_{r}^{a}\Sigma = 1,

-\]

-so that

-\[

-S_{r} = 1.

-\]

-Such a relation would reduce the group to a cyclical group of

-order~$3$. This is not admissible, if the group is actually to be

-generated by two distinct operations $S_{1}$ and~$S_{2}$.

-

-A relation

-\[

-(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_2)^{c} = 1,

-\]

-gives, on transformation by~$S_{2}^{-1}$,

-\[

-(S_{2} S_{1} S_{2})^{b} (S_{2}^{2} S_{1})^{c} = 1.

-\]

-

-Now

-\[

-S_{2}^{2} S_{1} = (S_{1} S_{2}^{2})^{-1} (S_{2} S_{1} S_{2})^{-1},

-\]

-so that

-\[

-(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b - c} = 1.

-\]

-

-If $m$~is the greatest common factor of $b$~and~$c$, so that

-\[

-b = b'm,\quad c = c'm,

-\]

-where $b'$~and~$c'$ are relatively prime; and if

-\[

-b'x - c'y = 1,

-\]

-the two relations

-\[

-(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,

-\]

-and

-\[

-(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b - c} = 1,

-\]

-lead to

-\[

-(S_{2} S_{1} S_{2})^{m(b'^{2} - b'c' + c'^{2})} = 1,

-\]

-and

-\[

-(S_{1} S_{2})^{m} = (S_{2} S_{1} S_{2})^{m\{(c' - b')y - c'x\}};

-\]

-and every operation of the group is contained, once and only

-once, in the form

-\[

-S_{1}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma},

-\]

-where

-\[

-\alpha = 0, 1, 2;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b'^{2} - b'c' + c'^{2}) - 1.

-\]

-Thus the group is of finite order $3(b^{2} - bc + c^{2})$. In this case

-then, unlike the previous one, a single additional relation is

-sufficient to ensure that the group is of finite order. Any

-further relation, which is independent, must of necessity reduce

-the group to a cyclical group of order~$3$ or to the identical

-operation.

-

-\Par{207.} The two remaining solutions may now be treated in

-less detail. The general group corresponding to the solution

-\[

-n = 3,\quad m_{1} = 2,\quad m_{2} = m_{3} = 4,

-\]

-%% -----File: 315.png---Folio 299-------

-is given by

-\begin{gather*}

-S_{1}^{2} = 1,\quad S_{2}^{4} = 1,\quad S_{3}^{4} = 1, \\

-S_{1} S_{2} S_{3} = 1,

-\end{gather*}

-and is represented graphically by fig.~16.

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-

-% [Illustration: Fig.~16.]

-\Figure{16}{fig16}

-

-All the translations of the group can be generated from the

-two operations

-\[

-S_{1} S_{2}^{2},\quad S_{2} S_{1} S_{2};

-\]

-and every operation of the group is given, once and only once,

-by the form

-\[

-S_{2}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma},\quad

-  (\alpha = 0, 1, 2, 3;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).

-\]

-

-An additional relation of the form

-\[

-S_{2}^{a} (S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,

-\]

-where $a$~is $1$,~$2$ or~$3$, leads either to

-\[

-S_{1} = 1,\quad S_{2} = 1 \quad\text{or}\quad S_{3} = 1,

-\]

-and is therefore inconsistent with the supposition that the

-group is generated by two distinct operations.

-

-An additional relation

-\[

-(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1

-\]

-gives

-\[

-(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b} = 1:

-\]

-and if

-\begin{gather*}

-b = b_{1}m,\quad c = c_{1}m, \\

-b_{1}x + c_{1}y = 1,

-\end{gather*}

-%% -----File: 316.png---Folio 300-------

-where $b_{1}$~and~$c_{1}$ are relatively prime, these relations are

-equivalent to

-\begin{gather*}

-(S_{2} S_{1} S_{2})^{m(b_{1}^{2} + c_{1}^{2})} = 1, \\

-(S_{1} S_{2}^{2})^{m} = (S_{2} S_{1} S_{2})^{m(b_{1}y - c_{1}x)}.

-\end{gather*}

-

-Every operation of the group is then contained, once and

-only once, in the form

-\begin{gather*}

-S_{2}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma}, \\

-(\alpha = 0, 1, 2, 3;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b_{1}^{2} + c_{1}^{2}) - 1);

-\end{gather*}

-and the order of the group is $4(b^{2} + c^{2})$.

-

-\Par{208.} Lastly, the general group corresponding to the

-solution

-\[

-n = 3,\quad m_{1} = 2,\quad m_{2} = 3,\quad m_{3} = 6,

-\]

-is given by

-\begin{gather*}

-S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{6} = 1, \\

-S_{1} S_{2} S_{3} = 1;

-\end{gather*}

-and it is represented graphically by fig.~17.

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-

-% [Illustration: Fig.~17.]

-\Figure{17}{fig17}

-%% -----File: 317.png---Folio 301-------

-

-Now it may again be verified, either from the generating

-relations or from the figure, that the two operations

-\[

-S_{2}^{2} S_{3}^{2} \quad\text{and}\quad S_{3} S_{2}^{2} S_{3},

-\]

-which are permutable with each other, generate all the operations

-which in the kinematical form of the group are translations;

-and that every operation of the group is represented, once and

-only once, by the form

-\begin{gather*}

-S_{3}^{\alpha} (S_{2}^{2} S_{3}^{2})^{\beta} (S_{3} S_{2}^{2} S_{3})^{\gamma}, \\

-(\alpha = 0, 1, \dots, 5;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).

-\end{gather*}

-

-Also as before, any further relation, which does not reduce

-the group to a cyclical group, is necessarily of the form

-\[

-(S_{2}^{2} S_{3}^{2})^{b} (S_{3} S_{2}^{2} S_{3})^{c} = 1.

-\]

-

-On transforming this relation by~$S_{3}^{-1}$ we obtain

-\[

-(S_{3} S_{2}^{2} S_{3})^{b} (S_{3}^{2} S_{2}^{2})^{c} = 1.

-\]

-

-Now

-\[

-S_{3}^{2} S_{2}^{2} = (S_{2}^{2} S_{3}^{2})^{-1} (S_{3} S_{2}^{2} S_{3});

-\]

-so that

-\[

-(S_{2}^{2} S_{3}^{2})^{-c} (S_{3} S_{2} S_{3})^{b + c} = 1.

-\]

-

-If then

-\begin{gather*}

-b = b_{1}m,\quad c = c_{1}m, \\

-b_{1}x + c_{1}y = 1,

-\end{gather*}

-where $b_{1}$~and~$c_{1}$ are relatively prime, it follows that

-\[

-(S_{3} S_{2}^{2} S_{3})^{m(b_{1}^{2} + b_{1}c_{1} + c_{1}^{2})} = 1,

-\]

-and

-\[

-(S_{2}^{2} S_{3}^{2})^{m} = (S_{3} S_{2} S_{3})^{m\{b_{1}y + c_{1}(y - x)\}}.

-\]

-

-Every operation of the group is then contained, once and

-only once, in the form

-\begin{gather*}

-S_{3}^{\alpha} (S_{2}^{2} S_{3}^{2})^{\beta} (S_{3} S_{2}^{2} S_{3})^{\gamma}, \\

-(\alpha = 0, 1, \dots, 5;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b_{1}^{2} + b_{1}c_{1} + c_{1}^{2}) - 1);

-\end{gather*}

-and the order of the group is $6(b^{2} + bc + c^{2})$.

-

-\Par{209.} There are thus four distinct classes of groups\footnote

-  {Dyck, ``Ueber Aufstellung und Untersuchung von Gruppe und Irrationalität

-  regulärer Riemann'scher Flächen,'' \textit{Math.\ Ann.}\ \VolNo{XVII}, (1880), pp.~501--509.}

-\index{Dyck@\Topic{Dyck}, quoted}%

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus unity}%

-of

-genus~$1$, which are defined in terms of their generating operations

-by the following sets of relations:---

-

-\Item{\Inum{I.}} $\begin{gathered}[t]

-S_{1}^{2} = 1,\quad S_{2}^{2} = 1,\quad S_{3}^{2} = 1, \quad (S_{1} S_{2} S_{3})^{2} = 1, \\

-(S_{1} S_{2})^{a} (S_{2} S_{3})^{b} = 1,\quad

-(S_{1} S_{2})^{a'} (S_{2} S_{3})^{b'} = 1,\quad (ab' - a'b > 0); \\

-N = 2(ab' - a'b).

-\end{gathered}$

-%% -----File: 318.png---Folio 302-------

-

-\Item{II.} $\begin{gathered}[t]

-S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad (S_{1} S_{2})^{3} = 1, \\

-(S_{1}^{2} S_{2}^{2})^{a} (S_{2} S_{1} S_{2})^{b} = 1; \\

-N = 3(a^{2} - ab + b^{2}).

-\end{gathered}$

-

-\Item{III.} $\begin{gathered}[t]

-S_{1}^{2} = 1,\quad S_{2}^{4} = 1,\quad (S_{1} S_{2})^{4} = 1, \\

-(S_{1} S_{2}^{2})^{a} (S_{2} S_{1} S_{2})^{b} = 1; \\

-N = 4 (a^{2} + b^{2}).

-\end{gathered}$

-

-\Item{IV.} $\begin{gathered}[t]

-S_{1}^{6} = 1,\quad S_{2}^{3} = 1,\quad (S_{1} S_{2})^{2} = 1, \\

-(S_{1}^{2} S_{2}^{2})^{a} (S_{1} S_{2}^{2} S_{1})^{b} = 1; \\

-N = 6 (a^{2} + ab + b^{2}).

-\end{gathered}$

-\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus unity}%

-\index{Groups@\Topic{Groups}!of genus one|)}%

-

-For special values of $a$~and~$b$, some of these groups may be

-groups of genus zero; for instance, in Class~\Inum{I}, if $ab' - a'b$ is a

-prime, the group is a dihedral group. It is left as an exercise

-to the reader to determine all such exceptional cases.

-

-\begin{Ex}

-Prove that the number of distinct types of group, of genus two, is

-\index{Groups@\Topic{Groups}!of genus two}%

-three; viz.\ the groups defined by

-\begin{alignat*}{4}

-&\Inum{(i)}      & A^{4} &= 1,\quad& B^{2} &= A^{2},\quad&& B^{-1}AB = A^{-1}; \\

-&\Inum{(ii)}     & A^{8} &= 1,     & B^{2} &= 1,         && B^{-1}AB = A^{3}; \\

-&\Inum{(iii)}\quad&A^{8} &= 1,     & B^{2} &= 1,         && (AB)^{3} = 1,\quad (A^{4}B)^{2} = 1.

-\end{alignat*}

-\end{Ex}

-

-\Par{210.} As a final illustration of the present method of

-\PageLabel{302}%

-graphical representation, we will consider the simple group of

-order~$168$ (§~146), given by

-\[

-\{(1236457),\ (234) (567),\ (2763) (45)\}.

-\]

-

-The operations of this group are of orders $7$,~$4$, $3$ and~$2$; and

-it is easy to verify that three operations of orders $2$,~$3$, and~$7$

-can be chosen such that their product is identity.

-

-In fact, if

-\[

-S_{2} = (16) (34),\quad S_{3} = (253) (476),\quad S_{7} = (1673524);

-\]

-then

-\[

-S_{2} S_{3} S_{7} = 1.

-\]

-

-Moreover, these three operations generate the group. The

-connectivity of the corresponding surface, by the regular division

-of which the group can be represented, is $2p + 1$; where

-\[

-2p - 2 = 168(3 - 2 - \tfrac{1}{2} - \tfrac{1}{3} - \tfrac{1}{7}).

-\]

-%% -----File: 319.png---Folio 303-------

-

-This gives

-\[

-p = 3;

-\]

-it follows from Theorem~II, §~197, that the genus of the

-group is~$3$.

-

-The figure for the general group, generated by $S_{2}$,~$S_{3}$, and~$S_{7}$,

-where

-\begin{gather*}

-S_{2}^{2} = 1,\quad S_{3}^{3} = 1,\quad S_{7}^{7} = 1, \\

-S_{2} S_{3} S_{7} = 1,

-\end{gather*}

-acquires as symmetrical a form as possible, by taking the

-centre of the orthogonal circle for that angular point of the

-triangle~$1$ at which the angle is~$\frac{1}{7}\pi$. In fig.~18 a portion of

-the general figure, which is contained between two radii of the

-orthogonal circle inclined at an angle~$\frac{2}{7}\pi$, is shewn. The

-remainder may be filled in by inversions at the different

-portions of the boundary.

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-

-%[Illustration: Fig. 18.]

-\Figure[0.8\textwidth]{18}{fig18}

-%% -----File: 320.png---Folio 304-------

-The operations, which correspond to the white triangles of

-the figure, are given by the following table:---

-\[

-\begin{array}{cr|cr|cr}

-1 &               1  & 9 &          S_{2}S_{7}^{5}S_{2} &17 & S_{2}S_{7}^{5}S_{2}S_{7}^{3}S_{2} \\

-2 &          S_{2} &10 &     S_{7}S_{2}S_{7}^{5}S_{2} &18 &      S_{7}^{6}S_{2}S_{7}^{3}S_{2} \\

-3 &     S_{7}S_{2} &11 & S_{7}^{2}S_{2}S_{7}^{5}S_{2} &19 &               S_{2}S_{7}^{3}S_{2} \\

-4 & S_{7}^{2}S_{2} &12 & S_{7}^{3}S_{2}S_{7}^{5}S_{2} &20 &          S_{7}S_{2}S_{7}^{3}S_{2} \\

-5 & S_{7}^{3}S_{2} &13 & S_{7}^{6}S_{2}S_{7}^{4}S_{2} &21 &               S_{2}S_{7}^{2}S_{2} \\

-6 & S_{7}^{4}S_{2} &14 &          S_{2}S_{7}^{4}S_{2} &22 &      S_{7}^{6}S_{2}S_{7}^{2}S_{2} \\

-7 & S_{7}^{5}S_{2} &15 &     S_{7}S_{2}S_{7}^{4}S_{2} &23 &      S_{7}^{5}S_{2}S_{7}^{2}S_{2} \\

-8 & S_{7}^{6}S_{2} &16 & S_{7}^{2}S_{2}S_{7}^{4}S_{2} &24 & S_{2}S_{7}^{4}S_{2}S_{7}^{2}S_{2}.

-\end{array}

-\]

-

-The representation of the special group is derived from

-this general figure by retaining only a set of $168$ white (and

-corresponding black) triangles, which are distinct when $S_{2}$,~$S_{3}$

-and~$S_{7}$ are replaced by the corresponding substitutions given on

-\PageRef{p.}{302}. When each white triangle is thus marked with the

-corresponding substitution, it is found that a complete set of

-$168$ distinct white (and black) triangles is given by the portion

-of the figure actually drawn and the six other distinct portions

-obtained by rotating it round the centre of the orthogonal

-circle through multiples of~$\frac{2}{7}\pi$.

-

-To complete the graphical representation of the group of

-order~$168$, it is necessary to determine the correspondence in

-pairs of the sides of the boundary. This is facilitated by

-noticing that the angular points $A_{1}$, $A_{2}$, $A_{3}$,~\Dots\ of the

-boundary must correspond in sets. Now the white triangle,

-which has an angle at~$A_{1}$ and lies inside the polygon, is

-given by

-\[

-S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2}.

-\]

-

-This must be equivalent to a white triangle, which lies

-outside the polygon, and has a side on the boundary and an

-angle at one of the points $A_{2}$, $A_{3}$,~\Dots.

-

-The triangle, which satisfies these conditions and has an

-angle at~$A_{2n + 1}$, is given by

-\[

-S_{7} S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2} S_{7}^{n};

-\]

-while the triangle, which satisfies the conditions and has an

-angle at~$A_{2n + 2}$, is given by

-\[

-S_{7}^{6} S_{2} S_{7}^{5} S_{2} S_{7}^{3} S_{2} S_{7}^{n}.

-\]

-%% -----File: 321.png---Folio 305-------

-

-When $(16) (34)$ and $(1673524)$ are written for $S_{2}$ and~$S_{7}$, we

-find that

-\[

-S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2}

-  = S_{7} S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2} S_{7}^{5}.

-\]

-

-The white triangle with an angle at~$A_{1}$ inside the polygon

-is therefore equivalent to the white triangle with an angle at~$A_{11}$,

-which lies outside the polygon and has a side on the

-boundary. It follows, from the continuity of the figure, that

-the arcs $A_{1}A_{2}$ and $A_{11}A_{10}$ of the boundary correspond. Since

-the operation~$S_{7}$ changes the figure and the boundary into

-themselves, it follows that $A_{3}A_{4}$,~$A_{13}A_{12}$; $A_{5}A_{6}$,~$A_{1}A_{14}$; $A_{7}A_{8}$,~$A_{3}A_{2}$;

-$A_{9}A_{10}$,~$A_{5}A_{4}$; $A_{11}A_{12}$,~$A_{7}A_{6}$; and $A_{13}A_{14}$,~$A_{9}A_{8}$; are pairs

-of corresponding sides. Hence the above single condition is

-sufficient to ensure that the general group shall reduce to the

-special group of order~$168$.

-

-By taking account of the relation

-\[

-(S_{7} S_{2})^{3} = 1,\quad \text{or}\quad (S_{2} S_{7})^{3} = 1,

-\]

-the form of the condition may be simplified. Thus it may be

-written

-\begin{align*}

-S_{7}^{4} S_{2} S_{7}^{2} S_{2}

-  &= S_{2} S_{7} S_{2} S_{7} · S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5} \\

-  &= S_{7}^{6} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5},

-\end{align*}

-or

-\[

-S_{2} S_{7}^{2} S_{2} = S_{7}^{2} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5}.

-\]

-

-Now

-\begin{align*}

-S_{2} S_{7} · S_{7} S_{2}

-  &= S_{7}^{6} S_{2} S_{7}^{6} S_{2} · S_{2} S_{7}^{6} S_{2} S_{7}^{6} \\

-  &= S_{7}^{6} S_{2} S_{7}^{5} S_{2} S_{7}^{6}.

-\end{align*}

-

-Hence

-\[

-S_{7}^{4} S_{2} S_{7}^{5} S_{2} S_{7} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},

-\]

-or

-\[

-S_{7}^{4} S_{2} S_{7}^{4} · S_{2} S_{7}^{6} S_{2} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},

-\]

-or

-\[

-S_{7}^{4} S_{2} S_{7}^{4} S_{2} S_{7}^{4} = S_{2} S_{7}^{3} S_{2},

-\]

-or finally

-\[

-(S_{7}^{4} S_{2})^{4} = 1.

-\]

-

-The simple group of order~$168$ is therefore defined abstractly

-\index{Defining relations@\Topic{Defining relations} of a group!for the simple group of order~$168$}%

-by the relations\footnote

-  {This agrees with the result as stated by Dyck, ``Gruppentheoretische

-  Studien,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XX}, (1882), p.~41.}

-\index{Dyck@\Topic{Dyck}, quoted}%

-\[

-S_{2}^{2} = 1,\quad S_{7}^{7} = 1,\quad (S_{7} S_{2})^{3} = 1,\quad (S_{7}^{4} S_{2})^{4} = 1.

-\]

-

-\begin{Ex} Shew that the symmetric group of degree five is a group

-\index{Defining relations@\Topic{Defining relations} of a group!for the symmetric group of degree~$5$}%

-of genus four; and that it is completely defined by the relations

-\[

-S_{2}^{2} = 1,\quad S_{5}^{5} = 1,\quad (S_{2} S_{5})^{4} = 1,\quad (S_{5}^{-1} S_{2} S_{5} S_{2})^{3} = 1.

-\]

-\end{Ex}

-%% -----File: 322.png---Folio 306-------

-

-\Par{211.} The regular division of a continuous surface into

-$2N$~black and white polygons is only one of many methods that

-may be conceived for representing a group graphically.

-

-We shall now describe shortly another such mode of

-\index{Colour-groups@\Topic{Colour-groups}|(}%

-representation, due to Cayley\footnotemark,

-\index{Cayley@\Topic{Cayley}, quoted}%

-\footnotetext{\textit{American Journal of Mathematics}, Vol.~\VolNo{I}, (1878), pp.~174--176, Vol.~\VolNo{XI},

-  (1889), pp.~139--157; \textit{Proceedings of the London Mathematical Society}, Vol.~\VolNo{IX},

-  (1878), pp.~126--133.}%

-who has called it the method

-of \emph{colour-groups}. As given by Cayley, this method is entirely

-independent of the one we have been hitherto dealing with;

-but there is an intimate relation between them, and the new

-method can be most readily presented to the reader by deriving

-it from the old one.

-

-Let

-\[

-1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1}

-\]

-be the operations of a group~$G$ of order~$N$. We may take the

-$N - 1$ operations other than identity as a set of generating

-operations. Their continued product

-\[

-S_{1} S_{2} \Dots S_{N - 1}

-\]

-is some definite operation of the group. If it is the identical

-operation, the only modification in the figure, which represents

-the group by the regular division of a continuous surface, will

-be that the $N$th~corner of the polygon has an angle of two right-angles.

-

-With this set of generating operations, the representation

-of the group is given by a regular division of a continuous

-surface into $N$~white and $N$~black polygons $A_{1}A_{2}\Dots A_{N}$, the

-angle at~$A_{r}$ being~$\dfrac{2\pi}{m_{r}}$, $m_{r}$~being the order of~$S_{r}$. Suppose

-now that in each white polygon we mark a definite point.

-From the marked point in the polygon~$\Sigma$, draw a line to

-the marked point in the polygon derived from it by a positive

-rotation round its angle~$A_{r}$. Call this line an $S_{r}$-line, and

-denote the direction in which it is drawn by means of an

-arrow. Carry out this construction for each polygon~$\Sigma$, and

-for each of its angles except~$A_{N}$. We thus form a figure

-which, disregarding the original surface, consists of $N$~points

-connected by $N(N - 1)$ directed lines, two distinct lines

-joining each pair of points. Now if the line drawn from~$a$

-%% -----File: 323.png---Folio 307-------

-to~$b$, where $a$~and~$b$ are two of the points, is an $S$-line, then

-the line drawn from $b$ to~$a$ is from the construction an $S^{-1}$-line.

-We may then at once modify our diagram, in the direction

-of simplification, by dropping out one of the two lines

-between $a$ and~$b$, say the $S^{-1}$-line, on the understanding that

-the remaining line, with the arrow-head reversed, will give the

-line omitted. If $S$~is an operation of order~$2$, $S$~and $S^{-1}$ are

-identical, and two arrow-heads may be drawn on such a line in

-opposite directions. The modified figure will now consist of

-$N$~points connected by $\frac{1}{2}N(N - 1)$ lines. From the construction

-it follows at once that, for every value of~$r$, a single $S_{r}$-line ends

-at each point of the figure and a single $S_{r}$-line begins at each

-point of the figure; these two lines being identical when the

-order of~$S_{r}$ is~$2$.

-

-We may pass from one point of the figure to another along

-the lines in various ways; but any path between two points of

-the figure will be specified completely by such directions as:

-follow first an $S_{r}$-line, then an $S_{s}$-line, then an $S_{t}^{-1}$-line, and

-so on. Such a set of directions is said to define a route.

-It is an immediate consequence of the construction that, if

-starting from some one particular point a given route leads back

-to the starting point, then it will lead back to the starting point

-from whatever point we begin. In fact, a route will be specified

-symbolically by a symbol

-\[

-S_{u} \Dots S_{t}^{-1}S_{s}S_{r},

-\]

-and if

-\[

-S_{u} \Dots S_{t}^{-1}S_{s}S_{r}\Sigma = \Sigma,

-\]

-then

-\[

-S_{u} \Dots S_{t}^{-1}S_{s}S_{r} = 1,

-\]

-and therefore

-\[

-S_{u} \Dots S_{t}^{-1}S_{s}S_{r}\Sigma' = \Sigma',

-\]

-whatever operation $\Sigma'$ may be.

-

-\Par{212.} If the diagram of $N$~points connected by $\frac{1}{2}N(N - 1)$

-directed lines is to appeal readily to the eye, some method must

-be adopted of easily distinguishing an $S_{r}$-line from an $S_{s}$-line.

-To effect this purpose, Cayley suggested that all the $S_{r}$-lines

-should be of one colour, all the $S_{s}$-lines of another, and so on.

-Suppose now that, independently of any previous consideration,

-we have a diagram of $N$~points connected by $\frac{1}{2}N(N - 1)$ coloured

-directed lines satisfying the following conditions:---

-%% -----File: 324.png---Folio 308-------

-

-\Inum{(i)} all the lines of any one colour have either \Inum{($a$)}~a single

-arrow-head denoting their directions: or \Inum{($b$)}~two opposed arrow-heads,

-in which case each may be regarded as equivalent to two

-coincident lines in opposite directions;

-

-\Inum{(ii)} there is a single line of any given colour leading to every

-point in the diagram, and a single line of the colour leading from

-every point: if the colour is one with double arrow-heads, the

-two lines are a pair of coincident lines;

-

-\Inum{(iii)} every route which, starting from some one given point

-in the diagram, is closed, \ie\ leads back again to the given point,

-is closed whatever the starting point.

-

-Then, under these conditions, the diagram represents in

-graphical form a definite group of order~$N$.

-

-It is to be noticed that the first two conditions are necessary

-in order that the phrase ``a route'' used in the third shall have

-a definite meaning. Suppose that $R$~and~$R'$ are two routes

-leading from $a$ to~$b$. Then $RR'^{-1}$~is a closed route and will

-lead back to the initial point whatever it may be. Hence if $R$~leads

-from $c$ to~$d$, so also must~$R'$; and therefore $R$~and~$R'$ are

-equivalent routes in the sense that from any given starting

-point they lead to the same final point. There are then, with

-identity which displaces no point, just $N$ non-equivalent routes

-on the diagram, and the product of any two of these is a definite

-route of the set. The $N$~routes may be regarded as operations

-performed on the $N$~points; on account of the last property

-which has been pointed out, they form a group. Moreover, the

-diagram gives in explicit form the complete multiplication table

-of the group, for a mere inspection will immediately determine

-the one-line route which is equivalent to any given route; \ie\

-the operation of the group which is the same as the product of

-any set of operations in any given order.

-

-From a slightly different point of view, every route will give

-a permutation of the $N$~points, regarded as a set of symbols,

-among themselves; no symbol remaining unchanged unless

-they all do. To the set of $N$ independent routes, there will

-correspond a set of $N$ substitutions performed on $N$~symbols;

-and we can therefore immediately from the diagram represent

-the group as a transitive substitution group of degree~$N$.

-%% -----File: 325.png---Folio 309-------

-

-\Par{213.} It cannot be denied that, even for groups of small

-order, the diagram we have been describing would not be easily

-grasped by the eye. It may however still be considerably

-simplified since, so far as a graphical definition of the group

-is concerned, a large number of the lines are always redundant.

-

-If in the diagram consisting of $N$~points and $\frac{1}{2}N(N - 1)$

-coloured lines, which satisfies the conditions of §~212, all the

-lines of one or more colours are omitted, two cases may occur.

-We may still have a figure in which it is possible to pass

-along lines from one point to any other; or the points may

-break up into sets such that those of any one set are connected

-by lines, while there are no lines which enable us

-to pass from one set to another.

-

-Suppose, to begin with, that the first is the case. There will

-then, as before, be $N$~non-equivalent routes in the figure, which

-form a group when they are regarded as operations; it is

-obviously the same group as is given by the general figure.

-The sole difference is that there will not now be a one-line

-route leading from every point to every other point, and therefore

-the diagram will no longer give directly the result of the

-product of any number of operations of the group.

-

-If on the other hand the points break up into sets, the new

-diagram will no longer represent the same group as the original

-diagram. Some of the routes of the original diagram will

-not be possible on the new one, but every route on the new

-one will be a route on the original diagram. Hence the new

-diagram will give a sub-group; and since it is still the case that

-no route, except identity, can leave any point unchanged, the

-number of points in each of the sets must be the same. The

-reader may verify that the sub-group thus obtained will be self-conjugate,

-only if the omitted colours interchange these sets

-bodily among themselves.

-

-\Par{214.} The simplest diagram that will represent the group

-will be that which contains the smallest number of colours and

-at the same time connects all the points. To each colour

-corresponds a definite operation of the group (and its inverse).

-Hence the smallest number of colours is the smallest number

-of operations that will generate the group. It may be noticed

-%% -----File: 326.png---Folio 310-------

-that this simplified diagram can be actually constructed from the

-previously obtained representation of the group by the regular

-division of a surface, the process being exactly the same as

-that by which the general diagram was obtained. For if

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}

-\]

-are a set of independent generating operations, and if

-\[

-S_{1} S_{2} \Dots S_{n} S_{n + 1} = 1,

-\]

-we may represent the group by the regular division of a surface

-into $2N$ black and white $(n + 1)$-sided polygons. When we

-draw on this surface the $S_{1}$-, $S_{2}$-,~\Dots, $S_{n}$-lines, the $N$~points

-will be connected by lines in a single set, since from

-\[

-S_{1},\ S_{2},\ \Dots,\ S_{n}

-\]

-every operation of the group can be constructed; and the set

-of points and directed coloured lines so obtained is clearly the

-diagram required.

-

-As an illustration of this form of graphical representation,

-\index{Colour-groups@\Topic{Colour-groups}!examples of}%

-\index{Graphical representation@\Topic{Graphical representation}!examples of}%

-we may consider the octohedral group (§~201), defined by

-\begin{gather*}

-S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{4} = 1, \\

-S_{1} S_{2} S_{3} = 1.

-\end{gather*}

-

-On the diagram already given (\PageRef{p.}{290}), we may at once draw

-$S_{1}$-,~$S_{2}$- and $S_{3}$-lines. These in the present figure are coloured

-respectively red, yellow, and green. On the lines which correspond

-to operations of order two (in this case the red lines),

-the double arrow-heads may be dispensed with. By omitting

-successively the red, the yellow, and the green lines we form

-from this the three simplest colour diagrams which will represent

-\index{Young@\Topic{Young}, quoted}%

-\index{Maschke@\Topic{Maschke}, quoted}%

-the group\footnotemark.

-\footnotetext{For further illustrations, the reader may refer to Young, \textit{Amer.\ Journal},

-  Vol.~\VolNo{XV}, (1893), pp.~164--167; Maschke, \textit{Amer.\ Journal}, Vol.~\VolNo{XVIII}, (1896),

-  pp. 156--188.}%

-\index{Colour-groups@\Topic{Colour-groups}|)}%

-%% -----File: 327.png----------

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Fractional linear group@\Topic{Fractional linear group}!definition of}%

-\index{Gierster@\Topic{Gierster}, quoted}%

-\index{Jordan@\Topic{Jordan}, quoted}%

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|(}%

-% [Plate 1] \textit{To face page}~310.

-\begin{figure}[hp!]

-\centering

-\Graphic[jpg]{\textwidth}{plate1}

-\end{figure}

-%% -----File: 328.png----------

-% [Blank Page]

-%% -----File: 329.png---Folio 311-------

-

-

-\Chapter{XIV.}{On the Linear Group\protect\footnotemark.}

-

-\footnotetext{The homogeneous linear group and its sub-groups forms the subject of the

-  greater part of Jordan's \textit{Traité des Substitutions}. The investigation of its

-  composition-series, given in the text, is due to Jordan.

-

-  The complete analysis of the fractional linear group, defined by

-  \[

-  y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p),

-  \]

-  where

-  \[

-  \alpha\delta - \beta\gamma \equiv 1,

-  \]

-  is due originally to Gierster, ``Die Untergruppen der Galois'schen Gruppe der

-  Modulargleichungen für den Fall eines primzahligen Transformationsgrades,''

-  \textit{Math.\ Ann.}\ Vol.~\VolNo{XVIII}, (1881), pp.~319--365. With a few unimportant

-  modifications, the investigation in the text follows the lines of Gierster's

-  memoir.

-

-  A similar analysis of the simple groups of order $2^{n}(2^{2n} - 1)$, which can be

-  expressed as triply transitive groups of degree~$2^{n} + 1$, has been given by the

-  author, ``On a class of groups defined by congruences,'' \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXV},

-  (1894), pp.~132--136.}

-

-\Par{215.} \First{We} shall now, in illustration of the general principles

-that have been developed in the preceding chapters, proceed to

-discuss and give an analysis of certain special groups. The first

-that we choose for this purpose is the group of isomorphisms

-of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.

-This group has been defined and its order determined in

-§§~171,~172. It is there shewn that the group is simply

-isomorphic with the homogeneous linear group defined by all

-sets of congruences

-\[

-\begin{aligned}

-y_{1} &\equiv a_{11} x_{1} + a_{12} x_{2} + \Dots + a_{1n} x_{n}, \\

-y_{2} &\equiv a_{21} x_{1} + a_{22} x_{2} + \Dots + a_{2n} x_{n}, \\

-\multispan{3}{\dotfill} \\

-y_{n} &\equiv a_{n1} x_{1} + a_{n2} x_{2} + \Dots + a_{nn} x_{n},

-\end{aligned}

-\quad(\mod p),

-\]

-%% -----File: 330.png---Folio 312-------

-whose determinants are not congruent to zero; and that its

-order is

-\[

-N = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n - 1}).

-\]

-The operation given by the above set of congruences will be

-denoted in future by the symbol

-%[** TN: Set on two lines in the original]

-\[

-(a_{11}x_{1} + \Dots + a_{1n}x_{n},\

-a_{21}x_{1} + \Dots + a_{2n}x_{n},\ \Dots,\

-a_{n1}x_{1} + \Dots + a_{nn}x_{n}).

-\]

-

-\Par{216.} It may be easily verified by direct calculation that, if

-$D$~and~$D'$ are the determinants of two operations $S$ and~$S'$ of

-the group, then $DD'$~is the determinant of the operation~$SS'$,

-all the numbers involved being reduced, $\pmod{p}$. Hence it

-immediately follows that those operations of the group, whose

-determinant is unity, form a sub-group. If this sub-group is

-denoted by~$\Gamma$, the group itself being~$G$, then $\Gamma$~is a self-conjugate

-sub-group of~$G$. For if $\Sigma$~is any operation of~$\Gamma$ and

-$S$~any operation of~$G$ whose determinant is~$D$, the determinant

-of~$S^{-1}\Sigma S$ is $D^{-1}D$ or unity; and therefore $S^{-1}\Sigma S$~belongs to~$\Gamma$.

-Suppose now that $S$~is an operation\footnote

-  {The operation $(zx_{1}, x_{2}, \dots, x_{n})$ has $z$~for its determinant.}

-of~$G$ whose determinant is~$z$,

-a primitive root of the congruence

-\[

-z^{p - 1} \equiv 1\ (\mod p).

-\]

-Then the determinant of every operation of the set

-\[

-S^{r}\Gamma

-\]

-is~$z^{r}$; and therefore, if $r$~and~$s$ are not congruent $(\mod p)$, the

-two sets

-\[

-S^{r}\Gamma\quad \text{and}\quad S^{s}\Gamma

-\]

-can have no operation in common. Moreover, if $S'$~is any

-operation of~$G$ whose determinant is~$z^{r}$, then $S^{-r}S'$~belongs to~$\Gamma$,

-and therefore $S'$~belongs to the set~$S^{r}\Gamma$. Hence finally, the

-sets

-\[

-\Gamma,\ S\Gamma,\ S^{2}\Gamma,\ \Dots,\ S^{p - 2}\Gamma

-\]

-are all distinct, and they include every operation of~$G$; so that

-\[

-G = \{S, \Gamma\}.

-\]

-

-The factor-group $\dfrac{G}{\Gamma}$ is therefore cyclical and of order~$p - 1$.

-%% -----File: 331.png---Folio 313-------

-

-\Par{217.} It may be very readily verified that the operations of

-the cyclical sub-group generated by

-\[

-(zx_{1}, zx_{2}, \Dots, zx_{n})

-\]

-are self-conjugate operations of~$G$. To prove that these are

-the only self-conjugate operations of~$G$, we will deal with the

-case $n = 3$: it will be seen that the method is perfectly

-general. Suppose then that

-\[

-T = (\alpha   x_{1} + \beta   x_{2} + \gamma   x_{3},\

-     \alpha'  x_{1} + \beta'  x_{2} + \gamma'  x_{3},\

-     \alpha'' x_{1} + \beta'' x_{2} + \gamma'' x_{3})

-\]

-is a self-conjugate operation of $G$, while

-\[

-S = (a  x_{1} + b  x_{2} + c  x_{3},\

-     a' x_{1} + b' x_{2} + c' x_{3},\

-     a''x_{1} + b''x_{2} + c''x_{3})

-\]

-is any operation. The relation

-\[

-ST = TS

-\]

-involves the nine simultaneous congruences\footnote

-  {These and all succeeding congruences are to be taken $\pmod{p}$, unless the

-  contrary is stated.}

-\begin{align*}

-a\alpha + b\alpha' + c\alpha'' &\equiv \alpha a + \beta a' + \gamma a'', \\

-a\beta  + b\beta'  + c\beta''  &\equiv \alpha b + \beta b' + \gamma b'', \\

-a\gamma + b\gamma' + c\gamma'' &\equiv \alpha c + \beta c' + \gamma c'', \\

-\PadTo{a\gamma + b\gamma' + c\gamma''}{\text{etc.},} &

-\PadTo{\equiv \alpha c + \beta c' + \gamma c''}{\text{etc.};}

-\end{align*}

-and these must be satisfied for all possible values of the

-coefficients of~$S$. Now

-\[

-b \equiv c \equiv a' \equiv 0

-\]

-is a possible relation between the coefficients of~$S$, whether

-regarded as an operation of~$G$ or~$\Gamma$; and therefore

-\[

-\gamma \equiv 0.

-\]

-In the same way, it may be shewn that

-\[

-\beta \equiv \alpha' \equiv \gamma' \equiv \alpha'' \equiv \beta'' \equiv 0,

-\]

-and that

-\[

-\alpha \equiv \beta' \equiv \gamma'';

-\]

-so that $T$~is a power of the operation

-\[

-(zx_{1}, zx_{2}, zx_{3}).

-\]

-The only self-conjugate operations of~$G$ are therefore the powers

-of~$A$, where $A$~denotes

-\[

-(zx_{1}, zx_{2}, \Dots, zx_{n});

-\]

-and the only self-conjugate operations of~$\Gamma$ are those operations

-%% -----File: 332.png---Folio 314-------

-of this cyclical sub-group which are contained in~$\Gamma$. Now the

-order of~$A$ is~$p - 1$ and its determinant is~$z^{n}$. Hence the self-conjugate

-operations of~$\Gamma$ form a cyclical sub-group~$D$ of order~$d$,

-where $d$~is the greatest common factor of $p - 1$ and~$n$; and

-this sub-group is generated by~$A^{\efrac{p - 1}{d}}$.

-

-\Par{218.} To determine completely the composition-series of~$G$,

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!composition series of|(}%

-it is necessary to find whether $\Gamma$ has a self-conjugate sub-group

-greater than and containing~$D$. A simple calculation

-will shew that, from

-\[

-(x_{1} + x_{s}, x_{2}, x_{3}, \Dots, x_{n})

-\]

-and its conjugate operations, all the operations of~$\Gamma$ may be

-generated; and hence no self-conjugate sub-group of~$\Gamma$ which

-is different from $\Gamma$ itself can contain an operation of this form.

-If then it is shewn that any self-conjugate sub-group of~$\Gamma$,

-distinct from~$D$, necessarily contains operations of this form, it

-follows that $D$~is a maximum self-conjugate sub-group of~$\Gamma$.

-

-We shall first deal with the case $n = 2$.

-

-If $p = 2$, the orders of $G$,~$\Gamma$ and~$D$ are $6$,~$6$ and~$1$. In this

-case, $\Gamma$~is simply isomorphic with the symmetric group of three

-symbols, which has a self-conjugate sub-group of order~$3$. The

-successive factor-groups of the composition-series of~$G$ are

-therefore cyclical groups of orders $2$ and~$3$.

-

-If $p = 3$, the orders of $G$,~$\Gamma$ and~$D$ are $48$,~$24$ and~$2$. The

-factor-group $\dfrac{\Gamma}{D}$ has $12$ for its order, and cannot therefore be a

-simple group. The reader will have no difficulty in verifying

-that, in this case, the successive factor-groups of~$G$ have orders

-$2$,~$3$, $2$,~$2$ and~$2$. We may therefore, in dealing with the case

-$n = 2$, assume that $p$~is not less than~$5$.

-

-Let us suppose now that $\Gamma$~has a self-conjugate sub-group~$I$

-that contains~$D$; and let $S$ or

-\[

-(ax_{1} + bx_{2},\ a'x_{1} + b'x_{2})

-\]

-be one of its operations, not contained in~$D$.

-

-If $b$~is different from zero, $\Gamma$~contains~$\Sigma$, where $\Sigma$~denotes

-\[

-\left(\alpha ax_{1} + \alpha bx_{2}, -\frac{1 + \alpha^{2}a^{2}}{\alpha b} x_{1} - \alpha ax_{2}\right),

-\]

-%% -----File: 333.png---Folio 315-------

-and therefore $I$~contains $\Sigma^{-1} S \Sigma S$, which is

-\[

-\left(-\alpha^{-2} x_{1}, -\frac{(1 + \alpha^{2})(b' + a\alpha^{2})}{b\alpha^{2}} x_{1} - \alpha^{2}x_{2}\right).

-\]

-

-If $b$~is zero, $b'$~is congruent with~$a^{-1}$; therefore, in any case,

-$I$~contains an operation~$S'$ of the form

-\[

-(cx_{1}, dx_{1} + c^{-1}x_{2}).

-\]

-

-Again, $\Gamma$~contains the operation~$T$, where $T$~denotes

-\[

-(x_{1}, x_{1} + x_{2});

-\]

-and $I$~therefore contains $S' T^{-1} S'^{-1} T$, which is

-\[

-(x_{1}, (1 - c^{2})x_{1} + x_{2}).

-\]

-

-Hence unless $1 - c{^2} \equiv 0$, $I$~must coincide with~$\Gamma$. Now,

-when $p > 5$, $c$~can always be chosen so that this congruence

-is not satisfied. If $p = 5$, the square of the above operation

-$\Sigma^{-1} S\Sigma S$, when unity is written for~$\alpha$, is

-\[

-\left(x_{1}, \frac{4}{b}(b' + a)x_{1} + x_{2}\right);

-\]

-unless $b' + a \equiv 0$, this again requires that $I$ coincides with~$\Gamma$.

-If finally, the condition $b' + a \equiv 0$ is satisfied in~$S$, it is not

-satisfied in~$ST^{-1} S^{-1}T$, another operation belonging to~$I$; and

-therefore again, in this case, $I$~coincides with~$\Gamma$.

-

-Hence finally, if $n = 2$, the factor-group $\dfrac{\Gamma}{D}$ is simple, except

-when $p$~is $2$ or~$3$.

-

-\Par{219.} When $n$~is greater than~$2$, it will be found that it is

-sufficient to deal in detail with the case $n = 3$, as the method

-will apply equally well for any greater value of~$n$. Suppose

-here again that $\Gamma$~has a self-conjugate sub-group~$I$ which

-contains~$D$; and let $S$, denoting

-\[

-(a  x_{1} + b  x_{2} + c  x_{3},\

- a' x_{1} + b' x_{2} + c' x_{3},\

- a''x_{1} + b''x_{2} + c''x_{3}),

-\]

-be one of the operations of~$I$ which is not contained in~$D$.

-$S$~cannot be permutable with all operations of the form

-$(x_{1}, x_{2}, x_{3} + x_{1})$, as it would then be permutable with every

-operation of~$\Gamma$. We may therefore suppose without loss of

-generality that $S$~and~$T$ are not permutable, $T$~denoting

-$(x_{1}, x_{2}, x_{3} + x_{1})$. Then $T^{-1}STS^{-1}$~is an operation, distinct from

-%% -----File: 334.png---Folio 316-------

-identity, belonging to~$I$. Now a simple calculation shews that

-this operation, say~$U$, is of the form

-\[

-\left(x_{1} - cX, x_{2} - c'X, Ax_{1} + Bx_{2} + Cx_{3}\right),

-\]

-where $X$~is the symbol with which $S^{-1}$ replaces~$x_{1}$.

-

-If $c$~and~$c'$ are both different from zero, $\Gamma$~will contain an

-operation~$V$ of the form

-\[

-\left(x_{1} - \frac{c}{c'} x_{2}, x_{2}, x_{3}\right);

-\]

-and $I$~contains $V^{-1}UV$, which is of the form

-\[

-(x_{1}, \alpha x_{1} + \beta x_{2} + \gamma x_{3}, \alpha' x_{1} + \beta' x_{2} + \gamma' x_{3}).

-\]

-

-Moreover, if either $c$~or $c'$ is zero, the operation $U$ itself

-leaves one symbol unaltered. Hence $I$~always contains operations

-by which one symbol is unaltered.

-

-This process may now be repeated to shew that $I$~necessarily

-contains operations of the form

-\[

-(x_{1}, x_{2}, \alpha'' x_{1} + \beta'' x_{2} + \gamma'' x_{3});

-\]

-and, since the determinant of the operation is unity, $\gamma''$~is

-necessarily congruent to unity. But it has been seen that

-the group~$\Gamma$ is generated from the last operation and the

-operations conjugate to it. Hence finally, if $n$~is greater than~$2$,

-the factor-group~$\dfrac{\Gamma}{D}$ is simple for all values of~$p$.

-

-\Par{220.} The composition-series of~$G$ is now, except as regards

-the constitution of the simple group~$\dfrac{\Gamma}{D}$, perfectly definite. It

-has, in fact, been seen that $\dfrac{G}{\Gamma}$~and $D$ are cyclical groups of orders

-$p - 1$ and~$d$; and therefore if $\alpha$,~$\beta$, $\gamma$,~\Dots\ are distinct primes

-whose product is~$p - 1$, and if $\alpha'$,~$\beta'$, $\gamma'$,~\Dots\ are distinct primes

-whose product is~$d$; the successive factor-groups of~$G$ are first,

-a series of simple groups of prime orders $\alpha$,~$\beta$, $\gamma$,~\Dots: then

-a simple group of composite order~$\dfrac{N}{(p - 1)d}$: and lastly, a series

-of simple groups of prime orders $\alpha'$,~$\beta'$, $\gamma'$,~\Dots.

-

-The sequence in which the set of simple groups of orders

-%% -----File: 335.png---Folio 317-------

-$\alpha$,~$\beta$, $\gamma$,~\Dots\ are taken in the composition-series may be clearly

-any whatever, and the same is true of the set of factor-groups

-of orders $\alpha'$,~$\beta'$, $\gamma'$,~\Dots; but it is to be noticed that, when $d$~is

-not equal to~$p - 1$, the composition-series is capable of further

-modifications. In this case, $\{A, \Gamma\}$~is a self-conjugate sub-group

-of~$G$ of order~$\dfrac{N}{d}$, which has a maximum self-conjugate sub-group~$\{A\}$

-of order~$p - 1$. The successive composition-factors of~$G$

-may therefore be taken in the sequence

-\[

-\alpha',\ \beta',\ \gamma',\ \Dots:\ \frac{N}{(p - 1) d}:\ \alpha,\ \beta,\ \gamma,\ \Dots:

-\]

-and their arrangement may be yet further changed by considering

-the self-conjugate sub-group $\{A^{m}, \Gamma\}$, where $m$~is a

-factor of $p - 1$ less than~$\dfrac{p - 1}{d}$.

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!composition series of|)}%

-

-\Par{221.} For every value of~$p^{n}$, except $2^{2}$~and~$3^{2}$, it thus

-appears that the linear group may be regarded as defining a

-simple group of composite order. We shall now proceed

-to a discussion of the constitution of the simple groups thus

-defined when $n = 2$, $p$~being greater than~$3$\footnotemark.

-\footnotetext{For the case $n = 3$ the reader may consult a paper by the author ``On a

-  class of groups defined by congruences,'' \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), pp.~58--106.}

-\index{Burnside@\Topic{Burnside}, quoted}%

-In this case, the

-group~$\Gamma$ is defined by the congruences

-\[

-\begin{aligned}

-y_{1} &\equiv \alpha x_{1} + \beta x_{2}, \\

-y_{2} &\equiv \gamma x_{1} + \delta x_{2}, \\

-\alpha\delta - \beta\gamma &\equiv 1,

-\end{aligned}

-\quad (\mod p);

-\]

-and since $p - 1$ is divisible by~$2$ when $p$~is an odd prime, $d$~is

-equal to~$2$. Hence the self-conjugate operations of~$\Gamma$ are

-\[

-(x_{1}, x_{2}) \quad\text{and}\quad (-x_{1}, -x_{2}).

-\]

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|)}%

-

-The order of~$\Gamma$ is~$p(p^{2} - 1)$, and therefore the order of the

-simple group,~$H$, which it defines is~$\frac{1}{2} p(p^{2} - 1)$. Suppose now,

-if possible, that $\Gamma$~contains a sub-group~$g$ simply isomorphic

-with~$H$. If $S$~is any operation of~$\Gamma$, not contained in~$g$, the

-whole of the operations of~$\Gamma$ are contained in the two sets

-\[

-g,\ Sg.

-\]

-%% -----File: 336.png---Folio 318-------

-

-Now $(x_{2}, -x_{1})$, whose square $(-x_{1}, -x_{2})$ is a self-conjugate

-operation, cannot be contained in the simple group~$g$. Hence

-both $(x_{2}, -x_{1})$ and $(-x_{1}, -x_{2})$ are contained in~$Sg$, an obvious

-contradiction. Therefore $\Gamma$~contains no sub-group simply

-isomorphic with~$H$.

-

-For a discussion of the properties of~$H$, some concrete

-representation of the group itself is necessary; this may be

-obtained in the following way. Instead of the pair of homogeneous

-congruences that define each operation of~$\Gamma$, let us,

-as in §~113, consider the single non-homogeneous congruence

-\[

-y \equiv \frac{\alpha x + \beta} {\gamma x + \delta},\ (\mod p),

-\]

-where

-\[

-\alpha\delta - \beta\gamma \equiv 1.

-\]

-Corresponding to every operation

-\[

-(\alpha x_{1} + \beta x_{2},\ \gamma x_{1} + \delta x_{2})

-\]

-of~$\Gamma$, there will be a single operation of this new set; namely

-that in which $\alpha$,~$\beta$, $\gamma$,~$\delta$ have respectively the same values. But

-since the operations

-\[

-y \equiv \frac{ \alpha x + \beta}{ \gamma x + \delta} \quad\text{and}\quad

-y \equiv \frac{-\alpha x - \beta}{-\gamma x - \delta}

-\]

-are identical, two operations

-\[

-( \alpha x_{1} + \beta x_{2},\  \gamma x_{1} + \delta x_{2}) \quad\text{and}\quad

-(-\alpha x_{1} - \beta x_{2},\ -\gamma x_{1} - \delta x_{2})

-\]

-of~$\Gamma$ will correspond to each operation

-\[

-y \equiv \frac{\alpha x + \beta}{\gamma x + \delta}

-\]

-of the new set; the two self-conjugate operations

-\[

-(x_{1}, x_{2}) \quad\text{and}\quad (-x_{1}, -x_{2}),

-\]

-in particular, corresponding to the identical operation of the

-new set. Moreover, direct calculation immediately verifies

-that, to the product of any two operations of~$\Gamma$, corresponds the

-product of the two corresponding operations of the new set.

-Hence the new set of operations forms a group of order~$\frac{1}{2}p(p^{2} - 1)$,

-with which $\Gamma$~is multiply isomorphic; the group of order~$2$

-formed by the self-conjugate operations of~$\Gamma$ corresponding to

-the identical operation of the new group.

-%% -----File: 337.png---Folio 319-------

-

-The simple group~$H$, of order~$\frac{1}{2}p(p^{2} - 1)$, which we propose

-to discuss, can therefore be represented by the set of operations

-\[

-y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p);

-\]

-where

-\[

-\alpha\delta - \beta\gamma \equiv 1,

-\]

-$\alpha$,~$\beta$, $\gamma$,~$\delta$ being integers reduced to modulus~$p$.

-

-\Par{222.} Since the order of~$H$ is divisible by~$p$ and not by~$p^{2}$,

-\index{Fractional linear group@\Topic{Fractional linear group}!analysis of|(}%

-the group must contain a single conjugate set of sub-groups of

-order~$p$. Now the operation

-\[

-y \equiv x + 1,

-\]

-or $(x + 1)$ as we will write it in future, is clearly an operation

-of order~$p$: for its $n$th~power is~$(x + n)$, and $p$~is the smallest

-value of~$n$ for which this is the identical operation. If

-$(x + 1)$ and $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ are represented by $P$ and~$S$, then

-\[

-S^{-1}PS = \left(\frac{(1 - \alpha\gamma)x + \alpha^{2}}{-\gamma^{2}x + 1 + \alpha\gamma}\right).

-\]

-

-This is identical with~$P$, only if

-\[

-\gamma \equiv 0,\quad \alpha^{2} \equiv 1;

-\]

-and therefore $P$~is permutable with no operations except its

-own powers. On the other hand, if

-\[

-\gamma \equiv 0,

-\]

-then

-\[

-S^{-1}PS = P^{a^{2}};

-\]

-and therefore every operation, for which $\gamma \equiv 0$, transforms the

-sub-group~$\{P\}$ into itself. These operations therefore form a

-sub-group: a result that may also be easily verified directly.

-The order of this sub-group is the number of distinct operations

-$\left(\dfrac{\alpha x + \beta}{\delta}\right)$ for which $\alpha\delta \equiv 1$. The ratio~$\dfrac{\alpha}{\delta}$ must be a quadratic

-residue, while $\beta$~may have any value whatever. Hence the

-order of the sub-group is~$\frac{1}{2}(p - 1)p$; and $H$~therefore contains

-$p + 1$ sub-groups of order~$p$. Since $H$~is a simple group, it

-follows (§~125) that it can be represented as a transitive

-substitution group of degree~$p + 1$.

-%% -----File: 338.png---Folio 320-------

-

-This representation of the group can be directly derived, as

-in §~113, from the congruences already used to define it. Thus

-if, in

-\[

-y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},

-\]

-we write for $x$ successively $0$, $1$, $2$,~\Dots, $p - 1$,~$\infty$, the $p + 1$

-values obtained for~$y$, when reduced $\pmod{p}$, will be the same

-$p + 1$ symbols in some other sequence. For if

-\[

-\frac{\alpha x_{1} + \beta}{\gamma x_{1} + \delta} \equiv

-\frac{\alpha x_{2} + \beta}{\gamma x_{2} + \delta},

-\]

-then

-\[

-(\alpha\delta - \beta\gamma)(x_{1} - x_{2}) \equiv 0,

-\]

-and therefore $x_{1} \equiv x_{2}$.

-

-Each operation of~$H$ gives therefore a distinct substitution

-performed on the symbols $0$, $1$,~\Dots, $p - 1$,~$\infty$; and the complete

-set of substitutions thus obtained gives the representation of~$H$

-as a transitive substitution group of degree~$p + 1$. Since $H$~contains

-operations of order~$p$, this substitution group must be

-doubly transitive. That this is the case may also be shewn

-directly. Thus

-\[

-\frac{y - a}{y - b} \equiv m \frac{x - a'}{x - b'}

-\]

-is an operation changing $a'$ into~$a$ and $b'$ into~$b$. This operation

-may be written

-\[

-y \equiv \frac{k(bm - a)x + k(ab' - ma'b)}

-              {k(m - 1)x + k(b' - ma')},

-\]

-and its determinant is

-\[

-k^{2}m(b - a)(b' - a').

-\]

-

-If now $(b - a)(b' - a')$ is a quadratic residue (or non-residue)

-$\pmod{p}$, $m$~may be any quadratic residue (or non-residue); and $k$~can

-always be chosen so that the determinant is unity. There

-are therefore $\frac{1}{2}(p - 1)$ substitutions in the group, changing any

-two symbols $a'$,~$b'$ into any other two given symbols $a$,~$b$.

-Further, if the operation $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ keeps $x$ unchanged in the

-substitution group, $x$~must satisfy the congruence

-\[

-x \equiv \frac{\alpha x + \beta}

-              {\gamma x + \delta},

-\]

-%% -----File: 339.png---Folio 321-------

-that is

-\[

-\gamma x^{2} + (\delta - a)x - \beta \equiv 0.

-\]

-Such a congruence cannot have more than two roots; and

-therefore every substitution displaces all, all but one, or all but

-two, of the $p + 1$ symbols.

-

-\Par{223.} The substitutions, which keep either one or two

-symbols fixed, must therefore be regular in the remaining $p$~or

-$p - 1$ symbols. Hence the order of every substitution which

-keeps just one symbol fixed must be~$p$; and the order of every

-substitution that keeps two symbols fixed must be equal to or

-be a factor of~$p - 1$. Now it was seen in the last paragraph

-that the order of the sub-group that keeps two symbols fixed is

-$\frac{1}{2}(p - 1)$. Moreover, if $z$~is a primitive root $\pmod{p}$, the

-sub-group that keeps $a$~and~$b$ fixed contains the operation

-\[

-\frac{y - a}{y - b} = z^{2} \frac{x - a}{x - b},

-\]

-and the order of this operation is $\frac{1}{2}(p - 1)$. Hence, the

-sub-group that keeps any two symbols fixed is a cyclical group

-of order~$\frac{1}{2}(p - 1)$; and every operation that keeps two symbols

-fixed is some power of an operation of order~$\frac{1}{2}(p - 1)$. Since the

-group is a doubly transitive group of degree~$p + 1$, there must

-be $\frac{1}{2}(p + 1)p$ sub-groups which keep two symbols fixed; and

-these must form a conjugate set. Each is therefore self-conjugate

-in a sub-group of order~$p - 1$. To determine the

-type of this sub-group, we may consider the sub-group keeping

-$0$~and~$\infty$ fixed: this is generated by~$Q$, where $Q$~denotes $\left(\dfrac{zx}{z^{-1}}\right)$.

-If $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ is represented by~$S$, then

-\[

-S^{-1}QS = \left(

-  \frac{(z\alpha\delta - z^{-1}\beta\gamma)x - (z - z^{-1})\alpha\beta}

-       {(z - z^{-1})\gamma\delta x + z^{-1}\alpha\delta - z\beta\gamma}

-\right),

-\]

-which can be a power of~$Q$ only if

-\[

-\alpha\beta \equiv 0,\quad \gamma\delta \equiv 0.

-\]

-

-Hence either

-\[

-\beta \equiv \gamma \equiv 0,

-\]

-in which case $S$~is a power of~$Q$: or

-\[

-\alpha \equiv \delta \equiv 0,\quad \gamma \equiv -\beta^{-1}.

-\]

-%% -----File: 340.png---Folio 322-------

-

-In the latter case, we have

-\[

-S = \left(\frac{-\beta}{\beta^{-1} x}\right),

-\]

-which is an operation of order~$2$; and then

-\[

-S^{-1} Q S = \left(\frac{z^{-1} x}{z}\right) = Q^{-1}.

-\]

-

-The group of order~$p - 1$, which contains self-conjugately a

-cyclical sub-group of order~$\frac{1}{2}(p - 1)$ that keeps two symbols fixed,

-is therefore a group of dihedral (§~202) type. Moreover, if $t$~is

-any factor of~$p - 1$, this investigation shews that $\{S, Q\}$~is the

-greatest sub-group that contains $\{Q^{t}\}$ self-conjugately.

-

-\Par{224.} A substitution that changes all the symbols must

-either be regular in the $p + 1$ symbols, or must be such that

-one of its powers keeps two symbols fixed. The latter case

-however cannot occur; for we have just seen that, if $Q$~is an

-operation, of order~$\frac{1}{2}(p - 1)$, which keeps two symbols fixed,

-the only operations permutable with~$Q^{r}$ are the powers of~$Q$.

-Hence the substitutions that change all the symbols must

-be regular in the $p + 1$ symbols, and their orders must be

-equal to or be factors of~$p + 1$.

-

-Suppose now that $i$ is a primitive root of the congruence

-\[

-i^{p^{2} - 1} - 1 \equiv 0\ (\mod p),

-\]

-so that $i$~and~$i^{p}$ are the roots of a quadratic congruence with

-real coefficients; and consider the operation~$K$, denoting

-\[

-\frac{y - i^{k}}{y - i^{kp}} \equiv i^{2(p - 1)} \frac{x - i^{k}}{x - i^{kp}},

-\]

-where $k$~is not a multiple of~$p + 1$. On solving with respect to~$y$,

-$K$~is expressed in the form

-\[

-y \equiv \frac{

-    \dfrac{i^{\left(\efrac{k}{2} + 1\right)(p - 1)} - i^{-\left(\efrac{k}{2} + 1\right)(p - 1)}}

-        {i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} x

-  - \dfrac{i^{(p - 1)} - i^{-(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} i^{\efrac{k}{2} (p + 1)} %[** End of top-level numerator]

- }{

-    \dfrac{i^{(p - 1)} - i^{-(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} x i^{-\efrac{k}{2} (p + 1)}

-  + \dfrac{i^{\left(\efrac{k}{2} - 1\right)(p - 1)} - i^{-\left(\efrac{k}{2} - 1\right)(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}}},

-\]

-%% -----File: 341.png---Folio 323-------

-an operation of determinant unity. It will be found, on writing

-$i^{p}$ for~$i$ in the coefficients of this operation, that they remain

-unaltered; therefore, since they are symmetric functions of

-$i$~and~$i^{p}$, they must be real numbers. The operation therefore

-belongs to~$H$. The $n$th~power of this operation is given by

-\[

-\frac{y - i^{k}}{y - i^{kp}} \equiv i^{2n(p - 1)} \frac{x - i^{k}}{x - i^{kp}},

-\]

-and therefore, since the first power of~$i$ which is congruent to

-unity, $\pmod{p}$, is the $(p^{2} - 1)$th, the order of the operation is

-$\frac{1}{2}(p + 1)$. If we write $kp$ for~$k$ in the operation~$K$, the new

-operation is~$K^{-1}$; but if $k$~is replaced by any other number~$k'$,

-which is not a multiple of $p + 1$, the new operation~$K'$, given by

-\[

-\frac{y - i^{k'}}{y - i^{k'p}} \equiv i^{2(p - 1)} \frac{x - i^{k'}}{x - i^{k'p}},

-\]

-generates a new sub-group of order~$\frac{1}{2}(p + 1)$, which has no

-operation except identity in common with~$\{K\}$. Now there are

-$p^{2} - p$ numbers less than $p^{2} - 1$ which are not multiples of

-$p + 1$; therefore $H$~contains $\frac{1}{2}(p^{2} - p)$ cyclical sub-groups of

-order~$\frac{1}{2}(p + 1)$, no two of which have a common operation

-except identity. The corresponding substitutions displace all

-the symbols.

-

-\Par{225.} A simple enumeration shews that the operations of

-the cyclical sub-groups of orders $\frac{1}{2}(p - 1)$,~$p$ and~$\frac{1}{2}(p + 1)$,

-exhaust all the operations of the group. Thus there are,

-omitting identity from each sub-group:

-

-\Inum{(i)} $\frac{1}{2}p(p + 1)$ sub-groups of order~$\frac{1}{2}(p - 1)$, containing

-$\frac{1}{4}p(p^{2} - 1) - \frac{1}{2}p^{2} - \frac{1}{2}p$ distinct operations;

-

-\Inum{(ii)} $\frac{1}{2}p(p - 1)$ sub-groups of order~$\frac{1}{2}(p + 1)$, containing

-$\frac{1}{4}p(p^{2} - 1) - \frac{1}{2}p^{2} + \frac{1}{2}p$ distinct operations;

-

-\Inum{(iii)} $p + 1$ sub-groups of order~$p$, containing

-$p^{2} - 1$ distinct operations; \\

-%

-and the sum of these numbers, with $1$ for the identical operation,

-gives $\frac{1}{2}p(p^{2} - 1)$, which is the order of the group.

-

-Every operation that displaces all the symbols is therefore

-the power of an operation of order~$\frac{1}{2}(p + 1)$.

-%% -----File: 342.png---Folio 324-------

-

-\Par{226.} We shall now further shew that the $\frac{1}{2}p(p - 1)$ sub-groups

-of order~$\frac{1}{2}(p + 1)$ form a single conjugate set, and that

-each is contained self-conjugately in a dihedral group of order

-$p + 1$. Let $S$~be any operation of~$H$, which is permutable with~$\{K\}$

-and replaces $i^{k}$ by some other symbol~$j$. Then $S^{-1}KS$~is

-an operation which leaves $j$ unaltered; it may therefore be expressed

-in the form

-\[

-\frac{y - j}{y - j'} = m \frac{x - j}{x - j'}.

-\]

-

-This can belong to the sub-group generated by~$K$, only if $j$

-and $j'$ are the same pair as $i^{k}$ and~$i^{kp}$. Hence $j$~must be either

-$i^{k}$ or~$i^{kp}$; and similarly, if $S$~replaces $i^{kp}$ by~$j'$, the latter must be

-either $i^{kp}$ or~$i^{k}$. Hence either $S$ must keep both the symbols

-$i^{k}$ and~$i^{kp}$ unchanged or it must interchange them; and conversely,

-every operation which either keeps both the symbols

-unchanged or interchanges them, must transform $\{K\}$ into

-itself. If $S$~keeps both of them unchanged, it is a power of~$K$.

-If $S$~interchanges them, it is of the form

-\[

-\frac{y - i^{k}}{y - i^{kp}} \equiv m \frac{x - i^{kp}}{x - i^{k}};

-\]

-and a simple calculation shews that

-\[

-S^{-1}KS = K^{-1}.

-\]

-If we take $m = 1$, $S$~becomes

-\[

-x + y \equiv i^{k} + i^{kp},

-\]

-an operation belonging to~$H$. Hence the cyclical sub-group~$\{K\}$

-is contained self-conjugately in the sub-group of~$\{S, K\}$ which is

-of dihedral type. If there were any other operation~$S'$, not contained

-in~$\{S, K\}$, which transformed~$K$ into its inverse, then

-$SS'$~would be an operation permutable with~$K$ and not contained

-in~$\{K\}$. It has just been seen that no such operation

-exists. Hence $\{S, K\}$, of order~$p + 1$, is the greatest sub-group

-that contains $\{K\}$ self-conjugately; and $\{K\}$~must be one of

-$\frac{1}{2}p(p - 1)$ conjugate sub-groups.

-

-\Par{227.} The distribution of the operations of~$H$ in conjugate

-sets is now known. A sub-group of order~$p$ is contained self-conjugately

-in a group of order~$\frac{1}{2}p(p - 1)$, while an operation

-%% -----File: 343.png---Folio 325-------

-of order~$p$ is permutable only with its own powers. There are

-therefore two conjugate sets of operations of order~$p$, each set

-containing $\frac{1}{2}(p^{2} - 1)$ operations. Again, each of the operations

-of a cyclical sub-group of order $\frac{1}{2}(p - 1)$ or $\frac{1}{2}(p + 1)$ is conjugate

-to its own inverse and to no other of its powers. Hence if

-$\frac{1}{2}(p + 1)$ is even and therefore $\frac{1}{2}(p - 1)$ odd, there are $\frac{1}{4}(p - 3)$

-conjugate sets of operations whose orders are factors of $\frac{1}{2}(p - 1)$,

-each set containing $p^{2} + p$ operations; $\frac{1}{4}(p - 3)$~conjugate sets

-of operations whose orders are factors of~$\frac{1}{2}(p + 1)$, other than

-the factor~$2$, each set containing $p^{2} - p$ operations; and a single

-set of operations of order~$2$, containing $\frac{1}{2}(p^{2} - p)$ operations. If

-$\frac{1}{2}(p - 1)$ is even and $\frac{1}{2}(p + 1)$~odd, there are $\frac{1}{4}(p - 1)$ conjugate

-sets of operations whose orders are factors of $\frac{1}{2}(p + 1)$, each

-containing $p^{2} - p$ operations; $\frac{1}{4}(p - 5)$ conjugate sets whose

-orders are factors of~$\frac{1}{2}(p - 1)$, other than the factor~$2$, each

-set containing $p^{2} + p$ operations; and a single set of $\frac{1}{2}(p^{2} + p)$

-conjugate operations of order~$2$. In either case, the group

-contains, exclusive of identity, $\frac{1}{2}(p + 3)$ conjugate sets of

-operations.

-

-\Par{228.} Since $p - 1$ and $p + 1$ can have no common factor except~$2$,

-it follows that, if $q^{m}$~denote\DPnote{** [sic]} the highest power of an odd

-prime, other than~$p$, which divides the order of~$H$, $q^{m}$~must be a

-factor of~$\frac{1}{2}(p - 1)$ or of~$\frac{1}{2}(p + 1)$; and the sub-groups of order~$q^{m}$

-must be cyclical. Moreover, since no two cyclical sub-groups

-of order~$\frac{1}{2}(p - 1)$, or~$\frac{1}{2}(p + 1)$, have a common operation except

-identity, the same must be true of the sub-groups of order~$q^{m}$.

-

-If $2^{m}$~is the highest power of~$2$ that divides $\frac{1}{2}(p - 1)$ or

-$\frac{1}{2} (p + 1)$, $2^{m + 1}$~will be the highest power of~$2$ that divides the

-order of~$H$. Moreover, a sub-group of order~$2^{m + 1}$ must contain

-a cyclical sub-group of order~$2^{m}$ self-conjugately, and it must

-contain an operation of order~$2$ that transforms every operation

-of this cyclical sub-group into its own inverse; in other words,

-the sub-groups of order~$2^{m + 1}$ are of dihedral type.

-

-Suppose now that two sub-groups of order~$2^{m + 1}$ ($m > 1$) have

-a common sub-group of order~$2^{r}$ ($r > 2$). Such a sub-group

-must be either cyclical or dihedral: in the latter case, it

-contains self-conjugately a single cyclical sub-group of order~$2^{r - 1}$.

-%% -----File: 344.png---Folio 326-------

-Hence, on the supposition made, a cyclical sub-group of

-order~$4$ at least would be contained self-conjugately in two

-distinct cyclical sub-groups of order~$2^{m}$. It has been seen that

-this is not the case; and therefore the greatest sub-group, that

-two sub-groups of order~$2^{m + 1}$ can have in common, must be a

-sub-group of order~$4$, whose operations, except identity, are all

-of order~$2$. Now every group of order~$2^{m + 1}$ contains one self-conjugate

-operation of order~$2$, and $2^{m}$~operations of order~$2$

-falling into $2$~conjugate sets of $2^{m - 1}$ each. Moreover, the group

-of order~$p ± 1$, which has a cyclical sub-group of order~$2^{m}$ and

-contains the operation~$A$ of order~$2$ of this cyclical group

-self-conjugately, has $\frac{1}{2}(p ± 1)$ other operations of order~$2$; and

-therefore it contains $\dfrac{p ± 1}{2^{m + 1}}$ sub-groups of order~$2^{m + 1}$, each of

-which has $A$ for its self-conjugate operation. If now $B$~is

-any operation of order~$2$ of this sub-group of order~$p ± 1$, and

-if it is distinct from~$A$, then $B$~enters into a sub-group of

-order~$2^{m + 1}$ that contains $A$ self-conjugately. But since $A$~is

-permutable with~$B$, $A$~must belong to the sub-group of order

-$p ± 1$, which contains $B$ self-conjugately; hence $A$~enters into

-a sub-group of order~$2^{m + 1}$ which contains $B$ self-conjugately.

-The sub-group $\{A, B\}$ is therefore common to two distinct

-sub-groups of order~$2^{m + 1}$. Now no group of order~$2^{r}$ ($r > 2$)

-can be common to two sub-groups of order~$2^{m + 1}$; and therefore

-$\{A, B\}$~must (§~80) be permutable with some operation~$S$ whose

-order~$s$ is prime to~$2$. If $s$~is not~$3$, $S$~must be permutable with

-$A$ and~$B$: and then $\{A, S\}$ and $\{B, S\}$ would be two distinct

-sub-groups of orders~$2s$, whose operations are permutable with

-each other. It has been seen that $H$~does not contain such

-sub-groups. Hence $s = 3$; and $S$~transforms $A$,~$B$ and~$AB$ cyclically,

-or $\{S, A, B\}$ is a sub-group of tetrahedral type (§~202).

-

-The number of quadratic\footnote

-  {A non-cyclical group of order~$4$ is called a \emph{quadratic} group.}

-sub-groups contained in~$H$ may

-\index{Group@\Topic{Group}!quadratic}%

-be directly enumerated. A group of order~$2^{m + 1}$ contains $2^{m - 1}$

-such sub-groups, which fall into $2$ conjugate sets of $2^{m - 2}$ each;

-a single group of order~$8$ containing each quadratic group self-conjugately.

-The quadratic groups, contained in the $\dfrac{p ± 1}{2^{m + 1}}$

-sub-groups of order~$2^{m + 1}$ of a sub-group of order~$p ± 1$, are

-%% -----File: 345.png---Folio 327-------

-clearly all distinct, and each quadratic group belongs to just $3$

-groups of order~$p ± 1$; thus $\{A, B\}$ belongs to the $3$~groups

-which contain $A$,~$B$ and~$AB$ respectively as self-conjugate

-operations. Hence the total number of quadratic groups contained

-in~$H$ is

-\[

-\tfrac{1}{2} p (p \mp 1) \frac{p ± 1}{2^{m + 1}} 2^{m - 1} \tfrac{1}{3}

-  = \frac{\frac{1}{2}p(p^{2} - 1)}{12}.

-\]

-

-\Par{229.} The greatest sub-group of a group of order~$2^{m + 1}$, that

-contains a quadratic group self-conjugately, is a group of order~$8$

-and dihedral type; and it has been shewn that $3$~is the

-only factor, prime to~$2$, that occurs in the order of the sub-group

-containing a quadratic group self-conjugately. Hence

-finally, the order of the greatest group containing a quadratic

-group self-conjugately is~$24$, and the $\dfrac{\frac{1}{2}p(p^{2} - 1)}{12}$ quadratic

-groups fall into two conjugate sets of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$ each. The

-group of order~$24$, that contains a quadratic group self-conjugately,

-contains also a self-conjugate tetrahedral sub-group,

-while the sub-groups of order~$8$ are dihedral. Hence (§~84) this

-group must be of octohedral type.

-

-Since every tetrahedral sub-group of~$H$ contains a quadratic

-sub-group self-conjugately, and every octohedral sub-group

-contains a tetrahedral sub-group self-conjugately, there must

-also be two conjugate sets of tetrahedral sub-groups and two

-conjugate sets of octohedral sub-groups, the number in each set

-being $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$.

-

-\Par{230.} We have hitherto supposed $m > 1$, or what is the

-same thing, $p \equiv ±1\ (\mod 8)$. If now $m = 1$, so that $p \equiv ±3\ (\mod 8)$,

-the highest power of~$2$ that divides the order of~$H$ is~$2^{2}$;

-and, since $2^{2}$~is not a factor of~$\frac{1}{2}(p ± 1)$, the sub-groups of

-order~$2^{2}$ are quadratic. Moreover, since $2^{2}$~is the highest power

-of~$2$ dividing the order of~$H$, the quadratic sub-groups form

-a single conjugate set. Each sub-group of order~$p ± 1$, which

-has a self-conjugate operation of order~$2$, contains $\frac{1}{4}(p ± 1)$

-%% -----File: 346.png---Folio 328-------

-sub-groups of order~$4$, and each of the latter belongs to $3$ of the

-former. The total number is as before

-\[

-\frac{\frac{1}{2}p(p^{2} - 1)}{12},

-\]

-and since they form a single conjugate set, each quadratic group

-is self-conjugate in a group of order~$12$. Also, for the same

-reason as in the previous case, this sub-group is of tetrahedral

-type.

-

-Finally, since every sub-group of~$H$ of tetrahedral type

-must contain a quadratic sub-group self-conjugately, $H$~must

-contain a single conjugate set of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{12}$ tetrahedral sub-groups.

-In this case, the order of~$H$ is not divisible by~$24$, and

-therefore the question of octohedral sub-groups does not arise.

-

-\Par{231.} The group~$H$ always contains tetrahedral sub-groups;

-when its order is divisible by~$24$, it contains also octohedral

-sub-groups. Now if $p \equiv ±1\ (\mod 5)$, the order of~$H$ is

-divisible by~$60$: it may be shewn as follows that, in these cases,

-$H$~contains sub-groups of icosahedral type.

-

-Let us suppose, first, that $p \equiv 1\ (\mod 5)$; and let $j$ be a

-primitive root of the congruence

-\[

-j^{5} \equiv 1\ (\mod p).

-\]

-Then $\left(\dfrac{jx}{j^{-1}}\right)$, which we will denote by~$A$, is an operation of order~$5$.

-The operations of order~$2$ of~$H$ are all of the form~$B$, where

-$B$~denotes $\left(\dfrac{\alpha x + \beta}{\gamma x - \alpha}\right)$, since each is its own inverse. Now

-\[

-AB = \left(\frac{\alpha jx + \beta j^{-1}}{\gamma jx - \alpha j^{-1}}\right);

-\]

-and (§~203) if $A$~and~$B$ generate an icosahedral group,

-\[

-(AB)^{3} = 1.

-\]

-A simple calculation shews that, if this condition is satisfied,

-then

-\[

-\alpha^{2}(j - j^{-1})^{2} \equiv 1.

-\]

-Also, since the determinant of~$B$ is unity,

-\[

-\alpha^{2} + \beta\gamma \equiv -1.

-\]

-%% -----File: 347.png---Folio 329-------

-These two congruences have just $p - 1$ distinct solutions,

-the solutions $\alpha$,~$\beta$,~$\gamma$ and $-\alpha$,~$-\beta$,~$-\gamma$ being regarded as

-identical. There are therefore $p - 1$ operations of order two in~$H$,

-namely the operations

-\[

-\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta}

-           {\gamma x - \dfrac{1}{j - j^{-1}}}

-\right),

-\]

-where

-\[

-\beta\gamma \equiv -1 - \frac{1}{(j - j^{-1})^{2}},

-\]

-which with $A$~generate an icosahedral sub-group.

-

-The group generated by

-\[

-\left(\frac{jx}{j^{-1}}\right)

-\quad\text{and}\quad

-\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta_{0}}

-           {\gamma_{0} x - \dfrac{1}{j - j^{-1}}}

-\right),

-\]

-contains $5$ of the $p - 1$ operations of order~$2$ of the form

-\[

-\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta}

-           {\gamma x - \dfrac{1}{j - j^{-1}}}

-\right),

-\]

-viz.\ those for which

-\[

-\beta  \equiv \beta_{0} j^{n},\quad \gamma \equiv \gamma_{0} j^{-n},\quad (n = 0, 1, 2, 3, 4).

-\]

-Hence the sub-group~$\{A\}$, of order~$5$, belongs to $\frac{1}{5}(p - 1)$

-distinct icosahedral sub-groups. Now each icosahedral sub-group

-has $6$~sub-groups of order~$5$; and $H$~contains $\frac{1}{2}p(p + 1)$

-sub-groups of order~$5$ forming a single conjugate set. The

-number of icosahedral sub-groups in~$H$ is therefore

-\[

-\tfrac{1}{6} \frac{p - 1}{5}\, \tfrac{1}{2} p(p + 1)

-  = \frac{\frac{1}{2}p(p^{2} - 1)}{30}.

-\]

-The group of isomorphisms of the icosahedral group is the

-symmetric group of degree~$5$ (§~173). Now $H$~can contain no

-sub-group simply isomorphic with the symmetric group of

-degree~$5$. For if it contained such a sub-group, an operation of

-%% -----File: 348.png---Folio 330-------

-order~$5$ would be conjugate to its own square; and this is not

-the case.

-

-Hence (§~165), if an icosahedral sub-group~$K$ of~$H$ is contained

-self-conjugately in a greater sub-group~$L$, then $L$~must be

-the direct product of~$K$ and some other sub-group. This also is

-impossible; for the greatest sub-group of~$H$ in which any cyclical

-sub-group, except those of order~$p$, is contained self-conjugately,

-is of dihedral type. Hence $L$~must coincide with~$K$, and $K$~must

-be one of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ conjugate sub-groups. The icosahedral

-sub-groups of~$H$ therefore fall into two conjugate sets of

-$\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ each.

-

-In a similar manner, when $p \equiv -1\ (\mod 5)$, we may take as

-a typical operation~$A$, of order~$5$,

-\[

-\frac{y - i}{y - i^{p}} \equiv i^{\efrac{p^{2} - 1}{5}} \frac{x - i}{x - i^{p}};

-\]

-and it may be shewn, the calculation being rather more

-cumbrous than in the previous case, that there are just $p + 1$

-operations~$B$, of the form $\left(\dfrac{\alpha x + \beta}{\gamma x - \alpha}\right)$, such that

-\[

-(AB)^{3} = 1,

-\]

-and that five of these belong to the icosahedral group generated

-by~$A$ and any one of them. It follows, exactly as

-in the previous case, that $H$~contains $\dfrac{\frac{1}{2}p(p^{2} - 1)}{30}$ icosahedral

-sub-groups, which fall into two conjugate sets, each set containing

-$\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ groups.

-

-\Par{232.} Finally, we proceed to shew that $H$ has no other

-sub-groups than those which have been already determined.

-Suppose, first, that a sub-group~$h$ of~$H$ contains two distinct

-sub-groups of order~$p$. These must, by Sylow's theorem, form

-part of a set of $kp + 1$ sub-groups of order~$p$ conjugate within~$h$.

-Now $H$~contains only $p + 1$ sub-groups of order~$p$, and

-%% -----File: 349.png---Folio 331-------

-therefore $k$~must be unity and $h$~must contain all the sub-groups

-of order~$p$; or since $H$~is simple, $h$~must contain and therefore

-coincide with~$H$. Hence the only sub-groups of~$H$, whose orders

-are divisible by~$p$, are those that contain a sub-group of order~$p$

-self-conjugately. They are of known types.

-

-Suppose next that $g$~is a sub-group of~$H$, whose order~$n$

-is not divisible by~$p$, and let $S_{1}$ be an operation of~$g$ whose

-order~$q_{1}$ is not less than the order of any other operation of~$g$.

-In~$H$ the sub-group~$\{S_{1}\}$ is self-conjugate in a dihedral group of

-order~$p ± 1$; and the greatest sub-group of this group, which

-contains no operation of order greater than~$q_{1}$, is a dihedral

-group of order~$2q_{1}$. Hence in~$g$ the sub-group~$\{S_{1}\}$ is self-conjugate

-in a group of order $q_{1}$ or~$2q_{1}$, and therefore it forms

-one of~$\dfrac{n}{q{_1}}$ or of~$\dfrac{n}{2q{_1}}$ conjugate sub-groups. Moreover, no two of

-these sub-groups contain a common operation except identity;

-and they therefore contain, excluding identity, $\dfrac{n(q_{1} - 1)}{\epsilon_{1} q_{1}}$ distinct

-operations, where $\epsilon_{1}$~is either $1$ or~$2$.

-

-Of the remaining operations of~$g$, let $S_{2}$ be one whose

-order~$q_{2}$ is not less than that of any of the others. The

-operation~$S_{2}$ cannot be permutable with any of the $\dfrac{n(q_{1} - 1)}{\epsilon_{1} q_{1}}$

-operations already accounted for, since $S_{2}$~is not a power of any

-one of these operations. Hence, exactly as before, $\{S_{2}\}$~must

-form one of~$\dfrac{n}{\epsilon_{2}q_{2}}$ conjugate sub-groups in~$g$, $\epsilon_{2}$~being either $1$ or~$2$;

-and these sub-groups contain $\dfrac{n(q_{2} - 1)}{\epsilon_{2} q_{2}}$ operations which are

-distinct from identity, from each other, and from those of the

-previous set. This process may be continued till the identical

-operation only remains. Hence, finally, $n$~being the total

-number of operations of~$g$, we must have

-\[

-n = 1 + \sum_{\nu} \frac{n(q_{\nu} - 1)}{\epsilon_{\nu}q_{\nu}}

-\]

-or

-\[

-\frac{1}{n} = 1 - \sum_{\nu}  \frac{q_{\nu} - 1}{\epsilon_{\nu}q_{\nu}}.

-\]

-%% -----File: 350.png---Folio 332-------

-

-\Par{233.} In this equation, let $r$ of the $\epsilon$'s be~$1$ and $s$~of them be~$2$,

-so that $r + s$ is their total number, say~$m$. Then

-\begin{align*}

-\frac{1}{n}

-  &= 1 - \sum \left(1 - \frac{1}{q_{\lambda}}\right)

-       - \sum \left(\tfrac{1}{2} - \frac{1}{2 q_{\mu}}\right) \\

-  &= 1 - r + \sum \frac{1}{q_{\lambda}} - \tfrac{1}{2} s + \sum \frac{1}{2q_{\mu}} \\

-  &\leq 1 - r + \tfrac{1}{2} r - \tfrac{1}{2} s + \tfrac{1}{4} s \\

-  &\leq 1 - \tfrac{1}{2} r - \tfrac{1}{4} s \\

-  &\leq 1 - \tfrac{1}{4} m.

-\end{align*}

-

-Hence, since $n$~is a positive integer, there cannot be more

-than three terms under the sign of summation. Moreover, since

-\[

-\frac{1}{n} \leq 1 - \tfrac{1}{2} r - \tfrac{1}{4} s,

-\]

-$r$ cannot be greater than~$1$, and therefore not more than one of

-the~$\epsilon$'s can be unity. Also, when one of the~$\epsilon$'s is unity, we have

-\begin{align*}

-\frac{1}{n} &= \frac{1}{q} - \tfrac{1}{2} s + \sum \frac{1}{2 q_{\mu}} \\

-  &\leq \tfrac{1}{2} - \tfrac{1}{2} s + \tfrac{1}{4} s \\

-  &\leq \tfrac{1}{4} (2 - s),

-\end{align*}

-so that, in this case, $s$~cannot be greater than unity. The

-solutions are now easily obtained by trial.

-

-\Class{\Inum{(i)}}{For one term in the sum, the only possible solution is}

-\[

-\epsilon_{1} = 1,\quad n = q_{1},

-\]

-and the corresponding group is cyclical.

-

-\Class{\Inum{(ii)}}{For two terms in the sum, the solutions are}

-\begin{align*}

-&\Inum{($\alpha$)}\quad \epsilon_{1} = \epsilon_{2} = 2,\quad n = \frac{2q_{1}q_{2}}{q_{1} + q_{2}}; \\

-&\Inum{($\beta$)}\quad  \epsilon_{1} = 2,\quad \epsilon_{2} = 1,\quad q_{2} = 2,\quad n = 2q_{1}; \\

-&\Inum{($\gamma$)}\quad \epsilon_{1} = 1,\quad \epsilon_{2} = 2,\quad q_{1} = 3,\quad q_{2} = 2,\quad n = 12.

-\end{align*}

-

-To the solution~\Inum{($\alpha$)} there corresponds no sub-group; for

-$n < 2q_{1}$, and the values $q_{1} = q_{1}$\DPnote{** [sic]}, $\epsilon_{1} = 2$ imply that $g$~has a sub-group

-of order~$2q_{1}$.

-

-To the solution~\Inum{($\beta$)} correspond the sub-groups of order~$2q_{1}$

-of dihedral type, for which $q_{1}$~is odd, so that the operations of

-order~$2$ form a single conjugate set.

-%% -----File: 351.png---Folio 333-------

-

-To the solution~\Inum{($\gamma$)} corresponds a sub-group of order~$12$

-containing $8$~operations of order~$3$ and $3$~operations of order~$2$,

-\ie\ a tetrahedral sub-group.

-

-\Class{\Inum{(iii)}}{For three terms in the sum, the solutions are}

-\begin{alignat*}{6}

-&\Inum{($\alpha$)}\quad && \epsilon_{1} = \epsilon_{2} = \epsilon_{3} = 2,\quad

-  &&& q_{2} &= 2,\quad & q_{3} &= 2,\quad & n &= 2q_{1}; \\

-%

-&\Inum{($\beta$)} && \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}

-  & q_{1} &= 3,\quad & q_{2} &= 3, & q_{3} &= 2, & n &= 12; \\

-&\Inum{($\gamma$)}&& \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}

-  & q_{1} &= 4,      & q_{2} &= 3, & q_{3} &= 2, & n &= 24; \\

-&\Inum{($\delta$)}&& \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}

-  & q_{1} &= 5,      & q_{2} &= 3, & q_{3} &= 2, & n &= 60.

-\end{alignat*}

-

-To the solution~\Inum{($\alpha$)} correspond the sub-groups of order~$2q_{1}$

-of dihedral type, in which $q_{1}$~is even, so that the operations of

-order~$2$, which do not belong to the cyclical sub-group of order~$q_{1}$,

-fall into two distinct conjugate sets.

-

-To the solution~\Inum{($\beta$)} would correspond a group of order~$12$

-containing $3$~operations of order~$2$ and $4$~sub-groups of order~$3$

-which fall into two conjugate sets of $2$ each. Sylow's theorem

-shews that such a group cannot exist; and therefore there is no

-sub-group of~$H$ corresponding to this solution.

-

-Solution~\Inum{($\gamma$)} gives a group of order~$24$, with $3$~conjugate

-cyclical sub-groups of order~$4$, $4$~conjugate cyclical sub-groups

-of order~$3$, and $6$~other operations of order~$2$ forming a single

-conjugate set. No operation of this group is permutable with

-each of the $4$ sub-groups of order~$3$; and therefore, if the group

-exists, it can be represented as a transitive group of $4$~symbols.

-On the other hand, the order of the symmetric group of~$4$~symbols,

-which (§~203) is simply isomorphic with the octohedral

-group, is~$24$; and its cyclical sub-groups are distributed as

-above. Hence to this solution there correspond the octohedral

-sub-groups of~$H$.

-

-Solution~\Inum{($\delta$)} gives a group of order~$60$, with $6$~conjugate

-sub-groups of order~$5$, $10$~conjugate sub-groups of order~$3$, and

-a conjugate set of~$15$ operations of order~$2$. It has been shewn,

-in §~85, that there is only one type of group of order~$60$ that

-has $6$~sub-groups of order~$5$; viz.\ the alternating group of

-degree~$5$: and that, in this group, the distribution of sub-groups

-in conjugate sets agrees with that just given. Moreover, the

-alternating group of degree~$5$ is simply isomorphic with the

-\index{Fractional linear group@\Topic{Fractional linear group}!analysis of|)}%

-%% -----File: 352.png---Folio 334-------

-icosahedral group. Hence to this solution there correspond the

-icosahedral sub-groups of~$H$.

-

-\Par{234.} When $p > 11$, then $\frac{1}{2}p(p - 1) > 60$; and, when $p > 3$,

-\index{Fractional linear group@\Topic{Fractional linear group}!generalization of}%

-$\frac{1}{2}p(p - 1) > p + 1$. Hence when $p > 11$, the order of the

-greatest sub-group of~$H$ is $\frac{1}{2}p(p - 1)$, and the least number of

-symbols in which $H$ can be expressed as a transitive substitution

-group is $p + 1$.

-

-When $p$~is $5$,~$7$ or~$11$, however, $H$~can be expressed as a

-transitive substitution group of $p$~symbols\footnotemark.

-\index{Galois@\Topic{Galois}, quoted}%

-\footnotetext{This is another of the results stated in the letter of Galois referred to in

-  the footnote on \PageRef{p.}{192}.}

-

-For, when $p = 5$, $H$~contains a tetrahedral sub-group of

-order~$12$, forming one of $5$ conjugate sub-groups; therefore $H$~can

-be expressed as a transitive group of $5$~symbols. It is to be

-noticed that in this case $H$~is an icosahedral group.

-

-When $p = 7$, $H$ contains an octohedral sub-group of order~$24$,

-which is one of $7$ conjugate sub-groups; and $H$~can therefore

-be expressed as a transitive group of $7$~symbols. Similarly,

-when $p = 11$, $H$~contains an icosahedral sub-group of order~$60$,

-which is one of $11$ conjugate sub-groups; and the group can be

-expressed transitively in $11$~symbols.

-

-\Par{235.} The simple groups, of the class we have been discussing

-in the foregoing sections, are self-conjugate sub-groups

-of the triply transitive groups of degree~$p + 1$, defined by

-\[

-y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p),

-\]

-the existence of which was demonstrated in §~113. In fact,

-since $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$

-and $\left(\dfrac{k\alpha x + k\beta}{k\gamma x + k\delta}\right)$ represent the same transformation,

-the determinant, $\alpha\delta - \beta\gamma$, of any transformation may

-always be taken as either unity or a given non-residue; and it

-follows at once that the transformations of determinant unity

-form a self-conjugate sub-group of the whole group of transformations.

-

-If, as in §~113, $\alpha$,~$\beta$, $\gamma$,~$\delta$, are powers of~$i$, where $i$~is a primitive

-root of the congruence

-\[

-i^{p^{n} - 1} \equiv 1,\ (\mod p),

-\]

-%% -----File: 353.png---Folio 335-------

-the triply transitive group~$G$ of degree~$p^{n} + 1$, which is defined

-by the transformations, has again, when $p$~is an odd prime, a

-self-conjugate sub-group~$H$ of order~$\frac{1}{2}p^{n}(p^{2n} - 1)$, which is

-given by the transformations of determinant unity. It follows

-from Theorem~IX, §~134, that $G$, being a triply transitive group

-of degree~$p^{n} + 1$, must have, as a self-conjugate sub-group, a

-doubly transitive simple group; and it is easy to shew that $H$

-is this sub-group.

-

-In fact, if a simple group~$h$ is a self-conjugate sub-group of~$G$

-it must be contained in~$H$. Also, since $h$~is a doubly

-transitive group of degree~$p^{n} + 1$, it must contain every operation

-of order~$p$ that occurs in~$G$. Now we may shew that these

-operations generate~$H$. Thus $\left(\dfrac{2x + 1}{-x}\right)$ and $(x + 2 - i - i^{-1})$ are

-operations of order~$p$ belonging to~$G$. Therefore $\left(\dfrac{(i + i^{-1})x + 1}{-x}\right)$

-belongs to~$h$. But this operation is transformed into $\left(\dfrac{ix}{i^{-1}}\right)$ by

-$\left(\dfrac{x + i^{-1}}{x + i}\right)$. Hence $\left(\dfrac{ix}{i^{-1}}\right)$ belongs to~$h$; and a sub-group of~$h$

-which keeps one symbol unchanged is the group of order

-$\frac{1}{2}p^{n}(p^{n} - 1)$ generated by $(x + 1)$ and~$\left(\dfrac{ix}{i^{-1}}\right)$. The order of~$h$

-therefore is not less than $\frac{1}{2}p^{n}(p^{2n} - 1)$; in other words $h$~is

-identical with~$H$.

-

-When $p = 2$, every power of~$i$ is a quadratic residue, and the

-determinant of every transformation is unity. In this case it

-may be shewn, by an argument similar to the above, that the

-group~$G$ of order $2^{n}(2^{2n} - 1)$ is itself a simple group.

-

-We are thus led to recognize the existence of a doubly-infinite

-series of simple groups of orders $2^{n}(2^{2n} - 1)$ and

-$\frac{1}{2}p^{n}(p^{2n} - 1)$, which are closely analogous to the groups of order

-$\frac{1}{2}p(p^{2} - 1)$ already discussed. For an independent proof of the

-existence of these simple groups and for an investigation of

-their properties, the reader is referred to the memoirs mentioned

-below\footnotemark.

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Fractional linear group@\Topic{Fractional linear group}!generalization of}%

-\index{Moore@\Topic{Moore}, quoted}%

-\footnotetext{Moore: ``On a doubly-infinite series of simple groups,'' \textit{Chicago Congress

-  Mathematical Papers}, (1893); Burnside: ``On a class of groups defined by congruences,''

-  \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXV}, (1894), pp.~113--139.}%

-%% -----File: 354.png---Folio 336-------

-

-\Par{236.} We will now return to the linear homogeneous group~$G$

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|(}%

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!represented as a transitive substitution group}%

-of transformations of $n$~symbols, taken to a prime modulus~$p$;

-and consider it more directly as the group of isomorphisms of

-an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.

-As in §~156, it may be expressed in the form of a substitution

-group performed on the $p^{n} - 1$ symbols of the operations, other

-than identity, of the Abelian group. In this form it is clearly

-transitive, since there are isomorphisms changing any operation

-of the Abelian group into any other operation. If $P$~is any

-operation of the Abelian group, an isomorphism which changes

-any one of the $p - 1$ operations

-\[

-P,\ P^{2},\ \Dots,\ P^{p - 1},

-\]

-into any other, will certainly interchange the set among themselves.

-Hence, when expressed as a group of degree~$p^{n} - 1$, $G$~is

-imprimitive; and the symbols forming an imprimitive system

-are those of the operations, other than identity, of any sub-group

-of order~$p$ of the Abelian group. If

-\[

-P_{1},\ P_{2},\ \Dots,\ P_{n}

-\]

-are a set of generating operations of the Abelian group, an

-isomorphism, which changes each of the sub-groups

-\[

-\{P_{1}\},\ \{P_{2}\},\ \Dots,\ \{P_{n}\}

-\]

-into itself, must be of the form

-\[

-\Sub{\PadTo{P_{1}^{\alpha_{1}}}{P_{1}},

-     \PadTo{P_{1}^{\alpha_{2}}}{P_{2}}, \Dots,

-     \PadTo{P_{1}^{\alpha_{n}}}{P_{n}}}

-    {P_{1}^{\alpha_{1}}, P_{2}^{\alpha_{2}}, \Dots, P_{n}^{\alpha_{n}}}.

-\]

-This isomorphism changes $P_{1}P_{2}$ into $(P_{1} P_{2}^{\efrac{\alpha_{2}}{\alpha_{1}}})^{\alpha_{1}}$; therefore it

-will only transform the sub-group $\{P_{1}P_{2}\}$ into itself when

-$\alpha_{1} \equiv \alpha_{2},\ (\mod p)$. If then the given isomorphism changes every

-sub-group of order~$p$ into itself, we must have

-\[

-\alpha_{1} \equiv \alpha_{2} \equiv \Dots \equiv \alpha_{n},\ (\mod p).

-\]

-

-Hence the only operations of~$G$, which interchange the

-symbols of each imprimitive system among themselves, are

-those given by the powers of

-\[

-\Sub{\PadTo{P_{1}^{\alpha}}{P_{1}},

-     \PadTo{P_{2}^{\alpha}}{P_{2}}, \Dots,

-     \PadTo{P_{n}^{\alpha}}{P_{n}}}

-    {P_{1}^{\alpha}, P_{2}^{\alpha}, \Dots, P_{n}^{\alpha}},

-\]

-where $\alpha$ is a primitive root of~$p$. This operation is the same as

-%% -----File: 355.png---Folio 337-------

-that denoted by $A$ in §~217. It follows immediately that the

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!represented as a transitive substitution group}%

-factor-group $\dfrac{G}{\{A\}}$ can be represented as a transitive group in

-$\dfrac{p^{n} - 1}{p - 1}$ symbols. In fact, the operations of~$\{A\}$ are the only

-operations of~$G$ which transform each of the $\dfrac{p^{n} - 1}{p - 1}$ sub-groups

-of order~$p$ in itself; and these $\dfrac{p^{n} - 1}{p - 1}$ sub-groups must be

-permuted among themselves by every operation of~$G$. The

-substitution group thus obtained is doubly transitive; for if $P$~and~$P'$

-are any two operations of the Abelian group such that

-$P'$~is not a power of~$P$, and if $Q$~and~$Q'$ are any other two

-operations of the Abelian group subject to the same condition,

-there certainly exists an isomorphism of the form

-\[

-\Sub{P, P', \Dots}{Q, Q', \Dots},

-\]

-and this isomorphism changes the sub-groups $\{P\}$ and $\{P'\}$ into

-the sub-groups $\{Q\}$ and~$\{Q'\}$.

-

-These results will still hold if, instead of considering $G$~the

-total group of isomorphisms, we take $\Gamma$~the group of isomorphisms

-of determinant unity. Thus the determinant of

-\[

-\Sub{\PadTo{Q^{\alpha}}{P}, P', \Dots}{Q^{\alpha}, Q', \Dots}

-\]

-is $\alpha$~times the determinant of

-\[

-\Sub{P, P', \Dots}{Q, Q', \Dots}.

-\]

-It is therefore possible always to choose $\alpha$ so that the determinant

-of

-\[

-\Sub{\PadTo{Q^{\alpha}}{P}, P', \Dots}{Q^{\alpha}, Q', \Dots}

-\]

-shall be unity; and this isomorphism still changes the sub-groups

-$\{P\}$ and $\{P'\}$ into $\{Q\}$ and $\{Q'\}$ respectively.

-

-The lowest power of~$A$ contained in~$\Gamma$ is (§~217) $A^{\efrac{p - 1}{d}}$.

-Hence the group $\dfrac{\Gamma}{\{A^{\efrac{p - 1}{d}}\}}$ can be represented as a doubly

-%% -----File: 356.png---Folio 338-------

-transitive group of degree $\dfrac{p^{n} - 1}{p - 1}$. This group is (§~220) simply

-isomorphic with the simple group of order $\dfrac{N}{(p - 1)d}$, which is

-defined by the composition-series of~$G$.

-

-We may sum up these results as follows:---

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!simple groups defined by}%

-

-\begin{Theorem}

-The homogeneous linear group of order

-\[

-N = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n - 1})

-\]

-when $p^{n}$~is neither $2^{2}$ nor~$3^{2}$, defines, by its composition-series, a

-simple group of order~$\dfrac{N}{(p - 1)d}$, where $d$~is the greatest common

-factor of $p - 1$ and~$n$. This simple group can be represented as

-a doubly transitive group of degree $p^{n - 1} + p^{n - 2} + \Dots + p + 1$.

-\end{Theorem}

-

-\Par{237.} The $\dfrac{p^{n} - 1}{p - 1}$ symbols, permuted by one of these doubly

-transitive simple groups, may be regarded as the sub-groups of

-order~$p$ of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to

-$n$ units})$. Now every pair of sub-groups of such an Abelian

-group enters in one, and only in one, sub-group of order~$p^{2}$;

-and every sub-group of order~$p^{2}$ contains $p + 1$ sub-groups of

-order~$p$. Hence from the $\dfrac{p^{n} - 1}{p - 1}$ symbols permuted by the

-doubly transitive group, $\dfrac{p^{n} - 1 · p^{n-1} - 1}{p - 1 · p^2 - 1}$\DPnote{** TN: No () in original} sets of $p + 1$ symbols

-each may be formed, such that every pair of symbols occurs in

-one set and no pair in more than one set, while the sets are

-permuted transitively by the operations of the group. These

-groups therefore belong to the class of groups referred to in

-§~148. The sub-group, that leaves two symbols unchanged,

-permutes the remaining symbols in two transitive systems of

-$p - 1$ and $p^{n - 1} + p^{n - 2} + \Dots + p^{2}$; and the sub-group, that leaves

-unchanged each of the symbols of one of the sets of $p + 1$, is

-contained self-conjugately in a sub-group whose order is

-$(p + 1)p$~times that of a sub-group leaving two symbols unchanged.

-This latter sub-group permutes the symbols in two

-transitive systems of $p + 1$ and $p^{n - 1} + p^{n - 2} + \Dots + p^{2}$. It may

-be pointed out that, when $n$~is~$3$, such a sub-group is simply

-\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|)}%

-%% -----File: 357.png---Folio 339-------

-isomorphic with, but is not conjugate to, the sub-groups that

-leave one symbol unchanged: this may be seen at once by

-noticing that an Abelian group, of order~$p^{3}$ and type $(1, 1, 1)$, has

-the same number of sub-groups of orders $p$ and~$p^{2}$.

-

-\begin{Remark}

-\Par{238.} Some special cases may be noticed. First, when $p = 2$,

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!simple groups defined by}%

-both $p - 1$ and~$d$ are unity, and the homogeneous linear group is

-itself a simple group.

-

-If $n = 3$, then $N = 168$; so that the group of isomorphisms of a

-group of order~$8$, whose operations are all of order~$2$, is the simple

-group of order~$168$ (§~146).

-

-If $n = 4$, then $N = 2^{6} · 3^{2} · 5 · 7$. This is the order of the

-alternating group of $8$~symbols; and it may be shewn that this

-group is simply isomorphic with the group of isomorphisms.

-

-The Abelian group of order~$16$ contains $35$~sub-groups of order~$4$;

-and it may be shewn that, from these $35$~sub-groups, sets of~$5$ can

-be formed in $56$~distinct ways, so that each set of~$5$ contains every

-operation of order~$2$ of the Abelian group once, and once only. If

-\[

-P_{1},\ P_{2},\ P_{3},\ P_{4}

-\]

-are a set of generating operations of the Abelian group, such a set

-of $5$~sub-groups of order~$4$ is given by

-\[

-\{P_{1}, P_{2}\},\ \{P_{3}, P_{4}\},\ \{P_{1}P_{3}, P_{2}P_{4}\},\

-\{P_{1}P_{4}, P_{1}P_{2}P_{3}\},\

-\{P_{2}P_{3}, P_{1}P_{3}P_{4}\}.

-\]

-Now

-\[

-\Sub{P_{1}, \PadTo{P_{2}P_{3}}{P_{2}}, P_{3}, \PadTo{P_{2}P_{4}}{P_{4}}}

-    {P_{1}, P_{2}P_{3}, P_{4}, P_{2}P_{4}}

-\]

-is an isomorphism of order~$7$ of the Abelian group, and $P_{1}$~is the

-only operation of the group, except identity, which is left unchanged

-by this isomorphism. It may be directly verified that the $7$~sets of

-$5$~groups of order~$4$, into which the given set is transformed by the

-powers of this isomorphism, contain every sub-group of order~$4$ of

-the Abelian group. The $7$~sets, being interchanged among themselves

-by this isomorphism of order~$7$ which leaves only $P_{1}$ unchanged,

-must be interchanged among themselves by isomorphisms of order~$7$

-which leave any other single operation of the Abelian group

-unchanged. There are therefore at least $15$~isomorphisms of order~$7$

-which interchange the $7$~sets among themselves. Now the

-isomorphisms, which interchange the $7$~sets among themselves, form a

-sub-group of the group of isomorphisms, which is isomorphic with a

-group of degree~$7$; and the only groups of degree~$7$, which contain at

-least $15$ operations of order~$7$, are the symmetric and the alternating

-groups. The group of isomorphisms must therefore contain a sub-group

-which is isomorphic with the symmetric or with the alternating

-group of degree~$7$. Hence at once, since the group of isomorphisms

-is simple, it must contain a sub-group which is simply isomorphic

-with the alternating group of degree~$7$. Since this must be one of

-%% -----File: 358.png---Folio 340-------

-$8$~conjugate sub-groups, the group of substitutions itself is simply

-isomorphic with the alternating group of degree~$8$.

-

-If $p^{n} = 3^{3}$, then $p^{n - 1} + \Dots + p + 1 = 13$, $d = 1$, and $N = 2^{4} · 3^{3} · 13$.

-There is therefore a doubly transitive simple group of degree~$13$ and

-order~$2^{4} · 3^{3} · 13$ (§§~145,~149).

-\end{Remark}

-

-\Par{239.} The homogeneous linear group may be generalized by

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!generalization of|(}%

-taking for the coefficients powers of a primitive root of

-\[

-i^{p^{\nu} - 1} \equiv 1,\ (\mod p),

-\]

-instead of powers of a primitive root of

-\[

-i^{p - 1} \equiv 1,\ (\mod p).

-\]

-When the coefficients are thus chosen, the order of the group

-$G_{p, n, \nu}$, defined by all sets of transformations

-\[

-\begin{aligned}

-  x_{1}^{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\

-  x_{2}^{1} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\

-  \multispan{3}{\dotfill} \\

-  x_{n}^{1} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},

-\end{aligned}

-\quad (\mod p),

-\]

-whose determinant differs from zero, may be shewn, as in §~172,

-to be

-\[

-N = (p^{n\nu} - 1) (p^{n\nu} - p^{\nu}) (p^{n\nu} - p^{2 \nu}) \Dots (p^{n\nu} - p^{(n - 1)\nu});

-\]

-and the order of the sub-group~$\Gamma$, formed of those transformations

-whose determinant is unity, is~$\dfrac{N}{p^{\nu} - 1}$. The only

-self-conjugate operations of~$\Gamma$ are the operations of the sub-group

-generated by $(ix_{1}, ix_{2}, \Dots, ix_{n})$, which are contained in~$\Gamma$.

-If $\delta$~is the greatest common factor of $p^{\nu} - 1$ and~$n$, these

-self-conjugate operations of~$\Gamma$ form a cyclical sub-group~$\gamma$ of

-order~$\delta$. Finally, the argument of §~219 maybe repeated to

-shew that $\dfrac{\Gamma}{\gamma}$~is a simple group.

-

-The homogeneous linear group $G_{p, n, \nu}$, when values of~$\nu$

-greater than unity are admitted, thus defines a triply infinite

-system of simple groups; it may be proved that these groups

-can, for all values of~$\nu$, be expressed as doubly transitive groups

-of degree~$\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$.

-%% -----File: 359.png---Folio 341-------

-

-\Par{240.} We may shew, in conclusion, that the group $G_{p, n, \nu}$ is

-simply isomorphic with a sub-group of $G_{p, n\nu, 1}$. For this purpose,

-we consider the group defined by

-\begin{gather*}

-x_{1}^{1} \equiv x_{1} + i^{r_{1}},\quad

-x_{2}^{1} \equiv x_{2} + i^{r_{2}},\quad \Dots,\quad

-x_{n}^{1} \equiv x_{n} + i^{r_{n}}, \\

-(r_{1}, r_{2}, \Dots, r_{n} = 0, 1, 2, \Dots, p^{\nu} - 1);

-\end{gather*}

-the congruences being taken to modulus~$p$. This is an Abelian group

-of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n\nu$ units})$. Moreover, the operation

-\begin{align*}

-  x_{1}^{1} &\equiv a_{11}x_{1} + \Dots + a_{1n}x_{n}, \\

-  \multispan{3}{\dotfill} \\

-  x_{n}^{1} &\equiv a_{n1}x_{1} + \Dots + a_{nn}x_{n},

-\end{align*}

-of $G_{p, n, \nu}$ transforms the given operation of the Abelian group into

-\[

-x_{1}^{1} \equiv x_{1} + i^{s_{1}},\quad

-x_{2}^{1} \equiv x_{2} + i^{s_{2}},\quad \Dots,\quad

-x_{n}^{1} \equiv x_{n} + i^{s_{n}};

-\]

-where

-\begin{align*}

-  i^{s_{1}} &\equiv a_{11}i^{r_{1}} + a_{12}i^{r_{2}} + \Dots + a_{1n}i^{r_{n}}, \\

-  \multispan{3}{\dotfill} \\

-  i^{s_{n}} &\equiv a_{n1}i^{r_{1}} + a_{n2}i^{r_{2}} + \Dots + a_{nn}i^{r_{n}}.

-\end{align*}

-

-Every operation of~$G_{p, n, \nu}$ as defined in §~239, is therefore

-permutable with the Abelian group, and gives a distinct isomorphism

-of it; or in other words, as stated above, $G_{p, n, \nu}$~is simply isomorphic

-with a sub-group of~$G_{p, n\nu, 1}$.

-

-Further, the sub-group

-\begin{gather*}

-x_{1}^{1} \equiv x_{1} + i^{r},\quad

-x_{2}^{1} \equiv x_{2},\quad \Dots,\quad

-x_{n}^{1} \equiv x_{n}, \\

-(r = 0, 1, 2, \Dots, p^{\nu} - 1),

-\end{gather*}

-is transformed by the given operation of~$G_{p, n, \nu}$ into the sub-group

-\begin{gather*}

-x_{1}^{1} \equiv x_{1} + a_{11}i^{r},\quad

-x_{2}^{1} \equiv x_{2} + a_{21}i^{r},\quad \Dots,\quad

-x_{n}^{1} \equiv x_{n} + a_{n1}i^{r}, \\

-(r = 0, 1, 2, \Dots, p^{\nu} - 1).

-\end{gather*}

-If

-\[

-a_{21} \equiv a_{31} \equiv \Dots \equiv a_{n1} \equiv 0,

-\]

-the two sub-groups are identical; but if these conditions are not

-satisfied, they have no operation in common except identity.

-Moreover,

-\[

-a_{11},\ a_{21},\ \Dots,\ a_{n1}

-\]

-may each have any value from $0$ to~$i^{p^{\nu} - 1}$, simultaneous zero values

-alone excluded. Hence the sub-group of order~$p^{\nu}$ defined by

-\begin{gather*}

-x_{1}^{1} \equiv x_{1} + i^{r},\quad

-x_{2}^{1} \equiv x_{2},\quad \Dots,\quad

-x_{n}^{1} \equiv x_{n}, \\

-(r = 0, 1, \Dots, p^{\nu} - 1),

-\end{gather*}

-is one of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ conjugate sub-groups in the group formed by combining

-the Abelian group with~$G_{p, n, \nu}$; and no two of these conjugate

-sub-groups have a common operation except identity.

-%% -----File: 360.png---Folio 342-------

-

-\begin{Remark}

-The $p^{n\nu} - 1$ operations, other than identity, of an Abelian group

-of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n\nu$ units})$, can therefore be

-divided into $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ sets of $p^{\nu} - 1$ each, such that each set, with

-identity, forms a sub-group of order~$p^{\nu}$; and the group~$G_{p, n, \nu}$ is isomorphic

-with a group of isomorphisms of the Abelian group which permutes

-among themselves such a set of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ sub-groups of order~$p^{\nu}$.

-\end{Remark}

-

-\begin{Ex}[1.] Shew that the $\dfrac{p^{n\nu} - 1 · p^{n\nu} - p\Dots p^{n\nu} - p^{\nu}}{p - 1 · p^{2} - 1\Dots p^{\nu} - 1}$\DPnote{** TN: No () in original} sub-groups of

-order~$p^{\nu}$ of an Abelian group of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to

-$n\nu$ units})$ can be divided into sets of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ each, such that each set

-contains every operation of the group, other than identity, once and

-once only; and discuss in how many distinct ways such a division

-may be carried out.

-\end{Ex}

-

-\begin{Ex}[2.] Shew that the simple group, defined by the group of

-isomorphisms of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$,

-admits a class of contragredient isomorphisms, which change the

-operations of the simple group, that correspond to isomorphisms

-leaving a sub-group of order~$p$ of the Abelian group unaltered, into

-operations that correspond to isomorphisms leaving a sub-group of

-order~$p^{n - 1}$ of the Abelian group unaltered.

-\end{Ex}

-\index{Homogeneous linear group@\Topic{Homogeneous linear group}!generalization of|)}%

-%% -----File: 361.png---Folio 343-------

-

-

-\Chapter{XV.}{On Soluble and Composite Groups.}

-

-\Par{241.} \First{The} most general problem of pure group-theory, so

-far as it is concerned with groups of finite order, is the determination

-and analysis of all distinct types of group whose order

-is a given integer. The solution of this problem clearly involves

-the previous determination of all types of simple groups whose

-orders are factors of the given integer.

-

-Now there is no known criterion by means of which we can

-say whether, corresponding to an arbitrarily given composite

-integer~$N$ as order, there exists a simple group or not. For

-certain particular forms of~$N$, this question can be answered in

-the affirmative. For instance, when $N = \frac{1}{2}n!$, previous investigation

-enables us to state that there is a simple group of

-order~$N$; but even in these cases we cannot, in general, say

-how many distinct types of simple group of order~$N$ there are.

-

-On the other hand, we have seen that, for certain forms of~$N$,

-it is possible to state that there is no simple group of order~$N$.

-Thus, when $N$~is the power of a prime, or is the product of two

-distinct primes, there is no simple group whose order is~$N$; and

-further, every group of order~$N$ is soluble. Again, if $N$~is

-divisible by a prime~$p$, and contains no factor of the form

-$1 + kp$, a group of order~$N$ cannot be simple; for, by Sylow's

-theorem, it must contain a self-conjugate sub-group having a

-power of~$p$ for its order.

-

-We propose, in the present Chapter, to prove a series of

-theorems which enable us to state, in a considerable variety of

-%% -----File: 362.png---Folio 344-------

-cases, that a group, which has a number of given form for its

-order, is either soluble or composite. If the results appear

-fragmentary, it must be remembered that this branch of the

-subject has only recently received attention: it should be

-regarded rather as a promising field of investigation than as

-one which is thoroughly explored.

-

-The symbols $p_{1}$,~$p_{2}$, $p_{3}$,~\Dots\ will be used throughout the

-Chapter to denote distinct primes in ascending order of magnitude;

-while distinct primes, without regard to their magnitude,

-will be denoted by $p$,~$q$, $r$,~\Dots.

-

-\Par{242.} \begin{Theorem}[I.]

-If $H$ is one of $n$ conjugate sub-groups

-of a group~$G$ of order~$N$, and if $N$~is not a factor of~$n!$, $G$~cannot

-be simple.

-\end{Theorem}

-

-The group~$G$ is isomorphic with the group of substitutions

-given on transforming the set of $n$ conjugate sub-groups by

-each of the operations of~$G$. This is a transitive group of

-degree~$n$, and its order therefore is equal to or is a factor of~$n!$.

-Hence, if $N$~is not a factor of~$n!$, the isomorphism cannot be

-simple; $G$~is therefore not simple.

-

-\begin{CorollaryFN}

-If a prime~$p$ divides~$N$, and if $m$~is the

-\index{Holder@\Topic{Hölder}, quoted}%

-greatest factor of~$N$ which is congruent to unity~$(\mod p)$, $G$~will

-be composite unless $N$~is a factor of~$m!$.

-\end{CorollaryFN}

-\footnotetext

-  {Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), p.~57.}

-

-There cannot, in fact, be more than $m$ conjugate sub-groups

-whose order is the highest power of~$p$ that divides~$N$; so that

-this result follows from the theorem.

-

-\begin{Remark}

-In dealing with a given integer as order, it may happen that,

-though no single application of this theorem will prove the corresponding

-group to be composite, a repeated application of Sylow's

-theorem, on which the theorem depends, will lead to that result.

-As an example, let us consider groups whose order is $3^{2} · 5 · 11$. Any

-such group must contain either $1$ or $45$ sub-groups of order~$11$, and

-either $1$ or $11$ sub-groups of order~$5$. Hence, if the group is simple,

-it must contain $45$~sub-groups of order~$11$, and $11$~sub-groups of

-order~$5$; and each one of the latter must be contained self-conjugately

-in a sub-group of order~$45$. Now the $45$~sub-groups of order~$11$

-would contain $450$~distinct operations of order~$11$, leaving only

-$45$~others; and therefore a sub-group of order~$45$, if the group contain

-such a sub-group, must in this case be self-conjugate. This is,

-%% -----File: 363.png---Folio 345-------

-however, in direct contradiction to the assumption that the group

-contains $11$~sub-groups of order~$5$; therefore the group cannot be

-simple.

-\end{Remark}

-

-\Par{243.} \begin{Theorem}[II.]

-Every group whose order is the power

-\index{Soluble groups@\Topic{Soluble groups}!special classes of|(}%

-of a prime is soluble.

-\end{Theorem}

-

-This has already been proved in §~54.

-

-\begin{Theorem}[III\protect\footnotemark.]

-A group~$G$ whose order is~$p^{\alpha} q^{\beta}$, where $\alpha$~is

-less than~$2m$, $m$~being the index to which $p$ belongs~$(\mod q)$, is

-soluble.

-\end{Theorem}

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, \textit{Berliner Sitzungsberichte} 1895, p.~190, and Burnside, \textit{Proc.\

-  L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~209; for the case where $\alpha \ngtr m$.}

-

-If $\alpha < m$, $G$~contains a self-conjugate sub-group~$H$ of order~$q^{\beta}$.

-

-If $\alpha = m$, $G$~contains either $1$ or $p^{m}$ sub-groups of order~$q^{\beta}$.

-In the latter case, if $q^{r}$~is the order of the greatest sub-group

-common to two groups of order~$q^{\beta}$, and if $h$~is such a sub-group,

-$h$~must (§~80) be contained self-conjugately in a

-sub-group~$k$ of order~$p^{x}q^{r + s}$, $x > 0$, $s > 0$; and this sub-group

-must contain more than one sub-group of order~$q^{r + s}$. Hence $x$~must

-be~$m$; and therefore $h$~must be common to all the $p^{m}$

-sub-groups of order~$q^{\beta}$, so that $h$~is self-conjugate, and $s$~is

-$\beta - r$. If $r$~is zero, no two sub-groups of order~$q^{\beta}$ have a

-common operation except identity, and the $p^{m}$~sub-groups

-contain $p^{m}(q^{\beta} - 1)$ distinct operations, so that a sub-group

-of order~$p^{m}$ is self-conjugate. Hence when $\alpha$~is equal to~$m$, $G$~contains

-either a self-conjugate sub-group of order~$q^{r}$ ($r > 0$) or

-else one of order~$p^{m}$.

-

-If $m < \alpha < 2m$, $G$~contains $1$ or $p^{m}$ sub-groups of order~$q^{\beta}$.

-If there is only one, it is self-conjugate.

-

-If there are $p^{m}$~sub-groups of order~$q^{\beta}$, and if~$q^{r}$ ($r > 0$) is

-the order of the greatest group common to any two of them,

-it may be shewn, exactly as in the preceding case, that $G$~contains

-a self-conjugate sub-group of order~$q^{r}$.

-

-If $r = 0$, the $q^{\beta} - 1$ operations, other than identity, of each

-of the $p^{m}$~sub-groups of order~$q^{\beta}$ are all distinct. Now $G$~is

-isomorphic with a group of degree~$p^{m}$. If the isomorphism

-is multiple, $G$~must have a self-conjugate sub-group whose order

-is a power of~$p$. If the isomorphism is simple, we may regard~$G$

-%% -----File: 364.png---Folio 346-------

-as a transitive group of degree~$p^{m}$; and every operation,

-except identity, of a sub-group of order~$q^{\beta}$ will leave one symbol

-only unchanged. Hence $q^{\beta}$~is equal to or is a factor of~$p^{m} - 1$.

-Moreover, when $G$~is thus regarded, the sub-groups of order~$p^{\alpha}$

-must be transitive; for all the sub-groups of order~$q^{\beta}$ will be

-given on transforming any one of them by the operations of a

-sub-group of order~$p^{\alpha}$.

-

-Consider now a sub-group of order~$p^{\alpha - m} q^{\beta}$, which keeps one

-symbol fixed and contains a sub-group of order $q^{\beta}$ self-conjugately.

-It will contain $q^{\gamma}$ ($\gamma \leq \beta$) sub-groups of order~$p^{\alpha - m}$.

-Any one of these must keep $p^{x}$~symbols fixed; for it is the

-sub-group of a transitive group of order~$p^{\alpha}$ and degree~$p^{m}$, which

-keeps one symbol fixed. If $\gamma < \beta$, there are operations of order~$q$

-which transform one of these sub-groups into itself, and

-therefore interchange among themselves the $p^{x}$~symbols unchanged

-by the sub-group. Hence $p^{x} - 1$ must be divisible

-by~$q$. This requires that $x = m$, which is impossible. Hence

-$\gamma = \beta$, and the sub-group of order~$p^{\alpha - m} q^{\beta}$ contains $q^{\beta}$~sub-groups

-of order~$p^{\alpha - m}$. It is obvious that no two of these can enter in

-the same sub-group of order~$p^{\alpha}$; for if they did, they would

-generate a group of order~$p^{\alpha - m + x}$ ($x > 0$), a sub-group

-of this order cannot enter in a group of order~$p^{\alpha - m} q^{\beta}$. Therefore

-$G$~contains $q^{\beta}$~sub-groups of order~$p^{\alpha}$.

-

-Since the sub-groups of~$G$, of order~$p^{\alpha}$, are transitive, their

-self-conjugate operations must (§~106) displace all the symbols,

-and they cannot therefore be permutable with any operation

-whose order is a power of~$q$. Suppose now that every

-operation of a sub-group~$Q$, of order~$q^{x}$, is permutable with

-an operation~$P$ of order~$p^{y}$, and that there is no sub-group

-of order~$q^{x + 1}$ of which this is true. If $S$~is a self-conjugate

-operation of a sub-group of order~$p^{\alpha}$ to which $P$~belongs,

-$P$~is transformed into itself and $Q$~into a new sub-group~$Q'$

-by~$S$. Hence the sub-group which contains $P$ self-conjugately

-has two, and therefore $p^{m}$, sub-groups of order~$q^{x}$.

-No two of these can belong to the same sub-group of order~$q^{\beta}$,

-for if they did they would generate a group of order~$q^{x + x'}$

-($x' > 0$). Now since $P$~is permutable with~$Q$, it must

-leave unchanged the symbol which $Q$~leaves unchanged.

-Hence $P$~must leave every symbol unchanged; \ie\ it must

-%% -----File: 365.png---Folio 347-------

-be the identical operation. The order of every operation of~$G$

-is therefore either a power of~$p$ or a power of~$q$.

-

-Suppose now that $p^{r}$~is the order of the greatest sub-group

-that is common to any two sub-groups of order~$p^{\alpha}$, and that $h$~is

-such a sub-group. Then (§~80) $h$~must be permutable with

-an operation of order~$q$. Since no operation of~$h$ is permutable

-with any operation of order~$q$, it follows that $p^{r} - 1$ is divisible

-by~$q$, and therefore that $r = m$.

-

-Let $k$, of order~$p^{m + t}q^{\gamma}$, be the greatest sub-group that

-contains $h$ self-conjugately. If $\gamma = \beta$, $h$~is common to each

-of the $q^{\beta}$ sub-groups of order~$p^{\alpha}$, and it is therefore a self-conjugate

-sub-group of~$G$.

-

-If $\gamma < \beta$, let $Q$~be a sub-group, of order~$q^{\gamma}$, of~$k$. In~$G$

-there must be a group, whose order is divisible by~$q^{\gamma + 1}$,

-containing $Q$ self-conjugately. Hence there must be operations

-in~$G$, which transform $Q$ into itself and $h$~into a conjugate

-sub-group~$h'$. If $h$~and~$h'$ have a common sub-group, it must

-be transformed into itself by every operation of~$Q$. This is

-impossible, since $p^{s} - 1$ is not divisible by~$q$ when $s < m$.

-

-In the group $\{h', Q\}$, $Q$~is one of $p^{m}$~conjugate sub-groups,

-and therefore no operation of~$h'$ can transform $Q$ into itself.

-Hence if an operation~$P'$ of~$h'$ transforms $h$ into itself, it must

-transform $Q$ into a group~$Q'$, which is conjugate to~$Q$ in~$k$.

-Hence, $k'$~being the greatest sub-group that contains $h'$

-self-conjugately, $\{Q, Q'\}$~must be common to $k$~and~$k'$. But

-$\{Q, Q'\}$ containing two sub-groups of order~$q^{\gamma}$, must contain $p^{m}$

-such sub-groups; and therefore the $p^{m}$~sub-groups of order~$q^{\gamma}$

-that enter in~$k$ are identical with those that enter in~$k'$.

-This is impossible if $H$~and~$H'$ are distinct; for the $p^{m}$~sub-groups

-of~$k$, of order~$q^{\gamma}$, generate $\{h, Q\}$. Hence no operation

-of~$h'$, except identity, transforms $h$ into itself, and the

-number of sub-groups in the conjugate set to which $h$~belongs

-must not be less than~$p^{m}$; so that

-\[

-p^{m} \ngtr p^{\alpha - m - t} q^{\beta - \gamma}.

-\]

-

-Now we have seen that

-\begin{align*}

-p^{m} &\equiv 1\ (\mod q^{\beta}), \\

-\lintertext{and}

-q^{\beta} &\equiv 1\ (\mod p^{\alpha - m});

-\end{align*}

-%% -----File: 366.png---Folio 348-------

-from these congruences, and the inequality $\alpha - m < m$, it follows

-that

-\[

-p^{m} \nless p^{\alpha - m} q^{\beta} - q^{\beta} + 1.

-\]

-

-This inequality is inconsistent with the previous one, which

-follows directly from the supposition that $\gamma$~is less than~$\beta$.

-Hence $\gamma = \beta$, and $h$~is a self-conjugate sub-group of~$G$.

-

-It follows therefore that, on every possible supposition, $G$~must

-have a self-conjugate sub-group. If this sub-group be

-represented by~$G_{1}$, the same reasoning may be repeated with

-respect to the groups $\dfrac{G}{G_{1}}$ and~$G_{1}$. Hence $G$~is soluble.

-

-\begin{Corollary}

-All groups whose orders are $p_{1} p_{2}^{\alpha}$, $p_{1}^{2} p_{2}^{\alpha}$, $p_{1}^{3} p_{2}^{\alpha}$,

-$p_{1}^{4} p_{2}^{\alpha}$, $p_{1}^{5} p_{2}^{\alpha}$, $p_{1}^{\alpha} p_{2}$ are soluble.

-\end{Corollary}

-

-Since the congruence

-\[

-p_{1}^{2} \equiv 1\ (\mod p_{2})

-\]

-is satisfied only by $p_{1} = 2$, $p_{2} = 3$, the results stated follow

-immediately from the theorem, except for the cases $2^{4}3^{\alpha}$ and

-$2^{5}3^{\alpha}$.

-

-If in these cases there are $16$ sub-groups of order~$3^{\alpha}$ ($\alpha > 1$),

-there must (§~78) be sub-groups of order~$3^{\alpha - 1}$ common to two

-sub-groups of order~$3^{\alpha}$; and, in the groups of order $2^{4}3^{\alpha}$~or

-$2^{5}3^{\alpha}$, such a sub-group, if not self-conjugate, must be one of

-either $4$ or~$8$ conjugate sub-groups. The groups must therefore

-be isomorphic with groups of degree $4$ or~$8$; from this it follows

-immediately that they must be soluble (§~146).

-

-\Par{244.} \begin{Theorem}[IV.]

-Groups of order~$p_{1}^{\alpha} p_{2}^{2}$ are soluble.

-\end{Theorem}

-

-If a group~$G$, of order~$p_{1}^{\alpha} p_{2}^{2}$, contains only $p_{2}$~sub-groups of

-order~$p_{1}^{\alpha}$, it cannot be simple, for a group of order~$p_{2}^{2}$ cannot be

-expressed as a substitution-group of $p_{2}$~symbols. Similarly, if $G$~contains

-a single group of order~$p_{1}^{\alpha}$, it is not simple.

-

-If $G$~contains $p_{2}^{2}$~sub-groups of order~$p_{1}^{\alpha}$, and if these sub-groups

-have no common operations, except identity, there are

-$p_{2}^{2}(p_{1}^{\alpha} - 1)$ operations in~$G$ whose orders are powers of~$p_{1}$; and

-therefore a group of order~$p_{2}^{2}$ is self-conjugate. If the operations

-of the $p_{2}^{2}$~sub-groups of order~$p_{1}^{\alpha}$ are not all distinct, let $p_{1}^{r}$ be

-the order of the greatest sub-group common to any two of them;

-%% -----File: 367.png---Folio 349-------

-and let $h$ be such a sub-group. Then (§~80) $h$~must be self-conjugate

-in a group~$k$ of order~$p_{1}^{r + s}p_{2}^{\beta}$ ($s > 0$, $\beta = 1$ or~$2$). If

-$\beta = 2$, $h$~is self-conjugate in~$G$; and if $\beta = 1$, $k$~must contain $p_{2}$~sub-groups

-of order~$p_{1}^{r + s}$, and therefore

-\[

-p_{2} \equiv 1\ (\mod p_{1}).

-\]

-

-Now (§~78)

-\[

-p_{2}^{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),

-\]

-and therefore

-\[

-p_{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),

-\]

-unless $p_{1} = 2$; but, if $p_{1} = 2$, then

-\[

-p_{2} \equiv ± 1\ (\mod p_{1}^{\alpha - r - 1}).

-\]

-

-Again, if  $s = \alpha - r$, $h$~is one of $p_{2}$~conjugate sub-groups; as

-before, $G$~cannot then be simple.

-

-Suppose now that $H$~is a sub-group of order~$p_{1}^{\alpha}$, and that

-the self-conjugate operations of~$H$ form a sub-group~$\Gh$. This

-must be one of $1$,~$p_{2}$ or~$p_{2}^{2}$ conjugate sub-groups; in the first

-two cases, $G$~cannot be simple. Moreover, if any two sub-groups

-of the conjugate set to which $\Gh$~belongs have a common

-sub-group, it must be self-conjugate in a group containing

-more than one sub-group of order~$p_{1}^{\alpha}$; again, $G$~cannot be

-simple.

-

-Let

-\[

-H,\ H_{1},\ \Dots,\ H_{p_{2} - 1}

-\]

-be the $p_{2}$~sub-groups (§~80) of order~$p_{1}^{\alpha}$ that contain~$h$; and

-suppose that $\Gh$~is not contained in~$h$. Every operation of each

-of the $p_{2}$~sub-groups

-\[

-\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1},

-\]

-is permutable with every operation of~$h$. But these sub-groups,

-since they occur in~$k$ and not in~$h$, generate a sub-group of~$k$,

-whose order is divisible by~$p_{2}$. Hence there must be an operation~$Q$,

-of order~$p_{2}$, which is permutable with every operation of~$h$.

-This operation would be permutable with every operation of a

-group of order $p_{1}^{r} p_{2}^{2}$ at least, and it would therefore be one of

-$p_{1}^{\alpha - r}$ conjugate operations at most. Now, if $p_{1} > 2$, $p_{1}^{\alpha - r} < p_{2}$; and

-as a group of order~$p_{2}^{2}$ cannot be represented in terms of less

-than $p_{2}$~symbols, $G$~would in this case have a self-conjugate

-sub-group. If $p_{1} = 2$, then either $p_{1}^{\alpha - r}$ or $p_{1}^{\alpha - r} = 2(p_{2} + 1)$.

-A group of order~$p_{2}^{2}$ cannot be expressed in less than $2p_{2}$

-symbols; a cyclical group of order~$p_{2}^{2}$ cannot be expressed in

-%% -----File: 368.png---Folio 350-------

-$2(p_{2} + 1)$~symbols; and a group of degree~$2(p_{2} + 1)$, which

-contains a non-cyclical sub-group of order~$p_{2}^{2}$, must (§~141)

-contain the alternating group. Hence again, in this case, $G$~must

-have a self-conjugate sub-group.

-

-Finally, suppose that $h$~contains the $p_{2}$~sub-groups

-\[

-\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1}.

-\]

-

-If $h$~contains further sub-groups of the conjugate set to

-which $\Gh$~belongs, they must be permuted among themselves in

-sets of~$p_{2}$ when $h$~is transformed by an operation~$Q$, of order~$p_{2}$,

-belonging to~$k$. Hence we may assume that

-\[

-\begin{array}{*{4}{l}}

-\Gh,&         \Gh_{1},& \Dots,& \Gh_{p_{2} - 1}, \\

-\Gh_{p_{2}},& \Gh_{p_{2} + 1},& \Dots,& \Gh_{2p_{2} - 1}, \\

-\multicolumn{4}{c}{\dotfill} \\

-\Gh_{(x-1)p_{2}},&  \Gh_{(x-1)p_{2} + 1},& \Dots,& \Gh_{xp_{2} - 1},

-\end{array}

-\]

-is the complete set of sub-groups of the conjugate set to which

-$\Gh$~belongs, contained in~$h$; and that, when transformed by~$Q$,

-those in each line are permuted among themselves. If $\{Q, Q'\}$

-is a sub-group\footnote

-  {If this sub-group is cyclical, then $Q = Q'^{p}$.}

-of order~$p_{2}^{2}$, the $p_{2}^{2}$~sub-groups

-\[

-\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2}^{2} - 1},

-\]

-are permuted transitively among themselves, when transformed

-by the operations of~$\{Q, Q'\}$. Hence if $Q'$~change\DPnote{** [sic]} $\Gh$ into $\Gh_{yp_{2}}$, it

-must change the set

-\[

-\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1},

-\]

-into the set

-\[

-\Gh_{yp_{2}},\ \Gh_{yp_{2} + 1},\ \Dots,\ \Gh_{(y + 1)p_{2} - 1};

-\]

-and therefore

-\[

-H_{yp_{2}},\ H_{yp_{2} + 1},\ \Dots,\ H_{(y + 1)p_{2} - 1},

-\]

-is a set of $p_{2}$~sub-groups of order~$p_{2}^{\alpha}$, which have the common

-sub-group $Q'^{-1} hQ'$.

-

-Now, since by supposition $h$~is not a self-conjugate sub-group

-of~$H$, there must (§~55) be in~$H$ a sub-group~$h'$ conjugate

-to and permutable with~$h$. Let this be the sub-group of order~$p_{1}^{r}$,

-common to

-\[

-H,\ H'_{1},\ \Dots,\ H'_{p_{2} - 1}.

-\]

-No one of the sets

-\[

-H_{yp_{2}},\ H_{yp_{2} + 1},\ \Dots,\ H_{(y + 1)p_{2} - 1},\quad (y = 0, 1, \Dots\Add{,} x - 1),

-\]

-%% -----File: 369.png---Folio 351-------

-can have more than one group in common with this set, as

-otherwise $p_{1}^{r}$ would not be the order of the greatest group

-common to two groups of order~$p_{1}^{\alpha}$. Hence, of the set of groups

-\[

-\Gh,\ \Gh_{1}',\ \Dots,\ \Gh_{p_{2} - 1}',

-\]

-at least $p_{2} - x$ are distinct from the groups, conjugate to~$\Gh$,

-which enter in~$h$. Every group of order~$p_{1}^{r + s}$ occurring in~$k$ will

-have a similar set of $p_{2} - x$ sub-groups of the set~$\Gh$, which do

-not occur in~$h$. Hence, since the operations of~$h$ are the only

-operations common to two sub-groups of order~$p_{1}^{r + s}$ in~$k$, there

-must be in~$k$ not less than $p_{2}(p_{2} - x) + p_{2}x$, \ie~$p_{2}^{2}$, sub-groups

-of the set to which $\Gh$~belongs.

-

-In other words, $k$~must contain a self-conjugate sub-group

-of~$G$.

-

-On every supposition that can be made, we have shewn that

-$G$~must have a self-conjugate sub-group. If this sub-group is~$G_{1}$,

-and if the order of either $G_{1}$ or $\dfrac{G}{G_{1}}$ contains~$p_{2}^{2}$, we may

-repeat the same reasoning; while if the orders of $\dfrac{G}{G_{1}}$ and $G_{1}$ are

-both divisible by~$p_{2}$, we have already seen that they must be

-soluble. Finally, then, $G$~itself must be soluble.

-

-\Par{245.}

-\begin{Theorem}[V.]

-A group of order~$p_{1}^{\alpha} p_{2}^{\beta}$, in which the

-sub-groups of orders $p_{1}^{\alpha}$ and $p_{2}^{\beta}$ are Abelian, is soluble.

-\end{Theorem}

-

-Suppose, first, that there are $p_{2}^{\gamma}$ ($\gamma < \beta$) sub-groups of order~$p_{1}^{\alpha}$.

-The group is then isomorphic with a substitution group

-of degree~$p_{2}^{\gamma}$. Now the operations of a sub-group of order~$p_{2}^{\beta}$

-transform the $p_{2}^{\gamma}$~sub-groups of order~$p_{1}^{\alpha}$ transitively among

-themselves; and therefore, in the isomorphism between the

-given group and the substitution group of degree~$p_{2}^{\gamma}$, there

-corresponds to an Abelian sub-group of order~$p_{2}^{\beta}$ a transitive

-substitution group of degree~$p_{2}^{\gamma}$. But (§~124) an Abelian group

-can only be represented as a transitive substitution group of a

-number of symbols equal to its order. Hence the isomorphism

-between the given group and the substitution group is multiple;

-and the given group is not simple.

-

-Suppose, next, that there are $p_{2}^{\beta}$~sub-groups of order~$p_{1}^{\alpha}$, and

-therefore (§~81) $p_{1}^{\alpha} - 1$ distinct sets of conjugate operations

-%% -----File: 370.png---Folio 352-------

-whose orders are powers of~$p_{1}$. Let $P$~be an operation whose

-order is a power of~$p_{1}$; and let $H$, of order~$p_{1}^{\alpha} p_{2}^{\delta}$, be the greatest

-group which contains $P$ self-conjugately. Then $H$~contains at

-least $p_{2}^{\delta}$~operations whose orders are powers of~$p_{2}$, and therefore

-at least $p_{2}^{\delta}$ distinct operations of the form~$PQ$, where $P$~and~$Q$

-are permutable and

-\[

-Q^{p_{2}^{\delta}} = 1.

-\]

-Now $P$~is one of $p_{2}^{\beta - \delta}$ conjugate operations; corresponding

-to each of these, there is a similar set of $p_{2}^{\delta}$~operations of the

-form~$PQ$; while no two such operations can be identical. The

-group therefore contains $p_{2}^{\beta}$~operations of the form~$PQ$; and

-similarly, it contains an equal number for each set of conjugate

-operations whose order is a power of~$p_{1}$. Hence, finally, the

-group contains $p_{2}^{\beta} (p_{1}^{\alpha} - 1)$ operations whose orders are divisible

-by~$p_{1}$. There is therefore in this case a self-conjugate sub-group

-of order~$p_{2}$.

-

-The group therefore always contains a self-conjugate sub-group.

-A repetition of the same reasoning shews that it is

-soluble.

-

-\Par{246.}

-\begin{Theorem}[VI.]

-A group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots\ p_{n}^{\alpha_{n}}$, in

-which the sub-groups of order $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n - 1}^{\alpha_{n - 1}}$ are cyclical, is

-soluble\footnotemark.

-\end{Theorem}

-\index{Burnside@\Topic{Burnside}, quoted}%

-\index{Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical}%

-\footnotetext{\textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~199.}%

-

-If the number of operations of the group, whose orders divide

-$p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$, is exactly equal to this number, it follows

-from Theorem~VII, Cor.~I, §~87, that the same is true for the

-number of operations of the group whose orders divide

-$p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$. Now, when $r = 1$, this relation obviously holds;

-and therefore it is true for all values of~$r$. The group therefore

-contains just $p_{n}^{\alpha_{n}}$ operations whose orders divide~$p_{n}^{\alpha_{n}}$; hence

-a sub-group of order~$p_{n}^{\alpha_{n}}$ is self-conjugate. Any sub-group of

-order~$p_{n - 1}^{\alpha_{n - 1}}$ must transform this self-conjugate sub-group of

-order~$p_{n}^{\alpha_{n}}$ into itself, so that the group must contain a sub-group

-of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$. If there were more than one sub-group

-of this order, the group would contain more than

-$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$ operations whose orders divide this number. Hence

-%% -----File: 371.png---Folio 353-------

-the group contains a single sub-group of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$, and

-this sub-group must be self-conjugate. By continuing this

-reasoning it may be shewn that, for each value of~$r$, the group

-contains a single sub-group of order $p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$, such a

-sub-group being necessarily self-conjugate. The group is therefore

-soluble; for we have already seen that a group whose

-order is the power of a prime is always soluble.

-

-The self-conjugate sub-group of order $p_{r}^{\alpha_{r}} \Dots\ p_{n}^{\alpha_{n}}$ must

-contain the complete set of sub-groups of order~$p_{r}^{\alpha_{r}}$. If

-then there are $m$ of these sub-groups, $m$~must be a factor of

-$p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$; and a sub-group of order~$p_{r}^{\alpha_{r}}$ must be contained

-self-conjugately in a sub-group of order

-\[

-p_{1}^{\alpha_{1}} \Dots p_{r}^{\alpha_{r}}

-  \frac{p_{r + 1}^{\alpha_{r + 1}} \Dots p_{n}^{\alpha_{n}}}{m}.

-\]

-Hence if $i$~and~$j$ are any two indices between $1$ and~$n$, the group

-contains sub-groups of order~$p_{i}^{\alpha_{i}} p_{j}^{\alpha_{j}}$. Now the above sub-group,

-which contains a sub-group of order~$p_{r}^{\alpha_{r}}$ self-conjugately, is of

-the same nature as the original group. Hence, $i$~and~$j$ being

-any two indices less than~$r$, it contains a sub-group of order

-$p_{i}^{\alpha_{i}} p_{j}^{\alpha_{j}} p_{r}^{\alpha_{r}}$. This process may be continued to shew that, if $\mu$~is

-any factor of $p_{1}^{\alpha_{1}} \Dots\ p_{n}^{\alpha_{n}}$ which is relatively prime to

-$\dfrac{p_{1}^{\alpha_{1}} \Dots\ p_{n}^{\alpha_{n}}}{\mu}$, then the group contains sub-groups of order~$\mu$.

-

-\begin{Ex}

-Shew that, when the groups of order~$p_{n}^{\alpha_{n}}$ are also cyclical,

-\index{Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical}%

-there are sub-groups whose orders are any factors whatever of the

-order of the group.

-\end{Ex}

-

-\begin{Remark}

-\Par{247.} A special case of the class of groups under consideration

-is that in which the order contains no repeated prime factor. These

-groups have formed the subject of a memoir by Herr O.~Hölder\footnotemark.

-\index{Holder@\Topic{Hölder}, quoted}%

-\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no squared factor}%

-\footnotetext{``Die Gruppen mit quadratfreier Ordnungszahl'' \textit{Göttingen Nachrichten},

-  1895: pp.~211--229. Compare also Frobenius, \textit{Berliner Sitzungsberichte}, 1895,

-  pp.~1043, 1044.}%

-He shews that they are capable of a specially simple form of representation,

-which we shall now consider.

-

-Let

-\[

-N = p_{1} p_{2} \Dots\ p_{n}

-\]

-be the order of a group~$G$. It has been seen above that a sub-group

-of order~$p_{n}$ is self-conjugate. Suppose then that $p_{l}$, $p_{m}$,~\Dots, $p_{n}$ are

-%% -----File: 372.png---Folio 354-------

-the orders of those cyclical sub-groups of prime order which are

-\index{Defining relations@\Topic{Defining relations} of a group!for groups whose orders contain no square factor}%

-\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no squared factor}%

-contained self-conjugately in~$G$. They generate a self-conjugate

-sub-group~$H$ of order $p_{l} p_{m} \Dots p_{n}$, say~$\mu$. Each of the cyclical

-sub-groups $\{P_{l}\}$, $\{P_{m}\}$,~\Dots, $\{P_{n}\}$ of~$H$, generated by an operation of

-prime order, is permutable with all the rest; and therefore (§~34)

-the operations $P_{l}$, $P_{m}$,~\Dots, $P_{n}$ are all permutable. Hence, since

-their orders are distinct primes, the sub-group~$H$ is cyclical. Let

-now $K$~be a sub-group of~$G$ of order~$\nu$, where $N = \mu\nu$. No self-conjugate

-sub-group of~$G$ can be contained in~$K$. For if it contained

-such a sub-group of order $p_{\alpha} p_{\beta} \Dots p_{\gamma}$ ($\alpha < \beta < \Dots < \gamma$), this sub-group

-would contain a single cyclical sub-group of order~$p_{\gamma}$; and $G$~would

-contain a self-conjugate sub-group of order~$p_{\gamma}$, contrary to

-supposition. Since then $K$~contains no self-conjugate sub-group of~$G$,

-it follows (§~123) that $G$~can be expressed transitively in $\mu$~symbols;

-when it is so expressed, an operation~$M$ of order~$\mu$, that

-generates~$H$, will be a circular substitution of $\mu$~symbols. The

-only substitutions performed on the $\mu$~symbols, which are permutable

-with~$M$, are its own powers; and therefore no operation of~$G$

-is permutable with~$M$ except its own powers. Every operation

-of~$K$ must therefore give a distinct isomorphism of~$\{M\}$. Now

-(§~169) the group of isomorphisms of a cyclical group is Abelian.

-Hence $K$~must be Abelian; and since its order contains no repeated

-prime factor, it must be cyclical. The group~$G$ is therefore completely

-defined by the relations

-\[

-M^{\mu} = 1,\quad N^{\nu} = 1,\quad N^{-1}MN = M^{\alpha},

-\]

-where $\alpha$~belongs to index $\nu, \pmod{\mu}$.

-\end{Remark}

-

-\Par{248.} The theorem of §~246 is a particular case of the

-following more general one, due to Herr Frobenius\footnotemark,

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{``Über auflösbare Gruppen,~II'': \textit{Berliner Sitzungsberichte}, 1895, p.~1035.}%

-from which

-a number of interesting and important results may be deduced.

-

-\begin{Theorem}[VII.]

-If $G$~is a group, of order $N = mn$, where $m$

-and~$n$ are relatively prime, and where

-\[

-m = p^{\alpha} q^{\beta} \Dots r^{\gamma};

-\]

-and if sub-groups $P$, $Q$,~\Dots, $R$, of orders $p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$, are

-Abelian, while $\theta(P)$ and~$\dfrac{N}{p^{\alpha}}$, $\theta(Q)$ and $\dfrac{N}{p^{\alpha} q^{\beta}}$,~\Dots, $\theta(R)$ and

-$\dfrac{N}{m}$, are relatively prime; then $G$~contains \Inum{(i)}~exactly $n$~operations

-whose orders divide~$n$, and \Inum{(ii)}~a sub-group of order~$m$ which

-has a self-conjugate sub-group of order~$r^{\gamma}$.

-\end{Theorem}

-

-The theorem will be proved inductively, by shewing that, if

-%% -----File: 373.png---Folio 355-------

-it is true for any similar group of order~$m'n'$, where $m'$~is a

-factor of~$m$, it is also true for~$G$.

-

-Suppose that the greatest group, which contains $R$ self-conjugately,

-is a group~$H'$ of order~$p^{\alpha'} q^{\beta'} \Dots r^{\gamma}n'$, where $n'$~is a

-factor of~$n$. Let $N'$~be any operation of this sub-group, whose

-order~$\nu'$ is a factor of~$n'$. Then (§~177) since $\theta(R)$ and~$\nu'$ are

-relatively prime, $N'$~is permutable with every operation of~$R$.

-Hence every operation of~$H'$, whose order divides~$n'$, is permutable

-with every operation of~$R$.

-

-Let now $S$~be any operation of~$R$; and suppose that the

-order of~$K$, the greatest sub-group which contains $S$ self-conjugately,

-is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}n_{1}$. If the order of~$S$ is~$r^{\delta}$, the

-order of~$\dfrac{K}{\{S\}}$ is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma - \delta} n_{1}$, say~$m_{1}n_{1}$. We have seen in

-§~176 that, if $P_{1}$~is a sub-group of~$P$, then $\theta(P_{1})$~is equal to or

-is a factor of~$\theta(P)$. Hence if $P_{1}$~is a sub-group of order~$p^{\alpha_{1}}$ of~$\dfrac{K}{\{S\}}$,

-$\theta(P_{1})$~and $\dfrac{N}{p^{\alpha}}$ are relatively prime; therefore, \DPtypo{à~fortiori}{a~fortiori},

-$\theta(P_{1})$ and $\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}}$ are relatively prime. Similarly $\theta(Q_{1})$ and

-$\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}q^{\beta_{1}}}$ are relatively prime, and so on. Moreover $m_{1}$~is a factor

-of~$m$. We may therefore assume that the theorem holds for~$\dfrac{K}{\{S\}}$.

-Hence this group contains exactly $n_{1}$ operations whose

-orders divide~$n_{1}$, and it also contains a sub-group of order~$m_{1}$

-in which a sub-group of order~$r^{\gamma - \delta}$ is self-conjugate. Therefore

-$K$~contains a sub-group of order $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}$ in which a

-sub-group of order~$r^{\gamma}$ is self-conjugate, and exactly $n_{1}$ operations

-of the form~$SN$, where $S$~and~$N$ are permutable and the order

-of~$N$ divides~$n$.

-

-Now $S$~is one of~$p^{\alpha - \alpha_{1}}q^{\beta - \beta_{1}} \Dots \dfrac{n}{n_{1}}$ conjugate operations in~$G$;

-corresponding to each of these, there is such a set of $n_{1}$

-operations of the form~$SN$, while (§~16) no two of these

-operations can be identical. Hence $G$~contains $p^{\alpha - \alpha_{1}} q^{\beta - \beta_{1}} \Dots n$

-operations of the form~$SN$, arising from the conjugate set to

-which $S$~belongs. If then we sum for the distinct conjugate

-sets of operations of~$G$ whose orders are powers of~$r$, the

-%% -----File: 374.png---Folio 356-------

-number of operations of~$G$, whose orders divide~$r^{\gamma}n$ and do not

-divide~$n$, is $n\sum p^{\alpha - \alpha_{1}}q^{\beta - \beta_{1}}\Dots$.

-

-Now since $H'$~is the greatest group that contains $R$ self-conjugately,

-the greatest common sub-group of $K$~and~$H'$ is

-the greatest sub-group of~$K$ that contains $R$ self-conjugately.

-The order of this group certainly contains $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}$ as

-a factor; and if the order is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}\nu$, then $S$~is one of

-$p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots \dfrac{n'}{\nu}$ conjugate operations in~$H'$. Every operation

-of~$H'$ however, whose order divides~$n'$, is permutable with every

-operation of~$R$; and therefore $\nu = n'$. Hence the number of

-operations of~$H'$, excluding identity, whose orders are powers of~$r$,

-is $\sum' p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots$, where the summation is extended to all the

-distinct conjugate sets of operations of~$H'$ whose orders are

-powers of~$r$. It has been shewn (Theorem~IV, Corollary, §~81)

-that the number of distinct conjugate sets of operations in~$G$,

-whose orders are powers of~$r$, is the same as the number in~$H'$.

-Hence the symbols $\sum$ and $\sum'$ represent summations of the same

-number of terms.

-

-Finally, in accordance with the induction we are using, we

-may assume that the number of operations of~$G$ whose orders

-divide~$r^{\gamma}n$ is exactly equal to~$r^{\gamma}n$. For the order of~$G$ may be

-separated into the factors

-\[

-m' = p^{\alpha} q^{\beta} \Dots,

-\]

-and

-\[

-n' = r^{\gamma}n,

-\]

-where $m'$~is a factor of~$m$, while the conditions of the theorem

-hold for this separation.

-

-Hence the number of operations of~$G$ whose orders divide~$n$,

-is equal to

-\[

-n\bigl(r^{\gamma} - \sum p^{\alpha - \alpha_{1}} q^{\beta - \beta_{1}} \Dots\bigr),

-\]

-while at the same time

-\[

-r^{\gamma} - 1 = \sum p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots.

-\]

-Unless $\alpha' = \alpha$, $\beta' = \beta$,~\Dots, the former of these numbers is

-negative, which is impossible; these conditions must therefore

-be satisfied. The number of operations, whose orders

-divide~$n$, is therefore exactly equal to~$n$; and the order of

-the greatest group~$H'$ that contains $R$ self-conjugately is

-%% -----File: 375.png---Folio 357-------

-$p^{\alpha} q^{\beta} \Dots r^{\gamma}n'$. The factor-group $\dfrac{H'}{R}$ has for its order $p^{\alpha} q^{\beta} \Dots n'$,

-and it satisfies the same conditions as~$G$. Hence it contains a

-sub-group of order~$p^{\alpha} q^{\beta} \Dots$; and $H'$~therefore contains a sub-group

-of order~$p^{\alpha}q^{\beta} \Dots r^{\gamma}$.

-

-\begin{Remark}

-The free use of the inductive process that has been made in the

-preceding proof may possibly lead the reader to doubt its validity.

-He will find it instructive to verify the truth of the theorem

-directly in the simpler cases. In fact, the direct verification for

-the case in which $m$~is a power of a prime is essential to the formal

-proof; it has been omitted for the sake of brevity.

-\end{Remark}

-

-\begin{Corollary}[I.]

-If $m = m_{1}m_{2}$, where $m_{1}$~and~$m_{2}$ are relatively

-prime, $G$~has a single conjugate set of sub-groups of order~$m_{1}$.

-\end{Corollary}

-

-That $G$~contains groups of order~$m_{1}$ may be shewn inductively

-at once. For since it is clearly true when $m$~contains

-only one, or two, distinct prime factors, we may assume it true

-when $m_{1}$~does not contain $r^{\gamma}$ as a factor. Now $G$~has a sub-group~$g$

-of order~$p^{\alpha} q^{\beta} \Dots r^{\gamma}$, which contains a sub-group~$R$ of

-order~$r^{\gamma}$ self-conjugately. The factor-group $\dfrac{g}{R}$ contains then a

-sub-group of order~$m_{1}'$ where $m_{1}'$~is any factor of $\dfrac{m}{r^{\gamma}}$ which is

-relatively prime to $\dfrac{m}{m_{1}'r^{\gamma}}$; and therefore $g$~contains a sub-group

-of order~$m_{1}'r^{\gamma}$.

-

-If now $I$~and~$I'$ are two sub-groups of~$G$ of order~$m_{1}'r^{\gamma}$,

-they may be assumed to contain the same sub-group~$R$ of

-order~$r^{\gamma}$. For the sub-groups of~$G$ of order~$r^{\gamma}$ form a single

-conjugate set; and therefore, if $I'$~contained a sub-group~$R'$ of

-this set, $S^{-1}I'S$~would contain $S^{-1}R'S$, which may be taken

-to be~$R$. Now $R$~is a self-conjugate sub-group of both

-$I$~and~$I'$, and therefore the factor-groups $\dfrac{I}{R}$ and $\dfrac{I'}{R}$ of order~$m_{1}'$

-are sub-groups of the factor-group~$\dfrac{H'}{R}$. Hence if the result is

-true when $m_{1}$~does not contain the factor~$r^{\gamma}$, it is also true

-when $m_{1}$~does contain~$r^{\gamma}$. But when $m_{1}$~is equal to~$p^{\alpha}$, the

-result is obviously true; and therefore it is true generally.

-%% -----File: 376.png---Folio 358-------

-

-\begin{Corollary}[II.]

-If the sub-groups $P$, $Q$,~\Dots, $R$ of~$G$

-contain characteristic sub-groups $P_{0}$, $Q_{0}$,~\Dots, $R_{0}$ of orders

-$p^{\alpha_{0}}$, $q^{\beta_{0}}$,~\Dots, $r^{\gamma_{0}}$; then $G$~contains a sub-group of order

-$m_{0} = p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma_{0}}$, which has a self-conjugate sub-group of

-order~$r^{\gamma_{0}}$.

-\end{Corollary}

-

-Assuming the truth of this statement when $m_{0}$~does not

-contain the factor~$r$, the factor-group $\dfrac{H'}{R}$ must contain a sub-group

-of order~$\dfrac{m_{0}}{r^{\gamma_{0}}}$; and therefore $G$~contains a sub-group~$J$ of

-order $p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma}$ which has $R$ for a self-conjugate sub-group.

-Hence, since $R_{0}$~is a characteristic sub-group of~$R$, $J$~contains a

-sub-group of order~$p^{\alpha_{0}} q^{\beta_{0}} \Dots$ and a self-conjugate sub-group

-of order~$r^{\gamma_{0}}$. It therefore contains a sub-group of order

-$p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma_{0}}$, which has a self-conjugate sub-group of order~$r^{\gamma_{0}}$.

-

-\Par{249.} The reader will have no difficulty in seeing, as has

-already been stated, that Theorem~VI is a direct result of the

-theorem proved in the last paragraph. We shall proceed at

-once to further applications of it.

-

-\begin{Theorem}[VIII.]

-A group~$G$ of order

-\[

-N = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}},

-\]

-in which the sub-groups of every order~$p_{r}^{\alpha_{r}}$ \($r = 1, 2, \Dots, n - 1$\)

-are Abelian, with either one or two generating operations, is

-generally soluble; the special case $p_{1} = 2$, $p_{2} = 3$, may constitute

-an exception\footnotemark.

-\end{Theorem}

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, \textit{loc.~cit.}, p.~1041.}%

-

-If an Abelian sub-group~$P_{r}$ of order~$p_{r}^{\alpha_{r}}$ is generated by a

-single operation, it is cyclical and $\theta(P_{r})$~is (§~176) equal to $p_{r} - 1$.

-If $P_{r}$~is generated by two independent and permutable operations,

-$\theta(P_{r})$~is equal to~$(p_{r} - 1)(p_{r}^{2} - 1)$. Now no prime greater

-than~$p_{r}$ can divide $(p_{r} - 1)(p_{r}^{2} - 1)$ unless $p_{r} + 1$ be a prime;

-and this is only possible when $p_{r}$~is equal to~$2$. Hence unless

-$P_{1} = 2$, $p_{2} = 3$, $\theta(P_{r})$~and $\dfrac{N}{p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{r}^{\alpha_{r}}}$ are relatively prime for

-all values of~$r$ from $1$ to~$n - 1$. Omitting for the present this

-exceptional case, the conditions of Theorem~VII are satisfied

-by~$G$; it therefore contains exactly $p_{n}^{a_{n}}$ operations whose

-%% -----File: 377.png---Folio 359-------

-orders are divisible by~$p_{n}$. Therefore $G$~has a self-conjugate

-sub-group~$P_{n}$ of order~$p_{n}^{\alpha_{n}}$, and hence also one or more sub-groups

-of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$. There are however only $p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$

-operations in~$G$ whose orders are divisible by no prime smaller

-than~$p_{n - 1}$, and therefore the sub-group of order~$p_{n - 1} ^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$ must

-be self-conjugate. This process clearly may be continued to

-shew that $G$~has a self-conjugate sub-group of every order

-$p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots p_{n}^{\alpha_{n}}$ ($r = 2, 3, \Dots, n$); from which it follows immediately

-that $G$~is soluble.

-

-Returning now to the exceptional case, suppose that a group

-of order~$2^{\alpha_{1}}$ is contained self-conjugately in a maximum group

-of order~$2^{\alpha_{1}} 3^{\beta} m$. If every operation of the group of order~$2^{\alpha_{1}}$ is

-self-conjugate within this sub-group, $G$~contains $2^{\alpha_{1}} - 1$ distinct

-conjugate sets of operations whose orders are powers of~$2$;

-hence it follows that there are just $\dfrac{N}{2^{\alpha_{1}}}$ operations in~$G$ whose

-orders are not divisible by~$2$. In this case, the conditions of

-Theorem~VII are satisfied by~$G$; and it is still soluble.

-

-If, lastly, the operations of the sub-groups of order~$2^{\alpha_{1}} 3^{\beta} m$,

-whose orders are powers of~$2$, are not all self-conjugate, there

-must be an operation~$B$, whose order is a power of~$3$, in this

-sub-group which is not permutable with every operation of

-the sub-group of order~$2^{\alpha_{1}}$. Let now

-\[

-A,\ A_{1},\ \Dots,\ A_{r},\ A_{r + 1},\ \dots

-\]

-be a characteristic series of the group~$A$ of order~$2^{\alpha_{1}}$; and

-suppose that $A_{r + 1}$~is the greatest of these groups with every

-one of whose operations~$B$ is permutable. Then (§~175) $B$~is

-not permutable with every operation of~$\dfrac{A_{r}}{A_{r + 1}}$. Hence $\dfrac{A_{r}}{A_{r + 1}}$~is

-a quadratic group; and its three operations of order~$2$ must

-be permuted cyclically when transformed by~$B$. It follows

-that $\{A_{r}, B\}$ is multiply isomorphic with a tetrahedral group.

-Hence finally, under the conditions of the theorem, $G$~is

-certainly soluble unless it contains a sub-group which is

-isomorphic with a tetrahedral group.

-

-\begin{Corollary}

-A group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \dots p_{n}^{\alpha_{n}}$ in which no

-one of the indices $\alpha_{1}$, $\alpha_{2}$,~\dots, $\alpha_{n - 1}$ is greater than~$2$, is soluble

-\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no cubed factor}%

-\index{Soluble groups@\Topic{Soluble groups}!special classes of|)}%

-%% -----File: 378.png---Folio 360-------

-unless it has a sub-group which is isomorphic with a tetrahedral

-group. For a group of order~$p_{r}^{2}$ is necessarily an Abelian group

-which is generated by either one or two independent operations.

-\end{Corollary}

-

-\Par{250.} We shall next consider certain groups of even order

-\index{Composite groups@\Topic{Composite groups}!of even order|(}%

-in which the operations of odd order form a self-conjugate sub-group.

-If $G$~is a group of order~$N$, where

-\[

-N = 2^{\alpha}n,\quad \text{($n$ odd),}

-\]

-we shall suppose, in this and the following paragraphs, that a

-sub-group~$\scrQ$ of order~$2^{\alpha}$ is Abelian, and that $\theta(\scrQ)$ and $n$ are

-relatively prime. Let us take first the case in which $\scrQ$~is

-cyclical, so that $\theta(\scrQ)$~is unity and the latter condition is satisfied

-for all values of~$n$. If such a group is represented in regular

-form as a substitution group of $2^{\alpha}n$~letters, the substitution

-corresponding to an operation of order~$2^{\alpha}$ will consist of $n$~cycles

-of $2^{\alpha}$~symbols each. This is an odd substitution; and therefore

-$G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. In this sub-group

-there are cyclical operations of order~$2^{\alpha - 1}$. Hence the

-same reasoning will apply to it, and it contains a self-conjugate

-sub-group of order~$2^{\alpha - 2}n$. Now (Theorem~VII, Cor.~I. §~87) $G$~contains

-exactly $2^{\alpha - 2}n$ operations whose orders are not divisible

-by~$2^{\alpha - 1}$. Hence the sub-group of order~$2^{\alpha - 2}n$ must be self-conjugate

-in~$G$. Proceeding thus, it may be shewn that $G$~contains

-self-conjugate (and characteristic) sub-groups of every

-order~$2^{\alpha - r}n$ ($r = 1, 2, \Dots, \alpha$)\footnotemark.

-\index{Frobenius@\Topic{Frobenius}, quoted}%

-\footnotetext{Frobenius, ``Über auflösbare \DPtypo{Grüppen}{Gruppen},~II'' \textit{Berliner Sitzungsberichte},

-  1895, p.~1039.}%

-

-\Par{251.} Suppose, secondly, that, under the same conditions, $G$~contains

-a self-conjugate sub-group~$\scrQ$ of order~$2^{\alpha}$. Since $\theta(\scrQ)$

-and $n$ are relatively prime, every operation of~$\scrQ$ is permutable

-with every operation of~$G$ (§~177). If $\scrQ'$~is a sub-group of~$\scrQ$

-of order~$2^{\alpha - 1}$, $\scrQ'$~must therefore be self-conjugate in~$G$. Now

-the order of the factor group $\dfrac{G}{\scrQ'}$ is~$2n$; hence, by the preceding

-result, it has a self-conjugate sub-group of order~$n$. It follows

-that $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. This

-group, again, has a self-conjugate sub-group of order~$2^{\alpha - 2}n$,

-and so on. Hence $G$~has a sub-group of order~$n$. Now by

-Theorem~VII (§~248), $G$~has just $n$~operations whose

-%% -----File: 379.png---Folio 361-------

-divide~$n$. Hence the sub-group~$H$, of order~$n$, must be self-conjugate;

-and $G$~is the direct product of the two groups $\scrQ$

-and~$H$.

-

-\Par{252.} Suppose next that $\scrQ$~is not self-conjugate in~$G$; and

-let~$I$, of order~$2^{\alpha}n'$, be the greatest group that contains $\scrQ$ self-conjugately.

-Every operation of~$I$ is permutable with every

-operation of~$\scrQ$; and therefore (§~81) $G$~contains $2^{\alpha - 1}$ distinct

-sets of conjugate operations whose orders are powers of~$2$.

-The case, in which $\scrQ$~is cyclical, has been already dealt with

-and will now be excluded; we may thus assume that $\scrQ$~contains

-$2^{\beta} - 1$ ($\beta \nless 2$) operations of order~$2$, and that $G$~contains an equal

-number of sets of conjugate operations of order~$2$.

-

-If possible, let no two sub-groups of order~$2^{\alpha}$ have a common

-operation except identity. Then if two operations $S$ and~$T$,

-of order~$2$, are chosen, belonging to distinct conjugate sets and

-to different sub-groups of order~$2^{\alpha}$, they cannot be permutable

-with each other. They will therefore generate a dihedral sub-group

-of order~$2m$. If $m$~were odd, $S$~and~$T$ would be conjugate

-operations. Hence $m$~must be even; and the dihedral sub-group

-must contain a self-conjugate operation~$U$ of order~$2$. Since

-$U$~is permutable with both $S$~and~$T$, it must occur in at least

-two different sub-groups of order~$2^{\alpha}$; and therefore the supposition,

-that no two sub-groups of order~$2^{\alpha}$ have a common

-operation except identity, is impossible.

-

-Let now $\scrQ'$ of order~$2^{\alpha'}$ be a sub-group common to $\scrQ$ and~$\scrQ_{1}$,

-two sub-groups of order~$2^{\alpha}$; and suppose that no two sub-groups

-of order~$2^{\alpha}$ have a common sub-group which contains~$\scrQ'$ and is

-of greater order. Since every operation of~$\scrQ$ is self-conjugate

-in~$I$, $\scrQ'$~must be self-conjugate in~$I$. If then $J$, of order~$2^{\alpha}n'r$,

-is the greatest sub-group of~$G$ that contains $\scrQ'$ self-conjugately,

-$J$~contains $r$~sub-groups of order~$2^{\alpha}$. Hence the factor-group

-$\dfrac{J}{\scrQ'}$ contains $r$~sub-groups of order~$2^{\alpha - \alpha'}$, and no two of these

-have a common sub-group; for if they had, some two sub-groups

-of order~$2^{\alpha}$ contained in~$J$ would have a common sub-group

-greater than and containing~$\scrQ'$, contrary to supposition. The

-$r$~sub-groups of order~$2^{\alpha - \alpha'}$ of~$\dfrac{J}{\scrQ'}$ must therefore be cyclical.

-%% -----File: 380.png---Folio 362-------

-

-\Par{253.} We will apply the results of the last paragraph to

-the case in which all the operations of~$\scrQ$ are of order~$2$. Then

-$\scrQ'$~must be of order~$2^{\alpha - 1}$, since $\dfrac{\scrQ}{\scrQ'}$ is cyclical. Suppose now that

-$R$~is an operation of~$J$ of odd order, which is permutable with~$\scrQ'$

-but not with~$\scrQ$. If $R$~were self-conjugate in a sub-group

-whose order is divisible by~$2^{\alpha}$, this sub-group would contain~$\scrQ'$

-and therefore one or more sub-groups of order~$2^{\alpha}$ containing~$\scrQ'$.

-But $R$~is not permutable with any sub-group of order~$2^{\alpha}$

-that contains~$\scrQ'$; and therefore the highest power of~$2$ that

-divides the order of the group in which $R$ is self-conjugate is~$2^{\alpha - 1}$,

-so that $R$~is one of~$2\mu$ ($\mu$~odd) conjugate operations. If

-now $A$~is an operation of order~$2$ that belongs to~$\scrQ$ and not to~$\scrQ'$,

-no operation conjugate to~$R$ can be permutable with~$A$.

-Hence in the substitution group of degree~$2\mu$, that results on

-transforming the set of operations conjugate to~$R$ among themselves

-by all the operations of~$G$, the substitution corresponding

-to~$A$ is an odd substitution. This substitution group has

-therefore a self-conjugate sub-group whose order is half its own,

-and therefore $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$.

-In the same way it may be shewn that this sub-group has a

-self-conjugate sub-group of order~$2^{\alpha - 2}n$; and so on. Hence $G$~has

-a sub-group $H$ of order~$n$. But it follows from Theorem~VII,

-§~248, that $G$~has exactly $n$ operations whose orders divide~$n$;

-and therefore $H$~is a self-conjugate sub-group. Moreover,

-since $\dfrac{G}{H}$~is an Abelian group of order~$2^{\alpha}$, it must contain self-conjugate

-sub-groups of every order~$2^{r}$ ($r = 1, 2, \Dots, \alpha - 1$);

-and $G$~therefore has self-conjugate sub-groups of every order~$2^{r}n$.

-In general however these sub-groups are not characteristic,

-as is the case when $\scrQ$~is cyclical.

-

-\Par{254.} We will consider next the case where $\scrQ$~is generated

-by two operations $A$~and~$B$, of orders $2^{\alpha - 1}$ and~$2$. There are in~$\scrQ$

-three operations of order~$2$, namely, $A^{2^{\alpha - 2}}$,~$A^{2^{\alpha - 2}}B$, and~$B$; and

-in~$\scrQ$ there are no operations, of order greater than~$2$, of which

-the two latter are powers. Suppose that $G$~contains an operation~$B'$,

-conjugate to~$B$, with which $A^{2^{\alpha - 2}}$~is not permutable.

-Then $\{A^{2^{\alpha - 2}}, B'\}$ is a dihedral group, and if $R$~is an operation

-%% -----File: 381.png---Folio 363-------

-of odd order of this group, no power of~$A$ is permutable with~$R$.

-Since $A^{2^{\alpha - 2}}$ and $B'$ are not conjugate operations in~$G$, there

-must be an operation of order~$2$, conjugate to~$A^{2^{\alpha - 2}}B$, and

-permutable with~$R$. The sub-group, in which $\{R\}$ is self-conjugate,

-therefore contains representatives of each of the

-three sets of conjugate operations of order~$2$ that belong to~$G$;

-and in this sub-group these representatives form three distinct

-conjugate sets. Hence no operation conjugate to~$A^{2^{\alpha - 2}}$ can be

-permutable with~$R$; and therefore no power, except identity,

-of an operation conjugate to~$A$ is permutable with~$R$. Hence

-$R$~is one of~$2^{\alpha - 1}\mu$ ($\mu$~odd) conjugate operations, with none of

-which is $A$ permutable. It follows that, in the substitution

-group of degree~$2^{\alpha - 1}\mu$, with which $G$~is isomorphic, $A$~is an odd

-substitution. Hence $G$~contains a self-conjugate sub-group of

-order~$2^{\alpha - 1}n$; and since this is of the same type as~$G$, the

-same reasoning may be applied to it. Exactly as before, $G$~will

-contain self-conjugate sub-groups of every order~$2^{r}n$

-($r = 0, 1, \Dots, \alpha - 1$).

-

-We have assumed that there is an operation~$B'$, conjugate

-to~$B$, with which $A^{2^{\alpha - 2}}$ is not permutable. If $A^{2^{\alpha - 2}}$~were permutable

-with every operation that is conjugate to~$B$, the

-sub-group in which $A^{2^{\alpha - 2}}$ is permutable would contain a self-conjugate

-sub-group~$G'$ of~$G$ whose order is divisible by~$2$. In

-this case, we may deal with~$\dfrac{G}{G'}$ exactly as we have been

-dealing with~$G$.

-

-The results of §§~250--254 may be summed up in the

-following form:---

-

-\begin{Theorem}[IX.]

-If in a group~$G$ of order~$2^{\alpha}n$, where $n$~is

-odd, a sub-group~$\scrQ$ of order~$2^{\alpha}$ is an Abelian group of type

-$(\alpha)$, $(\alpha - 1, 1)$, or $(1, 1, \Dots, 1)$, and if $\theta(\scrQ)$ and $n$ are relatively

-prime, then $G$~contains self-conjugate sub-groups of each order~$2^{\beta}n$

-\($\beta = 0, 1, \Dots, \alpha - 1$\).

-\end{Theorem}

-

-\Par{255.} Still representing the order of~$G$ by~$2^{\alpha}n$, where $n$~is

-odd, there are two cases, in which the sub-groups of order~$2^{\alpha}$

-are not Abelian, where it may be shewn without difficulty that

-$G$~contains a self-conjugate sub-group of order~$n$.

-%% -----File: 382.png---Folio 364-------

-

-\begin{Theorem}[X.]

-If the order of~$G$ is~$2^{\alpha}n$, where $n$~is odd and

-not divisible by~$3$, and if a sub-group~$\scrQ$ of order~$2^{\alpha}$ is of the type

-\begin{alignat*}{4}

-&\Inum{(i)}  & A^{2^{\alpha - 1}} &= 1, & B^{2} &= A^{2^{\alpha - 2}},\quad & B^{-1}AB &= A^{-1}, \\

-\lintertext{or}

-&\Inum{(ii)}\quad & A^{2^{\alpha - 1}} &= 1,\quad & B^{2} &= 1, & B^{-1}AB &= A^{-1},

-\end{alignat*}

-{\Loosen $G$~contains a self-conjugate sub-group of each order~$2^{\beta}n$

-\($\beta = 0, 1, \dots, \alpha - 1$\).}

-\end{Theorem}

-

-For each of these types, $\vartheta(\scrQ)$~is $3$~or~$1$; and if $\scrQ'$~is any

-sub-group of~$\scrQ$, $\vartheta(\scrQ')$~is either $3$~or~$1$. Hence if any operation

-of~$G$ of odd order is permutable with $\scrQ$~or~$\scrQ'$, it must be

-permutable with every operation of $\scrQ$~or~$\scrQ'$. In each type,

-the self-conjugate operations form a cyclical sub-group of order~$2$,

-namely $\{A^{2^{\alpha - 2}}\}$.

-

-We will consider first the case where $\scrQ$~is of type~\Inum{(i)}. In

-this case, $\{A^{2{\alpha - 3}}\}$~and $\{B\}$ are sub-groups of~$G$ of order~$4$, the

-former being self-conjugate in~$\scrQ$ while the latter is not. If

-they are conjugate sub-groups in~$G$, we have seen (§~82)

-that $G$~must contain an operation of odd order~$S$, such that

-the sub-groups $S^{-n}\{B\}S^{n}$ ($n = 0, 1, 2, \dots$) are permutable with

-each other. This is impossible, since every operation that

-is permutable with a sub-group~$\scrQ'$ is permutable with all its

-operations. Hence $\{A^{2^{\alpha - 3}}\}$~and $\{B\}$ are not conjugate sub-groups;

-and $A^{2^{\alpha - 3}}$~and~$B$ are not conjugate operations.

-

-Now $A^{2^{\alpha - 3}}$ is self-conjugate in~$\{A\}$ and is one of two conjugate

-operations in~$\scrQ$; therefore in~$G$ it must be one of $2\mu$ conjugate

-operations, where $\mu$~is odd. When these $2\mu$ conjugate

-operations are transformed by~$A^{2^{\alpha - 3}}$, two only, namely, $A^{2^{\alpha - 3}}$ and

-$A^{-2^{\alpha - 3}}$, remain unchanged. Let us suppose that, in the resulting

-substitution, there are $x$~cycles of $2$~symbols and $y$~cycles of $4$~symbols.

-

-When the $2\mu$ conjugate operations are transformed by~$B$,

-none can remain unchanged; and we may suppose that the

-resulting substitution has $x'$~cycles of $2$~symbols and $y'$~cycles of

-$4$~symbols.

-

-Now since

-\begin{gather*}

-A^{2^{\alpha - 2}} = B^{2}, \\

-y = y',

-\end{gather*}

-and therefore

-\[

-1 + x = x'.

-\]

-%% -----File: 383.png---Folio 365-------

-

-Hence one of the two substitutions $A^{2^{\alpha - 3}}$ and $B$ must be odd;

-$G$~has therefore a self-conjugate sub-group of order~$2^{\alpha - 1}n$. The

-groups of order~$2^{\alpha - 1}$ contained in this self-conjugate sub-group

-are either of type~\Inum{(i)}, with $\alpha - 1$ written for~$\alpha$: or they are

-cyclical; for, like~$\scrQ$, they can only contain a single operation of

-order~$2$. If they are cyclical, the reasoning of §~250 may

-be applied; and if they are of type~\Inum{(i)} the same reasoning

-will apply to the self-conjugate sub-group of order~$2^{\alpha - 1}n$

-that has been used for~$G$. Finally, then, $G$~must contain

-sub-groups of orders $n$,~$2n$, $2^{2}n$,~\dots, $2^{\alpha - 1}n$, each of which is

-contained self-conjugately in the next; and therefore the sub-group~$H$

-of order~$n$ must be self-conjugate.

-

-If $\scrQ$~is of type~\Inum{(ii)}, $A$~is one of $2\mu$ conjugate operations

-in~$G$, where $\mu$~is odd. It may be shewn, as in the previous

-case, that $B$~and $A^{2^{\alpha - 2}}$ cannot be conjugate in~$G$, so that no one

-of the operations conjugate to $A$~has~$B$ for one of its powers.

-If now $B$~were permutable with any one of these $2\mu$~conjugate

-operations, the group would contain an Abelian sub-group of

-order~$2^{\alpha}$, which is not the fact. Hence the substitution, given

-on transforming the $2\mu$~operations by~$B$, consists of $\mu$~transpositions

-and is an odd substitution. Therefore $G$~contains a self-conjugate

-sub-group of order~$2^{\alpha - 1}n$, in which the sub-groups of

-order~$2^{\alpha - 1}$ are cyclical. Hence, again, there is a self-conjugate

-sub-group of order~$n$.

-

-\Par{256.} The only non-Abelian groups of order~$2^{3}$ are those of

-types \Inum{(i)}~and~\Inum{(ii)} of §~255, when $\alpha = 3$. The Abelian groups

-of order $2^{2}$~and~$2^{3}$ are all included in the types considered in

-Theorem~IX, §~254. For an Abelian group~$\scrQ$, of order~$2^{3}$ and

-type~$(2, 1)$, $\theta(\scrQ)$~is~$3$; and for one of order~$2^{3}$ and type $(1, 1, 1)$,

-$\theta(\scrQ)$~is~$21$. Hence Theorems IX~and~X shew that, if $2$,~$2^{2}$

-or~$2^{3}$ divide the order of a group, but not~$2^{4}$, then the

-operations of odd order form a self-conjugate sub-group, with

-possible exceptions when $12$~or $56$ is a factor of the order.

-Hence:---

-

-\begin{Theorem}[XI.]

-A group of even order cannot be simple

-unless $12$,~$16$ or~$56$ is a factor of the order\footnotemark.

-\end{Theorem}

-\index{Burnside@\Topic{Burnside}, quoted}%

-\footnotetext{Burnside, \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~332.}%

-%% -----File: 384.png---Folio 366-------

-\index{Composite groups@\Topic{Composite groups}!of even order|)}%

-

-\begin{Remark}

-\Par{257.} In further illustration of the methods of the preceding

-paragraphs, we will deal with another case in which the sub-groups

-of order~$p^{\alpha}$ are not Abelian.

-

-{\Loosen Let $G$~be a group of order~$p^{\alpha}m$, where $m$~is relatively prime to

-$p(p - 1) \Dots (p^{\alpha} - 1)$; and suppose that a sub-group~$H$ of order~$p^{\alpha}$ is

-such that within it every operation is either self-conjugate or one

-of $p$ conjugate operations\footnotemark.}

-\PageLabel{366}%

-\footnotetext{The reader will easily verify that, when this condition is satisfied, $\dfrac{H}{\Gh}$

-  is Abelian, $\Gh$~being the sub-group formed by the self-conjugate operations of~$H$.}%

-Let $h$ be a sub-group of~$G$ whose order

-is a power of~$p$; and let $S$ be an operation of~$G$ whose order is not

-divisible by~$p$. Then since $m$~is relatively prime to $p(p - 1) \Dots

-(p^{\alpha} - 1)$, if $S$~is permutable with~$h$, it is permutable with every

-operation of~$h$. Hence it follows from §~82 that no operation of~$H$,

-which is not self-conjugate in~$H$, can be conjugate in~$G$ to

-a self-conjugate operation of~$H$. Suppose now that $S_{1}$~and $S_{2}$ are

-two operations of~$H$, each of which in~$H$ is one of $p$ conjugate

-operations; and that while $S_{1}$~and $S_{2}$ are conjugate in~$G$, they

-are not conjugate in~$I$, the greatest group that contains $H$ self-conjugately.

-Let $H_{1}$~and $H_{2}$ be the sub-groups of~$H$, of orders\DPnote{** [sic]}~$p^{\alpha - 1}$,

-which contain $S_{1}$ and $S_{2}$ self-conjugately. Since $S_{1}$ and $S_{2}$

-are conjugate in~$G$, all the sub-groups of order~$p^{\alpha - 1}$ which contain

-either of them self-conjugately belong to the same conjugate

-set. If then $H_{1}$ and $H_{2}$ are identical, there must be an operation,

-of order prime to~$p$, which will transform $S_{1}$ into~$S_{2}$ and $H_{1}$

-into itself. This, we have seen, is impossible. If $H_{1}$~and $H_{2}$ are

-not identical, there must be an operation which will transform $S_{1}$

-into $S_{2}$ and $H_{1}$~into $H_{2}$. Now $H_{1}$ and $H_{2}$ both contain the sub-group~$\Gh$

-formed of the self-conjugate operations of~$H$; and since no self-conjugate

-operation of~$H$ is conjugate in~$G$ with any operation of~$H$

-which is not self-conjugate, the operation in question, which

-transforms $S_{1}$ into $S_{2}$ and $H_{1}$~into $H_{2}$, must transform $\Gh$ into itself.

-Hence $S_{1}$ and $S_{2}$ are conjugate operations in that sub-group,~$G'$, of~$G$

-which contains $\Gh$ self-conjugately. Now in $\dfrac{G'}{\Gh}$ the operations, that

-correspond to $S_{1}$~and~$S_{2}$, are self-conjugate operations of~$\dfrac{H}{\Gh}$. Since

-then these operations are conjugate in~$\dfrac{G'}{\Gh}$, they must (§~82) be

-conjugate in~$\dfrac{I}{\Gh}$. This however is impossible, since every operation

-of~$\dfrac{I}{\Gh}$ whose order is a power of~$p$ is self-conjugate. Finally, then, no

-two operations of~$H$, which are not conjugate in~$H$, can be conjugate

-in~$G$; and the number of distinct sets of conjugate operations in~$G$,

-whose orders are powers of~$p$, is equal to the number in~$H$. From this

-%% -----File: 385.png---Folio 367-------

-it follows, as in previous cases, that the number of operations of~$G$

-whose orders divide~$m$ is equal to~$m$.

-\end{Remark}

-

-\Par{258.} \begin{Theorem}[XII.]

-The only simple groups, whose orders

-are the products of four or of five primes, are groups of orders

-$60$,~$168$, $660$, and~$1092$: and no group, whose order contains

-less than four prime factors, is simple.

-\end{Theorem}

-

-Groups, whose orders are $p_{1}$,~$p_{1} p_{2}$, $p_{1}^{2}$, $p_{1}^{3}$, $p_{1}^{2} p_{2}$, $p_{1} p_{2}^{2}$ or~$p_{1} p_{2} p_{3}$

-are all proved to be soluble by previous theorems in the present

-chapter.

-

-If the order of a group contains four prime factors, it must

-be of one of the forms $p_{1}^{4}$,~$p_{1}^{3} p_{2}$, $p_{1}^{2} p_{2}^{2}$, $p_{1}^{2} p_{2} p_{3}$, $p_{1} p_{2}^{3}$, $p_{1} p_{2}^{2} p_{3}$,

-$p_{1} p_{2} p_{3}^{2}$ or~$p_{1} p_{2} p_{3} p_{4}$. If $p_{1}$~is an odd prime, groups of any one of

-these orders have already been shewn to be soluble; while if $p_{1}$~is~$2$,

-the only case which can give a simple group is~$2^{2} p_{2} p_{3}$. If

-a group of this order is simple, it follows from Theorem~VIII,

-Cor.\ (§~249), that $p_{2}$~must be~$3$; and the order of the group~$12p_{3}$.

-A cyclical sub-group of order~$p_{3}$ must be one of $1 + kp_{3}$ conjugate

-sub-groups. Hence $1 + kp_{3}$ must be a factor of~$12$, so that $p_{3}$~is

-either $5$~or~$11$. If $p_{3}$~were~$11$, the $12$~conjugate sub-groups of

-order~$11$ would contain $120$~distinct operations of order~$11$, and

-the tetrahedral sub-group, which the group (if simple) must

-contain, would be self-conjugate. Hence $p_{3}$~must be~$5$ and the

-order of the group is~$60$. We have already seen that a simple

-group of order~$60$ actually exists, namely, the icosahedral group;

-and that there is only one type for such a group (§~85).

-

-If the order of a group contains five prime factors, it must

-be of one of the forms:---

-\begin{gather*}

-p_{1}^{5},\ p_{1}^{4} p_{2},\ p_{1}^{3} p_{2}^{2},\ p_{1}^{3} p_{2} p_{3},\ p_{1}^{2} p_{2}^{3},\ p_{1}^{2} p_{2}^{2} p_{3},\ p_{1}^{2} p_{2} p_{3}^{2},\ p_{1}^{2} p_{2} p_{3} p_{4},\ p_{1} p_{2}^{4}, \\

-p_{1} p_{2}^{3} p_{3},\ p_{1} p_{2}^{2} p_{3}^{2},\ p_{1} p_{2}^{2} p_{3} p_{4},\ p_{1} p_{2} p_{3}^{3},\ p_{1} p_{2} p_{3}^{2} p_{4},\ p_{1} p_{2} p_{3} p_{4}^{2},\ p_{1} p_{2} p_{3} p_{4} p_{5}.

-\end{gather*}

-

-If $p_{1}$~is an odd prime, it follows from previous theorems that

-none of these forms, except $p_{1}^{3} p_{2} p_{3}$ and $p_{1} p_{2}^{3} p_{3}$, can give simple

-groups.

-

-Taking first the order~$p_{1} p_{2}^{3} p_{3}$, let us suppose (without limitation

-to the particular case) that a group of order~$p_{1} p_{2}^{m} p_{3}$ is

-simple. It must contain $p_{3}$~or $p_{1} p_{3}$ conjugate sub-groups of

-order~$p_{2}^{m}$. In either case, the operations of these sub-groups

-must be all distinct, or else they must all have a common

-%% -----File: 386.png---Folio 368-------

-\index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|(}%

-sub-group, for the order of the group contains no factor

-congruent to unity $(\mod p_{2})$, except $p_{3}$~or $p_{1} p_{3}$. Now the group

-contains (§~248) just $p_{2}^{m} p_{3}$ operations whose orders are not

-divisible by~$p_{1}$; and if there are $p_{3}$~sub-groups of order~$p_{2}^{m}$

-whose operations are all distinct, there are $(p_{2}^{m} - 1)p_{3}$ operations

-whose orders are powers of~$p_{2}$. Hence, in this case, a sub-group

-of order~$p_{3}$ is self-conjugate. On the other hand, if there are

-$p_{1} p_{3}$~sub-groups of order~$p_{2}^{m}$, their operations cannot be all

-distinct; and the group has a self-conjugate sub-group whose

-order is a power of~$p_{2}$. A repetition of this reasoning shews

-that a group of order~$p_{1} p_{2}^{m} p_{3}$ is always soluble.

-

-We consider, next, a group whose order is~$p_{1}^{3} p_{2} p_{3}$. If $p_{1}$~is

-odd, $p_{1}^{2}$~cannot be congruent to unity, $\pmod{p_{2}}$ or $\pmod{p_{3}}$. The

-same is true, when $p_{1} = 2$, if $p_{2}$~is not equal to~$3$. We shall

-therefore first deal with a group of order~$p_{1}^{3} p_{2} p_{3}$ on the supposition

-that $p_{2}$~is not~$3$.

-

-If neither $p_{2}$~nor $p_{3}$ divides~$p_{1}^{3} - 1$, it follows from §§~248,~257

-that there are just $p_{2} p_{3}$ operations in the group whose orders

-divide~$p_{2} p_{3}$. When this is the case, the group clearly cannot

-be simple.

-

-If $p_{3}$~divides $p_{1}^{3} - 1$, and if there are more than $p_{2} p_{3}$

-operations whose orders divide~$p_{2} p_{3}$, a sub-group of order~$p_{1}^{3}$

-must be self-conjugate in a sub-group of order~$p_{1}^{3} p_{3}$, so that the

-group is isomorphic with a group of degree~$p_{2}$. In this case,

-again, the group cannot be simple.

-

-Finally, then, we have only to deal with the case in which a

-sub-group of order~$p_{1}^{3}$ is self-conjugate in a sub-group of order~$p_{1}^{3} p_{2}$,

-while in this sub-group an operation of order~$p_{2}$ is

-\PageLabel{368}%

-one of $p_{1}^{3}$ conjugate operations. Now the congruences

-\begin{align*}

-p_{1}^{3} &\equiv 1\ (\mod p_{2}), \\

-p_{2}     &\equiv 1\ (\mod p_{1}),

-\end{align*}

-are inconsistent; therefore, if a sub-group of order~$p_{2}$ is self-conjugate

-in a sub-group of order~$p_{1}^{\alpha} p_{2}$, the latter must be

-Abelian. Hence the group contains $p_{1}^{3}(p_{2} - 1)p_{3}$ operations

-whose orders are divisible by~$p_{2}$. Now if the sub-groups of

-order~$p_{1}^{3}$ have a common sub-group, it must be common to all of

-the sub-groups of order~$p_{1}^{3}$, and it is a self-conjugate sub-group.

-%% -----File: 387.png---Folio 369-------

-If, however, no two have a common sub-group, the group

-contains $(p_{1}^{3} - 1)p_{3}$ operations of order~$p_{1}$; and there remain

-only $p_{3}$~operations whose orders divide~$p_{3}$. The group is

-therefore in any case composite.

-

-If now $p_{2} = 3$, a group of order~$2^{3}3p_{3}$ has, by Sylow's theorem,

-a self-conjugate sub-group of order~$p_{3}$ unless $p_{3}$~is $5$,~$7$,~$11$ or~$23$.

-If $p_{3} = 23$, the group (if simple) would have just $24$~operations

-whose orders divide~$24$; it is thence easily seen to be non-existent.

-If $p = 11$, the group (if simple) could be expressed as

-a doubly transitive group of degree~$12$. In this form, however,

-each operation, which transforms an operation of order~$11$ into

-its own inverse, would be a product of $5$~transpositions and

-therefore an odd substitution. This group therefore cannot be

-simple. If $p_{3} = 5$, the group could be expressed as a doubly

-transitive group of degree~$6$. The sub-group of order~$4$ which

-transforms a sub-group of order~$5$ into itself would be cyclical;

-and the corresponding substitution being odd, the group could

-not be simple.

-

-Hence, finally, the only possibility is a group of order~$168$.

-That a simple group of this order actually exists is shewn in

-§~146; also, there is only one type of such group.

-

-When $p_{1}$~is equal to~$2$, the only case, that requires discussion

-in addition to those we have dealt with, is $p_{1}^{2} p_{2} p_{3} p_{4}$.

-A group of this order can only be simple (§~249) when it

-contains a tetrahedral sub-group. In this case, the operations

-of order~$2$ form a single conjugate set, and the group contains

-just $3p_{3} p_{4}$ operations whose orders are divisible by~$2$. If a sub-group

-of order~$2^{\alpha}3q$ is the greatest that contains a sub-group of

-order~$3$ self-conjugately, the group will contain $2^{2 - \alpha}(3 - 1)p_{3}p_{4}$

-operations whose orders are divisible by~$3$ and not by~$2$. Hence

-the group must contain either $p_{3}p_{4}$,~$5p_{3}p_{4}$, or~$7p_{3}p_{4}$, operations

-whose orders divide~$p_{3}p_{4}$. On the other hand, the number of

-these operations may be expressed in the form

-\[

-x(p_{3} - 1)p_{4} + y(p_{4} - 1) + 1,

-\]

-where $x$~is a factor of~$12$, and $y$~is a factor of~$12p_{3}$ which is

-congruent to unity, $\pmod{p_{4}}$.

-

-Hence

-\[

-x(p_{3} - 1)p_{4} + y(p_{4} - 1) = zp_{3}p_{4} - 1,\quad (z = 1, 5, 7).

-\]

-%% -----File: 388.png---Folio 370-------

-

-The case $z = 1$ leads to $y = 1$, so that the sub-group of order~$p_{4}$

-\index{Simple groups@\Topic{Simple groups}!whose orders do not exceed~$660$|(}%

-is self-conjugate.

-

-The case $z = 5$ gives no solution; but when $z = 7$, it will be

-found that there are two solutions, namely,

-\begin{alignat*}{4}

-x &= 6,\quad &y &= 12,\quad &p_{3} &= 5,\quad &p_{4} &= 11, \\

-\lintertext{and}

-x &= 6,\quad &y &= 14,\quad &p_{3} &= 7,\quad &p_{4} &= 13.

-\end{alignat*}

-

-That simple groups actually exist corresponding to the

-two orders $660$ and $1092$ thus arrived at, is shewn in §~221.

-There is also in each case a single type; the verification

-of this statement is left to the reader.

-

-\Par{259.} As has already been stated at the beginning of the

-present chapter, the solution of the general problem of pure

-group-theory, namely, the determination of all possible types of

-group of any given order, depends essentially on the previous

-determination of all possible simple groups. A complete solution

-of this latter problem is not to be expected; but for orders, which

-do not exceed some given limit, the problem may be attacked

-directly. The first determination of this kind was due to

-Herr O.~Hölder\footnotemark,

-\index{Holder@\Topic{Hölder}, quoted}%

-\footnotetext{\textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), pp.~55--88.}%

-who examined all possible orders up to~$200$:

-he proved that none of them, except $60$~and $168$, correspond to

-a simple group. Mr~Cole\footnote

-  {\textit{American Journal of Mathematics}, Vol.~\VolNo{XV}, (1893), pp.~303--315.}

-continued the investigation, by

-\index{Cole@\Topic{Cole}, quoted}%

-examining all orders from $201$ to~$660$, with the result of shewing

-that in this interval the only orders, which have simple groups

-corresponding to them, are $360$,~$504$ and~$660$. The existence

-of a simple group of order~$504$ had not been recognised

-before Mr~Cole's investigation.

-

-The author\footnote

-  {\textit{Proc.\ London Math.\ Soc.}, Vol.~\VolNo{XXVI}, (1895), pp.~333--338.}

-has carried on the examination from $661$ to~$1092$,

-\index{Burnside@\Topic{Burnside}, quoted}%

-with the result of shewing that $1092$~is the only number

-in this interval which is the order of a simple group.

-

-As the limit of the order is increased, such investigations

-as these rapidly become more laborious, since a continually increasing

-number of special cases have to be dealt with. There

-is little doubt however but that, with the aid of the theorems

-proved in the present chapter, the investigation might be

-continued without substantial difficulty up to~$2000$.

-%% -----File: 389.png---Folio 371-------

-

-We shall here be content with verifying the result of Herr

-Hölder's and Mr~Cole's investigations for orders up to~$660$. The

-method used in dealing with particular numbers may suggest

-to the reader how the determination might be continued.

-

-\Par{260.} It follows from Theorem~XII that, if an odd number

-can be the order of a simple group, it must be the product of at

-least $6$~prime factors. Moreover, by previous theorems, we have

-seen that $3^{6}$,~$3^{5}p$ and~$3^{4}p^{2}$ cannot be orders of simple groups.

-Hence certainly no odd number less than $3^{4} · 5 · 7$ or~$2835$\footnote

-  {It is not, of course, here suggested that $2835$~is the order of a simple

-  group. It may, in fact, be shewn with little difficulty that no number of the

-  form $3^{4}pq$,~$3^{3}p^{2}q$, or $3^{2}p^{3}q$, where $p$~and~$q$ are odd primes, can be the order of a

-  simple group. A much higher limit than~$2835$ may therefore be given for the

-  order, if odd, of a simple group.}

-can be

-the order of a simple group. Therefore, by Theorem~XI, we need

-only examine numbers less than~$660$ which are divisible by

-$12$,~$16$, or~$56$. When each number less than~$660$ which is

-divisible by $12$,~$16$, or~$56$ is written in the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}}$\dots, it

-will be found that, with eleven exceptions, Theorems II to

-XII of the present chapter immediately shew that there are no

-simple groups corresponding to them. The exceptions are $60$,~$168$,

-$240$, $336$, $360$, $480$, $504$, $528$, $540$, $560$,~$660$. We will first

-deal with such of these numbers as do not actually correspond

-to simple groups.

-

-\Subsection{$240 = 2^{4} · 3 · 5$.}

-

-%[** TN: No indentation in the original]

-A simple group of this order would contain $6$~or $16$ sub-groups

-of order~$5$. If it contained $6$~sub-groups of order~$5$, the group

-could be expressed as a transitive substitution group of degree~$6$;

-and there is no group of degree~$6$ and order~$240$. If there

-were $16$~sub-groups of order~$5$, each would be self-conjugate in

-a group of order~$15$, which must be cyclical. The group then

-would be a doubly transitive group of degree~$16$ and order

-$16 · 15$. Such a group (§~105) contains a self-conjugate sub-group

-of order~$16$.

-

-\Subsection{$336 = 2^{4} · 3 · 7$.}

-

-%[** TN: No indentation in the original]

-A simple group of this order would contain $8$~sub-groups of

-order~$7$, each self-conjugate in a group of order~$42$. We have

-seen (§~146) that there is no simple group of degree~$8$ and order

-$8 · 7 · 6$.

-\index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|)}%

-%% -----File: 390.png---Folio 372-------

-

-\Subsection{$480 = 2^{5} · 3 · 5$.}

-

-Since $2^{5}$~is not a factor of~$5!$, a group of order~$480$ must, if

-simple, contain $15$~sub-groups of order~$2^{5}$. Now $2^{5}$~is not congruent

-to unity, $\pmod{4}$, and therefore (§~78) some two sub-groups

-of order~$2^{5}$ must have a common sub-group of order~$2^{4}$. Such

-a sub-group, of order~$2^{4}$, must (§~80) either be self-conjugate, or

-it must be contained self-conjugately in a sub-group of order

-$2^{5} · 3$~or $2^{5} · 5$. In either case the group is composite.

-

-\Subsection{$528 = 2^{4} · 3 · 11$.}

-

-There must be $12$~sub-groups of order~$11$, each being

-self-conjugate in a group of order~$44$. A group of order~$44$

-necessarily contains operations of order~$22$, and such an operation

-cannot be represented as a substitution of $12$~symbols.

-The group is therefore not simple.

-

-\Subsection{$540 = 2^{2} · 3^{3} · 5$.}

-

-There must be $10$~sub-groups of order~$3^{3}$, each self-conjugate

-in a group of order~$2 · 3^{3}$. There must also be $36$~sub-groups of

-order~$5$, each self-conjugate in a group of order~$15$. Now when

-the group is expressed as transitive in $10$~symbols, the substitutions

-of order~$5$ consist of $2$~cycles of $5$~symbols each, and

-no such substitution can be permutable with a substitution of

-order~$3$. Hence the group is not simple.

-

-\Subsection{$560 = 2^{4} · 5 · 7$.}

-

-There must be $8$~sub-groups of order~$7$. A transitive group

-of degree~$8$ and order~$8 · 7 · 10$ does not exist (§~146).

-

-That there are actually simple groups of orders $60$,~$168$, $360$,

-and~$660$, has already been seen; the verification that there is

-only a single type of simple group of order~$360$ may be left

-to the reader. It remains then to consider the order~$504$.

-

-\Subsection{$504 = 2^{3} · 3^{2} · 7$.}

-

-A simple group of this order must contain $8$~or $36$ sub-groups

-of order~$7$. There is no group of degree~$8$ and order~$504$ (§~146).

-Hence there must be $36$~sub-groups of order~$7$.

-%% -----File: 391.png---Folio 373-------

-

-Again, there must be $7$~or $28$ sub-groups of order~$3^{2}$; and

-since there is no group of degree~$7$ and order~$504$ (\lc), there

-are $28$~sub-groups of order~$3^{2}$. If two of these have a common

-operation~$P$ of order~$3$, it must (§~80) be self-conjugate in a group

-containing more than one sub-group of order~$3^{2}$. If the order

-of this sub-group is~$2^{2} · 3^{2}$, it must contain a sub-group of order

-$2^{2}$ self-conjugately; and therefore this sub-group of order~$2^{2}$ must

-be self-conjugate in a group of order~$2^{3} · 3^{2}$ at least. Such a

-group would be one of $7$~conjugate sub-groups, and this is

-impossible. Similarly, $P$~cannot be self-conjugate in a group

-whose order is greater than~$2^{2} · 3^{2}$. Hence no two sub-groups

-of order~$3^{2}$ have a common operation, except identity.

-

-There are therefore in the group $216$~operations of order~$7$,

-and $224$~operations whose orders are powers of~$3$, leaving just

-$64$ other operations.

-

-Suppose now that $A$~is an operation of order~$2$, self-conjugate

-in a sub-group of order~$2^{3}$. If $A$~is self-conjugate in a

-group of order~$2^{3}x$ ($x$~being a factor of~$3^{2} · 7$), this group must

-contain at least $x$~operations of odd order and therefore at least $x$~operations

-of the form~$AS$, where $S$~is an operation of odd order

-permutable with~$A$. There are also $x$~operations of this form,

-corresponding to each of the $\dfrac{63}{x}$ operations conjugate to~$A$;

-so that the whole group contains $63$~operations of the form~$AS$,

-where $S$~is of odd order and permutable with~$A$, while $A$~is one

-of a set of conjugate operations of order~$2$, each of which is self-conjugate

-in a sub-group of order~$2^{3}$. Hence taking the identical

-operation with these $63$~operations of even order, all the

-operations of the group are accounted for; and there are therefore

-no operations whose orders are powers of~$2$ except those of

-the conjugate set to which $A$~belongs. The sub-groups of order~$2^{3}$

-are therefore Abelian groups whose operations are all of order~$2$;

-and since the $7$~operations of order~$2$, in a group of order~$2^{3}$,

-are all conjugate in the group of order~$504$, they must (§~81) be

-conjugate in the sub-group within which the group of order~$2^{3}$

-is self-conjugate. Hence, finally, there must be $9$~sub-groups of

-order~$2^{3}$, each self-conjugate in a group of order~$2^{3} · 7$, in which

-the $7$~operations of order~$2$ form a single conjugate set.

-%% -----File: 392.png---Folio 374-------

-

-A simple group of order~$504$, if it exists, must therefore be

-expressible as a triply transitive group of degree~$9$, in which

-the sub-group of order~$56$ that keeps one symbol fixed is of

-known type. From this, several inferences can be drawn.

-Firstly, all the $63$~operations of order~$2$ of the $9$~sub-groups

-of order~$2^{3}$ must be distinct, since each operation of order~$2$

-keeps just one symbol fixed. There is therefore a conjugate

-set of $63$~operations of order~$2$, and there are no other operations

-of even order in the group. Secondly, the $224$~operations whose

-orders are powers of~$3$ cannot all be operations of order~$3$. For,

-if they were, there would be $8$~operations of order~$3$ containing

-any three given symbols in a cycle; and therefore the group

-would contain operations other than identity which keep $3$~symbols

-fixed. This is not the fact; the sub-groups of order~$3^{2}$

-are therefore cyclical.

-

-Conversely, we may now shew that such a triply transitive

-group of degree~$9$, if it exists, is certainly simple. The distribution

-of its operations in conjugate sets has, in fact, been

-determined as follows. There are:---

-\begin{align*}

-&3 \text{ conjugate sets of operations of order } 7 \text{ each containing } 72 \text{ operations} \\

-&3 \Ditto{ conjugate sets of } \Ditto{operations of order } 9 \Ditto{ each containing } 56 \Ditto{ operations} \\

-&1 \Ditto{ conjugate sets of } \Ditto{operations of order } 3 \Ditto{ each containing } 56 \Ditto{ operations} \\

-&1 \Ditto{ conjugate sets of } \Ditto{operations of order } 2 \Ditto{ each containing } 63 \Ditto{ operations}

-\end{align*}

-A self-conjugate sub-group, that contains a single operation

-of order~$7$, must contain all of them; and such a sub-group,

-that contains a single operation of order~$9$, must contain all

-the operations of orders $3$~and~$9$. Hence if a self-conjugate sub-group

-contains operations of order~$9$, its order must be of the form

-\[

-1 + 224 + 216x + 63y;

-\]

-and if it contains no operations of order~$9$, its order must be of

-the form

-\[

-1 + 56x' + 216y' + 63z';

-\]

-each of the symbols $x$,~$y$, $x'$,~$y'$,~$z'$ being either zero or unity.

-The order must at the same time be a factor of~$504$. A very

-brief consideration will shew that all these conditions cannot be

-satisfied, and therefore that no self-conjugate sub-group exists.

-

-That there is a triply transitive group of degree~$9$ and order

-$9 · 8 · 7$, has already been seen in §~113; and the actual formation

-%% -----File: 393.png---Folio 375-------

-of the substitutions generating this group will verify that it

-satisfies the conditions just obtained.

-

-We may however very simply construct the group by the

-method of §~109; and this process has the advantage of shewing

-at the same time that there is only one type.

-

-There is no difficulty in constructing the sub-group of order~$56$

-as a doubly transitive group of $8$ symbols. If we denote by

-$s$~and~$t$ the two substitutions

-\[

-(1254673)\quad \text{and}\quad (12)(34)(56)(78),

-\]

-$\{s, t\}$ is, in fact, a group of the desired type. If now $A$~is any

-substitution of order~$2$ in the symbols $1$, $2$,~\Dots, $8$,~$9$ which

-contains the transposition~$(89)$, and satisfies the conditions

-\[

-AsA = S_{1},\quad AtA = S_{2}AS_{3},

-\]

-where $S_{1}$,~$S_{2}$,~$S_{3}$ belong to~$\{s, t\}$: then (§~109) it follows that

-$\{A, s, t\}$ is a triply transitive group of degree~$9$ and order~$504$.

-

-Now every operation of order~$2$ in the group, which interchanges

-$8$~and~$9$, must transform~$s$ into its inverse. Also from

-the $9$~symbols only $7$~operations of order~$2$, satisfying these

-conditions, can be formed; and if one of them belongs to the

-group, they must all belong to it. Hence there cannot be more

-than one type of group satisfying the conditions.

-

-Finally, if

-\[

-A = (89)(23)(46)(57),

-\]

-then

-\[

-AsA = s^{-1},

-\]

-and $AtA$~can be expressed in the form

-\[

-(1635842) A (1685437),

-\]

-where the two operations $(1635842)$ and $(1685437)$ belong

-to $\{s, t\}$. It follows, then, that $\{A, s, t\}$ is a triply transitive

-group of degree~$9$ and order~$504$, which satisfies all the conditions,

-and that there is only a single type of such group.

-\index{Simple groups@\Topic{Simple groups}!whose orders do not exceed~$660$|)}%

-

-\Par{261.} The order of a non-soluble group must be divisible by

-the order of some non-cyclical simple group. Moreover, if $N$~is

-the order of a non-cyclical simple group and if $n$~is any other

-number, there is always at least one non-soluble group of order~$Nn$;

-for, $G$~being a simple group of order~$N$ and $H$~any group

-of order~$n$, the direct product of $G$~and~$H$ is a non-soluble group

-%% -----File: 394.png---Folio 376-------

-of order~$Nn$. Herr Hölder\footnote

-  {``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}., (1895),

-  pp. 321--422; in particular, p.~420.}

-\index{Holder@\Topic{Hölder}, quoted}%

-\index{Composite groups@\Topic{Composite groups}!non-soluble|(}%

-has determined all distinct types of

-non-soluble groups whose orders are less than~$480$. He shews

-that, besides the three simple groups of orders $60$,~$168$, and

-$360$, there are $22$~non-soluble groups whose orders are given by

-the following table.

-\[

-\begin{array}{|*{7}{c|}}

-\hline

-\Strut

-120 &  180  & 240  & 300  & 336 & 360 & 420 \\

-\hline

-\Strut

- 3  &   1   &  8   &  1   &  3  &  5  &  1 \\

-\hline

-\end{array}

-\]

-

-For the proof of all these results except the first, and for

-the very interesting and suggestive methods that lead to them,

-the reader is referred to Herr Hölder's memoir. The case

-of a non-soluble group of order~$120$ is susceptible of simple

-treatment; and we will consider it here as exemplifying how

-composite groups may be constructed when their factor-groups

-and the isomorphisms of the latter are completely known.

-

-\begin{Remark}

-\Par{262.} The composition-factors of a non-soluble group~$G$ of order~$120$

-are $2$~and~$60$. If these may be taken in either order, then $G$~contains

-a self-conjugate sub-group of order~$2$ and a self-conjugate

-sub-group of order~$60$. The latter, being simple, must be an

-icosahedral group and cannot contain the former. Hence, in this

-case, $G$~must (§~34) be the direct product of an icosahedral group

-and a group of order~$2$.

-

-Next, suppose that the composition-factors can only be taken in

-the order $2$,~$60$. Then $G$~contains a self-conjugate sub-group~$H$ of

-icosahedral type and no self-conjugate sub-group of order~$2$. Hence

-if $S$~is any operation of~$G$ which is not contained in~$H$, the

-isomorphism of~$H$ which arises on transforming its operations by~$S$

-must (§~165) be contragredient. It follows that $G$~is simply isomorphic

-with a group of isomorphisms of~$H$. Now $H$~is isomorphic

-with the alternating group of degree five, and its group of isomorphisms

-is therefore (§~173) simply isomorphic with the symmetric

-group of degree five. Hence, in this case, $G$~is simply isomorphic

-with the symmetric group of degree five.

-

-Lastly, suppose that the composition-factors can only be taken

-in the order $60$,~$2$. Then $G$~has a self-conjugate sub-group of order~$2$

-and no self-conjugate sub-group of order~$60$. Hence (§~35) $G$~has

-%% -----File: 395.png---Folio 377-------

-no sub-group of order~$60$. Suppose that $A$~is the self-conjugate

-operation of order~$2$, and arrange the operations of~$G$ in the sets

-\[

-1,\ A;\quad S_{1},\ S_{1}A;\quad S_{2},\ S_{2}A;\ \Dots;\quad S_{59},\ S_{59}A.

-\]

-Since $\dfrac{G}{\{A\}}$ is simply isomorphic with an icosahedral group, it must

-(§~203) be possible to choose three sets

-\[

-S',\ S'A;\quad S'',\ S''A;\quad S''',\ S'''A;

-\]

-such that either operation of the first, multiplied by either operation

-of the second, belongs to the third set; while at the same time the

-cube of either operation of the first set, the fifth power of either

-operation of the second set, and the square of either operation of the

-third set, belong to the set $1$,~$A$. Now if

-\[

-S'^{3} = 1,

-\]

-then

-\[

-(S'A)^{3} = A.

-\]

-Hence we may assume, without loss of generality, that

-\[

-S'^{3} = 1,\quad S''^{5} = 1,\quad S'S'' = S''' \quad\text{or}\quad S'''A.

-\]

-

-If $S'''$~is of order~$2$, so also is~$S'''A$, and $G$~would then contain an

-icosahedral sub-group. Hence

-\[

-S'''^{4} = 1,

-\]

-and

-\[

-S'''^{2} = A.

-\]

-

-Now

-\[

-S_{1}^{3} = 1,\quad S_{2}^{5} = 1,\quad (S_{1}S_{2})^{2} = 1,

-\]

-is a complete set of defining relations for the icosahedral group; so

-that, from the symbols $S_{1}$~and~$S_{2}$, exactly $60$ distinct products can be

-formed. Hence from $S'$,~$S''$, and~$A$, where

-\[

-S'^{3} = 1,\quad S''^{5} = 1,\quad (S'S'')^{2} = A,\quad A^{2} = 1,

-\]

-$A$~being permutable both with $S'$ and~$S''$, exactly $120$ distinct

-products can be formed. It follows that the relations just given are

-the defining relations of a non-soluble group of order~$120$, which has

-a self-conjugate sub-group of order~$2$ and no sub-group of order~$60$;

-and that there is only one type of such group. We have already

-seen (§~221) that there must be at least one such type of group; viz.\

-the group defined by

-\[

-\begin{aligned}

-x_{1}' &\equiv \alpha x_{1} + \beta x_{2}, \\

-x_{2}' &\equiv \gamma x_{1} + \delta x_{2}, \\

-\alpha \delta - \beta \gamma &\equiv 1,

-\end{aligned}

-\quad (\mod 5).

-\]

-\end{Remark}

-

-\begin{Ex} If $p$~is an odd prime, shew that a group, whose composition-factors

-are $60$~and~$p$, must be the direct product of an icosahedral

-group and a group of order~$p$. \Attrib{Hölder.}

-\end{Ex}

-

-\Par{263.} Let $G$~be a composite group; and suppose that the

-factor-groups of the composition-series of~$G$ are the two non-cyclical

-simple groups $H'$~and~$H$ in the order given. Then

-%% -----File: 396.png---Folio 378-------

-(§~165) if $H_{1}$~is the group formed of all the operations of~$G$

-which are permutable with every operation of~$H$, the direct

-product $\{H, H_{1}\}$ of $H$~and $H_{1}$ is a self-conjugate sub-group of~$G$.

-Now the composition-factors of $\{H, H_{1}\}$ are composition-factors

-of~$G$. Hence $H_{1}$~must either be isomorphic with~$H'$, or it must

-reduce to the identical operation. The latter alternative can

-only occur if $\dfrac{L}{H}$~has a sub-group simply isomorphic with~$H'$,

-$L$~being the group of isomorphisms of~$H$. In the cases of the

-simple groups whose groups of isomorphisms have been investigated

-in Chapter~XI, \ie\ the alternating groups, the doubly

-transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$, and the

-triply transitive groups of degree~$2^{n} + 1$ and order $2^{n}(2^{2n} - 1)$,

-$\dfrac{L}{H}$~was found to be an Abelian group. (These groups include

-all the simple groups whose orders do not exceed~$660$.) If $H$~and

-$H'$ belong to these classes of groups, then $G$~must be the

-direct product of simple groups simply isomorphic with $H$~and~$H'$.

-For instance, a group of order~$3600$, whose composition-factors

-are $60$~and~$60$, must be the direct product of two icosahedral

-groups.

-

-This result may clearly be extended to the case of a group,

-the factor-groups of whose composition-series are a number of

-non-cyclical simple groups $H_{1}$, $H_{2}$,~\Dots, $H_{n}$, such that no two

-of them are of the same type, while for each of them $\dfrac{L}{H}$~is

-soluble. Such a group must be the direct product of $n$~simple

-groups which are isomorphic with $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

-

-If, however, several of these groups are of the same type, the

-inference is no longer necessarily true. Thus if $G$~is the direct

-product of five icosahedral groups, and if $L$~is the group of

-isomorphisms of~$G$, the order of~$\dfrac{L}{G}$ is~$2^{5} · 5!$, and this group

-contains icosahedral sub-groups. A group of order~$(60)^{6}$, the

-factor-groups of whose composition-series are all of order~$60$,

-is not therefore necessarily the direct product of $6$~icosahedral

-groups.

-\PageLabel[endpar]{263}

-\index{Composite groups@\Topic{Composite groups}!non-soluble|)}%

-%% -----File: 397.png---Folio 379-------

-

-\Note{\textit{Note to} §~257.}

-

-\begin{Remark}

-The property stated in the footnote on \PageRef{p.}{366} may be proved as

-follows. Let $S_{1}$~and~$S_{2}$ be two operations of the group which are

-not permutable with each other; and suppose that

-\[

-S_{2}^{-1} S_{1}S_{2} = S_{1}\Sigma.

-\]

-The sub-group of order~$p^{n - 1}$, which contains $S_{1}$ self-conjugately, also

-contains $S_{1}\Sigma$ self-conjugately (§~55); therefore it contains $\Sigma$ self-conjugately.

-Similarly the sub-group, which contains $S_{2}$ self-conjugately,

-contains $S_{2}\Sigma^{-1}$ and therefore $\Sigma$ self-conjugately. Hence $\Sigma$,

-being contained self-conjugately in two distinct sub-groups of order~$p^{n - 1}$,

-is a self-conjugate operation. From this it follows at once that

-$\dfrac{H}{\Gh}$~is Abelian.

-\end{Remark}

-

-

-\Note{\textit{Note to} §~258.}

-

-\begin{Remark}

-The statement, on \PageRef{p.}{368}, that the congruences $p_{1}^{3} \equiv 1\ (\mod p_{2})$

-and $p_{2} \equiv 1\ (\mod p_{1})$ are inconsistent, is subject to an exception

-if $p_{1}^{2} + p_{1} + 1$ is a prime. When this is the case and when

-$p_{2} = p_{1}^{2} + p_{1} + 1$, a group of order~$p_{1}^{2} p_{2} p_{3}$ can only be simple, under

-the conditions assumed in the text, if it contains $(p_{1}^{3} - 1)p_{3}$ operations

-of order~$p_{1}$, $p_{1}^{2}(p_{2} - 1)p_{3}$ operations of order~$p_{2}$, $p_{1}^{2} p_{2}(p_{3} - 1)$

-operations of order~$p_{3}$, and no other operation except identity. The

-relation

-\[

-p_{1}^{3} p_{2} p_{3}

-  = (p_{1}^{3} - 1)p_{3} + p_{1}^{2}(p_{2} - 1)p_{3} + p_{1}^{2} p_{2}(p_{3} - 1) + 1

-\]

-cannot however be satisfied; the group is therefore composite.

-\end{Remark}

-

-

-\Note{\textit{Note to} §~260.}

-

-\begin{Remark}

-No simple group of odd order is at present known to exist. An

-investigation as to the existence or non-existence of such groups

-would undoubtedly lead, whatever the conclusion might be, to

-results of importance; it may be recommended to the reader as

-well worth his attention. Also, there is no known simple group

-whose order contains fewer than three different primes. This

-suggests that Theorems III~and~IV, §§~243,~244, may be capable of

-generalisation. Investigation in this direction is also likely to lead

-to results of interest and importance.

-\end{Remark}

-%% -----File: 398.png---Folio 380-------

-\BackMatter

-

-

-\Section{Appendix.}

-

-\begin{Remark}

-\First{The} technical phraseology that has been used in this book is

-borrowed almost entirely from French or German; and far the

-greater number of important memoirs on the subject of finite groups

-are written in one or the other of those languages. To enable the

-reader to refer, with as little trouble as possible, to the writings of

-foreign mathematicians, a table is here given of the French and the

-German equivalents of the more important technical terms. Even

-abroad, the phraseology of the subject has not yet arrived at that

-settled state in which every writer uses a technical term in the

-same sense; but the variations of usage are not very serious. In

-his recently published \textit{Lehrbuch der Algebra}, however, Herr Weber

-has introduced or adopted several deviations from ordinary usage;

-the chief of these are noted, by the addition of his name after the

-term, in the subjoined table.

-

-\begin{longtable}{*{2}{>{\hangindent1em}p{0.45\textwidth}}}

-Group & Groupe \\

-      & Gruppe \\

-Abelian group & Groupe des opérations échangeables \\

-      & Abel'sche Gruppe \\

-Alternating group & Groupe alterné \\

-      & Alternierende Gruppe \\

-Complete group & Vollkommene Gruppe \\

-Composite group & Groupe composé \\

-      & Zusammengesetzte Gruppe \\

-Factor-group & Groupe facteur \\

-      & Factorgruppe \\

-Primitive or imprimitive group & Groupe primitif ou non-primitif \\

-      & Primitive oder imprimitive Gruppe \\

-Transitive or intransitive group & Groupe transitif ou intransitif \\

-      & Transitive oder intransitive Gruppe \\

-Simple group & Groupe simple \\

-      & Einfache Gruppe \\

-Soluble group & Groupe résoluble \\

-      & Auflösbare Gruppe \\

-      & Metacyclische Gruppe\footnotemark\ (\textit{Weber})

-\footnotetext{This term is used by other German writers to denote the holomorph of a

-group of prime order.} \\

-Substitution group & Groupe des substitutions \textit{or} système

-des substitutions conjuguées \\

-      & Substitutionengruppe \\

-Symmetric group & Groupe symétrique \\

-      & Symmetrische Gruppe \\

-Group of isomorphisms & Gruppe der Isomorphismen \\

-%% -----File: 399.png---Folio 381-------

-Sub-group & Sousgroupe \\

-      & Untergruppe \\

-      & Theiler (\textit{Weber}) \\

-Conjugate sub-groups & Sousgroupes conjugués \\

-      & Gleichberechtigte Untergruppen \\

-      & Conjugirte Theiler (\textit{Weber}) \\

-Self-conjugate sub-group & Sousgroupe invariant \\

-      & Ausgezeichnete Untergruppe \textit{or} invariante Untergruppe \\

-      & Normaltheiler (\textit{Weber}) \\

-Maximum self-conjugate sub-group & Sousgroupe invariant maximum \\

-      & Ausgezeichnete \textit{or} invariante Maximaluntergruppe \\

-      & Grösster Normaltheiler (\textit{Weber}) \\

-Characteristic sub-group & Charakteristische Untergruppe \\

-\hline

-\Strut

-Composition-series & Suite de composition \\

-      & Reihe der Zusammensetzung \\

-      & Compositionsreihe (\textit{Weber}) \\

-Composition factors & Facteurs de composition \\

-      & Factoren der Zusammensetzung \\

-Chief series & Hauptreihe \\

-Characteristic series & Lückenlose Reihe charakteristischer

-  Untergruppen \\

-\hline

-\Strut

-Simple isomorphism & Isomorphisme holoédrique \\

-      & Holoëdrischer \textit{or} einstufiger Isomorphismus \\

-Multiple isomorphism\footnotemark & Isomorphisme meriédrique

-\footnotetext{It is to be noticed that, if $G$~is multiply isomorphic with~$H$, then $H$~is

-  ``meriédriquement isomorphe'' with~$G$.} \\

-      & Meriëdrischer \textit{or} mehrstufiger Isomorphismus \\

-Isomorphism of a group with itself & Isomorphismus der Gruppe in sich \\

-Cogredient or contragredient isomorphism & Cogredient oder contragredient Isomorphismus \\

-\hline

-\Strut

-Degree (of a group) & Degré \\

-      & Grad \\

-Order \Ditto{(of a} \Ditto{group)} & Ordre \\

-      & Ordnung \\

-      & Grad (\textit{Weber}) \\

-Genus \Ditto{(of a} \Ditto{group)} & Geschlecht \\

-\hline

-\Strut

-%% -----File: 400.png---Folio 382-------

-Substitution & Substitution \\

-      & Substitution \textit{or} Buchstabenvertauschung \\

-Circular substitution & Substitution circulaire \\

-      & Cirkularsubstitution \\

-Even and odd substitutions & Substitutions positives et négatives \\

-      & Gerade und ungerade Substitutionen \\

-Regular substitution & Substitution régulière \\

-      & Reguläre Substitution \\

-Similar substitutions & Substitutions semblables \\

-      & Ähnliche Substitutionen \\

-Transposition & Transposition \\

-      & Transposition \\

-\hline

-\end{longtable}

-

-One other term, introduced by Herr Weber, for which no simple

-equivalent is used in this book, must be mentioned. The greatest

-sub-group which is common to two groups he calls the ``Durchschnitt''

-of the two groups.

-

-The ratio of the order of a sub-group~$H$, to the order of the group~$G$

-containing it, is called by French writers the ``indice,'' by German

-the ``Index,'' of~$H$ in~$G$. The phrase is most commonly used of a

-substitution group in relation to the symmetric group of the same

-degree.

-

-The smallest number of symbols displaced by any substitution,

-except identity, of a substitution group is called by French writers

-the ``classe,'' by Germans the ``Klasse,'' of the group.

-\end{Remark}

-%% -----File: 401.png---Folio 383-------

-

-

-% [** TN: \printindex macro prints "(The numbers refer to pages.)"]

-\printindex

-

-\iffalse

-Abel@\Topic{Abel}, quoted#Abel 46

-

-

-Abelian group@\Topic{Abelian group}

-  definition of 46

-  sub-groups of 47, 48, 55-59

-  existence of independent generating operations of 52

-  invariance of the orders of a set of independent generating operations of 54

-  symbol for, of given type 55

-

-

-Abelian group@\Topic{Abelian group}

-  of order~$p^{m}$ and type $(1, 1, \dots, 1)$#Abelian group 58

-  number of sub-groups of, whose order is given 60

-  number of distinct ways of choosing a set of independent generating operations of 59

-  group of isomorphisms of 243, 244

-    (|seealso \Topic{Homogeneous linear group})

-  holomorph of 245

-

-

-Alternating group@\Topic{Alternating group}

-  definition of 139

-  group of isomorphisms of 246

-  is simple, except for degree~$4$#simple 154

-

-

-Bochert@\Topic{Bochert}, quoted#Bochert 153

-

-

-Bolza@\Topic{Bolza}, quoted#Bolza 162

-

-

-Burnside@\Topic{Burnside}, quoted#Burnside 65, 97, 157, 311, 317, 335, 345, 352, 365, 370

-

-

-Cauchy@\Topic{Cauchy}, quoted#Cauchy 90

-

-

-Cayley@\Topic{Cayley}, quoted#Cayley 306

-

-

-Characteristic series@\Topic{Characteristic series}

-  definition of 232

-  invariance of 233

-  of a group whose order is a power of a prime 233-235

-

-

-Characteristic sub-group@\Topic{Characteristic sub-group}

-  definition of 232

-  groups with no, are either simple or the direct product of simply isomorphic simple groups 232

-

-

-Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}

-  definition of 123

-  invariance of 123

-

-

-Cole@\Topic{Cole}, quoted#Cole 196, 370

-

-

-Cole and Glover@\Topic{Cole} and \Topic{Glover}, quoted#Cole and Glover 137

-

-

-Colour-groups@\Topic{Colour-groups} 306-310

-  examples of 310

-

-

-Complete group@\Topic{Complete group}

-  definition of 236

-  group of isomorphisms of a simple group of composite order is a 238

-  holomorph of an Abelian group of order~$p^{m}$ and type $(1, 1, \dots, 1)$ is a#holomorph 239

-  holomorph of a cyclical group of odd order is a 241

-  symmetric group is a, except for degree~$6$#symmetric 246

-  groups which contain a, self-conjugately are direct products 236

-%% -----File: 402.png---Folio 384-------

-

-

-Composite groups@\Topic{Composite groups}

-  definition of 29

-  non-soluble 376-378

-  of even order 360-365

-

-

-Composition factors@\Topic{Composition factors}

-  definition of 118

-

-

-Composition-series@\Topic{Composition-series}

-  definition of 118

-  invariance of 122

-  examples of 128, 129

-

-

-Conjugate operations@\Topic{Conjugate operations}

-  definition of 27

-  complete set of 31

-

-

-Conjugate sub-groups@\Topic{Conjugate sub-groups}

-  definition of 29

-  complete set of 32

-  operations common to or permutable with a complete set of, form a self-conjugate sub-group 33

-

-

-Dedekind@\Topic{Dedekind}, quoted#Dedekind 89

-

-

-Defining relations@\Topic{Defining relations} of a group

-  definition of 21

-  limitation on the number of, when the genus is given 283

-  for groups of genus zero 291

-  for groups of genus unity 301, 302

-  for groups of orders $p^{2}$, $p^{3}$, $p^{4}$#groups 87, 88, 89

-  for groups of order~$pq$#groups 100

-  for groups of order~$p^{2}q$#groups 133-137

-  for groups whose orders contain no square factor#groups 354

-  for the holomorph of a cyclical group#holomorph 240, 241

-  for the symmetric group of degree~$5$#symmetric 305

-  for the simple group of order~$168$#simple 305

-

-

-Degree@\Topic{Degree} of a substitution group

-  definition of 138

-  is a factor of the order, if the group is transitive 140

-

-

-Direct product@\Topic{Direct product} of two groups

-  definition of 40

-  represented as a transitive group 190

-

-

-Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$#product 146, 147

-

-

-Doubly transitive groups@\Topic{Doubly transitive groups}

-  generally contain simple self-conjugate sub-groups 192

-  the sub-groups of, which keep two symbols fixed 212

-  with a complete set of triplets 214

-  of degree~$n$ and order~$n(n - 1)$#degree 155-157

-

-

-Dyck@\Topic{Dyck}, quoted#Dyck 22, 44, 172, 195, 255, 292, 301, 305

-

-

-Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$#Dyck's 22-24

-

-

-Factor groups@\Topic{Factor groups}

-  definition of 38

-  set of, for a given group 118

-  invariance of 118

-

-

-Forsyth@\Topic{Forsyth}, quoted#Forsyth 280, 283

-

-

-Fractional linear group@\Topic{Fractional linear group}

-  definition of 311

-  analysis of 319-333

-  generalization of 334, 335

-

-

-Frobenius@\Topic{Frobenius}, quoted#Frobenius 39, 45, 65, 71, 90, 97, 108, 110, 114, 166, 232, 250, 251, 345

-  354, 358, 360

-

-

-Frobenius and Stickelberger@\Topic{Frobenius and Stickelberger}, quoted#Frobenius and Stickelberger 46

-

-

-Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$#Frobenius 110-112

-

-

-Galois@\Topic{Galois}, quoted#Galois 192, 334

-

-

-General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations 256

-  relation of special groups to 257, 258

-

-

-Genus@\Topic{Genus} of a group, definition of 280

-

-

-Gierster@\Topic{Gierster}, quoted#Gierster 311

-

-

-Graphical representation@\Topic{Graphical representation}

-  of a cyclical group 260, 261, 288

-  of a general group 262-266

-  of a special group 266-269

-%% -----File: 403.png---Folio 385-------

-  of a group of finite order 273-278

-  examples of 269-272, 279, 288-290, 294, 296, 299, 300, 303, 310

-

-

-Group@\Topic{Group}

-  definition of 11, 12

-  continuous, discontinuous, or mixed 13, 14

-  order of 14

-  multiplication table of 20, 49

-  fundamental or generating operations of 21

-  defining relations of 21

-  cyclical 25

-  simple and composite 29

-  Abelian 46

-  soluble 130

-  symmetric 139

-  alternating 139

-  complete 236

-  group of isomorphisms of a 223

-  general 256

-  special 257

-  quadratic 326

-  dihedral 287

-  tetrahedral 289

-  octohedral 289

-  icosahedral 289

-  holomorph of a 228

-  genus of a 280

-  symbol for a 27

-  (|seealso \Topic{Substitution group}).

-

-

-Group of isomorphisms@\Topic{Group of isomorphisms}

-  contains the group of cogredient isomorphisms self-conjugately 226

-  of a cyclical group 239-242

-  of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$, 243, 244, 311-317#Abelian 336-338

-  of the alternating group 246

-  of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$#doubly 246-248

-

-

-Groups@\Topic{Groups}

-  of genus zero 286-292

-  of genus one 295-302

-  of genus two 302

-

-

-Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime 61

-  always contain self-conjugate operations 62

-  every sub-group of, is contained self-conjugately in a greater sub-group 65

-  number of sub-groups of given order is congruent to unity, $\pmod{p}$ 71

-  case in which there is only one sub-group of a given order 72-75

-  number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$#self-conjugate 76-81

-  determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime#types 81-88

-  table of distinct types of orders $8$ and~$16$#table 88, 89

-

-

-Groups whose orders contain@\Topic{Groups} whose orders contain

-  no squared factor 353, 354

-  no cubed factor 359

-

-

-Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical#cyclical 352, 353

-

-

-Heffter@\Topic{Heffter}, quoted#Heffter 215

-

-

-Holder@\Topic{Hölder}, quoted#Hölder 38, 87, 119, 137, 224, 236, 246, 249, 344, 353, 370, 376

-

-

-Holomorph@\Topic{Holomorph}

-  definition of 228

-  of a cyclical group 240-242

-  of an Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$#Abelian 245

-

-

-Homogeneous linear group@\Topic{Homogeneous linear group}

-  definition of 244

-  composition series of 314-317

-  represented as a transitive substitution group 336, 337

-%% -----File: 404.png---Folio 386-------

-

-

-Homogeneous linear group@\Topic{Homogeneous linear group}

-  simple groups defined by 338, 339

-  generalization of 340-342

-

-

-Hurwitz@\Topic{Hurwitz}, quoted#Hurwitz 24, 280, 282

-

-

-Identical operation@\Topic{Identical operation}

-  definition of 12

-

-

-Imprimitive groups@\Topic{Imprimitive groups}

-  definition of 171

-  of degree~$6$#degree 181, 182

-

-

-Imprimitive systems@\Topic{Imprimitive systems}

-  definition of 172

-  of a regular group 173

-  of any transitive group 176

-  properties of 185, 186

-

-

-Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group 193, 194

-

-

-Intransitive groups@\Topic{Intransitive groups}

-  definition of 140

-  properties of 159-162

-  transitive sets of symbols in 159, 166

-  of degree~$7$ with transitive sets of $3$ and~$4$#degree 163, 164

-

-

-Isomorphism of two groups@\Topic{Isomorphism of two groups}

-  simple, definition of 32

-  multiple, definition of 36

-  general, definition of 41

-

-

-Isomorphisms@\Topic{Isomorphisms} of a group with itself

-  definition of 222

-  cogredient and contragredient, definition of 224

-  class of, definition of 224

-  limitation on the order of 252

-

-

-Isomorphism@\Topic{Isomorphism} of general and special groups 257, 258

-

-

-Jordan@\Topic{Jordan}, quoted#Jordan 46, 119, 143, 146, 149, 150, 153, 189, 196, 198, 204, 311

-

-

-Klein@\Topic{Klein}, quoted#Klein 224, 289, 292

-

-

-Klein and Fricke@\Topic{Klein} and \Topic{Fricke}, quoted#Klein and Fricke 162

-

-

-Limitation@\Topic{Limitation}

-  on the order of a group of given genus 282

-  on the number of defining relations of a group of given genus 283

-

-

-Maillet@\Topic{Maillet}, quoted#Maillet 108

-

-

-Marggraff@\Topic{Marggraff}, quoted#Marggraff 198

-

-

-Maschke@\Topic{Maschke}, quoted#Maschke 310

-

-

-Mathieu@\Topic{Mathieu}, quoted#Mathieu 155

-

-

-Maximum sub-group@\Topic{Maximum sub-group}, definition of 35

-

-

-Maximum self-conjugate sub-group@\Topic{Maximum self-conjugate sub-group}

-  definition of 35

-

-

-Miller@\Topic{Miller}, quoted#Miller 105, 196, 259

-

-

-Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}

-  definition of 124

-  is a simple group or the direct product of simply isomorphic simple groups 127

-

-

-Moore@\Topic{Moore}, quoted#Moore 215, 335

-

-

-Multiplication table@\Topic{Multiplication table} of a group 20, 49

-

-

-Multiply isomorphic groups@\Topic{Multiply isomorphic groups}

-  definition of 36

-

-

-Netto@\Topic{Netto}, quoted#Netto 141, 198, 215

-

-

-Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group 165, 166

-

-

-Operations@\Topic{Operations}

-  common to two groups form a group 27

-  of a group, which are permutable with a given operation or sub-group, form a group 29, 31

-  common to or permutable with each of a complete set of conjugate sub-groups form a self-conjugate sub-group 33

-

-

-Order@\Topic{Order}

-  of a group, definition of 14

-  of an operation, definition of 14

-

-

-Permutable operations@\Topic{Permutable operations}

-  definition of 12

-

-

-Permutable groups@\Topic{Permutable groups}

-  definition of 41

-

-

-Permutability@\Topic{Permutability} of an operation with a group

-  definition of 29

-

-

-Primitive groups@\Topic{Primitive groups}

-  definition of 171

-  when soluble, have a power of a prime for degree 192

-  limit to the order of, for a given degree 199

-  with a transitive sub-group of smaller degree 197, 198

-  of prime degree 201

-  of degrees $3$,~$4$ and~$5$#degrees 205

-  of degree~$6$#degree 205, 206

-%% -----File: 405.png---Folio 387-------

-  of degree~$7$#degree 206-208

-  of degree~$8$#degree 209-211

-

-

-Primitivity@\Topic{Primitivity}, test of 184

-

-

-Regular division@\Topic{Regular division} of a surface, representation of a group by means of 278

-

-

-Representation of a group@\Topic{Representation of a group}

-  in transitive form; \ie\ as a transitive substitution group 22, 173-179

-  in primitive form 177

-

-

-Representation@\Topic{Representation}, graphical (|see \Topic{Graphical representation}).

-

-

-Self-conjugate operation@\Topic{Self-conjugate operation}

-  definition of 28

-  of a group whose order is the power of a prime 62

-  of a transitive substitution group must be a regular substitution 144

-

-

-Self-conjugate sub-group@\Topic{Self-conjugate sub-group}

-  definition of 29

-  generated by a complete set of conjugate operations 34

-  of a primitive group must be transitive 187

-  of an imprimitive group 187, 188

-  of a $k$-ply transitive group is, in general, $(k - 1)$-ply transitive#transitive 189

-

-

-Simple groups@\Topic{Simple groups}

-  definition of 29

-  whose orders are the products of not more than $5$~primes#products 367-370

-  whose orders do not exceed~$660$#exceed 370-375

-

-

-Simply isomorphic groups@\Topic{Simply isomorphic groups}

-  definition of 22

-  said to be of the same type 22

-

-

-Soluble groups@\Topic{Soluble groups}

-  definition of 130

-  properties of 131, 132

-  special classes of 345-359

-

-

-Sub-group@\Topic{Sub-group}

-  definition of 25

-  order of a, divides order of group 26

-  (|seealso, \Topic{conjugate}, \Topic{self-conjugate} and \Topic{characteristic sub-group}).

-

-

-Substitution@\Topic{Substitution}

-  definition of 1

-  cycles of 2

-  identical 4

-  inverse 4

-  order of 6

-  circular 7

-  regular 7

-  similar 7

-  permutable 8

-  even and odd 10

-  symbol for the product of two or more 4

-

-

-Substitutions@\Topic{Substitutions} which are permutable

-  with a given substitution 215, 216

-  with every substitution of a given group 145, 146, 217

-  with every substitution of a group, whose degree is equal to its order, form a simply isomorphic group 146

-

-

-Substitution group@\Topic{Substitution group}

-  degree of, definition of 138

-  regular, definition of 24

-  transitive and intransitive, definition of 140

-  primitive and imprimitive, definition of 171

-  multiply transitive, definition of 148

-  representation of any group as a regular 22

-  degree of transitive, is a factor of its order 140

-  order of a $k$-ply transitive, whose degree is~$n$, is a multiple of $n(n - 1)\Dots (n - k + 1)$#order 148

-  construction of multiply transitive 150

-  limit to the degree of transitivity of 152

-  transitive, whose substitutions, except identity, displace all or all but one of the symbols 141-144

-  doubly transitive, of degree~$n$ and order~$n(n - 1)$#doubly 155-157

-  triply transitive, of degree~$n$ and order~$n(n - 1)(n - 2)$#triply 158

-  quintuply transitive, of degree~$12$#quintuply 170

-  conjugate substitutions of, are similar 144

-

-

-Substitution groups@\Topic{Substitution groups} whose orders are powers of primes 218, 219

-%% -----File: 406.png---Folio 388-------

-

-

-Sylow@\Topic{Sylow}, quoted#Sylow 62, 90

-

-

-Sylow's theorem@\Topic{Sylow's theorem} 92-95

-  some direct consequences of 97-100

-  extension of 110

-

-

-Symbol@\Topic{Symbol}

-  for the product of two or more substitutions 4

-  for the product of two or more operations 12

-  for a group generated by given operations 27

-  theta@$\vartheta(P)$ and $\theta(P)$, definition of#definition 250, 251

-

-

-Symmetric group@\Topic{Symmetric group}

-  definition of 139

-  of degree~$n$ has a single set of conjugate sub-groups of order~$(n - 1)!$ except when $n$ is~$6$#degree 200

-  of degree~$6$ has $12$~simply isomorphic sub-groups of order~$5!$ which form two distinct conjugate sets#degree 200

-  is a complete group, except for degree~$6$#complete 246

-

-

-Transforming an operation@\Topic{Transforming an operation}

-  definition of 27

-

-

-Transitive group@\Topic{Transitive group}

-  definition of 140

-  representation of any group as a 175-179

-  number of distinct modes of representing the alternating group of degree~$5$ as a#modes 179, 180

-

-

-Transpositions@\Topic{Transpositions}

-  definition of 9

-  representation of a substitution by means of 9

-  number of, which enter in the representation of a substitution is either always even or always odd 10

-

-

-Triply transitive@\Topic{Triply transitive} groups of degree~$n$ and order $n(n - 1)(n - 2)$#triply 158

-

-

-Type of a group@\Topic{Type of a group} (|see \Topic{Simply isomorphic groups}).

-

-

-Types of group@\Topic{Types of group}, distinct, whose order is

-  $p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime#types 87, 88

-  $8$ or~$16$#types 88, 89

-  $pq$, where $p$~and $q$ are different primes#types 100, 101

-  $p^{2}q$, where $p$~and $q$ are different primes#types 133-137

-  $24$#types 101-104

-  $60$#types 105-108

-

-

-Weber@\Topic{Weber}, quoted#Weber 46

-

-

-Young@\Topic{Young}, quoted#Young 63, 68, 87, 310

-\fi % ** End of index text

-% [** TN: Text printed by \printindex macro]

-% CAMBRIDGE: PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS.

-%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%%

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-

-\PGLicense

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-% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %

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-% End of the Project Gutenberg EBook of Theory of Groups of Finite Order, by 

-% William Burnside                                                        %

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