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diff --git a/.gitattributes b/.gitattributes new file mode 100644 index 0000000..6833f05 --- /dev/null +++ b/.gitattributes @@ -0,0 +1,3 @@ +* text=auto +*.txt text +*.md text diff --git a/930-pdf.zip b/930-pdf.zip Binary files differnew file mode 100644 index 0000000..3c0f27b --- /dev/null +++ b/930-pdf.zip diff --git a/9930-pdf.pdf b/9930-pdf.pdf Binary files differnew file mode 100644 index 0000000..6d5b654 --- /dev/null +++ b/9930-pdf.pdf diff --git a/9930-pdf.zip b/9930-pdf.zip Binary files differnew file mode 100644 index 0000000..ae96c1d --- /dev/null +++ b/9930-pdf.zip diff --git a/9930-t.tex b/9930-t.tex new file mode 100644 index 0000000..18ab013 --- /dev/null +++ b/9930-t.tex @@ -0,0 +1,3705 @@ +\documentclass[oneside]{article} +\usepackage{enumerate} +\usepackage[leqno]{amsmath} +\allowdisplaybreaks[1] +\begin{document} + +\thispagestyle{empty} +\small +\begin{verbatim} + +The Project Gutenberg EBook of Groups of Order p^m Which Contain Cyclic +Subgroups of Order p^(m-3), by Lewis Irving Neikirk + +Copyright laws are changing all over the world. Be sure to check the +copyright laws for your country before downloading or redistributing +this or any other Project Gutenberg eBook. + +This header should be the first thing seen when viewing this Project +Gutenberg file. Please do not remove it. Do not change or edit the +header without written permission. + +Please read the "legal small print," and other information about the +eBook and Project Gutenberg at the bottom of this file. Included is +important information about your specific rights and restrictions in +how the file may be used. You can also find out about how to make a +donation to Project Gutenberg, and how to get involved. + + +**Welcome To The World of Free Plain Vanilla Electronic Texts** + +**eBooks Readable By Both Humans and By Computers, Since 1971** + +*****These eBooks Were Prepared By Thousands of Volunteers!***** + + +Title: Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3) + +Author: Lewis Irving Neikirk + +Release Date: February, 2006 [EBook #9930] +[Yes, we are more than one year ahead of schedule] +[This file was first posted on November 1, 2003] + +Edition: 10 + +Language: English + +Character set encoding: US-ASCII + +*** START OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M *** + + + + +Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, +John Hagerson, and the Online Distributed Proofreading Team. + +\end{verbatim} +\normalsize +\newpage + + + +\begin{center} +\noindent \Large GROUPS OF ORDER $p^m$ WHICH CONTAIN CYCLIC +SUBGROUPS OF ORDER $p^{m-3}$ + +\bigskip +\normalsize\textsc{by} +\bigskip + +\large LEWIS IRVING NEIKIRK + +\footnotesize\textsc{sometime harrison research fellow in +mathematics} + +\bigskip +\large 1905 +\end{center} + +\newpage +\begin{center} +\large \textbf{INTRODUCTORY NOTE.} +\end{center} +\normalsize + +This monograph was begun in 1902-3. Class I, Class II, Part I, and +the self-conjugate groups of Class III, which contain all the groups with +independent generators, formed the thesis which I presented to the Faculty +of Philosophy of the University of Pennsylvania in June, 1903, in partial +fulfillment of the requirements for the degree of Doctor of Philosophy. + +The entire paper was rewritten and the other groups added while the +author was Research Fellow in Mathematics at the University. + +I wish to express here my appreciation of the opportunity for scientific +research afforded by the Fellowships on the George Leib Harrison Foundation +at the University of Pennsylvania. + +I also wish to express my gratitude to Professor George H.\ +Hallett for his kind assistance and advice in the preparation of +this paper, and especially to express my indebtedness to Professor +Edwin S.\ Crawley for his support and encouragement, without which +this paper would have been impossible. + +\begin{flushright} +\textsc{Lewis I.\ Neikirk.} +\end{flushright} + +\footnotesize \textsc{ University Of Pennsylvania,} \textit{May, +1905.} \normalsize + +\newpage + +\begin{center} +\large GROUPS OF ORDER $p^m$, WHICH CONTAIN CYCLIC SUBGROUPS OF +ORDER $p^{(m-3)}$\footnote{Presented to the American Mathematical +Society April 25, 1903.} + +\bigskip \normalsize \textsc{by} + +\bigskip \textsc{lewis irving neikirk} + +\bigskip\textit{Introduction.} +\end{center} + +The groups of order $p^m$, which contain self-conjugate cyclic +subgroups of orders $p^{m-1}$, and $p^{m-2}$ respectively, have +been determined by \textsc{Burnside},\footnote{\textit{Theory of +Groups of a Finite Order}, pp.\ 75-81.} and the number of groups of +order $p^m$, which contain cyclic non-self-conjugate subgroups of +order $p^{m-2}$ has been given by +\textsc{Miller}.\footnote{Transactions, vol.\ 2 (1901), p.\ 259, and +vol.\ 3 (1902), p.\ 383.} + +Although in the present state of the theory, the actual tabulation +of all groups of order $p^m$ is impracticable, it is of importance +to carry the tabulation as far as may be possible. In this paper +\textit{all groups of order} $p^m$ ($p$ being an odd prime) +\textit{which contain cyclic subgroups of order $p^{m-3}$ and none +of higher order} are determined. The method of treatment used is +entirely abstract in character and, in virtue of its nature, it is +possible in each case to give explicitly the generational +equations of these groups. They are divided into three classes, +and it will be shown that these classes correspond to the three +partitions: $(m-3,\, 3)$, $(m-3,\, 2,\, 1)$ and $(m-3,\, 1,\, 1,\, 1)$, of +$m$. + +We denote by $G$ an abstract group $G$ of order $p^m$ containing +operators of order $p^{m-3}$ and no operator of order greater than +$p^{m-3}$. Let $P$ denote one of these operators of $G$ of order +$p^{m-3}$. The $p^3$ power of every operator in $G$ is contained +in the cyclic subgroup $\{P\}$, otherwise $G$ would be of order +greater than $p^m$. The complete division into classes is effected +by the following assumptions: +\begin{enumerate}[I.] +\item There is in $G$ at least one operator $Q_1$, such that +$Q{}_1^{p^2}$ is not contained in $\{P\}$. +\item The $p^2$ power of every operator in $G$ is contained in +$\{P\}$, and there is at least one operator $Q_1$, such that +$Q{}_1^p$ is not contained in $\{P\}$. +\item The $p$th power of every operator in $G$ is +contained in $\{P\}$. +\end{enumerate} + +\newpage +The number of groups for Class I, Class II, and Class III, +together with the total number, are given in the table below: +\bigskip + +\begin{tabular}{|c|c|c|c|c|c|c|c|} +\hline + & I & II$_1$ & II$_2$ & II$_3$ & II & III & Total \\ \hline +$p>3$ & & & & & & & \\ +$m>8$ & 9 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $64+5p$ \\ \hline +$p>3$ & & & & & & & \\ +$m=8$ & 8 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $63+5p$ \\ \hline +$p>3$ & & & & & & & \\ +$m=7$ & 6 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $61+5p$ \\ \hline +$p=3$ & & & & & & & \\ +$m>8$ & 9 & 23 & 12 & 12 & 47 & 16 & 72 \\ \hline +$p=3$ & & & & & & & \\ +$m=8$ & 8 & 23 & 12 & 12 & 47 & 16 & 71 \\ \hline +$p=3$ & & & & & & & \\ +$m=7$ & 6 & 23 & 12 & 12 & 47 & 16 & 69 \\ \hline +\end{tabular} + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} I.\normalsize +\end{center} + +1. \textit{General notations and relations.}---The group $G$ is +generated by the two operators $P$ and $Q_1$. For brevity we +set\footnote{With J.~W.\ \textsc{Young}, \textit{On a certain +group of isomorphisms}, American Journal of Mathematics, vol.\ 25 +(1903), p.\ 206.} +\begin{equation*} +Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots = [a,\, b,\, c,\, d,\, \cdots]. +\end{equation*} + +Then the operators of $G$ are given each uniquely in the form +\begin{equation*} +[y,\, x] \quad \left( \begin{aligned}y &= 0,\, 1,\, 2,\, \cdots,\, p^3-1 \\ + x &= 0,\, 1,\, 2,\, \cdots,\, p^{m-3}-1 + \end{aligned} \right) . +\end{equation*} + +We have the relation +\begin{equation} +Q{}_1^{p^3} = P^{hp^3}. %% 1 +\end{equation} + +\noindent There is in $G$, a subgroup $H_1$ of order $p^{m-2}$, which +contains $\{P\}$ self-con\-ju\-gate\-ly.\footnote{\textsc{Burnside}: +\textit{Theory of Groups}, Art.\ 54, p.\ 64.} The subgroup $H_1$ is +generated by $P$ and some operator $Q{}_1^y P^x$ of $G$; it then +contains $Q{}_1^y$ and is therefore generated by $P$ and +$Q{}_1^{p^2}$; it is also self-conjugate in $H_2 = \{Q{}_1^p, P\}$ +of order $p^{m-1}$, and $H_2$ is self-conjugate in $G$. + +From these considerations we have the +equations\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{align} +Q{}_1^{-p^2}\,P\,Q{}_1^{p^2} &= P^{1+kp^{m-4}}, \\ %% 2 +Q{}_1^{-p}\,P\,Q{}_1^p &= Q{}_1^{\beta p^2}\,P^{\alpha_1}, \\ %% 3 +Q{}_1^{-1}\,P\,Q_1 &= Q{}_1^{bp}\,P^{a_1}. %% 4 +\end{align} + +\medskip +2. \textit{Determination of $H_1$. Derivation of a formula for +$[yp^2, x]^s$.}---From (2), by repeated multiplication we obtain +\begin{gather*} +[-p^2,\, x,\, p^2] = [0,\, x(1 + kp^{m-4})]; \\ +\intertext{and by a continued use of this equation we have} +[-yp^2,\, x,\, yp^2] = [0,\, x(1 + kp^{m-4})^y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4) +\end{gather*} +\noindent and from this last equation, +\begin{equation} +[yp^2,\, x]^s = \bigl[syp^2,\, x\{s + k \tbinom{s}{2}yp^{m-4}\}\bigr]. %% 5 +\end{equation} + +\medskip +3. \textit{Determination of $H_2$. Derivation of a formula for +$[yp,\, x]^s$.}---It follows from (3) and (5) that +\begin{equation*} +[-p^2,\, 1,\, p^2] = \left[\beta\frac{\alpha_1^p-1}{\alpha_1-1}p^2,\, + \alpha_1^p\left \{ 1+\frac{\beta k}{2} + \frac{\alpha_1^p-1}{\alpha_1-1} p^{m-4}\right \} \right] \quad (m > 4). +\end{equation*} +\noindent Hence, by (2), +\begin{gather*} +\beta\frac{\alpha_1^p - 1}{\alpha_1 - 1}p^2 \equiv 0 \pmod{p^3}, \\ +\alpha{}_1^p \left \{ 1 + \frac{\beta k}{2} + \frac{\alpha{}_1^p-1}{\alpha_1 - 1} p^{m-4} \right \} + + \beta\frac{\alpha{}_1^p - 1}{\alpha_1 - 1}hp^2 + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}. \\ +\intertext{From these congruences, we have for $m > 6$} +\alpha{}_1^p \equiv 1 \pmod{p^3}, \qquad \alpha_1 \equiv 1 \pmod{p^2}, \\ +\intertext{and obtain, by setting} +\alpha_1 = 1 + \alpha_2 p^2, \\ +\intertext{the congruence} +\frac{(1 + \alpha_2 p^2)^p - 1}{\alpha_2 p^3}(\alpha_2 + h\beta)p^3 + \equiv kp^{m-4} \pmod{p^{m-3}}; \\ +\intertext{and so} +(\alpha_2 + h\beta)p^3 \equiv 0 \pmod{p^{m-4}}, \\ +\intertext{since} +\frac{(1+\alpha_2 p^2)^p-1}{\alpha_2 p^3} \equiv 1 \pmod{p^2}. +\end{gather*} +\noindent From the last congruences +\begin{gather} +(\alpha_2 + h\beta)p^3 \equiv kp^{m-4} \pmod{p^{m-3}}. \\ %% 6 +\intertext{Equation (3) is now replaced by} +Q{}_1^{-p}\,P\, Q{}_1^{-p} = Q{}_1^{\beta p^2} P^{1 + \alpha_2 p^2}. %% 7 +\end{gather} +\noindent From (7), (5), and (6) +\begin{equation*} +[-yp,\, x,\, yp] = \left[\beta xyp^2,\, x\{1 + \alpha_2 yp^2\} + + \beta k \tbinom{x}{2}yp^{m-4}\right]. +\end{equation*} +\noindent A continued use of this equation gives +\begin{multline} +[yp,\, x]^s = [syp + \beta \tbinom{s}{2}xyp^2, \\ + xs + \tbinom{s}{2} \{\alpha_2 xyp^2 + \beta k\tbinom{x}{2}yp^{m-4}\} + + \beta k\tbinom{s}{3}x^2yp^{m-4}]. %% 8 +\end{multline} + +\medskip +4. \textit{Determination of $G$.}---From (4) and (8), +\begin{gather*} +[-p,\, 1,\, p] = [Np,\, a{}_1^p + Mp^2]. \\ +\intertext{From the above equation and (7),} +a{}_1^p \equiv 1 \pmod{p^2}, \qquad a_1 \equiv 1 \pmod{p}. +\end{gather*} + +Set $a_1 = 1 + a_2 p$ and equation (4) becomes +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q{}_1^{bp}\, P^{1 + a_2 p}. %% 9 +\end{equation} +\noindent From (9), (8) and (6) +\begin{gather*} +[-p^2,\, 1,\, p^2] = \left[\frac{(1 + a_2 p)^{p^2}-1}{a_2 p}bp, + (1 + a_2 p)^{p^2}\right], \\ +\intertext{and from (1) and (2)} +\frac{(1 + a_2 p)^{p^2} - 1}{a_2 p}bp \equiv 0 \pmod{p^3}, \\ +(1 + a_2 p)^{p^2} + bh\frac{(1 + a_2 p)^{p^2} - 1}{a_2 p}p + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}. +\end{gather*} +\noindent By a reduction similar to that used before, +\begin{equation} +(a_2 + bh)p^3 \equiv kp^{m-4} \pmod{p^{m-3}}. %% 10 +\end{equation} + +The groups in this class are completely defined by (9), (1) and (10). + +These defining relations may be presented in simpler form by a suitable +choice of the second generator $Q_1$. From (9), (6), (8) and (10) +\begin{gather*} +[1,\, x]^{p^3} = [p^3,\, xp^3] = [0,\, (x + h)p^3] \quad (m > 6), \\ +\intertext{and, if $x$ be so chosen that} +x + h \equiv 0 \pmod{p^{m-6}}, \\ +\intertext{$Q_1\, P^x$ is an operator of order $p^3$ whose $p^2$ +power is not contained in $\{P\}$. Let $Q_1\, P^x = Q$. The group +$G$ is generated by $Q$ and $P$, where} +Q^{p^3} = 1, \quad P^{p^{m-3}} = 1. \\ +\intertext{Placing $h = 0$ in (6) and (10) we find} +\alpha_2 p^3 \equiv a_2 p^3 \equiv k p^{m-4} \pmod{p^{m-3}}. +\end{gather*} +\noindent Let $\alpha_2 = \alpha p^{m-7}$, and $a_2 = ap^{m-7}$. +Equations (7) and (9) are now replaced by +\begin{equation} +\left. + \begin{aligned} + Q^{-p}\, P\, Q^p &= Q^{\beta p^2} P^{1 + \alpha p^{m-5}},\\ + Q^{-1}\, P\, Q &= Q^{bp} P^{1 + ap^{m-6}}. + \end{aligned} +\right. %% 11 +\end{equation} + +As a direct result of the foregoing relations, the groups in this +class correspond to the partition $(m-3,\, 3)$. From (11) we +find\footnote{For $m = 8$ it is necessary to add +$a^2\binom{y}{2}p^4$ to the exponent of $P$ and for $m = 7$ the +terms $a(a + \frac{abp}{2})\binom{y}{2}p^2 + a^3\binom{y}{3}p^3$ +to the exponent of $P$, and the term $ab\binom{y}{2}p^2$ to the +exponent of $Q$. The extra term $27ab^2 k\binom{y}{3}$ is to be +added to the exponent of $P$ for $m = 7$ and $p = 3$.} +\begin{equation*} +[-y,\, 1,\, y] = [byp,\, 1 + ayp^{m-6}] \qquad (m > 8). +\end{equation*} + +It is important to notice that by placing $y = p$ and $p^2$ in the +preceding equation we find that\footnote{For $m = 7,\, +ap^2-\frac{a^2p^3}{2} \equiv ap^2 \pmod{p^4},\, ap^3 \equiv kp^3 +\pmod{p^4}$. For $m = 7$ and $p = 3$ the first of the above +congruences has the extra terms $27(a^3 + ab\beta k)$ on the left +side.} +\begin{equation*} +b \equiv \beta \pmod{p}, \qquad a \equiv \alpha \equiv k \pmod{p^3} + \qquad (m > 7). +\end{equation*} + +A combination of the last equation with (8) yields\footnote{For $m += 8$ it is necessary to add the term $a\binom{y}{2}xp^4$ to the +exponent of $P$, and for $m = 7$ the terms $x\{a(a + +\frac{abp}{2})\binom{y}{2}p^2 + a^3\binom{y}{3}p^3\}$ to the +exponent of $P$, with the extra term $27ab^2 k\binom{y}{3}x$ for +$p = 3$, and the term $ab\binom{y}{2}xp^2$ to the exponent of +$Q$.} +\begin{multline} +[-y,\, x,\, y] = [bxyp + b^2\tbinom{x}{2}yp^2, \\ + x(1 + ayp^{m-6}) + ab\tbinom{x}{2}yp^{m-5} + + ab^2\tbinom{x}{3}yp^{m-4}] \qquad (m > 8). %% 12 +\end{multline} + +\newpage +From (12) we get\footnote{For $m = 8$ it is necessary to add the +term $\frac{1}{2} axy \binom{s}{2}[\frac{1}{3}y(2s - 1) - 1]p^4$ +to the exponent of $P$, and for $m=7$ the terms +\begin{multline*} + x \Bigl\{ \frac{a}{2} \bigl( a + \frac{ab}{2} p \bigr) + \bigl(\frac{2s-1}{3} y - 1 \bigr) \tbinom{s}{2}yp^2 + + \frac{a^{3}}{3!} \bigl(\tbinom{s}{2}y^2 - (2s - 1)y + 2 \bigr) yp^3 \\ + + \frac{a^2 bxy^2}{2} \tbinom{s}{3} \frac{3s-1}{2}p^3 + \frac{a^2 b}{2} + \bigl( \frac{s(s - 1)^2 (s - 4)}{4!}y - \tbinom{s}{3} \bigr) yp^3 \Bigr\} +\end{multline*} +\noindent with the extra terms +\begin{equation*} + 27abxy \Bigl\{ \frac{bk}{3!}\bigl[\tbinom{s}{2}y^2 - (2s - 1)y + 2\bigr] \tbinom{s}{3} + + x(b^2 k + a^2)(2y^2 + 1)\tbinom{s}{3} \Bigr\}, +\end{equation*} +\noindent for $p=3$, to the exponent of $P$, and the terms +$\frac{ab}{2} \bigl\{ 2s - \frac{1}{3}y - 1 \bigr\} \tbinom{s}{2}xyp^2$ +to the exponent of $Q$.} %% END OF FOOTNOTE +\begin{multline} +[y,\, x]^s = \bigl[ys + by\bigl\{(x +b\tbinom{x}{2}p)\tbinom{s}{2} + x\tbinom{s}{3}p\bigr\}p, \\ + xs + ay\bigl\{(x+b\tbinom{x}{2}p + b^2\tbinom{x}{3}p^2)\tbinom{s}{2} \\ + + (bx^2 p + 2b^2 x\tbinom{x}{2}p^2)\tbinom{s}{3} + bx^2\tbinom{s}{4}p^2\bigr\}p^{m-6}\bigr] + \qquad (m > 8). %% 13 +\end{multline} + +\medskip +5. \textit{Transformation of the Groups.}---The general group $G$ +of Class I is specified, in accordance with the relations (2) (11) +by two integers $a$, $b$ which (see (11)) are to be taken mod +$p^3$, mod $p^2$, respectively. Accordingly setting +\begin{gather*} +a = a_1 p^\lambda, \quad b = b_1 p^\mu, +\intertext{where} +dv[a_1,\, p] = 1, \quad dv[b_1,\, p] = 1 \qquad (\lambda = 0,\, 1,\, 2,\, 3;\; \mu = 0,\, 1,\, 2), +\intertext{we have for the group $G = G(a,\, b) = G(a,\, b)(P,\, Q)$ the +generational determination:} +G(a,\, b):\; \left \{ + \begin{gathered} + Q^{-1}\, P\, Q = Q^{b_1 p^{\mu + 1}} P^{1 + a_1 p^{m + \lambda - 6}} \\ + Q^{p^3} = 1, \quad P^{p^{m-3}} = 1. + \end{gathered} \right. +\end{gather*} + +Not all of these groups however are distinct. Suppose that +\begin{gather*} +G(a,\, b)(P,\, Q) \sim G(a',\, b')(P',\, Q'), +\intertext{by the correspondence} +C = \left[\begin{array}{cc} + Q, & P \\ + Q'_1, & P'_1 \\ + \end{array} \right], +\intertext{where} +Q'_1 = Q'^{y'} P'^{x'p^{m-6}}, \qquad \hbox{ and } \qquad P'_1 = Q'^y P'^x, +\end{gather*} +\noindent with $y'$ and $x$ prime to $p$. + +Since +\begin{gather*} +Q^{-1}\, P\, Q = Q^{bp} P^{1 + ap^{m-6}}, \\ +\intertext{then} +{Q'}_1^{-1}\, P'_1\, Q'_1 = {Q'}_1^{bp} {P'}_1^{1 + ap^{m-6}}, \\ +\intertext{or in terms of $Q'$, and $P'$} +\begin{aligned} + \bigl[y + b'xy'p &+ b'^2\tbinom{x}{2}y'p^2, x(1 + a'y'p^{m-6}) + a'b'\tbinom{x}{2}y'p^{m-5} \\ + &+ a'b'^2\tbinom{x}{3}y'p^{m-4}\bigr] = [y + by'p, x + (ax + bx'p)p^{m-6}] \qquad (m > 8) +\end{aligned} +\end{gather*} +\noindent and +\begin{gather} +by' \equiv b'xy' + b'^2\tbinom{x}{2}y'p \pmod{p^2}, \\ %% 14 +ax + bx'p \equiv a'y'x + a'b'\tbinom{x}{2}y'p + a'b'^2\tbinom{x}{3}y'p^2 \pmod{p^3}. %% 15 +\end{gather} +\noindent The necessary and sufficient condition for the simple +isomorphism of these two groups $G(a,\, b)$ and $G(a',\, b')$ is, that +the above congruences shall be consistent and admit of solution +for $x$, $y$, $x'$ and $y'$. The congruences may be written +\begin{gather*} +b_1 p^\mu \equiv b'_1 xp^{\mu'} + {b'}_1^2\tbinom{x}{2}p^{2\mu' + 1} \pmod{p^2}, \\ +\begin{aligned} + a_1 xp^{\lambda} + b_1 x'p^{\mu + 1} &\equiv \\ + y'\{a'_1 xp^{\lambda'} &+ a'_1 b'_1\tbinom{x}{2}p^{\lambda'+\mu'+1} + + a'_1 {b'}_1^2\tbinom{x}{3}p^{\lambda'+2\mu'+2}\} \pmod{p^3}. +\end{aligned} +\end{gather*} +\noindent Since $dv[x,\, p] = 1$ the first congruence gives $\mu = +\mu'$ and $x$ may always be so chosen that $b_1 = 1$. + +We may choose $y'$ in the second congruence so that $\lambda = +\lambda'$ and $a_1 = 1$ except for the cases $\lambda' \ge \mu + 1 += \mu' + 1$ when we will so choose $x'$ that $\lambda = 3$. + +The type groups of Class I for $m > 8$\footnote{For $m = 8$ the +additional term $ayp$ appears on the left side of the congruence +(14) and $G(1,\, p^2)$ and $G(1,\, p)$ become simply isomorphic. The +extra terms appearing in congruence (15) do not effect the result. +For $m = 7$ the additional term $ay$ appears on the left side of +(14) and $G(1,\, 1)$, $G(1,\, p)$, and $G(l,\, p^2)$ become simply +isomorphic, also $G(p,\, p)$ and $G(p,\, p^2)$.} are then given by +\begin{multline} +G(p^\lambda,\, p^\mu):\; Q^{-1}\, P\, Q = Q^{p^{1+\mu}} + P^{1+p^{m-6+\lambda}},\, Q^{p^3} = 1,\, P^{p^{m-3}} = 1 \\ +\left( + \begin{aligned} + \mu = 0,\, 1,\, 2;\;& \lambda = 0,\, 1,\, 2;\; \lambda \ge \mu; \\ + \mu = 0,\, 1,\, 2;\;& \lambda = 3 + \end{aligned} \right) +\tag{I}. +\end{multline} + +Of the above groups $G(p^\lambda,\, p^\mu)$ the groups for $\mu = 2$ have +the cyclic subgroup $\{P\}$ self-conjugate, while the group $G(p^3,\, p^2)$ +is the abelian group of type $(\mbox{$m-3$},\, 3)$. + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} II. \normalsize +\end{center} +\setcounter{equation}{0} +1. \textit{General relations.} + +There is in $G$ an operator $Q_1$ such that $Q{}_1^{p^2}$ is contained in +$\{P\}$ while $Q{}_1^p$ is not. +\begin{equation} +Q{}_1^{p^2} = P^{hp^2}. %% 1 +\end{equation} + +The operators $Q_1$ and $P$ either generate a subgroup $H_2$ of order +$p^{m-1}$, or the entire group $G$. + +\bigskip +\begin{center} +\large\textit{Section} 1. \normalsize +\end{center} + +2. \textit{Groups with independent generators.} + +Consider the first possibility in the above paragraph. There is in +$H_2$, a subgroup $H_1$ of order $p^{m-2}$, which contains $\{P\}$ +self-conjugately.\footnote{\textsc{Burnside}, \textit{Theory of +Groups}, Art.\ 54, p.\ 64.} $H_1$ is generated by $Q{}_1^p$ and $P$. +$H_2$ contains $H_1$ self-conjugately and is itself self-conjugate +in $G$. + +From these considerations\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{align} +Q{}_1^{-p}\, P\, Q{}_1^p &= P^{1 + kp^{m-4}}, \\ %% 2 +Q{}_1^{-1}\, P\, Q &= Q{}_1^{\beta p} P^{\alpha_1}. %% 3 +\end{align} + +\medskip +3. \textit{Determination of $H_1$ and $H_2$.} + +From (2) we obtain +\begin{equation} +[yp,\, x]^s = \bigl[syp,\, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\bigr] \quad (m > 4), %% 4 +\end{equation} +\noindent and from (3) and (4) +\begin{equation*} +[-p,\, 1,\, p] = \left[\frac{\alpha{}_1^p - 1}{\alpha_1 - 1}\beta p,\, + \alpha{}_1^p\left\{1 + \frac{\beta k}{2} + \frac{\alpha{}_1^p-1}{\alpha_1-1} p^{m-4} \right\} \right]. +\end{equation*} + +A comparison of the above equation with (2) shows that +\begin{gather*} +\frac{\alpha{}_1^p - 1}{\alpha_1 - 1} \beta p \equiv 0 \pmod{p^2}, \\ +\alpha{}_1^p \left\{1 + \frac{\beta k}{2}\frac{\alpha{}_1^p-1}{a_1-1} + p^{m-4} \right\} + \frac{\alpha{}_1^p - 1}{\alpha_1 - 1} \beta hp + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}, +\end{gather*} +\noindent and in turn +\begin{equation*} +\alpha{}_1^p \equiv 1 \pmod{p^2}, \qquad \alpha_1 \equiv 1 \pmod{p} \qquad (m > 5). +\end{equation*} + +Placing $\alpha_1 = 1 + \alpha_2 p$ in the second congruence, we obtain +as in Class I +\begin{equation} +(\alpha_2 + \beta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}} \qquad (m > 5). %% 5 +\end{equation} + +Equation (3) now becomes +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q^\beta P^{1 + \alpha_2 p}. %% 6 +\end{equation} +\noindent The generational equations of $H_2$ will be simplified +by using an operator of order $p^2$ in place of $Q_1$. + +From (5), (6) and (4) +\begin{gather*} +[y,\, x]^s = [sy + U_s p,\, sx + W_s p] +\intertext{in which} +\begin{aligned} +U_s &= \beta \tbinom{s}{2}xy, \\ +W_s &= \alpha_2 \tbinom{s}{2}xy + \Bigl\{ \beta k \bigl[\tbinom{s}{2}\tbinom{x}{2} + + \tbinom{s}{3}x^2 y\bigr] \\ + & \qquad \qquad \qquad + \frac{1}{2}\alpha k\bigl[\frac{1}{3!}s(s - 1)(2s - 1)y^2 + - \tbinom{s}{2}y\bigr]x \Bigr\} p^{m-5}. +\end{aligned} +\end{gather*} + +Placing $s = p^2$ and $y = 1$ in the above +\begin{gather*} +[Q_1\, P^x]^{p^2} = Q{}_1^{p^2} P^{xp^2} = P^{(x+h)p^2}. +\intertext{If $x$ be so chosen that} +(x + h) \equiv 0 \pmod{p^{m-5}} \qquad (m > 5) +\end{gather*} +\noindent $Q_1 P^x$ will be the required $Q$ of order $p^2$. + +Placing $h = 0$ in congruence (5) we find +\begin{equation*} +\alpha_2 p^2 \equiv kp^{m-4} \pmod{p^{m-3}}. +\end{equation*} + +Let $\alpha_2 = \alpha p^{m-6}$. $H_2$ is then generated by +\begin{equation*} +Q^{p^2} = 1, \quad P^{p^{m-3}} = 1. +\end{equation*} +\begin{equation} +Q^{-1}\, P\, Q = Q^{\beta p} P^{1 + \alpha p^{m-5}}. %% 7 +\end{equation} + +Two of the preceding formul\ae\ now become +\begin{gather} +[-y,\, x,\, y] = \bigl[\beta xyp,\, x(1 + \alpha yp^{m-5}) + \beta k\tbinom{x}{2}yp^{m-4}\bigr], \\ %% 8 +[y,\, x]^s = [sy + U_s p,\, xs + W_s p^{m-5}], %% 9 +\end{gather} +\noindent where +\begin{equation*} +U_s = \beta \tbinom{s}{2}xy +\end{equation*} +\noindent and\footnote{For $m = 6$ it is necessary to add the terms +$\frac{ak}{2} \left \{ \frac{s(s - 1)(2s - 1)}{3!}y^2 - \tbinom{s}{2}y \right \}p$ +to $W_s$.} +\begin{equation*} +W_s = \alpha \tbinom{s}{2}xy + \beta k\bigl\{\tbinom{s}{2}\tbinom{x}{2} + + \tbinom{s}{3}x^2\bigr\}yp \quad (m > 6). +\end{equation*} + +\medskip +4. \textit{Determination of $G$.} + +Let $R_1$ be an operation of $G$ not in $H_2$. $R{}_1^p$ is in $H_2$. Let +\begin{equation} +R{}_1^p = Q^{\lambda p} P^{\mu p}. %% 10 +\end{equation} %% 10 + +Denoting $R{}_1^a\, Q^b\, P^c\, R{}_1^d\, Q^e\, P^f \cdots$ by the symbol $[a,\, b,\, c,\, +d,\, e,\, f,\, \cdots]$, all the operations of $G$ are contained in the set $[z,\, +y,\, x]$; $z = 0,\, 1,\, 2,\, \cdots,\, p - 1$; $y = 0,\, 1,\, 2,\, \cdots,\, p^2 - 1$; $x = 0,\, +1,\, 2,\, \cdots,\, p^{m-3} - 1$. + +The subgroup $H_2$ is self-conjugate in $G$. From +this\footnote{\textsc{Burnside}, \textit{Theory of Groups}, Art.\ +24, p.\ 27.} +\begin{gather} +R{}_1^{-1}\, P\, R_1 = Q^{b_1} P^{a_1}, \\ %% 11 +R{}_1^{-1}\, Q\, R_1 = Q^{d_1} P^{c_1 p^{m-5}}. %% 12 +\end{gather} +\noindent In order to ascertain the forms of the constants in (11) +and (12) we obtain from (12), (11), and (9) +\begin{gather*} +[-p,\, 1,\, 0,\, p] = [0,\, d{}_1^p + Mp,\, Np^{m-5}]. +\intertext{By (10) and (8)} +R{}_1^p\, Q\, R{}_1^p = P^{-\mu p}\, Q\, P^{\mu p} = Q\, P^{-a\mu p^{m-4}}. +\intertext{From these equations we obtain} +d{}_1^p \equiv 1 \pmod p \quad \hbox{ and } \quad d_1 \equiv 1 \pmod p . +\end{gather*} +\noindent Let $d_1 = 1 + dp$. Equation (12) is replaced by +\begin{equation} +R{}_1^{-1}\, Q\, R_1 = Q^{1+dp} P^{e_1 p^{m-5}}. %% 13 +\end{equation} +\noindent From (11), (13) and (9) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = \left[\frac{a{}_1^p - 1}{a_1 - 1}b_1 + Kp,\, a{}_1^p + b_1 Lp^{m-5}\right] +\intertext{in which} +K = a_1 b_1 \beta \sum_1^{p-1}\tbinom{a{}_1^y}{2}. +\intertext{By (10) and (8)} +R{}_1^{-p}\, P\, R{}_1^p = Q^{-\lambda p} P\, Q^{\lambda p} = P^{1 + a \lambda p^{m-4}}, +\intertext{and from the last two equations} +a{}_1^p \equiv 1 \pmod{p^{m-5}} +\intertext{and} +a_1 \equiv 1 \pmod{p^{m-6}} \quad (m > 6); \qquad a_1 \equiv 1 \pmod{p} \quad (m = 6). +\end{gather*} + +Placing $a_1 = 1 + a_2 p^{m-6} \quad (m > 6)$; \qquad $a_1 = 1 + a_2 p \quad (m=6)$. +\begin{equation*} +K \equiv 0 \pmod{p}, +\end{equation*} +\noindent and\footnote{$K$ has an extra term for $m = 6$ and $p = +3$, which reduces to $3b_1 c_1$. This does not affect the +reasoning except for $c_1 = 2$. In this case change $P^2$ to $P$ +and $c_1$ becomes $1$.} +\begin{gather*} +\frac{a{}_1^p - 1}{a_1 - 1}b_1 \equiv b_1 p \equiv 0 \pmod{p^2}, + \qquad b_1 \equiv 0 \pmod p. +\intertext{Let $b_1 = bp$ and we find} +a{}_1^p \equiv 1 \pmod{p^{m-4}}, \qquad a_1 \equiv 1 \pmod{p^{m-5}}. +\end{gather*} + +Let $a_1 = 1 + a_3 p^{m-5}$ and equation (11) is replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q^{bp} P^{1 + a_3 p^{m-5}}. %% 14 +\end{equation} +\noindent The preceding relations will be simplified by taking for +$R_1$ an operator of order $p$. This will be effected by two +transformations. + +From (14), (9) and (13)\footnote{The extra terms appearing in the +exponent of $P$ for $m=6$ do not alter the result.} +\begin{gather*} +[1,\, y]^p = \Bigl[p,\, yp,\, \frac{-c_1 y}{2} p^{m-4}\Bigr] + = \Bigl[0,\, (\lambda + y)p,\, \mu p - \frac{c_1 y}{2} p^{m-4}\Bigr], +\intertext{and if $y$ be so chosen that} +\lambda + y \equiv 0 \pmod{p}, +\end{gather*} +\noindent $R_2 = R_1\, Q^y$ is an operator such that $R{}_2^p$ is in +$\{P\}$. + +Let +\begin{gather*} +R{}_2^p = P^{lp}. +\intertext{Using $R_2$ in the place of $R_1$, from (15), (9) and (14)} +[1,\, 0,\, x]^p = \Bigl[p,\, 0,\, xp + \frac{ax}{2} p^{m-4}\Bigr] = + \Bigl[0,\, 0,\, (x + l)p + \frac{ax}{2} p^{m-4}\Bigr], +\intertext{and if $x$ be so chosen that} +x + l + \frac{ax}{2} p^{m-5} \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent then $R = R_2 P^x$ is the required operator of order $p$. + +$R^p = 1$ is permutable with both $Q$ and $P$. Preceding equations now +assume the final forms +\begin{align} +Q^{-1}\, P\, Q & = Q^{\beta p} P^{1 + ap^{m-5}}, \\ %% 15 +R^{-1}\, P\, R & = Q^{bp} P^{1 + ap^{m-4}}, \\ %% 16 +R^{-1}\, Q\, R & = Q^{1 + dp} P^{cp^{m-4}}, %% 17 +\end{align} +with $R^p = 1$, $Q^{p^2} = 1$, $P^{p^{m-3}} = 1$. + +The following derived equations are necessary\footnote{For $m=6$ +the term $a^2 \tbinom{x}{2} xp^2$ must be added to the exponent of +$P$ in (18).} +\begin{align} +[0,\, -y,\, x,\, 0,\, y] &= \bigl[0,\, \beta xyp,\, x(1 + \alpha yp^{m-5}) + \alpha + \beta \tbinom{x}{2}yp^{m-4}\bigr], \\ %% 18 +[-y,\, 0,\, x,\, -y] &= \bigl[0,\, bxyp,\, x(1 + ayp^{m-4}) + + ab\tbinom{x}{2} yp^{m-4}\bigr], \\ %% 19 +[-y,\, x,\, 0,\, y] &= [0,\, x(1 + dyp),\, cxyp^{m-4}]. %% 20 +\end{align} + +From a consideration of (18), (19) and (20) we arrive at the +expression for a power of a general operator of $G$. +\begin{equation} +[z,\, y,\, x]^s = [sz,\, sy + U_s p,\, sx + V_s p^{m-5}], %% 21 +\end{equation} +\noindent where\footnote{When $m = 6$ the following terms are to +be added to $V_s$: $\frac{a^2 x}{2} \left\{\frac{s(s - 1)(2s - 1)}{3!}y^2 + - \tbinom{s}{2}y\right\}p.$} +\begin{align*} +U_s &= \tbinom{s}{2} \{bxz +\beta xy + dyz \}, \\ +V_s &= \tbinom{s}{2} \Bigl\{\alpha xy + \bigl[axz + \alpha \beta \tbinom{x}{2}y + + cyz + ab\tbinom{x}{2}z\bigr]p\Bigr\} \\ + & \qquad \qquad \qquad + \alpha\tbinom{s}{3} \{bxz + \beta xy + dyz \} xp. +\end{align*} + +\medskip +5. \textit{Transformation of the groups.} All groups of this +section are given by equations (15), (16), and (17) with $a,\, b,\, +\beta,\, c,\, d = 0,\, 1,\, 2,\, \cdots ,\, p - 1$, and $\alpha = 0,\, 1,\, 2,\, +\cdots ,\, p^2 - 1$, independently. Not all these groups, however, +are distinct. Suppose that $G$ and $G'$ of the above set are +simply isomorphic and that the correspondence is given by +\begin{equation*} +C = \left[ + \begin{matrix} + R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ + \end{matrix} +\right], +\end{equation*} +\noindent in which +\begin{align*} +R'_1 &= R'^{z''} Q'^{y''p} P'^{x''p^{m-4}}, \\ +Q'_1 &= R'^{z'} Q'^{y'} P'^{x'p^{m-5}}, \\ +P'_1 &= R'^z Q'^y P'^x, +\end{align*} +\noindent where $x$, $y'$ and $z''$ \textit{are prime} to $p$. + +The operators $R'_1$, $Q'_1$, and $P'_1$ must be independent since +$R$, $Q$, and $P$ are, and that this is true is easily verified. +The lowest power of $Q'_1$ in $\{P'_1\}$ is $Q'{}_1^{p^2} = 1$ and +the lowest power of $R'_1$ in $\{Q'_1, P'_1\}$ is $R'{}_1^p = 1$. +Let $Q'{}_1^{s'} = P'{}_1^{sp^{m-5}}$. + +This in terms of $R'$, $Q'$, and $P'$ is +\begin{gather*} +\Bigl[s'z',\, y'\bigl\{s' + d'\tbinom{s'}{2}z'p\bigr\},\, s'x'p^{m-5} + +c'\tbinom{s'}{2}y'z'p^{m-4}\Bigr] = [0,\, 0,\, sxp^{m-5}]. \\ +\intertext{From this equation $s'$ is determined by} +s'z' \equiv 0 \pmod{p} \\ +y'\{s' + d'\tbinom{s}{2}z'p\} \equiv 0 \pmod{p^2}, +\intertext{which give} +s'y' \equiv 0 \pmod{p^2}. +\intertext{Since $y'$ is prime to $p$} +s' \equiv 0 \pmod{p^2} +\end{gather*} +\noindent and the lowest power of $Q'_1$ contained in $\{P'_1\}$ +is $Q'{}_1^{p^2} = 1$. + +Denoting by ${R'}_1^{s''}$ the lowest power of $R'_1$ contained in +$\{Q'_1, P'_1\}$. +\begin{equation*} +{R'}_1^{s''} = {Q'}_1^{s'p} {P'}_1^{sp^{m-4}}. +\end{equation*} + +This becomes in terms of $R'$, $Q'$, and $P'$ +\begin{gather*} +[s''z'',\, s''y''p,\, s''x''p^{m-4}] = [0,\, s'y'p,\, \{s'x' + sx\}p^{m-4}]. +\intertext{$s''$ is now determined by} +s''z'' \equiv 0 \pmod{p} +\intertext{and since $z''$ is prime to $p$} +s'' \equiv 0 \pmod{p}. +\end{gather*} +\noindent The lowest power of $R'_1$ contained in $\{Q'_1, P'\}$ is therefore +${R'}_1^p = 1$. + +Since $R$, $Q$, and $P$ satisfy equations (15), (16), and (17) $R'_1$, +$Q'_1$, and $P'_1$ also satisfy them. Substituting in these equations the +values of $R'_1$, $Q'_1$, and $P'_1$ and reducing we have in terms of +$R'$, $Q'$, and $P'$ +\begin{gather} +[z,\, y + \theta_1 p,\, x + \phi_1 p^{m-5}] = + [z,\, y + \beta y'p,\, x(1 + \alpha p^{m-5}) + \beta xp^{m-4}], \\ %% 22 +[z,\, y + \theta_2 p,\, x + \phi_2 p^{m-4}] = + [z,\, y + by'p,\, x(1 + ap^{m-4}) + bx'p^{m-4}], \\ %% 23 +[z',\, y' + \theta_3 p,\, (x' + \phi_3 p)p^{m-5}] = + [z',\, y'(1 + dp),\, x(1 + dp)p^{m-5} + cxp^{m-4}], %% 24 +\end{gather} +\noindent in which +\begin{align*} +\theta_1 &= d'(yz' - y'z) + x(b'z' + \beta'y'), \\ +\theta_2 &= d'yz'' + b'xz'', \\ +\theta_3 &= d'y'z'', \\ +\phi_1 &= \alpha'xy' + \bigl\{\alpha'(\beta'y' + b'z')\tbinom{x}{2} + + a'xz + c'(yz'-y'z)\bigr\}p, \\ +\phi_2 &= \alpha'xy'' + a'xz'' + \alpha'b'\tbinom{x}{2}z'' + c'yz'', \\ +\phi_3 &= c'yz''. +\end{align*} + +A comparison of the members of the above equations give six congruences +between the primed and unprimed constants and the nine indeterminates. + +\begin{align*} +\theta_1 &\equiv \beta y' \pmod{p}, \tag{I} \\ +\phi_1 &\equiv \alpha x + \beta x'p \pmod{p^2}, \tag{II} \\ +\theta_2 &\equiv by' \pmod{p}, \tag{III} \\ +\phi_2 &\equiv ax + bx' \pmod{p}, \tag{IV} \\ +\theta_3 &\equiv dy' \pmod{p}, \tag{V} \\ +\phi_3 &\equiv cx + dx' \pmod{p}. \tag{VI} +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of +the two groups $G$ and $G'$ is, \textit{that the above congruences shall be +consistent and admit of solution for the nine indeterminates, with the +condition that $x$, $y'$ and $z''$ be prime to $p$.} + +For convenience in the discussion of these congruences, the groups are +divided into six sets, and each set is subdivided into 16 cases. + +The group $G'$ is taken from the simplest case, and we associate with +this case all cases, which contain a group $G$, simply isomorphic with +$G'$. Then a single group $G$, in the selected case, simply isomorphic +with $G'$, is chosen as a type. + +$G'$ is then taken from the simplest of the remaining cases and we proceed +as above until all the cases are exhausted. + +Let $\kappa = \kappa_1 p^{\kappa_2}$, and $dv_1[\kappa_1 ,\, p] = 1$ +$(\kappa = a,\, b,\, \alpha ,\, \beta ,\, c,$ and $d)$. + +The six sets are given in the table below. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c||c|c|c|} +\hline + &$\alpha_2$&$d_2$& &$\alpha_2$&$d_2$ \\ \hline +$A$& 0 & 0 &$D$& 2 & 0 \\ \hline +$B$& 0 & 1 &$E$& 1 & 1 \\ \hline +$C$& 1 & 0 &$F$& 2 & 1 \\ \hline +\end{tabular} +\end{center} + +\medskip +The subdivision into cases and the results are given in Table II. + +\begin{center} +\large II. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} +\hline + &$a_2$&$b_2$&$\beta_2$&$c_2$& $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\ \hline + 1& 1 & 1 & 1 & 1 & & & & & & \\ \hline + 2& 0 & 1 & 1 & 1 &$A_1$& $B_1$ & &$C_2$& & $E_2$ \\ \hline + 3& 1 & 0 & 1 & 1 &$A_1$& &$C_1$&$D_1$& & \\ \hline + 4& 1 & 1 & 0 & 1 &$A_1$& &$C_1$&$D_1$& & $E_4$ \\ \hline + 5& 1 & 1 & 1 & 0 &$A_1$& &$C_1$&$D_1$& & $E_5$ \\ \hline + 6& 0 & 0 & 1 & 1 &$A_1$& $B_3$ &$C_2$&$C_2$& $E_3$ & $F_3$ \\ \hline + 7& 0 & 1 & 0 & 1 &$A_1$& $B_4$ &$C_2$&$C_2$& & $E_7$ \\ \hline + 8& 0 & 1 & 1 & 0 &$A_1$& $B_5$ &$C_2$&$C_2$& $E_5$ & $E_5$ \\ \hline + 9& 1 & 0 & 0 & 1 &$A_1$& $B_3$ &$C_1$&$D_1$& $E_3$ & $F_3$ \\ \hline +10& 1 & 0 & 1 & 0 &$A_1$& &$C_2$&$C_2$& &$E_{10}$ \\ \hline +11& 1 & 1 & 0 & 0 &$A_1$& & * &$C_1$& &$E_{11}$ \\ \hline +12& 0 & 0 & 0 & 1 &$A_1$& $B_3$ &$C_2$&$C_2$& * & $E_3$ \\ \hline +13& 0 & 0 & 1 & 0 &$A_1$&$B_{10}$& * & * &$E_{10}$&$E_{10}$ \\ \hline +14& 0 & 1 & 0 & 0 &$A_1$&$B_{11}$&$C_2$&$C_2$&$E_{11}$&$E_{11}$ \\ \hline +15& 1 & 0 & 0 & 0 &$A_1$&$B_{10}$&$C_2$&$C_2$&$E_{10}$&$E_{10}$ \\ \hline +16& 0 & 0 & 0 & 0 &$A_1$&$B_{10}$& * & * &$E_{10}$&$E_{10}$ \\ \hline +\end{tabular} + +\footnotesize The groups marked (*) divide into two or three parts. \normalsize +\end{center} + +\medskip +Let $ad - bc = \theta_1 p^{\theta_2}$, $\alpha_1 d - \beta c = +\phi_1 p^{\phi_2}$ and $\alpha_1 b - a\beta = \chi_1 p^{\chi_2}$ with +$\theta_1$, $\phi_1$, and $\chi_1$ prime to $p$. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c||c|c|c|c|c|} +\hline + * &$\theta_2$&$\phi_2$&$\chi_2$& & * &$\theta_2$&$\phi_2$&$\chi_2$& \\ \hline +$C_{11}$& & 1 & &$D_1$&$D_{13}$& 1 & & &$D_1$ \\ \hline +$C_{11}$& & 0 & &$C_1$&$D_{13}$& 0 & & &$C_2$ \\ \hline +$C_{13}$& 1 & & &$C_1$&$D_{16}$& 1 & & &$C_1$ \\ \hline +$C_{13}$& 0 & & &$C_2$&$D_{16}$& 0 & & &$C_2$ \\ \hline +$C_{16}$& 1 & 1 & &$D_1$&$E_{12}$& & & 1 &$F_3$ \\ \hline +$C_{16}$& 1 & 0 & &$C_1$&$E_{12}$& & & 0 &$E_3$ \\ \hline +$C_{16}$& 0 & & &$C_2$& & & & & \\ \hline +\end{tabular} + +\newpage +6. \textit{Types.} +\end{center} + +The type groups are given by equations (15), (16) and (17) with the +values of the constants given in Table IV. + +\begin{center} +\large IV. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c||c|c|c|c|c|c|c|} +\hline + & $a$ &$b$&$\alpha$&$\beta$& $c$ &$d$& &$a$&$b$&$\alpha$&$\beta$& $c$ &$d$ \\ \hline +$A_1$ & 0 & 0 & 1 & 0 & 0 & 1 &$E_1$ & 0 & 0 & $p$ & 0 & 0 & 0 \\ \hline +$B_1$ & 0 & 0 & 1 & 0 & 0 & 0 &$E_2$ & 1 & 0 & $p$ & 0 & 0 & 0 \\ \hline +$B_3$ & 0 & 1 & 1 & 0 & 0 & 0 &$E_3$ & 0 & 1 & $p$ & 0 & 0 & 0 \\ \hline +$B_4$ & 0 & 0 & 1 & 1 & 0 & 0 &$E_4$ & 0 & 0 & $p$ & 1 & 0 & 0 \\ \hline +$B_5$ & 0 & 0 & 1 & 0 & 1 & 0 &$E_5$ & 0 & 0 & $p$ & 0 & 1 & 0 \\ \hline +$B_{10}$& 0 & 1 & 1 & 0 &$\kappa$& 0 &$E_7$ & 1 & 0 & $p$ & 1 & 0 & 0 \\ \hline +$B_{11}$& 0 & 0 & 1 & 1 & 1 & 0 &$E_{10}$& 0 & 1 & $p$ & 0 &$\kappa$& 0 \\ \hline +$C_1$ & 0 & 0 & $p$ & 0 & 0 & 1 &$E_{11}$& 0 & 0 & $p$ & 1 & 1 & 0 \\ \hline +$C_2$ &$\omega$& 0 & $p$ & 0 & 0 & 1 &$F_1$ & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$D_1$ & 0 & 0 & 0 & 0 & 0 & 1 &$F_3$ & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +\end{tabular} + +\footnotesize +\begin{align*} +\kappa &= 1, \hbox{ and a non-residue } \pmod{p}, \\ +\omega &= 1, 2, \cdots, p - 1. +\end{align*} +\normalsize +\end{center} + +\medskip +The congruences for three of these cases are completely analyzed as +illustrations of the methods used. + +\medskip +\begin{equation*} B_{10}. \end{equation*} + +The congruences for this case have the special forms. +\begin{gather*} +b'xz' \equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha'y' \equiv \alpha \pmod{p}, \tag{II} \\ +b'xz'' \equiv by' \pmod{p}, \tag{III} \\ +\alpha'xy'' + \alpha'b'\tbinom{x}{2}z'' + c'yz'' \equiv ax + bx' \pmod{p}, \tag{IV} \\ +d \equiv 0 \pmod{p}, \tag{V} \\ +c'y'z'' \equiv cx \pmod{p}. \tag{VI} +\end{gather*} + +Since $z'$ is unrestricted (I) gives $\beta \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +From (II) since $y' \not\equiv 0, \alpha \not\equiv 0 \pmod{p}$. + +From (III) since $x, y', z'' \not\equiv 0$, $b \not\equiv 0 \pmod{p}$. + +In (IV) $b \not\equiv 0$ and $x'$ is contained in this congruence alone, +and, therefore, $a$ may be taken $\equiv 0$ or $\not\equiv 0 \pmod{p}$. + +(V) gives $d \equiv 0 \pmod{p}$ and (VI), $c \not\equiv 0 \pmod{p}$. + +Elimination of $y'$ between (III) and (VI) gives +\begin{equation*} +b'c'z''^{2} \equiv bc \pmod{p} +\end{equation*} +\noindent so that $bc$ is a quadratic residue or non-residue (mod $p$) +according as $b'c'$ is a residue or non-residue. + +The types are given by placing $a = 0$, $b = 1$, $\alpha = 1$, $\beta = 0$, +$c = \kappa$, and $d = 0$ where $\kappa$ has the two values, 1 and a +representative non-residue of $p$. + +\medskip +\begin{equation*} C_2. \end{equation*} + +The congruences for this case are +\begin{gather*} +d'(yz' - y'z) \equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha'_1 xy' + a'xz' \equiv \alpha_1 x + \beta x' \pmod{p}, \tag{II} \\ +d'yz'' \equiv by' \pmod{p}, \tag{III} \\ +a'xz'' \equiv ax + bx' \pmod{p}, \tag{IV} \\ +d'z'' \equiv d \pmod{p}, \tag{V} \\ +cx + dx' \equiv 0 \pmod{p}. \tag{VI} +\end{gather*} + +Since $z$ appears in (I) alone, $\beta$ can be either $\equiv 0$ or +$\not\equiv 0 \pmod{p}$. (II) is linear in $z'$ and, therefore, $\alpha +\equiv 0$ or $\not\equiv 0 \pmod{p}$, (III) is linear in $y$ and, +therefore, $b \equiv 0$ or $\not\equiv 0$. + +Elimination of $x'$ and $z''$ between (IV), (V), and (VI) gives +\begin{equation*} +a'd^2 \equiv d'(ad - bc) \pmod{p}. +\end{equation*} +\noindent Since $z''$ is prime to $p$, (V) gives $d \not\equiv 0 \pmod{p}$, so that +$ad - bc \not\equiv 0 \pmod{p}$. We may place $b = 0$, $\alpha = p$, +$\beta = 0$, $c = 0$, $d = 1$, then $a$ will take the values $1, 2, 3, \cdots, +p - 1$ giving $p - 1$ types. + +\medskip +\begin{equation*} D_1. \end{equation*} + +The congruences for this case are +\begin{align*} +d'(yz' - y'z) &\equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha_1 x + \beta x' &\equiv 0 \pmod{p}, \tag{II} \\ +d'yz'' &\equiv by' \pmod{p}, \tag{III} \\ +ax + bx' &\equiv 0 \pmod{p}, \tag{IV} \\ +d'z'' &\equiv d \pmod{p}, \tag{V} \\ +cx + dx' &\equiv 0 \pmod{p}. \tag{VI} +\end{align*} +\noindent $z$ is contained in (I) alone, and therefore $\beta \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +(III) is linear in $y$, and $b \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +(V) gives $d \not\equiv 0 \pmod{p}$. + +Elimination of $x'$ between (II) and (VI) gives $\alpha_1 d - \beta c +\equiv 0 \pmod{p}$, and between (IV) and (VI) gives $ad - bc \equiv 0 +\pmod{p}$. The type group is derived by placing $a = 0$, $b = 0$, $\alpha = 0$, +$\beta = 0$, $c = 0$ and $d = 1$. + +\bigskip +\begin{center} +\large\textit{Section} 2. \normalsize +\end{center} +\setcounter{equation}{0} + +1. \textit{Groups with dependent generators.} In this section, $G$ is generated +by $Q_1$ and $P$ where +\begin{equation} +Q{}_1^{p^2} = P^{hp^2}. %% 1 +\end{equation} +\noindent There is in $G$, a subgroup $H_1$, of order $p^{m-2}$, which contains $\{P\}$ +self-con\-ju\-gate\-ly.\footnote{\textsc{Burnside}, \textit{Theory of Groups}, +Art.\ 54, p.\ 64.} $H_1$ either contains, or does not contain $Q{}_1^p$. We will +consider the second possibility in the present section, reserving the first for +the next section. + +\medskip +2. \textit{Determination of $H_1$.} $H_1$ is generated by $P$ and some other operator +$R_1$ of $G$. $R{}_1^p$ is contained in $\{P\}$. Let +\begin{equation} +R{}_1^p = P^{lp}. %% 2 +\end{equation} +\noindent Since $\{P\}$ is self-conjugate in $H_1$,\footnote{\textsc{Burnside}, +\textit{Theory of Groups}, Art.\ 56, p.\ 66.} +\begin{equation} +R{}_1^{-1}\, P\, R_1 = P^{1 + kp^{m-4}} %% 3 +\end{equation} +\noindent Denoting $R{}_1^a\, P^b\, R{}_1^c\, P^d \cdots$ by the symbol $[a,\, b,\, c,\, d,\, +\cdots]$ we derive from (3) +\begin{gather} +[-y,\, x,\, y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4), %% 4 +\intertext{and} +[y,\, x]^s = \Bigl[sy,\, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\Bigr] %% 5 +\end{gather} +\noindent Placing $s = p$ and $y = 1$ in (5) we have, from (2) +\begin{gather*} +[R_1\, P^x]^p = R{}_1^p P^{xp} = P^{(l + x)p}. +\intertext{Choosing $x$ so that} +x + l \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent $R = R_1 P^x$ is an operator of order $p$, which will be used in the place +of $R_1$, and $H = \{R, P\}$ with $R^p = 1$. + +\medskip +3. \textit{Determination of $H_2$.} We will now use the symbol $[a,\, b,\, c,\, d,\, e,\, f,\, +\cdots]$ to denote $Q{}_1^a\, R^b\, P^c\, Q{}_1^d\, R^e\, P^f \cdots$. + +$H_1$ and $Q_1$ generate $G$ and all the operations of $G$ are given by +$[x,\, y,\, z]$ ($z = 0,\, 1,\, 2,\, \cdots,\, p^2 - 1$; $y = 0,\, 1,\, 2,\, \cdots,\, p - 1$; +$x = 0,\, 1,\, 2,\, \cdots,\, p^{m-3} - 1$), since these are $p^m$ in number and +are all distinct. There is in $G$ a subgroup $H_2$ of order $p^{m-1}$ +which contains $H_1$ self-conjugately. $H_2$ is generated by $H_1$ and +some operator $[z,\, y,\, x]$ of $G$. $Q{}_1^z$ is then in $H_2$ and $H_2$ +is the subgroup $\{Q{}_1^p, H_1\}$. Hence, +\begin{gather} +Q{}_1^{-p}\, P\, Q{}_1^p = R^\beta P^{\alpha_1}, \\ %% 6 +Q{}_1^{-p}\, P\, Q{}_1^p = R^{b_1} P^{ap^{m-4}}. %% 7 +\end{gather} +\noindent To determine $\alpha_1$ and $\beta$ we find from (6), (5) and (7) +\begin{multline*} +[-p^2,\, 0,\, 1,\, p^2] = \biggl[ 0,\, \frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1} + \beta,\, \alpha{}_1^p\Bigl\{1 + \frac{\beta k}{2}\frac{\alpha{}_1^p - 1} + {\alpha_1 - 1}p^{m-4} \Bigr\} \\ + + a\beta\Bigl\{ p\frac{\alpha{}_1^{p-1}}{\alpha_1 - b_1} - + \frac{\alpha{}_1^p - b{}_1^p}{(\alpha_1 - b_1)^2} + \Bigr\}p^{m-4} \biggr]. +\end{multline*} +\noindent By (1) +\begin{gather*} +Q{}_1^{-p^2}\, P\, Q{}_1^{p^2} = P, +\intertext{and, therefore,} +\frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta \equiv 0 \pmod{p}, \\ + \alpha{}_1^p \equiv 1 \pmod{p^{m-4}}, \qquad \hbox{ and } \qquad + \alpha_1 \equiv 1 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} + +Let $\alpha_1 = 1 + \alpha_2 p^{m-5}$ and equation (6) is replaced by +\begin{equation} +Q{}_1^{-p}\, P\, Q{}_1^p = R^\beta P^{1 + \alpha_2 p^{m-5}}. %% 8 +\end{equation} + +To find $a$ and $b_1$ we obtain from (7), (8) and (5) +\begin{gather*} +[-p^2,\, 1,\, 0,\, p^2] = \Bigl[0,\, b{}_1^p,\, a\frac{b{}_1^p - 1}{b_1 - 1}p^{m-4} \Bigr]. +\intertext{By (1) and (4)} +Q{}_1^{-p^2}\, R\, Q{}_1^{p^2} = P^{-lp^2} R\, P^{lp^2} = R, +\intertext{and, hence,} +b{}_1^p \equiv 1 \pmod{p}, \qquad a\frac{b{}_1^p - 1}{b_1 - 1} \equiv 0 \pmod{p}, +\end{gather*} +\noindent therefore $b_1 = 1$. + +Substituting $b_1 = 1$ and $\alpha_1 = 1 + \alpha_2 p^{m-5}$ in the +congruence determining $\alpha_1$ we obtain $(1 + \alpha_2 p^{m-5})^p +\equiv 1 \pmod{p^{m-3}}$, which gives $\alpha_2 \equiv 0 \pmod{p}$. + +Let $\alpha_2 = \alpha p$ and equations (8) and (7) are now replaced by +\begin{align} +Q{}_1^p\, P\, Q{}_1^p &= R^\beta P^{1 + \alpha p^{m-4}}, \\ %% 9 +Q{}_1^{-p}\, R\, Q{}_1^p &= RP^{ap^{m-4}}. %% 10 +\end{align} + +From these we derive +\begin{align} +[-yp,\, 0,\, x,\, yp] &= \Bigl[0,\, \beta xy,\, x + \bigl\{\alpha xy + + a\beta x\tbinom{y}{2} + \beta k\tbinom{x}{2}y\bigr\}p^{m-4}\Bigr], \\ %% 11 +[-yp,\, x,\, 0,\, yp] &= [0,\, x,\, axyp^{m-4}]. %% 12 +\end{align} + +A continued use of (4), (11), and (12) yields +\begin{equation} +[zp,\, y,\, x]^s = [szp,\, sy + U_s,\, sx + V_sp^{m-4}] %% 13 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta\tbinom{s}{2}xz, \\ +V_s &= \tbinom{s}{2}\Bigl\{\alpha xz + \beta k\tbinom{s}{2}z + kxy + + ayz\Bigr\} + \beta k\tbinom{s}{3}x^2 z \\ + & \qquad \qquad \qquad + \frac{1}{2}a\beta\Bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 + - \tbinom{s}{2}z\Bigr\}. +\end{align*} + +\medskip +4. \textit{Determination of $G$.} + +Since $H_2$ is self-conjugate in $G_1$ we have +\begin{align} +Q{}_1^{-1}\, P\, Q_1 &= Q{}_1^{\gamma p} R^\delta P^{\epsilon_1}, \\ %% 14 +Q{}_1^{-1}\, R\, Q_1 &= Q{}_1^{cp} R^d P^{ep^{m-4}}. %% 15 +\end{align} + +From (14), (15) and (13) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = [\lambda p,\, \mu,\, \epsilon{}_1^p + vp^{m-4}] +\intertext{and by (9) and (1)} +\lambda p \equiv 0 \pmod{p^2}, \qquad +\epsilon{}_1^p + \nu p^{m-4} + \lambda hp \equiv 1 + \alpha p^{m-4} + \pmod{p^{m-3}}, +\intertext{from which} +\epsilon{}_1^p \equiv 1 \pmod{p^2}, \quad \hbox{ and } \quad \epsilon_1 \equiv 1 \pmod{p} + \qquad (m > 5). +\end{gather*} + +Let $\epsilon_1 = 1 + \epsilon_2 p$ and equation (14) is replaced by +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q{}_1^{\gamma p} R^\delta P^{1 + \epsilon_2 p}. %% 16 +\end{equation} + +From (15), (16), and (13) +\begin{gather*} +[-p,\, 1,\, 0,\, p] = \left[c\frac{d^p - 1}{d - 1}p,\, d^p,\, Kp^{m-4} \right] +\intertext{where} +K = \frac{d^p - 1}{d - 1}e + \sum_{1}^{p-1} acd\frac{d^n(d^n - 1)}{2}. +\end{gather*} + +By (10) +\begin{gather*} +d^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad d = 1 +\intertext{and by (1)} +chp^2 \equiv ap^{m-4} \pmod{p^{m-3}}. +\end{gather*} + +Equation (15) is now replaced by +\begin{equation} +Q{}_1^{-1}\, R\, Q_1 = Q{}_1^{cp} R P^{ep^{m-4}}. %% 17 +\end{equation} + +A combination of (17), (16) and (13) gives +\begin{equation*} +[-p,\, 0,\, 1,\, p] = \Bigl[\bigl\{\gamma\frac{(1 + \epsilon_2 p)^p - 1} +{\epsilon_2 p^2} + c\delta\frac{p - 1}{2} \bigr\}p^2,\, 0,\, (1 + +\epsilon_2 p)^p \Bigr]. +\end{equation*} + +By (9) +\begin{equation*} +\Bigl\{\gamma\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} + c\delta +\frac{p - 1}{2}\Bigr\}hp^2 + (1 + \epsilon_2 p)^p \equiv 1 + \alpha p^{m-4} +\pmod{p^{m-3}}, +\end{equation*} +\noindent $\beta \equiv 0 \pmod{p}.$ + +A reduction of the first congruence gives +\begin{gather*} +\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon p^2}\bigl\{\epsilon_2 + \gamma h\bigr\}p^2 + \equiv \Bigl\{\alpha - a\delta\frac{p - 1}{2}\Bigr\}p^{m-4} \pmod{p^{m-3}} +\intertext{and, since} +\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} \equiv 1 \pmod{p}, \qquad +(\epsilon_2 + \gamma h)p^2 \equiv 0 \pmod{p^{m-4}} +\end{gather*} +\noindent and +\begin{equation} +(\epsilon_2 + \gamma h)p^2 \equiv \bigl(\alpha + \frac{a\delta}{2}\bigr)p^{m-4} + \pmod{p^{m-3}}. %% 18 +\end{equation} + +From (17), (16), (13) and (18) +\begin{align} +[-y,\, x,\, 0,\, y] &= \Bigl[cxyp,\, x,\, \bigl\{exy + ac\tbinom{x}{2}y\bigr\}p^{m-4}\Bigr], \\ %% 19 +[-y,\, 0,\, x,\, y] &= \Bigl[x\bigl\{\gamma y + c\delta\tbinom{y}{2}\bigr\}p,\, + \delta xy,\, x(1 + \epsilon_2 yp) + \theta p^{m-4}\Bigr] %% 20 +\end{align} +\noindent where +\begin{multline*} +\theta = \Bigl\{e\delta x + a\delta\gamma x + \epsilon_2 \left(\alpha + + \frac{a\delta}{2}\right)x\Bigr\}\tbinom{y}{2} \\ + + \frac{1}{2}ac \Bigl\{\frac{1}{3!}y(y-1)(2y-1)\delta^2 + - \tbinom{y}{2}\delta\Bigr\} \\ + + \bigl\{\alpha\gamma y + \delta ky + a\delta xy^2 + + (ac\delta^2 y + ac\delta) \tbinom{y}{2}\bigr\} \tbinom{x}{2}. +\end{multline*} + +From (19), (20), (4) and (18) +\begin{equation*} +\{Q_1\, P^x\}^{p^2} = Q{}_1^{p^2} P^{xp^2} = P^{(h+x)p^2}. +\end{equation*} + +If $x$ be so chosen that +\begin{equation*} +h + x \equiv 0 \pmod{p^{m-5}} +\end{equation*} +\noindent $Q = Q_1\, P^x$ is an operator of order $p^2$ which will be used in place +$Q_1$ and $Q^{p^2} = 1$. + +Placing $h = 0$ in (18) we get +\begin{equation*} +\epsilon_2 p^2 \equiv 0 \pmod{p^{m-4}}. +\end{equation*} + +Let $\epsilon_2 = \epsilon p^{m-6}$ and equation (16) is replaced by +\begin{equation} +Q^{-1}\, P\, Q = Q^{\gamma p} R^\delta P^{1 + \epsilon p^{m-5}} %% 21 +\end{equation} + +The congruence +\begin{gather*} +ap^{m-4} \equiv chp^2 \pmod{p^{m-3}} +\intertext{becomes} +ap^{m-4} \equiv 0 \pmod{p^{m-3}}, \qquad \hbox{ and } \qquad a \equiv 0 \pmod{p}. +\end{gather*} +\noindent Equations (19) and (20) are replaced by +\begin{align} +[-y,\, x,\, 0,\, y] &= [cxyp,\, x,\, exyp^{m-4}] \\ %% 22 +[-y,\, 0,\, x,\, y] &= \Bigl[ \bigl\{\gamma y + c\delta\tbinom{y}{2} \bigr\}xp,\, + \delta xy,\, x(1 + \epsilon yp^{m-5}) + \theta p^{m-4} \Bigr] %% 23 +\end{align} +\noindent where +\begin{equation*} +\theta = e\delta x\tbinom{y}{2} + \bigl\{\alpha\gamma y + \delta ky + + \alpha c\delta\tbinom{y}{2}\bigr\}\tbinom{x}{2}. +\end{equation*} + +A formula for any power of an operation of $G$ is derived from (4), +(22) and (23) +\begin{equation} +[z,\, y,\, x]^s = [sz + U_s p,\, sy + V_s,\, sx + W_s p^{m-5}] %% 24 +\end{equation} +\noindent where +\begin{align*} +U_s &= \tbinom{s}{2}\bigl\{\gamma xz + cyz\bigr\} + \frac{1}{2}c\delta x + \Bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\Bigr\}, \\ +V_s &= \delta \tbinom{s}{2}xz, \\ +W_s &= \tbinom{s}{2} \Bigl\{\epsilon xz + \bigl[(a\gamma + \delta k)\tbinom{x}{2}z + eyz + kxy\bigr]p\Bigr\} \\ + & \qquad \qquad + \tbinom{s}{3}\bigl\{\epsilon \gamma x + \epsilon y + \delta kx \bigr\}xzp + + \frac{1}{2}c \delta \epsilon \bigl\{\frac{1}{2}(s - 1)z^2 - z\bigr\} \tbinom{s}{3}xp \\ + & \qquad \qquad + \frac{1}{2}\bigl\{\delta ex + \alpha c \delta \tbinom{x}{2}\bigr\} + \bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\bigr\}p. +\end{align*} + +\medskip +5. \textit{Transformations of the groups.} Placing $y = 1$ and $x = -1$ in (22) +we obtain (17) in the form +\begin{equation*} +R^{-1}\, Q\, R = Q^{1-cp} P^{-ep^{m-4}}. +\end{equation*} +\noindent A comparison of the generational equations of the present section with +those of Section 1, shows that groups, in which $\delta \equiv 0 \pmod{p}$, +are simply isomorphic with those in Section 1, so we need consider only +those cases in which $\delta \not\equiv 0 \pmod{p}$. + +All groups of this section are given by +\begin{equation*} +G: \left\{ \begin{aligned} + R^{-1}\, P\, R &= P^{1 + kp^{m-4}}, \\ + Q^{-1}\, P\, Q &= Q^{\gamma p} R^\delta P^{1 + \epsilon p^{m-5}}, \\ + Q^{-1}\, R\, Q &= Q^{cp} R\, P^{\epsilon p^{m-4}}. \\ \end{aligned} +\right. \tag*{(25), (26), (27)} +\end{equation*} +\noindent $R^p = 1$, $Q^{p^2} = 1$, and $P^{p^{m-3}} = 1$, $(k,\, \gamma,\, c,\, e = 0,\, 1,\, +2,\, \cdots ,\, p - 1$; $\delta = 1,\, 2,\, \cdots,\, p - 1$; $\epsilon = 0,\, 1,\, 2,\, \cdots,\, +p^2 - 1)$. + +Not all these groups, however, are distinct. Suppose that $G$ and $G'$ of +the above set are simply isomorphic and that the correspondence is given by +\begin{equation*} +C = \left[\begin{matrix}R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right] . +\end{equation*} +\noindent Since $R^p = 1$, $Q^{p^2} = 1$, and $P^{p^{m-3}} = 1$, $R'{}_1^p = 1$, +$Q'{}_1^{p^2} = 1$ and $P'{}_1^{p^{m-3}}$. + +The forms of these operators are then +\begin{align*} +P'_1 &= Q'^z R'^y P'^x, \\ +R'_1 &= Q'^{z'p} R'^{y'} P'^{x'p^{m-4}}, \\ +Q'_1 &= Q'^{z''} R'^{y''}P'^{x''p^{m-5}}, \ +\end{align*} +\noindent where $dv[x,\, p] = 1$. + +Since $R$ is not contained in $\{P\}$, and $Q^p$ is not contained in +$\{R, P\}$ $R'_1$ is not contained in $\{P'_1\}$, and $Q'{}_1^p$ is not contained in +$\{R'_1, P'_1\}$. + +Let +\begin{gather*} +{R'}_1^{s'} = {P'}_1^{sp^{m-4}}. +\intertext{This becomes in terms of $Q'$, $R'$ and $P'$} +[s'z'p,\, s'y',\, s'x'p^{m-4}] = [0,\, 0,\, sxp^{m-4}], +\intertext{and} +s'y' \equiv 0 \pmod{p}, \qquad s'z' \equiv 0 \pmod{p}. +\end{gather*} +\noindent Either $y'$ or $z'$ is prime to $p$ or $s'$ may be taken $= 1$. + +Let +\begin{gather*} +{Q'}_1^{s''p} = {R'}_1^{s'} P'{}_1^{sp^{m-4}}, +\intertext{and in terms of $Q'$, $R'$ and $P'$} +[s''z''p,\, 0,\, s''x''p^{m-4}] = [s'z'p,\, s'y',\, (s'x' + sx)p^{m-4}], +\intertext{from which} +s''z'' \equiv s'z' \pmod{p}, \qquad \hbox{ and } \qquad s'y' \equiv 0 \pmod{p}. +\intertext{Eliminating $s'$ we find} +s''y'z'' \equiv 0 \pmod{p}, +\end{gather*} +\noindent $dv[y'z'',\, p] = 1$ or $s''$ may be taken $= 1$. We have then $z''$, $y'$ +and $x$ \textit{prime to} $p$. + +Since $R$, $Q$ and $P$ satisfy equations (25), (26) and (27) $R'_1$, $Q'_1$ +and $P'_1$ do also. These become in terms of $R'$, $Q'$ and $P'$. +\begin{align*} +[z + \Phi'_1 p,\, y,\, x + \Theta'_1 p^{m-4}] &= [z,\, y,\, x(1 + kp^{m-4})], \\ +[z + \Phi'_2 p,\, y + \delta'xz'',\, x + \Theta'_2 p^{m-5}] &= [z + \Phi_2 p,\, + y + \delta y',\, x + \Theta_2 p^{m-5}], \\ +[(z' + \Phi'_3)p,\, y',\, \Theta'_3 p^{m-4}] &= [(z' + \Phi_3)p,\, y,\, \Theta'_3 p^{m-4}], +\end{align*} +\noindent where +\begin{align*} +\Phi'_1 &= -c'yz', \quad \Theta'_1 = \epsilon'xz' + k'xy' - e'y'z, \\ +\Phi'_2 &= \Bigl\{\gamma'z'' + c'\delta'\tbinom{z}{2}\Bigr\}x + c'(yz'' - y''z), \\ +\Theta'_2 &= \epsilon'xz'' + \Bigl\{\tbinom{x}{2}\bigl[\alpha'\gamma'z'' + + \alpha'c'\delta'\tbinom{z''}{2} + \delta'k'z''\bigr] \\ + & \qquad \qquad \qquad + \delta'e'x \tbinom{z''}{2} + e'(yz'' - y''z) + k'xy''\Bigr\}p, \\ +\Phi_2 &= \gamma z'' + \delta z' + c'\delta y'z, \quad \Theta_2 \equiv + \epsilon x + (\gamma x'' + \delta x + e'\delta y'z)p, \\ +\Phi'_3 &= c'y'z'', \quad \Theta'_3 = e'y'z'', \quad \Phi_3 = cz'', \quad + \Theta_3 = ex + cx''. +\end{align*} + +A comparison of the members of these equations give seven congruences +\begin{align*} +\Phi'_1 &\equiv 0 \pmod{p}, \tag{I} \\ +\Theta'_1 &\equiv kx \pmod{p}, \tag{II} \\ +\Phi'_2 &\equiv \Phi_2 \pmod{p}, \tag{III} \\ +\delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ +\Theta'_2 &\equiv \Theta_2 \pmod{p^2}, \tag{V} \\ +\Phi_3' &\equiv cz'' \pmod{p}, \tag{VI} \\ +\Theta'_3 &\equiv \Theta_3 \pmod{p}. \tag{VII} +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of $G$ +and $G'$ is, \textit{that these congruences be consistent and admit of solution +for the nine indeterminants with $x$, $y'$, and $z''$ prime to $p$}. + +Let $\kappa = \kappa_1 p^{\kappa_2},\, dv[\kappa_1,\, p] = 1\; (\kappa = k,\, +\delta,\, \gamma,\, \epsilon,\, c,\, e)$. + +The groups are divided into three parts and each part is subdivided into +16 cases. + +The methods used in discussing the congruences are the same as those +used in Section 1. + +\medskip +6. \textit{Reduction to types.} The three parts are given by + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|} \hline + & $\epsilon_2$ & $\delta_2$ \\ \hline + $A$ & 0 & 0 \\ \hline + $B$ & 1 & 0 \\ \hline + $C$ & 2 & 0 \\ \hline +\end{tabular} +\end{center} + +The subdivision into cases and the results of the discussion of the +congruences are given in Table II. + +\medskip +\begin{center} +\large II. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline + &$k_2$&$\gamma_2$&$c_2$&$e_2$& $A$ & $B$ & $C$ \\ \hline + 1 & 1 & 1 & 1 & 1 & & & $B_1$ \\ \hline + 2 & 0 & 1 & 1 & 1 & & & $B_2$ \\ \hline + 3 & 1 & 0 & 1 & 1 & $A_2$ & $B_1$ & $B_1$ \\ \hline + 4 & 1 & 1 & 0 & 1 & & & $B_4$ \\ \hline + 5 & 1 & 1 & 1 & 0 & & & $B_5$ \\ \hline + 6 & 0 & 0 & 1 & 1 & * & $B_2$ & $B_2$ \\ \hline + 7 & 0 & 1 & 0 & 1 & $A_4$ & & $B_7$ \\ \hline + 8 & 0 & 1 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 9 & 1 & 0 & 0 & 1 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 10 & 1 & 0 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 11 & 1 & 1 & 0 & 0 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 12 & 0 & 0 & 0 & 1 & $A_4$ & $B_7$ & $B_7$ \\ \hline + 13 & 0 & 0 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 14 & 0 & 1 & 0 & 0 & $A_4$ & $B_7$ & $B_7$ \\ \hline + 15 & 1 & 0 & 0 & 0 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 16 & 0 & 0 & 0 & 0 & $A_4$ & $B_7$ & $B_7$ \\ \hline +\end{tabular} +\end{center} + +$A_6$ divides into two parts. + +The groups of $A_6$ in which $\delta k + \epsilon\gamma \equiv 0 \pmod{p}$ +are simply isomorphic with the groups of $A_1$ and those in which $\delta +k + \epsilon\gamma \not\equiv 0 \pmod{p}$ are simply isomorphic with the +groups of $A_2$. The types are given by equations (25), (26) and (27) where +the constants have the values given in Table III. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|} \hline + & $k$ & $\delta$ & $\gamma$ & $\epsilon$ & $c$ & $e$ \\ \hline + $A_1$ & 0 & 1 & 0 & 1 & 0 & 0 \\ \hline + $A_2$ & 1 & 1 & 0 & 1 & 0 & 0 \\ \hline + $A_4$ & 0 & 1 & 0 & 1 & 1 & 0 \\ \hline + $A_5$ & 0 & 1 & 0 & 1 & 0 & $\omega$ \\ \hline + $B_1$ & 0 & 1 & 0 & $p$ & 0 & 0 \\ \hline + $B_2$ & 1 & 1 & 0 & $p$ & 0 & 0 \\ \hline + $B_4$ & 0 & 1 & 0 & $p$ & 1 & 0 \\ \hline + $B_5$ & 0 & 1 & 0 & $p$ & 0 & $\kappa$ \\ \hline + $B_7$ & 1 & 1 & 0 & $p$ & $\omega$ & 0 \\ \hline +\end{tabular} + +\footnotesize +\begin{align*} +\kappa &= 1, \hbox { and a non-residue } \pmod{p}, \\ +\omega &= 1, 2, \cdots, p - 1. +\end{align*} +\normalsize +\end{center} + +A detailed analysis of several cases is given below, as a general +illustration of the methods used. + +\medskip +\begin{equation*} A_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{align*} + \epsilon'xz' &\equiv kx \pmod{p}, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ + \epsilon'xz'' &\equiv \epsilon x \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + ex &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} +\noindent Congruence (IV) gives $\delta \not\equiv 0 \pmod{p}$, from (II) $k$ can be +$\equiv 0$ or $\not\equiv 0 \pmod{p}$, (III) gives $\gamma \equiv 0$ +or $\not\equiv 0$, (V) gives $\epsilon \not\equiv 0$, (VI) and (VII) +give $c \equiv e \equiv 0 \pmod{p}$. Elimination of $x$, $z'$ and $z''$ +between (II), (III) and (V) gives $\delta k + \gamma\epsilon \equiv 0 +\pmod{p}$. If $k \equiv 0$, then $\gamma \equiv 0 \pmod{p}$ and +if $k \not\equiv 0$, then $\gamma \not\equiv 0 \pmod{p}$. + +\medskip +\begin{equation*} A_2. \end{equation*} + +The congruences for this case are +\begin{align*} + \epsilon'xz' + k'xy' &\equiv kx \pmod{p}, \tag{II} \\ +\gamma x'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ + \epsilon'xz'' &\equiv \epsilon x \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + ex &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} +\noindent Congruence (III) gives $\gamma \equiv 0$ or $\not\equiv 0$, (IV) gives +$\delta \not\equiv 0$, (V) $\epsilon \not\equiv 0$, (VI) and (VII) give $c +\equiv e \equiv 0 \pmod{p}$. Elimination of $x$, $z'$, and $z''$ between +(II), (III) and (V) gives +\begin{gather*} +\delta k + \gamma\epsilon \equiv k'\delta y' \pmod{p} +\intertext{from which} +\delta k + \gamma\epsilon \not\equiv 0 \pmod{p}. +\end{gather*} +\noindent If $k \equiv 0$, then $\gamma\not\equiv 0$, and if $\gamma \equiv 0$ then +$k \not\equiv 0 \pmod {p}$. + +Both $\gamma$ and $k$ can be $\not\equiv 0 \pmod{p}$ provided the above +condition is fulfilled. + +\medskip +\begin{equation*} A_5. \end{equation*} + +The congruences for this case are +\begin{align*} + \epsilon'xz'-e'y'z &\equiv kx \pmod p, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod p, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod p, \tag{IV} \\ + \epsilon'xz'' &\equiv ex \pmod p, \tag{V} \\ + cz'' &\equiv 0 \pmod p, \tag{VI} \\ + e'y'z'' &\equiv ex \pmod p. \tag{VII} +\end{align*} +\noindent (II) and (III) are linear in $z$ and $z'$ so $k$ and $\gamma$ are $\equiv +\hbox{ or } \not\equiv 0 \pmod{p}$ independently, (IV) gives $\delta \not +\equiv 0$, (V) give $\epsilon \not\equiv 0$, (VI) $c \equiv 0$, and (VII) +$e \not\equiv 0$. + +Elimination between (IV), (V), and (VII) gives +\begin{equation*} +\delta'e'\epsilon^2 \equiv \delta e \epsilon'^2 \pmod{p}. +\end{equation*} + +The types are derived by placing $\epsilon = \delta = 1$, and $e = 1, 2, +\cdots, p - 1$. + +\begin{equation*} B_5. \end{equation*} + +The congruences for this case are +\begin{align*} + -e'y'z &\equiv kx \pmod{p}, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ +\epsilon'_1 xz'' + \delta'e'x\tbinom{z''}{2} + e'yz'' + &\equiv e_1 x + \gamma x'' + \delta x' + \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + e'y'z'' &\equiv ex \pmod{p}. \tag{VII} +\end{align*} +\noindent (II), and (III) being linear in $z$ and $z'$ give $k \equiv 0 \hbox{ or } +\not\equiv 0$, and $\gamma \equiv 0 \hbox{ or } \not\equiv 0 \pmod{p}$, +(IV) gives $\delta \not\equiv 0$, (V) being linear in $x'$ gives +$\epsilon_1 \equiv 0 \hbox{ or } \not\equiv 0 \pmod{p}$, (VI) gives $c +\equiv 0$ and (VII) $e \not\equiv 0$. + +Elimination of $x$ and $y'$ from (IV) and (VII) gives +\begin{equation*} +\delta'e'z''^2 \equiv \delta e \pmod{p}. +\end{equation*} + +$\delta e$ is a quadratic residue or non-residue $\pmod{p}$ according as +$\delta'e'$ is a residue or non-residue. + +The two types are given by placing $\delta = 1$, and $e = 1$ and a +non-residue $\pmod{p}$. + +\bigskip +\begin{center} +\textit{Section} 3. +\end{center} +\setcounter{equation}{0} + +1. \textit{Groups with dependent generators continued.} As in Section 2, $G$ +is here generated by $Q_1$ and $P$, where +\begin{equation*} +Q{}_1^{p^2} = P^{hp^2}. +\end{equation*} +\noindent $Q{}_1^p$ is contained in the subgroup $H_1$ of order $p^{m-2}$, $H_1 += \{Q{}_1^p, P\}$. + +\medskip +2. \textit{Determination of $H_1$.} Since $\{P\}$ is self-conjugate in $H_1$ +\begin{equation} +Q{}_1^{-p}\, P\, Q{}_1^p = P^{1 + kp^{m-4}}. %% 1 +\end{equation} +\noindent Denoting $Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots$ by the symbol $[a, b, c, d, +\cdots]$, we have from (1) +\begin{equation} +[-yp,\, x,\, yp] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4). %% 2 +\end{equation} +\noindent Repeated multiplication with (2) gives +\begin{equation} +[yp, x]^s = \Bigl[syp, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\Bigr]. %% 3 +\end{equation} + +\medskip +3. \textit{Determination of $H_2$.} There is a subgroup $H_2$ of order $p^{m-1}$ +which contains $H_1$ self-conjugately.\footnote{\textsc{Burnside}, \textit{Theory of +Groups}, Art.\ 54, p.\ 64.} $H_2$ is generated by $H_1$ and some operator +$R_1$ of $G$. $R{}_1^p$ is contained in $H_1$, in fact in $\{P\}$, +since if $R{}_1^{p^2}$ is the first power of $R_1$ in $\{P\}$, then $H_2 += \{R_1, P\}$, which case was treated in Section 1. +\begin{equation} +R{}_1^p = P^{lp}. %% 4 +\end{equation} + +Since $H_1$ is self-conjugate in $H_2$ +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q{}_1^{\beta p} P^{\alpha_1}, \\ +R{}_1^{-1}\, Q^p\, R_1 &= Q{}_1^{bp} P^{\alpha_1 p}. +\end{align} + +Using the symbol $[a, b, c, d, e, f, \cdots]$ to denote $R{}_1^a\, Q{}_1^b\, +P^c\, R{}_1^d\, Q{}_1^e\, P^f \cdots$, we have from (5), (6) and (3) +\begin{equation} +[-p,\, 0,\, 1,\, p] = [0,\, \beta Np,\, \alpha{}_1^p + Mp], %% 7 +\end{equation} +\noindent and by (4) +\begin{equation*} +\alpha{}_1^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad \alpha_1 \equiv 1 \pmod{p}. +\end{equation*} + +Let $\alpha_1 = 1 + \alpha_2 p$ and (5) is now replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q{}_1^{\beta p} P^{1 + \alpha_2 p}. +\end{equation} + +From (6), (8) and (3) +\begin{gather*} +[-p,\, p,\, 0,\, p] = \bigl[0,\, b^p p,\, a_1 \frac{b^p - 1}{b - 1}p + a_1 Up^2\bigr], +\intertext{and by (4) and (2)} +R{}_1^{-p}\, Q{}_1^p\, R{}_1^p = Q{}_1^p +\end{gather*} +\noindent and therefore $b^p \equiv 1 \pmod{p}$, and $b = 1$. Placing +$b = 1$ in the above equation the exponent of $P$ takes the form +\begin{gather*} +a_1 p^2 (1 + U'p) = a_1 \frac{\left\{1 + (\alpha_2 + \beta h)p\right\}^p +- 1}{(\alpha_2 + \beta h)p^2}p^2 +\intertext{from which} +a_1 p^2 (1 + U'p) \equiv 0 \pmod{p^{m-3}} +\intertext{or} +a_1 \equiv 0 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} + +Let $a_1 = ap^{m-5}$ and (6) is replaced by +\begin{equation} +R{}_1^{-1}\, Q{}_1^p\, R_1 = Q{}_1^p\, P^{ap^{m-4}}. %% 9 +\end{equation} + +(7) now has the form +\begin{gather*} +[-p,\, 0,\, 1,\, p] = [0,\, \beta Np,\, (1 + \alpha_2 p)^p + Mp^2], +\intertext{where} +N = p \quad \hbox{ and } \quad M = \beta h \left\{ \frac{(1 + \alpha_2 p)^p - 1} + {\alpha_2 p^2} -1 \right\}, +\intertext{from which} +(1 + \alpha_2 p)^p + \frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2}\beta hp^2 + \equiv 1 \pmod{p^{m-3}} +\intertext{or} +\frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2}\{\alpha_2 + \beta h\}p^2 + \equiv 0 \pmod{p^{m-3}} +\intertext{and since} +\frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2} \equiv 1 \pmod{p} +\end{gather*} +\begin{equation} +(\alpha_2 + \beta h)p^2 \equiv 0 \pmod{p^{m-3}}. %% 10 +\end{equation} + +From (8), (9), (10) and (3) +\begin{align} +[-y,\, 0,\, x,\, y] &= [0,\, \beta xyp,\, x(1+\alpha_2 yp)+\theta p^{m-4}], \\ %% 11 +[-y,\, xp,\, 0,\, y] &= [0,\, xp,\, axyp^{m-4}], %% 12 +\end{align} +\noindent where +\begin{equation*} +\theta = a \beta x \tbinom{y}{2} + \beta k \tbinom{x}{2} y. +\end{equation*} + +By continued use of (11), (12), (2) and (10) +\begin{equation} +[z,\, yp,\, x]^s = [sz,\, (sy + U_s)p,\, xs+ V_s p], %% 13 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta \tbinom{s}{2} xz \\ +V_s &= \tbinom{s}{2} \Bigl\{ \alpha_2 xz + \bigl[ayz + kxy + \beta k\tbinom{x}{2}z \bigr] + p^{m-5} \Bigr\} \\ & \qquad \qquad +\Bigl\{\beta\tbinom{s}{3}x^2 z + \frac{1}{2}a\beta + \Bigl[\frac{1}{3!} s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\Bigr]x \Bigr\}p^{m-5}. +\end{align*} + +Placing in this $y = 0$, $z = 1$ and $s = p$,\footnote{Terms of the form +$(Ax^2 + Bx)p^{m-4}$ in the exponent of $P$ for $p = 3$ and $m > 5$ do not +alter the result.} +\begin{gather*} +(R_1\, P^x)^p = R{}_1^p\, P^{xp} = P^{(x+l)p}, +\intertext{determine $x$ so that} +x + l \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +then $R = R_1 P^x$ is an operator of order $p$ which will be used in the +place of $R_1$, $R^p = 1$. + +\medskip +4. \textit{Determination of $G$.} Since $H_2$ is self-conjugate in $G$ +\begin{align} +Q{}_1^{-1}\, P\, Q_1 &= R^\gamma\, Q{}_1^{\delta p}\, P^{\epsilon_1}, \\ %% 14 +Q{}_1^{-1}\, R\, Q_1 &= R^c\, Q{}_1^{dp}\, P^{e_1 p}. %% 15 +\end{align} + +From (15) +\begin{gather*} +(R^c\, Q{}_1^{dp}\, P^{e_1 p})^p = 1, +\intertext{by (13)} +Q{}_1^{dp^2}\, P^{e_1 p^2} = P^{(e_1 + dh)p^2} = 1, +\end{gather*} +\noindent and +\begin{equation} +(e_1+ dh)p^2 \equiv 0 \pmod{p^{m-3}}. %% 16 +\end{equation} + +From (14), (15) and (13) +\begin{equation} +[0,\, -p,\, 1,\, 0,\, p] = [L,\, Mp,\, \epsilon_1^p + Np]. %% 17 +\end{equation} + +By (1) +\begin{equation*} +\epsilon{}_1^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad \epsilon_1 \equiv 1 \pmod{p}. +\end{equation*} + +Let $\epsilon_1 = 1 + \epsilon_2 p$ and (14) is replaced by +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = R^\gamma\, Q{}_1^{\delta p}\, P^{1 + \epsilon_2 p}. %% 18 +\end{equation} + +From (15), (18), and (13) +\begin{equation*} +[0,\, -p,\, 0,\, 1,\, p] = \left[c_p,\, \frac{c^p - 1}{c - 1}dp,\, Kp \right]. +\end{equation*} + +Placing $x = 1$ and $y =-1$ in (12) we have +\begin{equation} +[0,\, -p,\, 0,\, 1,\, p] = [1,\, 0,\, -ap^{m-4}], %% 19 +\end{equation} +and therefore $c^p \equiv 1 \pmod{p}$, and $c = 1$. (15) is now replaced by +\begin{equation} +Q{}_1^{-1}\, R\, Q_1 = R\, Q{}_1^{dp}\, P^{e_1 p}. %% 20 +\end{equation} + +Substituting $1 + \epsilon_2 p$ for $\epsilon_1$ and 1 for $c$ in (17) +gives, by (16) +\begin{gather*} +[0,\, -p,\, 1,\, p] = [0,\, Mp^2,\, (1 + \epsilon_2 p)^p + Np^2], +\intertext{where} +M = \gamma d \frac{p-1}{2} + \delta\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} +\intertext{and} +N = \frac{e_1\gamma}{(\epsilon_2 + \delta h)p^2} \left\{\frac{[1 + (\epsilon_2 + +\delta h)p]^p - 1}{(\epsilon_2 + \delta h)p} - p \right\}. +\end{gather*} + +By (1) +\begin{gather*} +(1 + \epsilon_2 p)^p + (N + Mh)p^2 \equiv 1 + kp^{m-4} \pmod{p^{m-3}}, +\intertext{or reducing} +\psi(\epsilon_2 + \delta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}}, +\intertext{where} +\psi = \frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} + N - + e_1 \gamma \frac{p-1}{2}. +\end{gather*} + +Since +\begin{equation*} +\psi = 1 \pmod{p}. +\end{equation*} +\begin{equation} +(\epsilon_2 + \delta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}}. %% 21 +\end{equation} + +From (18), (20), (13), (16) and (21) +\begin{align} +[0,\, -y,\, x,\, 0,\, y] &= [\gamma xy,\, \theta_1 p,\, x + \phi_1 p], \\ %% 22 +[0,\, -y,\, 0,\, x,\, y] &= [x,\, dxyp,\, \phi_2 p], %% 23 +\end{align} +\noindent where +\begin{align*} +\theta_1 &= d\gamma x\tbinom{y}{2} + \delta xy + \beta\gamma\tbinom{x}{2}y, \\ +\phi_1 &= \epsilon_2 xy + \alpha_2\gamma\tbinom{x}{2}y + e_1\gamma + \tbinom{y}{2}x + \bigl\{x\tbinom{y}{2}(\epsilon_2 k + \delta \gamma) \\ +& \qquad + \frac{1}{2}ad\left[\frac{1}{3!}y(y - 1)(2y - 1)\gamma^2 - + \frac{y}{2}\gamma\right]x + a\gamma^2 dx\frac{1}{3!}y(y + 1)(y - 1) \\ +& \qquad + e_1\gamma k\tbinom{y}{3}x + \frac{1}{2}a\beta\left[\frac{1}{3!}x(x - 1)(2x - 1)\gamma^2 y^2 + - \tbinom{x}{2}\gamma y\right] \\ +& \qquad \qquad + \tbinom{x}{2}(a + k)\left[dy\tbinom{y}{2} + + \delta y\right] + \beta\gamma \tbinom{x}{3}\bigr\}p^{m-5}, \\ +\phi_2 &= e_1 xy + \left\{e_1 k\tbinom{y}{2} + ad\tbinom{x}{2}y \right\}p^{m-5}. +\end{align*} + +Placing $x = 1$ and $y = p$ in (23) and by (16) +\begin{gather*} +Q{}_1^{-p}\, R\, Q{}_1^p = R, +\intertext{and by (19)} +a \equiv 0 \pmod{p}. +\end{gather*} + +A continued multiplication, with (11), (22), and (23), gives +\begin{gather*} +(Q_1\, P^x)^{p^2} = Q{}_1^{p^2}\, P^{xp^2} = P^{(x + l)p^2}. +\intertext{Let $x$ be so chosen that} +(x + l) \equiv 0 \pmod{p^{m-5}}, +\end{gather*} +\noindent then $Q = Q_1\, P^x$ is an operator of order $p^2$ which will be used in +place of $Q_1$, $Q^{p^2} = 1$ and +\begin{equation*} +h \equiv 0 \pmod{p^{m-5}}. +\end{equation*} + +From (21), (10) and (16) +\begin{equation*} +\epsilon_2 p^2 \equiv kp^{m-4}, \qquad \alpha_2 p^2 \equiv 0 \qquad \hbox{ and } + \qquad e_1 p^2 \equiv 0 \pmod{p^{m-3}}. +\end{equation*} +\noindent Let $\epsilon_2 = \epsilon p^{m-6}$, $\alpha_2 = \alpha p^{m-5}$ and +$e_1 = ep^{m-5}$. Then equations (18), (20) and (8) are replaced by +\begin{gather*} +G: \left\{ \begin{aligned} +Q^{-1}\, P\, Q &= R^\gamma\, Q^{\delta p}\, P^{1 + \epsilon p^{m-5}}, \\ +Q^{-1}\, R\, Q &= R\, Q^{dp}\, P^{ep^{m-4}}, \\ +R^{-1}\, P\, R &= Q^{\beta p}\, P^{1 + \alpha p^{m-4}}, \\ \end{aligned} \right. +\tag*{(24), (25), (26)} \\ +R^p = 1, \qquad Q^{p^2} = 1, \qquad P^{p^{m-3}} = 1. +\end{gather*} + +\setcounter{equation}{26} +(11), (22) and (23) are replaced by +\begin{align} + [-y,\, 0,\, x,\, y] &= [0,\, \beta xyp,\, x + \phi p^{m-4}], \\ %% 27 +[0,\, -y,\, x,\, 0,\, y] &= [\gamma xy,\, \theta_1 p,\, x + \phi_1 p^{m-5}], \\ %% 28 +[0,\, -y,\, 0,\, x,\, y] &= [x,\, dxyp,\, \phi_2 p^{m-4}], %% 29 +\end{align} +\noindent where +\begin{gather*} +\phi = \alpha xy + \beta k\tbinom{x}{2}y, \quad + \theta_1 = d\gamma\tbinom{y}{2}x + \delta xy + \beta\gamma\tbinom{x}{2}y, \\ +\phi_1 = exy + \left\{e\gamma x\tbinom{y}{2} + \tbinom{x}{2}\left(\alpha\gamma y + + d\gamma k\tbinom{y}{2} + \delta ky\right) + \beta\gamma y\tbinom{x}{3}\right\}p, \\ +\phi_2 = exy. +\end{gather*} +\noindent A formula for a general power of any operator of $G$ is derived from (27), +(28) and (29) +\begin{equation} +[0,\, z,\, 0,\, y,\, 0,\, z]^s = [0,\, sz + U_s p,\, 0,\, sy + V_s,\, 0,\, sx + W_s p^{m-5}], %% 30 +\end{equation} +\noindent where +\begin{align*} +U_s &= \tbinom{s}{2}\left\{\delta xz + dyz + \beta xy + \beta\gamma \tbinom{x}{2}z \right\} \\ + & \qquad + \frac{1}{2}dx \left\{ \frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z \right\}x + + \beta\gamma\tbinom{s}{2}x^2 z, \\ +V_s &= \gamma\tbinom{s}{2}xz, \displaybreak \\ +W_s &= \tbinom{s}{2} \left\{\epsilon xz + \left[axy + eyz + (\beta ky + \alpha\beta + \gamma + \delta kz)\tbinom{x}{2}\right]p \right\} \\ + & \qquad + \tbinom{s}{3}\left\{\alpha\gamma x^2 z + dkxyz + \delta kx^2 z + + \beta kx^2 y + 2\beta\gamma k\tbinom{x}{2}xz\right\}p \\ + & \qquad + \beta yk\tbinom{s}{4}x^3 zp + \frac{1}{2}\left\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 + - \frac{s}{2}z \right\} \left\{e\gamma x + d\gamma k\tbinom{x}{2}\right\}p \\ + & \qquad + \frac{1}{2}d\gamma k\left[\frac{1}{2}(s - 1)z^2 - z\right]\tbinom{s}{3}x^2. +\end{align*} +\noindent A comparison of the generational equations of the present section with those +of Sections 1 and 2, shows that, $\gamma \equiv 0 \pmod{p}$ gives groups +simply isomorphic with those of Section 1, while $\beta \equiv 0 \pmod{p}$, +groups simply isomorphic with those of Section 2 and we need consider only +the groups in which $\beta$ and $\gamma$ are prime to $p$. + +\medskip +5. \textit{Transformation of the groups.} All groups of this section are given by +equations (24), (25), and (26), where $\gamma, \beta = 1, 2, \cdots, p - 1$; +$\alpha, \delta, d, e = 0, 1, 2, \cdots, p - 1$; and $\epsilon = 0, 1, +2, \cdots, p^2 - 1$. + +Not all of these, however are distinct. Suppose that $G$ is simply +isomorphic with $G'$ and that the correspondence is given by +\begin{equation*} +C = \left[\begin{matrix}R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right]. +\end{equation*} +\noindent An inspection of (30) gives +\begin{align*} +R'_1 &= Q'^{z''p}\, R'^{y''}\, P'^{x''p^{m-4}}, \\ +Q'_1 &= Q'^{z'}\, R'^{y'}\, P'^{x'p^{m-5}}, \\ +P'_1 &= Q'^z\, R'^y\, P'^x, +\end{align*} +\noindent with $dv[x,\, p] = 1$. Since $Q^p$ is not in $\{P\}$, and $R$ is not in +$\{Q^p, P\}$, ${Q'}_1^p$ is not in $\{P'_1\}$ and $R'_1$ is not in +$\{{Q'}_1^p, P'_1\}$. Let +\begin{gather*} +{Q'}_1^{s'p} = {P'}_1^{sp^{m-4}}. +\intertext{This is in terms of $R'$, $Q'$, and $P'$,} +[0,\, s'z'p,\, s'x'p^{m-4}] = [0,\, 0,\, sxp^{m-4}]. +\intertext{From which} +s'z'p \equiv 0 \pmod{p^2}, +\intertext{and $z'$ must be prime to $p$, since otherwise $s' \hbox{ can } = 1$. Let} +{R'}_1^{s''} = {Q'}_1^{s'p}\, {P'}_1^{sp^{m-4}}, +\intertext{or in terms of $R'$, $Q'$, and $P'$,} +[s''y'',\, s''z''p,\, s''x''p^{m-4}] = [0,\, s'z'p,\, (sx + s'x')p^{m-4}] +\intertext{and} +s''z'' \equiv s'z' \pmod{p}, \qquad s''y'' \equiv 0 \pmod{p}, +\end{gather*} +\noindent and $y''$ is prime to $p$, since otherwise $s''$ can $= 1$. Since $R$, +$Q$, and $P$ satisfy equations (24), (25) and (26), $R'_1$, $Q'_1$, and +$P'_1$ must also satisfy them. These become when reduced in terms of $R'$, +$Q'$ and $P'$ +\begin{align*} +[0,&\, z + \theta'_1 p,\, 0,\, y + \gamma'xz',\, 0,\, x + \psi'_1 p^{m-5}] \\ + & \qquad \qquad \qquad \qquad \qquad = [0,\, z + \theta_1 p,\, 0,\, y + \gamma y'',\, 0,\, x + \psi_1 p^{m-5}], \\ +[0,&\, (z'' + \theta'_2)p,\, 0,\, y'',\, 0,\, (x'' + \psi_2)p^{m-4}] \\ + & \qquad \qquad \qquad \qquad \qquad = [0,\, (z'' + \theta_2)p,\, 0,\, y'',\, 0,\, (x'' + \psi_2)p^{m-4}], \\ +[0,&\, z + \theta'_3p,\, 0,\, y,\, 0,\, x + \psi'_3 p^{m-4}] + = [0,\, z + \theta_3p,\, 0,\, y,\, 0,\, x + \psi_3 p^{m-4}], +\end{align*} +\noindent where +\begin{align*} +\theta'_1 &= d'(yz' - y'z) + x\left\{d'\gamma'\tbinom{z'}{2} + \delta'z' + +\beta'y'\right\} + \beta'\gamma'\tbinom{x}{2}z', \\ +\theta_1 &= \gamma z'' + \delta z' + d'\gamma y''z, \\ +\psi'_1 &= \epsilon'xz' + \bigl\{e'\gamma'x\tbinom{z'}{2} + \tbinom{x}{2}\left[\alpha' + \gamma'z' + \gamma'\epsilon'd'k'\tbinom{z'}{2} + \delta'\epsilon k'z' + + \beta'k'y'\right] \\ + & \qquad + \beta'\gamma'\tbinom{x}{3}z' + e'(yz' - y'z) + \alpha'xy'\bigr\}p, \\ +\psi_1 &= \epsilon x + \{\delta x' + \gamma x'' + e'\gamma y''z\}p, \\ +\theta'_2 &= d'y''z', \qquad \theta_2 = dz', \qquad \psi'_2 = e'y''z, \qquad + \psi_2 = dx' + ex, \\ +\theta'_3 &= \beta'xy'' - d'y''z, \qquad \theta_3 = \beta z', \\ +\psi_3 &= \epsilon'xz'' - e'y''z + \alpha'xy'' + \beta'\epsilon'\tbinom{x}{2}y'', + \qquad \psi_3 = \alpha x + \beta x'. +\end{align*} + +A comparison of the two sides of these equations give seven congruences +\begin{align*} +\theta'_1 &\equiv \theta_1 \pmod{p}, \tag{I} \\ +\gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\psi'_1 &\equiv \psi_1 \pmod{p^2}, \tag{III} \\ +\theta'_2 &\equiv \theta_2 \pmod{p}, \tag{IV} \\ +\psi'_2 &\equiv \psi_2 \pmod{p}, \tag{V} \\ +\theta'_3 &\equiv \theta_3 \pmod{p}, \tag{VI} \\ +\psi'_3 &\equiv \psi_3 \pmod{p}. \tag{VII} +\end{align*} + +(VI) is linear in $z$ provided $d' \not\equiv 0 \pmod{p}$ and $z$ may be +so determined that $\beta \equiv 0 \pmod{p}$ and therefore all groups in +which $d' \not\equiv 0 \pmod{p}$ are simply isomorphic with groups in +Section 2. + +Consequently we need only consider groups in which $d' \equiv 0 \pmod{p}$. + +As before we take for $G'$ the simplest case and associate with it all +simply isomorphic groups $G$. We then take as $G'$ the simplest case left +and proceed as above. + +Let $\kappa = \kappa_1 p^{\kappa_2}$ where $dv[\kappa_1, p] = 1, (\kappa = +\alpha, \beta, \gamma, \delta, \epsilon, d, e)$. + +For convenience the groups are divided into three sets and each set is +subdivided into eight cases. + +The sets are given by +\begin{equation*} +\begin{matrix} +A: & \epsilon_2 = 0, & \beta_2 = 0, & \gamma_2 = 0, \\ +B: & \epsilon_2 = 1, & \beta_2 = 0, & \gamma_2 = 0, \\ +C: & \epsilon_2 = 2, & \beta_2 = 0, & \gamma_2 = 0. +\end{matrix} +\end{equation*} + +The subdivision into cases and results of the discussion are given in +Table I. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|} \hline + & $\delta_2$ & $e_2$ & $\alpha_2$ & $A$ & $B$ & $C$ \\ \hline + 1 & 1 & 1 & 1 & & & $B_1$ \\ \hline + 2 & 0 & 1 & 1 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 3 & 1 & 0 & 1 & & & $B_3$ \\ \hline + 4 & 1 & 1 & 0 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 5 & 0 & 0 & 1 & $A_3$ & $B_3$ & $B_3$ \\ \hline + 6 & 0 & 1 & 0 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 7 & 1 & 0 & 0 & $A_3$ & $B_3$ & $B_3$ \\ \hline + 8 & 0 & 0 & 0 & $A_3$ & $B_3$ & $B_3$ \\ \hline +\end{tabular} +\end{center} + +\medskip +6. \textit{Reduction to types.} The types of this section are given by equations +(24), (25) and (26) with $\alpha = 0, \beta = 1, \lambda = 1$ or a +quadratic non-residue (mod $p$), $\delta \equiv 0; \epsilon = l, e = 0, 1, +2, \cdots, p - 1;$ and $\epsilon = p, e = 0, 1,$ or a +non-residue (mod $p$), $2p+6$ in all. + +The special forms of the congruences for these cases are given below. + +\medskip +\begin{equation*} A_1. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'xz' &\equiv \epsilon x \pmod{p}, \tag{III} \\ + dz' &\equiv 0 \pmod{p}, \tag{IV} \\ + ex &\equiv 0 \pmod{p}, \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\epsilon'xz'' + \beta'\epsilon'\tbinom{x}{2}y' + &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} +\end{align*} + +(I) is linear in $z''$ and $\delta \equiv 0$ or $\not\equiv 0$, (II) +gives $\gamma \not\equiv 0$, (III) $\epsilon \not\equiv 0$, (IV) and (V) +$d \equiv e \equiv 0$, (VI) $\beta \not\equiv 0$, (VII) is linear in $x'$ +and $\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +Elimination of $y''$ and $z'$ between (II) and (VI) gives +\begin{equation*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p} +\end{equation*} +\noindent and $\beta\gamma$ is a residue or non-residue $\pmod{p}$ according as +$\beta'\gamma'$ is a residue or non-residue. + +\medskip +\begin{equation*} A_3. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'z' &\equiv \epsilon \pmod{p}, \tag{III} \\ + d &\equiv 0 \pmod{p}, \tag{IV} \\ + e'y''z' &\equiv ex \pmod{p}, \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\epsilon'xz'' - e'y''z + \beta'\epsilon'\tbinom{x}{2}y' + &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} +\end{align*} + +(I) is linear in $z''$ and $\delta \equiv 0$ or $\not\equiv 0$. (II) +gives $\gamma \not\equiv 0$, (III) $\epsilon \not\equiv 0$, (V) $e \not +\equiv 0$ and (VI) $\beta \not\equiv 0$. (VII) is linear in $x'$ and +$\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +Elimination between (II) and (VI) gives +\begin{gather*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}, +\intertext{and between (II), (III), and (IV) gives} +\epsilon'^2 \gamma e \equiv \epsilon^2\gamma'e' \pmod{p}. +\end{gather*} + +$\beta\gamma$ is a residue, or non-residue, according as $\beta'\gamma'$ is +or is not, and if $\gamma$ and $\epsilon$ are fixed, $e$ must take the +$(p - 1)$ values $1, 2, \cdots, p - 1$. + +\medskip +\begin{equation*} B_1. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ +\gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'_1 xz' + \beta'xz'\tbinom{x}{3} + &\equiv \epsilon_1 x + \delta x' + \gamma x'' \pmod{p}, \tag{III} \\ + ex &\equiv 0 \pmod{p}, \tag{IV} \\ +\beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\alpha x + \beta x' &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} + +(I) gives $\delta \equiv 0$ or $\not\equiv 0$, (II) $\gamma \not +\equiv 0$, (III) is linear in $x''$ and gives $\epsilon_1 \equiv 0$ +or $\not\equiv 0$, (V) $e = 0$, (VI) $\beta \not\equiv 0$ and +(VII) is linear in $x'$ and gives $\alpha \equiv 0$ or $\not\equiv 0$. + +Elimination between (II) and (VI) gives +\begin{equation*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}. +\end{equation*} + +\newpage +\begin{equation*} B_3. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma\beta'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y' \pmod{p}, \tag{II} \\ + \epsilon'_1 xz' + e'\gamma'x\tbinom{z'}{2} + \beta'\gamma'\tbinom{x}{3} &+ e'(yz' - y'z) \\ + &\equiv \epsilon_1 x + \delta x' + \gamma x'' + e'\gamma zy'' \pmod{p}, + \tag{III} \\ + e'y''z' &\equiv ex \pmod{p}. \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ + -e'y''z &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} \\ +\end{align*} +(I) gives $\delta\equiv 0$ or $\not\equiv 0$, (II) $\gamma \not +\equiv 0$, (III) is linear in $x''$ and gives $\epsilon_1 \equiv 0$ or +$\not\equiv 0$, (V) $e \not\equiv 0$, (VI) $\beta \not\equiv 0$, (VII) is +linear in $x'$ and gives $\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. +Elimination of $y''$ and $z'$ between (II) and (VI) gives +\begin{gather*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}, +\intertext{and between (V) and (VI) gives} +\beta'e'y''^2 \equiv \beta e \pmod{p} +\end{gather*} +\noindent and $\beta\gamma$ and $\beta e$ are residues or non-residues, independently, +according as $\beta'\gamma'$ and $\beta'e'$ are residues or non-residues. + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} III.\normalsize +\end{center} +\setcounter{equation}{0} + +1. \textit{General relations.} In this class, the $p$th power of every operator of +$G$ is contained in $\{P\}$. There is in $G$ a subgroup $H_1$ of order +$p^{m-2}$, which contains $\{P\}$ self-conjugately.\footnote{\textsc{Burnside}, +\textit{Theory of Groups}, Art.\ 54, p.\ 64.} + +\medskip +2. \textit{Determination of $H_1$.} $H_1$ is generated by $P$ and some operator +$Q_1$ of $G$. +\begin{equation*} +Q{}_1^p = P^{hp}. +\end{equation*} +\noindent Denoting $Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots$ by the symbol $[a, b, c, d, +\cdots]$, all operators of $H_1$ are included in the set $[y, x]; (y = 0, +1, 2, \cdots, p - 1, x = 0, 1, 2, \cdots, p^{m-3} - 1)$. + +Since $\{P\}$ is self-conjugate in $H_1$\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{gather} +Q{}_1^{-1}\, P\, Q_1 = P^{1 + kp^{m-4}}. %% 1 +\intertext{Hence} +[-y,\, x,\, y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4). %% 2 +\intertext{and} +[y,\, x]^s = \left[sy, x\left\{s + ky \tbinom{s}{2}p^{m-4} \right\}\right]. %% 3 +\end{gather} +\noindent Placing $y = 1$ and $s = p$ in (3), we have, +\begin{gather*} +[Q_1\, P^x]^p = Q{}_1^p\, P^{xp} = P^{(x + h)p} +\intertext{and if $x$ be so chosen that} +(x + h) \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent $Q = Q_1\, P^x$ will be an operator of order $p$ which will be used in place +of $Q_1$, $Q^p = 1$. + +\medskip +3. \textit{Determination of $H_2$.} There is in $G$ a subgroup $H_2$ of order +$p^{m-1}$, which contains $H_1$ self-conjugately. $H_2$ is generated by +$H_1$, and some operator $R_1$ of $G$. +\begin{equation*} +R{}_1^p = P^{lp}. +\end{equation*} + +We will now use the symbol $[a, b, c, d, e, f, \cdots]$ to denote $R{}_1^a$ $Q^b$ +$P^c$ $R{}_1^d$ $Q^e$ $P^f$ $\cdots$. + +The operations of $H_2$ are given by $[z, y, x];$ $(z, y = 0, 1, \cdots, p - 1$; +$x = 0, 1, \cdots, p^{m-3} - 1)$. Since $H_1$ is self-conjugate in $H_2$ +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q{}_1^\beta P^{\alpha_1}, \\ %% 4 +R{}_1^{-1}\, Q\, R_1 &= Q{}_1^{b_1} P^{\alpha p^{m-4}}. %% 5 +\end{align} + +From (4), (5) and (3) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = \left[0,\, \frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta,\, + \alpha{}_1^p + \theta p^{m-4} \right] = [0,\, 0,\, 1], \\ +\intertext{where} +\theta = \frac{\alpha{}_1^p \beta k}{2}\frac{\alpha{}_1^p - 1}{\alpha_1-1} + + a\beta \left\{\frac{\alpha{}_1^p - 1}{\alpha_1 - b_1}p - + \frac{\alpha{}_1^p - b{}_1^p}{(\alpha_1 - b_1)^2}\right\}. +\end{gather*} +\noindent Hence +\begin{equation} +\frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta \equiv 0 \pmod{p}, \qquad +\alpha{}_1^p + \theta p^{m-4} \equiv 1 \pmod{p^{m-3}}, %% 6 +\end{equation} +\noindent and $\alpha{}_1^p \equiv 1 \pmod{p^{m-4}}$, or $\alpha_1 \equiv 1 +\pmod{p^{m-5}} \qquad (m > 5)$, $\alpha_1 = 1 + \alpha_2 p^{m-5}.$ Equation (4) +is replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q^\beta P^{1 + \alpha_2 p^{m-5}}, +\end{equation} + +From (5), (7) and (3). +\begin{equation*} +[-p,\, 1,\, 0,\, p] = \left[0,\, b{}_1^p,\, a\frac{b{}_1^p - 1}{b - 1} p^{m-4} \right]. +\end{equation*} +\noindent Placing $x = lp$ and $y = 1$ in (2) we have $Q^{-1} P^{lp} Q = P^{lp}$, and +\begin{equation*} +b{}_1^p \equiv 1 \pmod{p}, \qquad a\frac{b{}_1^p - 1}{b_1 - 1} \equiv 0 +\pmod{p}. +\end{equation*} +\noindent Therefore, $b_1 = 1$. + +Substituting 1 for $b_1$ and $1 + \alpha_2 p^{m-5}$ for $\alpha_1$ in +congruence (6) we find +\begin{equation*} +(1 + \alpha_2 p^{m-5})^p \equiv 1 \pmod{p^{m-3}}, \qquad \hbox{ or } \qquad +\alpha_2 \equiv 0 \pmod{p}. +\end{equation*} + +Let $\alpha_2 = \alpha p$ and equations (7) and (5) are replaced by +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q^\beta P^{1 + \alpha p^{m-4}}, \\ %% 8 +R{}_1^{-1}\, Q\, R_1 &= Q P^{\alpha p^{m-4}}. %% 9 +\end{align} + +From (8), (9) and (3) +\begin{align} +[-y,\, 0,\, x,\, y] &= \Bigl[0,\, \beta xy,\, x + \bigl\{\alpha xy + a\beta x \tbinom{y}{2} + + \beta ky\tbinom{x}{2}\bigr\}p^{m-4}\Bigr], \\ %% 10 +[-y,\, x,\, 0,\, y] &= [0,\, x,\, axyp^{m-4}]. %% 11 +\end{align} + +From (2), (10), and (11) +\begin{equation} +[z,\, y,\, x]^s = [sz,\, sy + U_s,\, sx + V_s p^{m-4}], %% 12 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta \tbinom{s}{2}xz, \\ +V_s &= \tbinom{s}{2}\left\{\alpha xz + kxy + ayz + \beta k\tbinom{x}{2}z\right\} \\ + & \qquad \qquad + \beta k\tbinom{s}{3}x^2 z + \frac{1}{2}a\beta\tbinom{s}{2} \left\{\frac{1}{3!} + (2s - 1)z - 1\right\}xz. +\end{align*} + +Placing $z = 1$, $y = 0$, and $s = p$ in (12)\footnote{The terms of the form +$(Ax + Bx^2)p^{m-4}$ which appear in the exponent of $P$ for $p = 3$ +do not alter the conclusion for $m > 5$.} +\begin{equation*} +[R_1\, P^x]^p = R{}_1^p\, P^{xp} = P^{(x+l)p}. +\end{equation*} + +If $x$ be so chosen that +\begin{equation*} +x + l \equiv 0 \pmod{p^{m-4}} +\end{equation*} +\noindent then $R =R_1 P^x$ is an operator of order $p$ which will be used in +place of $R_1$, and $R^p = 1$. + +\medskip +4. \textit{Determination of $G$.} $G$ is generated by $H_2$ and some operation +$S_1$. +\begin{equation*} +S{}_1^p = P^{\lambda p}. +\end{equation*} + +Denoting $S{}_1^a\, R^b\, Q^c\, P^d \cdots$ by the symbol $[a, b, c, d, \cdots]$ +all the operators of $G$ are given by +\begin{equation*} +[v,\, z,\, y,\, x];\, (v,\, z,\, y = 0, 1, \cdots, p - 1; x = 0, 1, \cdots, p^{m-3} - 1). +\end{equation*} + +Since $H_2$ is self-conjugate in $G$ +\begin{align} +S{}_1^{-1}\, P\, S_1 &= R^\gamma Q^s P^{\epsilon_1}, \\ %% 13 +S{}_1^{-1}\, Q\, S_1 &= R^c Q^d P^{ep^{m-4}}, \\ %% 14 +S_1\, R\, S_1 &= R^f Q^g P^{jp^{m-4}}. %% 15 +\end{align} + +From (13), (14), (15), and (12) +\begin{gather*} +[-p,\, 0,\, 0,\, 1,\, p] = [0,\, L,\, M,\, \epsilon{}_1^p + Np^{m-4}] = [0,\, 0,\, 0,\, 1] +\intertext{and} +\epsilon{}_1^p \equiv 1 \pmod{p^{m-4}} \qquad \hbox{ or } \qquad + \epsilon_1 \equiv 1 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} +\noindent Let $\epsilon_1 = 1 + \epsilon_2 p^{m-5}$. Equation (13) is +now replaced by +\begin{equation} +S{}_1^{-1}\, P\, S_1 = R^\gamma Q^\delta P^{1 + \epsilon_2 p^{m-5}}. %% 16 +\end{equation} + +If $\lambda = 0 \pmod{p}$ and $\lambda = \lambda'p,$ +\begin{multline*} +[1,\, 0,\, 0,\, 1]^p = \left[p,\, 0,\, 0,\, p + \epsilon\tbinom{p}{2}p^{m-5} + Wp^{m-4}\right] \\ += [0,\, 0,\, 0,\, p + \lambda'p^2 + W'p^{m-4}] +\end{multline*} +\noindent and for $m>5$ $S_1 P$ is of order $p^{m-3}$. We will take this in place +of $S_1$ and assume $dv [\lambda, p] = 1$. +\begin{equation*} +S{}_1^{p^{m-3}} = 1. +\end{equation*} +\noindent There is in $G$ a subgroup $H'_1$ of order $p^{m-2}$ which contains +$\{S_1\}$ self-conjugately. $H'_1 = \{S_1,\, S^v_1\, R^z\, Q^y\, P^x\}$ and the +operator $T= R^z\, Q^y\, P^x$ is in $H'_1$. + +There are two cases for discussion. + +\smallskip +$1^\circ$. Where $x$ is prime to $p$. + +$T$ is an operator of $H_2$ of order $p^{m-3}$ and will be taken as $P$. +Then +\begin{gather*} +H'_1 = \{S_1, P\}. +\intertext{Equation (16) becomes} +S{}_1^{-1}\, P\, S_1 = P^{1 + \epsilon p^{m-4}}. +\end{gather*} +\noindent There is in $G$ a subgroup $H'_2$ of order $p^{m-1}$ which contains $H'_1$ +self-conjugately. +\begin{equation*} +H'_2 = \{H'_1,\, S{}_1^{v'}\, R^{z'}\, Q^{y'}\, P^{x'}\}. +\end{equation*} +\noindent $T' = R^{z'}Q^{y'}$ is in $H'_2$ and also in $H_2$ and is taken as $Q$, +since $\{P, T'\}$ is of order $p^{m-2}$. + +$H'_2 = \{H'_1, Q\} = \{S_1, H_1\}$ and in this case $c$ may be taken +$\equiv 0 \pmod{p}$. + +\smallskip +$2^\circ$. \textit{Where $x = x_1 p$.} $P^p$ is in $\{S_1\}$ since $\lambda$ is +prime to $p$. In the present case $R^z\, Q^y$ is in $H'_1$ and also in $H_2$. +If $z \not\equiv 0 \pmod{p}$ take $R^z\, Q^y$ as $R$; if $z \equiv 0 +\pmod{p}$ take it as $Q$. +\begin{gather*} +H'_1 = \{S_1, R\} \quad \hbox{ or } \quad \{S_1,Q\}, +\intertext{and} +R^{-1}\, S_1\, R = S{}_1^{1 + k'p^{m-4}} \qquad \hbox{ or } \qquad +Q^{-1}\, S_1\, Q = S{}_1^{1 + k''p^{m-4}}. +\intertext{On rearranging these take the forms} +S{}_1^{-1}\, R\, S_1 = R\,S{}_1^{np^{m-4}} = R\,P^{jp^{m-4}} \quad \hbox{ or } + \quad S_1^{-1}\, Q\, S_1 = Q\,S{}_1^{n'p^{m-4}} = Q\,P^{ep^{m-4}}, +\end{gather*} +\noindent and either $c$ or $g$ may be taken $\equiv 0 \pmod{p}$, +\begin{equation} +cg \equiv 0 \pmod{p}. %% 17 +\end{equation} +\noindent From (14), (15), (16), (12) and (17) +\begin{equation*} +[-p,\, 0,\, 1,\, 0,\, p] = \left[0,\, c\frac{d^p - f^p}{d - f},\, d^p,\, Wp^{m-4}\right]. +\end{equation*} + +Place $x = \lambda p$ and $y = 1$ in (12) +\begin{gather*} +Q^{-1}\, P^{\lambda p}\,Q = P^{\lambda p} \qquad \hbox{ or } \qquad + S{}_1^p\, Q\, S{}_1^p = Q, +\intertext{and} +d^p \equiv 1 \pmod{p}, \qquad d = 1. +\end{gather*} +\noindent Equation (14) is replaced by +\begin{equation} +S{}_1^{-1}\, Q\, S_1 = R^c\, Q\, P^{ep^{m-4}}. %% 18 +\end{equation} + +From (15), (18), (17), (16) and (12) +\begin{equation*} +[-p,\, 1,\, 0,\, 0,\, p] = \left[0,\, f^p,\, \frac{d^p - f^p}{d - f}g, W'p^{m-4}\right]. +\end{equation*} +\noindent Placing $x = \lambda p,\, y = 1$ in (10) +\begin{equation*} +R^{-1}\, P^{\lambda p}\, R = P^{\lambda p}, +\end{equation*} +\noindent and $f^p \equiv 1 \pmod{p},\, f = 1$. Equation (15) is replaced by +\begin{equation} +S{}_1^{-1}\, R\, S_1 = R\, Q^g\, P^{jp^{m-4}}. %% 19 +\end{equation} +\noindent From (16), (18), (19) and (12) +\begin{equation*} +S{}_1^{-p}\, P\, S{}_1^p = P^{1 + \epsilon_2 p^{m-4}} = P +\end{equation*} +\noindent and $\epsilon_2 \equiv 0 \pmod{p}$. Let $\epsilon_2 = \epsilon p$ and (16) +is replaced by +\begin{equation} +S{}_1^{-1}\, P\, S_1 = R^\gamma\, Q^\delta\, P^{1 + \epsilon p^{m-4}}. %% 20 +\end{equation} +\noindent Transforming both sides of (1), (8) and (9) by $S_1$ +\begin{align*} +S{}_1^{-1} Q^{-1} S_1 \cdot S{}_1^{-1} P S_1 \cdot S{}_1^{-1} Q S_1 &= + S{}_1^{-1} P^{1 + kp^{m-4}} S_1, \\ +S{}_1^{-1} R^{-1} S_1 \cdot S{}_1^{-1} P S_1 \cdot S{}_1^{-1} R S_1 &= + S{}_1^{-1} Q^\beta S_1 \cdot S{}_1^{-1} P^{1 + \alpha p^{m-4}} S_1, \\ +S{}_1^{-1} R^{-1} S_1 \cdot S{}_1^{-1} Q S_1 \cdot S{}_1^{-1} R S_1 &= + S{}_1^{-1} Q S_1 \cdot S{}_1^{-1} P^{ap^{m-4}} S_1. +\end{align*} +\noindent Reducing these by (18), (19), (20) and (12) and rearranging +\begin{align*} +\bigl[0,&\, \gamma,\, \delta + \beta c,\, 1 + \left \{ \epsilon + \alpha c + k + ac\delta + + a\beta\tbinom{c}{2} - a\gamma \right \} p^{m-4} \bigr] \\ + & \qquad \qquad \qquad = [0,\, \gamma,\, \delta, 1 + (\epsilon + k)p^{m-4}]. \\ +[0,&\, \gamma,\, \beta + \delta,\, 1 + \{kg + \epsilon + \alpha + a\delta - + a\gamma g\}p^{m-4}] \\ + & \qquad \qquad \qquad = \Bigl[0,\, \gamma + \beta c,\, \beta + \delta,\, 1 + + \bigl\{\epsilon + \alpha + \beta e + \alpha\tbinom{\beta}{2}c + + a\beta\gamma\bigr\}p^{m-4}\Bigr], \\ +[0,&\, c,\, 1,\, (e + a)p^{m-4}] = [0,\, c,\, 1,\, (e + a)p^{m-4}]. +\end{align*} + +The first gives +\begin{gather} +\beta c \equiv 0 \pmod{p}, \\ %% 21 +ac + ac\delta - a\gamma \equiv 0 \pmod{p}. %% 22 +\intertext{Multiplying this last by $g$} +ag\gamma \equiv 0 \pmod{p}. %% 23 +\intertext{From the second equation above} +gk + \alpha\delta \equiv \beta e + a\beta\gamma \pmod{p}. %% 24 +\intertext{Multiplying by $c$} +ac\delta \equiv 0 \pmod{p}. %% 25 +\end{gather} + +These relations among the constants \textit{must be satisfied} in order that our +equations should define a group. + +From (20), (19), (18) and (12) +\begin{align} +[-y,\, 0,\, 0,\, x,\, y] &= [0,\, \gamma xy + \chi_1 (x, y),\, \delta xy + \phi_1 + (x, y),\, x + \Theta_1 (x, y)p^{m-4}], \\ %% 26 +[-y,\, 0,\, x,\, 0,\, y] &= [0,\, cxy,\, x,\, \Theta_2 (x, y)p^{m-4}], \\ %% 27 +[-y,\, x,\, 0,\, 0,\, y] &= [0,\, x,\, gxy,\, \Theta_3 (x, y)p^{m-4}], %% 28 +\end{align} +\noindent where +\begin{align*} +\chi_1 (x,\, y) &= c\delta x\tbinom{y}{2}, \\ +\phi_1 (x,\, y) &= \gamma gx\tbinom{y}{2} + \beta\gamma\tbinom{x}{2}y, \\ +\Theta_1 (x,\, y) &= \epsilon xy + \tbinom{y}{2}\left[\gamma jx + e\delta x + + a\delta\gamma + (\alpha\gamma + k\delta)\tbinom{x}{2}\right] \\ + & \qquad \qquad + \tbinom{y}{3} [c\delta j + eg\gamma]x + + \tbinom{x}{2}[\alpha\gamma y + \delta ky + a\delta\gamma y^2] + + \beta\gamma k\tbinom{x}{3}y^2, \\ +\Theta_2 (x, y) &= exy + cjx\tbinom{y}{2} + ac\tbinom{x}{2}y, \\ +\Theta_3 (x, y) &= jxy + egx\tbinom{y}{2} + ag\tbinom{x}{2}y. +\end{align*} + +Let a general power of any operator be +\begin{equation} +[v,\, z,\, y,\, x]^s = [sv,\, sz + U_s,\, sy + V_s,\, sx + W_s p^{m-4}]. %% 29 +\end{equation} + +Multiplying both sides by $[v,\, z,\, y,\, x]$ and reducing by (2), (10), (11), +(26), (27) and (28), we find +\begin{align*} +U_{s+1} &\equiv U_s + (cy + \gamma x)sv + c \delta\tbinom{sv}{2}x \pmod{p}, \\ +V_{s+1} &\equiv V_s + (gz + \delta x)sv + \gamma g\tbinom{sv}{2}x + + \beta\gamma\tbinom{x}{2}sv + \beta (sz + U_s)x \pmod{p}, \\ +W_{s+1} &\equiv W_s + \Theta_1 (x, sv) + \left\{ey + jz + a \gamma xy + + ac\tbinom{y}{2} + ag\tbinom{z}{2} \right\}sv \\ + & \qquad \qquad + \left\{\alpha x + \beta k\tbinom{x}{2} + + ay + a\delta sx + \alpha gsvz\right\}sz + ksxy \\ + & \qquad \qquad + \tbinom{sv}{2}\{cjy + egz\} + U_s \left\{\alpha x + + \beta k\tbinom{x}{2} + ay + a(\delta x + gz)sv\right\} \\ + & \qquad \qquad + a\beta\tbinom{sz + Us}{2}x + kV_s x \pmod{p}. +\end{align*} + +From (29) +\begin{equation*} +U_1 \equiv 0, \qquad V_1 \equiv 0, \qquad W_1 \equiv 0 \pmod{p}. +\end{equation*} + +A continued use of the above congruences give +\begin{align*} +U_s &\equiv (cy + \gamma x)\tbinom{s}{2}v + \frac{1}{2} c\delta xv + \{\frac{1}{3} (2s - 1)v - 1\}\tbinom{s}{2} \pmod{p}, \\ +V_s &\equiv \{[gz + \delta x + \beta\gamma\tbinom{x}{2}v + \beta xz\} + \tbinom{s}{2} \\ & \qquad + \frac{1}{2} \gamma gxv\{\frac13 (2s - 1)v -1\} + \tbinom{s}{2} + \beta\gamma\tbinom{s}{3}x^2 v \pmod{p}, \displaybreak \\ +%% +W_s &\equiv \tbinom{s}{2} + \Bigl\{ + \epsilon xv + egv + (\alpha\gamma + \delta kv + \beta kz)\tbinom{s}{2} + + \beta\gamma\ k\tbinom{x}{3}v + ac\tbinom{y}{2}v \\ + & \qquad + jvz + ag\tbinom{z}{2}v + \alpha xz + kxy + a\gamma xyv + ayz + \Bigr\} ++ \tbinom{s}{3} + \Bigl\{ + \alpha cxyv \\ & \qquad + \alpha\gamma x^2 v + 2\beta\gamma k\tbinom{x}{2} xv + + gkxzv + \delta kx^2 v + + \beta kx^2 z + acvy^2 \\ & \qquad + a\gamma xvy + \Bigr\} ++ \beta k \gamma\tbinom{s}{4}x^3 v + \tbinom{s}{2}\frac{2s-1}{3} + \Bigl\{ + a\delta\gamma\tbinom{x}{2}v^2 + a\delta xzv \\ + & \qquad + agvz^2 + \Bigr\} ++ \frac{1}{2}v\tbinom{s}{2} + \Bigl\{ + \frac13(2s-1)v - 1 + \Bigr\} + \Bigl\{ + \gamma jx + e\delta x + a\delta\gamma x \\ + & \qquad + \alpha c \delta\tbinom{x}{2} + \gamma gk \tbinom{x}{2} + cjy + egz + \Bigr\} ++ \frac{1}{6}\tbinom{s}{2} + \Bigl\{ + \tbinom{s}{2}v^2 - (2s-1)v \\ & \qquad + 2 + \Bigr\} + \bigl\{ + c\delta jx + eg\gamma x + \bigr\}v ++ \frac{1}{2}\tbinom{s}{3} + \Bigl\{ + \frac{1}{2}(s-1)v-1 + \Bigr\} + \bigl\{ + \alpha c \delta \\ & \qquad + \gamma gk + \bigr\} x^2 v ++ \frac12 a\beta x \tbinom{s}{2} + \Bigl\{ + \frac{1}{3}(2s-1)z - 1 + \Bigr\} z \\ +& \qquad + \frac{1}{2} a\delta\gamma x^2 v\tbinom{s}{3} \frac{1}{2}(3s-1) \pmod{p} +\end{align*} + +Placing $v = 1,\, z = y = s = p$ in (29)\footnote{For $p = 3$ and +$c\delta \equiv \gamma g \equiv \beta \gamma +\equiv 0 \pmod{p}$ there are terms of the form $(A + Bx + Cx^2 + Dx^3) +p^{m-4}$ in the exponent of $P$. For $m > 5$ these do not vitiate our +conclusion. For $p = 3$ and $c\delta$, $\gamma g$, or $\beta\gamma$ prime +to $p$, $[S_1\, P^x]^p$ is not contained in $\{P\}$ and the groups defined +belong to Class II.} +\begin{gather*} +[S_1\, P^x]^p = S{}_1^p P^{xp} = P^{(\lambda + x)p} \qquad (p > 3). +\intertext{If $x$ be so chosen that} +x + \lambda \equiv 0 \pmod{p^{m-4}}. +\intertext{$S = S_1\, P^x$ is an operator of order $p$ and is taken in place +of $S_1$.} +S^p = 1. +\end{gather*} + +The substitution of $S$ for $S_1$ leaves congruence (17) invariant. + +\medskip +5. \textit{Transformation of the groups.} All groups of this class are given by + +\begin{equation} +G: \begin{cases} +Q^{-1} P\, Q = P^{1 + kp^{m-4}}, \\ +R^{-1} P\, R = Q^\beta\, P^{1 + \alpha p^{m-4}}, \\ +R^{-1} Q\, R = Q\, P^{ap^{m-4}}, \\ +S^{-1} P\, S = R^\gamma\, Q^\delta P^{1 + \epsilon p^{m-4}}, \\ +S^{-1} Q\, S = R^c\, Q\, P^{ep^{m-4}}, \\ +S^{-1} R\, S = R\, Q^g\, P^{jp^{m-4}}, \\ +\end{cases} %% 30 +\end{equation} +\noindent with +\begin{equation*} +P^{p^{m-3}} = 1, \quad Q^p = R^p = S^p = 1, +\end{equation*} +\noindent$(k,\, \beta,\, \alpha,\, a,\, \gamma,\, \delta,\, \epsilon,\, c,\, +e,\, g,\, j = 0,\, 1,\, 2,\, \cdots,\, p - 1)$. + +These constants are however subject to conditions (17), (21), (22), (23), +(24) and (25). Not all these groups are distinct. Suppose that $G$ and +$G'$ of the above set are simply isomorphic and that the correspondence is +given by + +\begin{equation*} +C = \left[ \begin{matrix}S, & R, & Q, & P \\ + S'_1, & R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right]. +\end{equation*} + +Inspection of (29) gives +\begin{align*} +S'_1 &= S'^{v'''} R'^{z'''} Q'^{y'''} P'^{x'''p^{m-4}}, \\ +R'_1 &= S'^{v''} R'^{z''} Q'^{y''} P'^{x''p^{m-4}}, \\ +Q'_1 &= S'^{v'} R'^{z'} Q'^{y'} P'^{x'p^{m-4}}, \\ +P'_1 &= S'^v R'^z Q'^y P'^x, +\end{align*} +\noindent in which $x$ and one out of each of the sets $v'$, $z'$, $y'$, $x'$; $v''$, +$z''$, $y''$, $x''$; $v'''$, $z'''$, $y'''$, $x'''$ are prime to $p$. + +Since $S$, $R$, $Q$, and $P$ satisfy equations (30), $S'_1$, $R'_1$, $Q'_1$ +and $P'_1$ also satisfy them. Substituting these operators and reducing in +terms of $S'$, $R'$, $Q'$, and $P'$ we get the six equations +\begin{equation} +[V'_\kappa,\, Z'_\kappa,\, Y'_\kappa,\, X'_\kappa] = [V_\kappa,\, Z_\kappa,\, +Y_\kappa,\, X_\kappa] \qquad (\kappa = 1,\, 2,\, 3,\, 4,\, 5,\, 6), %% 31 +\end{equation} +\noindent which give the following twenty-four congruences +\begin{equation} + \begin{cases} + V'_\kappa \equiv V_\kappa \pmod{p}, \\ + Z'_\kappa \equiv Z_\kappa \pmod{p}, \\ + Y'_\kappa \equiv Y_\kappa \pmod{p}, \\ + X'_\kappa \equiv X_\kappa \pmod{p^{m-3}}, + \end{cases} %% 32 +\end{equation} +\noindent where +\begin{align*} +V'_1 &= v, \quad V_1 = v, \\ +Z'_1 &= Z + c'(yv' - y'v) + \gamma'xv' + c\delta x\tbinom{v'}{2}, \quad Z_1 = z, \\ +Y'_1 &= y + g'(zv' - z'v) + \delta'xv' + \gamma'g'x\tbinom{v}{2} + \beta'xz', \quad Y_1 = y, \\ +%% +X'_1 &= x + \Bigl\{\epsilon'xv' + (\gamma'j'x + e'\delta'x + a'\delta'\gamma'x) + \tbinom{v'}{2} + c'\delta'j'\tbinom{v'}{3} + (\alpha'\gamma'v' + \delta'k'v' \\ + &\quad+ a'\delta'\gamma'v^2 + \beta'k'z')\tbinom{x}{2} + j'(zv' - z'v) + + e'g'[z\tbinom{v'}{2} - z'\tbinom{v}{2}] \\ + &\quad+ a'g'[\tbinom{z}{2}v' + \tbinom{z'}{2}v - zz'v] + e'(yv' - y'v) + + c'j'[y\tbinom{v'}{2} - y'\tbinom{v}{2}] \\ + &\quad+ a'c'[\tbinom{y}{2}v' + v\tbinom{-y'}{2} - yy'v] + a'(yz' - y'z) + - a'\beta'xz'^2 + \alpha'xz' \\ + &\quad+ \alpha'\beta'x\tbinom{z'}{2} + a'\gamma'x(y - y')v' + k'xy'\Bigr\}p^{m-4}, \\ +%% +X_1 &= x + kxp^{m-4}, \\ +V'_2 &= v, \quad V_2 = v + \beta v', \\ +Z'_2 &= z + c'(yv'' - y''v) + \gamma'xv'' + e'\delta'\tbinom{v''}{2}, + \quad Z_2 = z + \beta z' + c'\beta y'v, \\ +Y'_2 &= y + g'(zv'' - z''v) + \delta'xv'' + \gamma'g'x\tbinom{v''}{2} + + \beta'\gamma'\tbinom{x}{2}v'' + \beta'xz'', \\ +%% +Y_2 &= y + \beta y' + g'\beta z'v, \\ +X'_2 &= x + \Bigl\{\Theta'_1(x, v'') + j'(zv'' - z''v) + e'g'[z\tbinom{v''}{2} - + z''\tbinom{v}{2}] + a'g'[\tbinom{x}{2}v'' \\ + &\quad+ \tbinom{-z''}{2}v - zz''v] + e'(yv'' - y''v) + c'j'[y\tbinom{v''}{2} + - y''\tbinom{v}{2}] + a'c'[\tbinom{y}{2}v'' \\ + &\quad+ \tbinom{-y''}{2}v - yy''v''] + a'g'(zv'' - z''v)z'' + a'(yz'' - y''z) + + a'\delta'v''z'' \\ + &\quad+ a'\gamma'(y - y'')v''x + \alpha'xz'' + a'\beta'x\tbinom{z''}{2} + + \beta'k'\tbinom{x}{2}z'' + k'xy''\Bigr\}p^{m-4}, \\ +%% +X_2 &= x + \Bigl\{\alpha x + \beta x' + a'\tbinom{\beta}{2}y'z' + e'\beta vy' + + (c'j'\beta + e'g'\beta z')\tbinom{v}{2} \\ + &+ a'c'\tbinom{\beta y'}{2}v + j'\beta vz' + a'g'\tbinom{\beta z'}{2} + + a'\beta(g'z'v + y')z\Bigr\}p^{m-4}, \\ +V'_3 &= v', \quad V_3 = v', \\ +Z'_3 &= z' + c'(y'v'' - y''v'), \quad Z_3 = z', \\ +Y'_3 &= y' + g'(z'v'' - z''v'), \quad Y_3 = y', \\ +%% +X'_3 &= \Bigl\{x' + j'(z'v'' - z''v') + e'g'[\tbinom{v''}{2}z' - + \tbinom{v'}{2}z''] + a'g'[\tbinom{z'}{2}v'' + \tbinom{-z''}{2}v' \\ + &\quad- z'z''v'] + e'(y'v'' - y''v') + c'j'[y'\tbinom{v''}{2} - y''\tbinom{v'}{2}] + + a'c'[\tbinom{y'}{2}v'' + \tbinom{-y''}{2}v' \\ + &\quad- y''y'v''] + a'(y'z'' - y''z')\Bigr\}p^{m-4}, \\ +X_4 &= (x' + a'x)p^{m-4}, \\ +%% +V'_4 &= v, \quad V_4 = v + \gamma v'' + \delta v', \\ +Z'_4 &= z + c'(yv''' - y'''v) + \gamma'xv''' + c'\delta'x\tbinom{v'''}{2}, \\ +Z_4 &= z + \gamma z'' + \delta z' + c'[\tbinom{\gamma}{2}v''y'' + + \tbinom{\delta}{2}v'y'] + c'(\gamma y'' + \delta y')v + c'\gamma\delta y''v, \\ +Y'_4 &= y + g'(zv''' - z'''v) + \delta' xv''' + \gamma' g'x\tbinom{v'''}{2} + + \beta'\gamma'\tbinom{x}{2}v''' + \beta'xz''', \\ +%% +Y_4 &= y + \gamma y'' + \delta y' + g'[\tbinom{\gamma}{2}v''z'' + + \tbinom{\delta}{2}v'z'] + g'(\gamma z'' + \delta z')v + g'\delta\gamma v'z'', \\ +X'_4 &= x + \Bigl\{\Theta'_1 (x, v''') + j'(zv''' - z'''v) + + e'g'[\tbinom{v'''}{2}z - \tbinom{v}{2}z'''] + a'g'\left[\tbinom{z}{2}v''' \right. \\ + &\quad \left. + \tbinom{-z'''}{2}v - zz'''v\right] + e'(yv''' - y'''v) + + c'j'[y\tbinom{v'''}{2} - y'''\tbinom{v}{2}] \\ + &\quad+ a'c'[\tbinom{y}{2}v''' + \tbinom{-y'''}{2}v - yy'''v'''] + + a'g'(v'''z - vz''')z''' \\ + &\quad+ a'(yz''' - y'''z) + a'\delta' xz'''v''' + a'\gamma' x(y - y''')v''' + \alpha' xz''' \\ + &\quad+ a'\beta' x\tbinom{z'''}{2} + \beta' k'z'''\tbinom{x}{2} + k'xy''' \Bigr\} p^{m-4}. \displaybreak \\ +%% +X_4 &= x + \Bigl\{\epsilon x + \delta x' + \gamma x'' + \tbinom{\gamma}{2} + [a'c'\tbinom{y''}{2}v'' + a'y''z'' + e'v''y'' + j'v''z'' \\ + &\quad+ a'g'\tbinom{z''}{2}v'' + (c'j'v''y'' + e'g'v''z'')(v + \delta v') + + a'(z + \delta z')v''z'' \\ + &\quad+ \frac{2\gamma - 1}{3}a'g'v''z''^2 + \frac{1}{2}[\frac{1}{3}(2\gamma - 1)v'' + - 1](c'j'y'' + e'g'z'')v''] \\ + &\quad+ \tbinom{\gamma}{3}a'c'v''y'' + \tbinom{\delta}{2}[a'c'\tbinom{y'}{2}v' + + a'y'z' + e'v'y' + j'v'z' \\ + &\quad+ a'g'\tbinom{z'}{2}v' + j'c'vv'y' + e'g'vv'z' + a'g'v'zz' + a'c'\gamma y'y''v' \\ + &\quad+ \frac{2\delta - 1}{3}a'g'v'z'^2 + \frac{1}{2}\{\frac{1}{3}(2\delta - 1)v' + - 1\}(c'j'y' + e'g'z')] \\ + &\quad+ \tbinom{\delta}{3}a'c'v'y'^2 + (v + \delta v')[j'\gamma z'' + + \tbinom{\gamma z''}{2}a'g' + e'\gamma y'' + \tbinom{\gamma y''}{2}a'c' \\ + &\quad+ a'g'(z + \delta z')] + \tbinom{v + \delta v'}{2}[e'g'\gamma z'' + + c'j'\gamma y''] + \delta[(e'g'z' \\ + &\quad+ c'j'y')\tbinom{v}{2} + e'vy' + j'z' + a'zy + a'g'vzz' + a'\gamma z'y'' + + a'c'\gamma vy'y''] \\ + &\quad+ a'g'\tbinom{\delta z'}{2}v + a'c'\tbinom{\delta y'}{2} v + + a'\gamma zy''\Bigr\}p^{m-4}, \\ +%% +V'_5 &= v', \quad V_5 = v' + cv'', \\ +Z'_5 &= z' + c'(y'v''' - y'''v'), \quad Z_5 = z' + cz'' + c'cy''v, \\ +Y'_5 &= y' + g'(z'v''' - z'''v'), \quad Y_5 = y' + cy'' + g'cv'z'', \\ +X'_5 &= \Bigl\{x' + j'(z'v''' - z'''v') + e'g'[\tbinom{v'''}{2}z' + - \tbinom{v'}{2}z'''] + a'g'[\tbinom{z'}{2}v''' \\ + &\quad+ \tbinom{-z'''}{2}v' - z'z'''v'] + c'(y'v''' - y'''v') + + c'j'[y'\tbinom{\delta'''}{2} - y'''\tbinom{v'}{2}] \\ + &\quad+ a'c'[\tbinom{y'}{2}v''' + \tbinom{-y'''}{2}v' - y'y'''v'''] + + a'(y'z''' - y'''z')\Bigr\}p^{m-4}, \\ +%% +X_5 &= \Bigl\{x' + ex + cx'' + a'\tbinom{c}{2}y''z'' + j'cv'z'' + + (e'g'cz'' + c'cj'y'')\tbinom{v'}{2} + e'cy''v \\ + &\quad+ a'cy''z' + a'g'z'v' + a'g'\tbinom{cz''}{2} + a'c'\tbinom{cy''}{2}\Bigr\}p^{m-4}, \\ +V'_6 &= v'', \quad V_6 = v'' + gv', \\ +Z'_6 &= z'' + c'(y''v''' - y'''v''), \quad Z_6 = z'' + gz', \\ +Y'_6 &= y'' + g'(z''v''' - z'''v''), \quad Y_6 = y'' + gy', \\ +%% +X'_6 &= \Bigl\{x'' + j'(z''v''' - z'''v'') + e'g'[\tbinom{v'''}{2}z'' + - \tbinom{v''}{2}z'''] + a'g'\left[\tbinom{z''}{2}v''' \right. \\ + &\quad+ \left. \tbinom{-z'''}{2}v'' - z''z'''v''\right] + e'(y''v''' - y'''v'') + + c'j'[y''\tbinom{v'''}{2} - y'''\tbinom{v''}{2}] \\ + &\quad+ a'c'[\tbinom{y''}{2}v''' + \tbinom{-y'''}{2}v'' - y''y'''v'''] + + a'(y''z''' - y'''z'')\Bigr\}p^{m-4}, \\ +X_6 &= \{x'' + jx + gx' + a'gy''z'\}p^{m-4}. +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of the +two groups $G$ and $G'$ is \textit{that congruences (32) shall be consistent and +admit of solution} subject to conditions derived below. + +\medskip +6. \textit{Conditions of transformation.} Since $Q$ is not contained in $\{P\}$, +$R$ is not contained in $\{Q, P\}$, and $S$ is not contained in $\{R, Q, +P\}$, then $Q'_1$ is not contained in $\{P'_1\}$, $R'_1$ is not contained +in $\{Q'_1, P'_1\}$, and $S'_1$ is not contained in $\{R'_1, Q'_1, P'_1\}$. + +Let +\begin{equation*} +{Q'}_1^{s'} = {P'}_1^{sp^{m-4}}. +\end{equation*} + +This equation becomes in terms of $S'$, $R'$, $Q'$ and $P'$ +\begin{gather*} +[s'v',\, s'z' + c'\tbinom{s'}{2}v'y',\, s'y' + g'\tbinom{s'}{2}v'z',\, Dp^{m-4}] + = [0,\, 0,\, 0,\, sxp^{m-4}], +\intertext{and} +s'v' \equiv s'z' \equiv s'y' \equiv 0 \pmod{p}. +\end{gather*} + +At least one of the three quantities $v'$, $z'$ or $y'$ is prime to $p$, +since otherwise $s'$ may be taken $= 1$. + +Let +\begin{equation*} +{R'}_1^{s''} = {Q'}_1^{s'} {P'}_1^{sp^{m-4}}, +\end{equation*} +\noindent or in terms of $S'$, $R'$, $Q'$ and $P'$ +\begin{multline*} +[s''v'',\, s''z'' + c'\tbinom{s''}{2}v''y'',\, s''y'' + g'\tbinom{s''}{2}v'' + z'',\, Ep^{m-4}] \\ = [s'v',\, s'z' + c'\tbinom{s'}{2}v'y',\, s'y' + + g'\tbinom{s'}{2}v'z',\, E_1 p^{m-4}], +\end{multline*} +\noindent and +\begin{align*} +s''v'' &\equiv s'v' \pmod{p}, \\ +s''z'' + c'\tbinom{s''}{2}v''y'' &\equiv s'z' + c'\tbinom{s'}{2}v'y' \pmod{p}, \\ +s''y'' + g'\tbinom{s''}{2}v''z'' &\equiv s'y' + g'\tbinom{s'}{2}v'z' \pmod{p}. +\end{align*} + +Since $c'g' \equiv 0 \pmod{p}$, suppose $g' \equiv 0 \pmod{p}$. Elimination +of $s'$ between the last two give by means of the congruence $Z'_3 \equiv +Z_3 \pmod{p}$, +\begin{equation*} +s''\{2(y'z'' - y''z') + c'y'y''(v' - v'')\} \equiv 0 \pmod{p}, +\end{equation*} +\noindent between the first two +\begin{equation*} +s''\{2(v'z'' - v''z') + c'v'v''(y' - y'')\} \equiv 0 \pmod{p}, +\end{equation*} +\noindent and between the first and last +\begin{equation*} +s''(y'v'' - y''v') \equiv 0 \pmod{p}. +\end{equation*} + +At least one of the three above coefficients of $s''$ is prime to $p$, +since otherwise $s''$ may be taken $= 1$. + +Let +\begin{equation*} +{S'}_1^{s'''} = {R'}_1^{s''} {Q'}_1^{s'} {P'}_1^{sp^{m-4}} +\end{equation*} +\noindent or, in terms of $S'$, $R'$, $Q'$, and $P'$ +\begin{multline*} +[s'''v''', s'''z''' + c'\tbinom{s'''}{2}v'''y''', s'''y''' + +g'\tbinom{s'''}{2}v'''z''', E_2 p^{m-4}] \\ = [s''v'' + s'v', s''z'' + s'z' + +c'\{\tbinom{s''}{2}v''y'' + \tbinom{s'}{2}v'y' + s's''y''v'\}, \\ s''y'' + +s'y' + g'\{\tbinom{s''}{2}v''z'' + \tbinom{s'}{2}v'z' + s's''v'z''\}, E_3 p^{m-4}] +\end{multline*} +\noindent and +\begin{align*} +s'''v''' & \equiv s''v'' + s'v' \pmod{p}, \\ +s'''z''' & + c'\tbinom{s'''}{2}v'''y''' \\ + & \qquad \equiv s''z'' + s'z' + c'\{\tbinom{s''}{2}v''y'' + \tbinom{s'}{2}v'y' + + s's''y''v'\} \pmod{p}, \\ +s'''y''' & + g'\tbinom{s'''}{2}v'''z''' \\ + & \qquad \equiv s''y'' + s'y' + g'\{\tbinom{s''}{2}v''z'' + \tbinom{s'}{2}v'z' + + s's''z''v'\} \pmod{p}. +\end{align*} + +If $g' \equiv 0$ and $c' \not\equiv 0 \pmod{p}$ the congruence $Z'_3 +\equiv Z_3 \pmod{p}$ gives +\begin{equation*} +(y'v'' - y''v') \equiv 0 \pmod{p}. +\end{equation*} + +Elimination in this case of $s''$ between the first and last congruences +gives +\begin{equation*} +s'''(y''v''' - y'''v'') \equiv 0 \pmod{p}. +\end{equation*} + +Elimination of $s''$ between the first and second, and between the second +and third, followed by elimination of $s'$ between the two results, gives +\begin{equation*} +s'''\left(z''^2 - c'y''z''v' + \frac{c'^2}{4}y''v''\right) +(y'v''' - y'''v') \equiv 0 \pmod{p}. +\end{equation*} + +Either $(y''v''' - y'''v'')$, or $(y'v''' - y'''v')$ is prime to $p$, +since otherwise $s'''$ may be taken $= 1$. + +A similar set of conditions holds for $c' \equiv 0$ and $g' \not\equiv 0 +\pmod{p}$. + +When $c' \equiv g' \equiv 0 \pmod{p}$ elimination of $s'$ and $s''$ between +the three congruences gives +\begin{equation*} +s'''\Delta \equiv s''' \left|\begin{matrix} + v' & v'' & v''' \\ + y' & y'' & y''' \\ + z' & z'' & z''' \\ \end{matrix}\right| +\equiv 0 \pmod{p} +\end{equation*} +\noindent and $\Delta$ is prime to $p$, since otherwise $s'''$ +may be taken $= 1$. + +\newpage +7. \textit{Reduction to types.} In the discussion of congruences (32), the +group $G'$ is taken from the simplest case and we associate with it all +simply isomorphic groups $G$. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|r|c|c|c|c|c|c||r|c|c|c|c|c|} +\multicolumn{7}{c}{A.}&\multicolumn{6}{c}{B.} \\ \hline + &$a_2$&$\beta_2$&$c_2$&$g_2$&$\gamma_2$&$\delta_2$& &$k_2$&$\alpha_2$&$\epsilon_2$&$e_2$&$j_2$ \\ \hline +\textbf{ 1}& 1 & 1 & 1 & 1 & 1 & 1 &\textbf{ 1}& 1 & 1 & 1 & 1 & 1 \\ \hline +\textbf{ 2}& 0 & 1 & 1 & 1 & 1 & 1 &\textbf{ 2}& 0 & 1 & 1 & 1 & 1 \\ \hline +\textbf{ 3}& 0 & 0 & 1 & 1 & 1 & 1 &\textbf{ 3}& 1 & 0 & 1 & 1 & 1 \\ \hline +\textbf{ 4}& 0 & 0 & 1 & 1 & 1 & 0 &\textbf{ 4}& 1 & 1 & 0 & 1 & 1 \\ \hline +\textbf{ 5}& 0 & 0 & 1 & 0 & 1 & 1 &\textbf{ 5}& 1 & 1 & 1 & 0 & 1 \\ \hline +\textbf{ 6}& 0 & 0 & 1 & 0 & 1 & 0 &\textbf{ 6}& 1 & 1 & 1 & 1 & 0 \\ \hline +\textbf{ 7}& 0 & 1 & 0 & 1 & 1 & 1 &\textbf{ 7}& 0 & 0 & 1 & 1 & 1 \\ \hline +\textbf{ 8}& 0 & 1 & 0 & 1 & 0 & 1 &\textbf{ 8}& 0 & 1 & 0 & 1 & 1 \\ \hline +\textbf{ 9}& 0 & 1 & 1 & 0 & 1 & 1 &\textbf{ 9}& 0 & 1 & 1 & 0 & 1 \\ \hline +\textbf{10}& 0 & 1 & 1 & 0 & 1 & 0 &\textbf{10}& 0 & 1 & 1 & 1 & 0 \\ \hline +\textbf{11}& 1 & 0 & 1 & 1 & 1 & 1 &\textbf{11}& 1 & 0 & 0 & 1 & 1 \\ \hline +\textbf{12}& 1 & 0 & 1 & 0 & 1 & 1 &\textbf{12}& 1 & 0 & 1 & 0 & 1 \\ \hline +\textbf{13}& 1 & 0 & 1 & 1 & 0 & 1 &\textbf{13}& 1 & 0 & 1 & 1 & 0 \\ \hline +\textbf{14}& 1 & 0 & 1 & 1 & 1 & 0 &\textbf{14}& 1 & 1 & 0 & 0 & 1 \\ \hline +\textbf{15}& 1 & 0 & 1 & 0 & 0 & 1 &\textbf{15}& 1 & 1 & 0 & 1 & 0 \\ \hline +\textbf{16}& 1 & 0 & 1 & 0 & 1 & 0 &\textbf{16}& 1 & 1 & 1 & 0 & 0 \\ \hline +\textbf{17}& 1 & 0 & 1 & 1 & 0 & 0 &\textbf{17}& 0 & 0 & 0 & 1 & 1 \\ \hline +\textbf{18}& 1 & 0 & 1 & 0 & 0 & 0 &\textbf{18}& 0 & 0 & 1 & 0 & 1 \\ \hline +\textbf{19}& 1 & 1 & 0 & 1 & 1 & 1 &\textbf{19}& 0 & 0 & 1 & 1 & 0 \\ \hline +\textbf{20}& 1 & 1 & 0 & 1 & 0 & 1 &\textbf{20}& 0 & 1 & 0 & 0 & 1 \\ \hline +\textbf{21}& 1 & 1 & 0 & 1 & 1 & 0 &\textbf{21}& 0 & 1 & 0 & 1 & 0 \\ \hline +\textbf{22}& 1 & 1 & 0 & 1 & 0 & 0 &\textbf{22}& 0 & 1 & 1 & 0 & 0 \\ \hline +\textbf{23}& 1 & 1 & 1 & 0 & 1 & 1 &\textbf{23}& 1 & 0 & 0 & 0 & 1 \\ \hline +\textbf{24}& 1 & 1 & 1 & 1 & 0 & 1 &\textbf{24}& 1 & 0 & 0 & 1 & 0 \\ \hline +\textbf{25}& 1 & 1 & 1 & 1 & 1 & 0 &\textbf{25}& 1 & 0 & 1 & 0 & 0 \\ \hline +\textbf{26}& 1 & 1 & 1 & 0 & 0 & 1 &\textbf{26}& 1 & 1 & 0 & 0 & 0 \\ \hline +\textbf{27}& 1 & 1 & 1 & 0 & 1 & 0 &\textbf{27}& 0 & 0 & 0 & 0 & 1 \\ \hline +\textbf{28}& 1 & 1 & 1 & 1 & 0 & 0 &\textbf{28}& 0 & 0 & 0 & 1 & 0 \\ \hline +\textbf{29}& 1 & 1 & 1 & 0 & 0 & 0 &\textbf{29}& 0 & 0 & 1 & 0 & 0 \\ \hline + & & & & & & &\textbf{30}& 0 & 1 & 0 & 0 & 0 \\ \hline + & & & & & & &\textbf{31}& 1 & 0 & 0 & 0 & 0 \\ \hline + & & & & & & &\textbf{32}& 0 & 0 & 0 & 0 & 0 \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &1 &2 &3 &4&5 &6 &7 &8 &9 &10 \\ \hline +\textbf{1} &$\times$&$\times$&$\times$& &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{2} &$\times$&$2_1$ &$3_1$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{3} &$1_2$ &$2_1$ &$3_1$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{4} &$1_2$ &$\times$&$\times$& &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{5} &$2_1$ &$2_1$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{6} &$2_1$ &$2_1$ &$3_1$ & &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{7} &$1_2$ &$2_1$ &$3_1$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{8} &$1_2$ &$2_4$ &$3_4$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{9} &$2_1$ &$2_1$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{10}&$2_4$ &$2_4$ &$3_4$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{11}&$1_2$ &$2_4$ &$3_4$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{12}&$2_4$ &$2_4$ & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{13}&$2_1$ &$2_1$ &$3_1$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{14}&$2_1$ &$2_4$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{15}&$2_1$ &$2_4$ &$3_4$ & &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{16}&$2_1$ &$2_1$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{17}&$1_2$ &$2_4$ &$3_4$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{18}&$2_4$ &$2_4$ & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{19}&$2_4$ &$2_4$ &$3_4$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{20}&$2_1$ &$2_4$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{21}&$2_4$ &* &* & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{22}&$2_4$ &$2_4$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{23}&$2_4$ &* & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{24}&$2_1$ &$2_4$ &$3_4$ & & &$19_6$& & &$19_6$&$19_6$ \\ \hline +\textbf{25}&$2_4$ &$2_4$ & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{26}&$2_1$ &$2_4$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{27}&$2_4$ &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{28}&$2_4$ &* &* & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{29}&* &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{30}&$2_4$ &* & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{31}&$2_4$ &* & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{32}&* &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize (continued) + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &11 &12 &13 &14 &15 &16 &17 &18 &19 \\ \hline +\textbf{1} &$\times$ &$19_1$&$\times$ &$11_1$ &$19_1$&$19_1$&$13_1$ &$19_1$&$\times$ \\ \hline +\textbf{2} &$25_2$ & &$\times$ &$25_2$ & & &$13_2$ & &$\times$ \\ \hline +\textbf{3} &$11_1$ &$19_2$&$13_1$ &$24_2$ &$21_2$&$19_2$&$13_1$ &$21_2$& \\ \hline +\textbf{4} &$24_2$ &$19_2$&$13_1$ &$24_2$ &$19_1$&$19_2$&$13_1$ &$19_1$&$19_2$ \\ \hline +\textbf{5} & & & & & & & & &$19_1$ \\ \hline +\textbf{6} &$\times$ &$19_2$&$\times$ &$11_6$ &$21_2$&$19_2$&$13_6$ &$21_2$&$\times$ \\ \hline +\textbf{7} &$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & & \\ \hline +\textbf{8} &$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & &$19_2$ \\ \hline +\textbf{9} & &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_2$ \\ \hline +\textbf{10}&$25_{10}$& &$\times$ &$25_{10}$& & &$13_{10}$& &$19_6$ \\ \hline +\textbf{11}&$24_2$ &$19_2$&$13_1$ &* &$21_2$&$19_2$&$13_1$ &$21_2$& \\ \hline +\textbf{12}& & & & & & & & & \\ \hline +\textbf{13}&$11_6$ &* &$13_6$ &$11_6$ &* &$19_2$&$13_6$ &* & \\ \hline +\textbf{14}& & & & & & & & &$19_2$ \\ \hline +\textbf{15}&$11_6$ &$19_2$&$13_6$ &$11_6$ &$21_2$&$19_2$&$13_6$ &$21_2$&$19_6$ \\ \hline +\textbf{16}& & & & & & & & &$19_6$ \\ \hline +\textbf{17}&$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & & \\ \hline +\textbf{18}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{19}&$25_{10}$& &$13_{10}$&$25_{10}$& & &$13_{10}$& & \\ \hline +\textbf{20}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_2$ \\ \hline +\textbf{21}&$25_{10}$& &$13_{10}$&$25_{10}$& & &$13_{10}$& &$19_6$ \\ \hline +\textbf{22}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_6$ \\ \hline +\textbf{23}& & & & & & & & & \\ \hline +\textbf{24}& &$11_6$&$19_2$ &$13_6$ &$11_6$&* &* &$13_6$&* \\ \hline +\textbf{25}& & & & & & & & & \\ \hline +\textbf{26}& & & & & & & & &$19_6$ \\ \hline +\textbf{27}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{28}&$25_{10}$& &$13_{10}$&$25_10$ & & &$13_{10}$& & \\ \hline +\textbf{29}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{30}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_6$ \\ \hline +\textbf{31}& & & & & & & & & \\ \hline +\textbf{32}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize (concluded) + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &20 &21 &22 &23 &24 &25 &26 &27 &28 &29 \\ \hline +\textbf{1} &$19_1$&$19_1$ &$19_1$&$19_1$&$11_1$ &$11_1$ &$19_1$ &$19_1$&$11_1$&$19_1$ \\ \hline +\textbf{2} &$19_2$&$\times$&$21_2$& &$\times$ &$\times$ & & &$25_2$& \\ \hline +\textbf{3} & & & &$19_2$&$25_2$ &$24_2$ &$21_2$ &$19_2$&$25_2$&$21_2$ \\ \hline +\textbf{4} &$19_2$&$19_1$ &$19_1$&$19_2$&$11_1$ &$11_1$ &$19_1$ &$19_2$&$11_1$&$19_1$ \\ \hline +\textbf{5} &$19_2$&$19_1$ &$19_1$&$19_6$&$11_6$ &$3_1$ &$21_6$ &$19_6$&$3_1$ &$21_6$ \\ \hline +\textbf{6} &$19_6$&$\times$&$21_6$&$19_1$&$3_1$ &$11_6$ &$19_1$ &$19_2$&$3_1$ &$19_1$ \\ \hline +\textbf{7} & & & & &$25_2$ &$25_2$ & & &* & \\ \hline +\textbf{8} &$19_2$&$21_2$ &$21_2$& &$24_2$ &$25_2$ & & &$25_2$& \\ \hline +\textbf{9} &$19_2$&$21_2$ &$21_2$& &$11_6$ &$3_1$ & & &$3_1$ & \\ \hline +\textbf{10}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$\times$ & & &$3_4$ & \\ \hline +\textbf{11}& & & &$19_2$&$25_2$ &$24_2$ &$21_2$ &$19_2$&$25_2$&$21_2$ \\ \hline +\textbf{12}& & & &$19_6$&$25_{10}$ &$3_4$ &$21_6$ &$19_6$&$3_4$ &$21_6$ \\ \hline +\textbf{13}& & & &$19_2$&$3_1$ &$11_6$ &$21_2$ &$19_2$&$3_1$ &$21_2$ \\ \hline +\textbf{14}&* &$19_1$ &$19_1$&$19_6$&$11_6$ &$3_1$ &$21_6$ &$19_6$&$3_1$ &$21_6$ \\ \hline +\textbf{15}&$19_6$&$21_6$ &$21_6$&$19_2$&$3_1$ &$11_6$ &$19_1$ &* &$3_1$ &$19_1$ \\ \hline +\textbf{16}&$19_6$&$21_6$ &$21_6$&$19_6$&$3_1$ &$3_1$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{17}& & & & &$25_2$ &$25_2$ & & &* & \\ \hline +\textbf{18}& & & & &$25_{10}$ &$3_4$ & & &$3_4$ & \\ \hline +\textbf{19}& & & & &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{20}&$19_2$&$21_2$ &$21_2$& &$11_6$ &$3_1$ & & &$3_1$ & \\ \hline +\textbf{21}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{22}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$3_4$ & & &* & \\ \hline +\textbf{23}& & & &$19_6$&$25_{10}$ &$3_4$ &$21_6$ &$19_6$&$3_4$ &$21_6$ \\ \hline +\textbf{24}& & & &$19_2$&$3_1$ &$11_6$ &$21_2$ &$19_2$&$3_1$ &$21_2$ \\ \hline +\textbf{25}& &$21_6$ &$21_6$&$19_6$&$3_4$ &$3_4$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{26}&$19_6$&$21_6$ &$21_6$&$19_6$&$3_1$ &$3_1$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{27}& & & & &$25_{10}$ &$3_4$ & & &$3_4$ & \\ \hline +\textbf{28}& & & & &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{29}& & & & &* &* & & &* & \\ \hline +\textbf{30}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$3_4$ & & &* & \\ \hline +\textbf{31}& & & &$19_6$&$3_4$ &$3_4$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{32}& & & & &* &* & & &* & \\ \hline +\end{tabular} +\end{center} + +For convenience the groups are divided into cases. + +The double Table I gives all cases consistent with congruences (17), (21), +(23) and (25). The results of the discussion are given in Table II. The +cases in Table II left blank are inconsistent with congruences (22) and +(24), and therefore have no groups corresponding to them. + +Let $\kappa = \kappa_1 p^{k_2}$ where $dv[\kappa_1,\, p] = 1\; (\kappa = a,\, +\beta,\, c,\, g,\, \gamma,\, d,\, k,\, \alpha,\, \epsilon,\, e,\, j)$. + +In explanation of Table II the groups in cases marked +$\boxed{r_s}$ are simply isomorphic with groups in $A_r B_s$. + +The group $G'$ is taken from the cases marked +$\boxed{\times}$. The types are also selected from these cases. + +The cases marked $\boxed{*}$ divide into two or more parts. Let +\begin{align*} +a\epsilon - \alpha e + jk &= I_1, & a\epsilon - jk &= I_2, \\ +a\delta(a - e) + 2I_1 &= I_3, & \alpha g - \beta j &= I_4, \\ +\alpha\delta - \beta\epsilon &= I_5, & \epsilon g - \delta j &= I_6, \\ +c\epsilon - e\gamma &= I_7, & \alpha e - jk &= I_8, \\ +\delta e + \gamma j &= I_9, & \alpha\gamma + \delta k &= I_{10}. +\end{align*} + +The parts into which these groups divide, and the cases with which they are +simply isomorphic, are given in Table III. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|l|l|c|l|c|} \hline +$A_{1,2}B^*$ &$dv[I_1,p]=p$ &$2_1$ &$dv[I_1,p]=1$ &$2_4$ \\ \hline +$A_{3}B^*$ &$dv[I_2,p]=p$ &$3_1$ &$dv[I_2,p]=1$ &$3_4$ \\ \hline +$A_{4}B^*$ &$dv[I_3,p]=p$ &$3_1$ &$dv[I_3,p]=1$ &$3_4$ \\ \hline +$A_{12}B_{13}$ &$dv[I_4,p]=p$ &$19_1$ &$dv[I_4,p]=1$ &$19_2$ \\ \hline +$A_{14}B_{11}$ &$dv[I_5,p]=p$ &$11_1$ &$dv[I_5,p]=1$ &$24_2$ \\ \hline +$A_{15,18}B^*$ &$dv[I_4,p]=p$ &$19_1$ &$dv[I_4,p]=1$ &$21_2$ \\ \hline +$A_{16}B_{24}$ &$dv[I_6,I_5,p]=p$ &$19_1$ &$dv[I_6,I_5,p]=1$ &$19_2$ \\ \hline +$A_{20}B_{14}$ &$dv[I_7,p]=p$ &$19_1$ &$dv[I_7,p]=1$ &$19_2$ \\ \hline +$A_{24,25}B^*$ &$dv[I_8,p]=p$ &$3_1$ &$dv[I_8,p]=1$ &$3_4$ \\ \hline +$A_{27}B_{15}$ &$dv[I_6,p]=p$ &$19_1$ &$dv[I_6,p]=1$ &$19_2$ \\ \hline +$A_{29}B_{7,17}$ &$dv[I_{10},p]=p$ &$24_2$ &$dv[I_{10},p]=1$ &$25_2$ \\ \hline +$A_{29}B_{16,26}$ &$dv[I_9,p]=p$ &$11_6$ &$dv[I_9,p]=1$ &$3_1$ \\ \hline +$A_{29}B_{22,25,30,31}$&$dv[I_9,p]=p$ &$25_{10}$&$dv[I_9,p]=1$ &$3_4$ \\ \hline +$A_{29}B_{29,32}$ &$dv[I_8,I_9,p]=p$ &$11_6$ &$[I_8,p]=p,[I_9,p]=1$&$3_1$ \\ \hline +$A_{29}B_{29,32}$ &$[I_8,p]=1,[I_9,p]=p$&$25_{10}$&$[I_8,p]=1,[I_9,p]=1$&$3_4$ \\ \hline +\end{tabular} +\end{center} + +\newpage +8. \textit{Types.} The types for this class are given by equations (30) where the +constants have the values given in Table IV. + +\begin{center} +\large IV. \normalsize + +\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline + & $a$ &$\beta$&$c$&$g$&$\gamma$&$\delta$&$k$&$\alpha$&$\epsilon$&$e$&$j$ \\ \hline +$1_1$ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$2_1$ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$3_1$ &$\kappa$& 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$11_1$ & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$*13_1$& 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$19_1$ & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$1_2$ & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$*13_2$& 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$19_2$ & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$*21_2$& 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ \hline +$24_2$ & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$25_2$ & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ \hline +$2_4$ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline +$3_4$ &$\kappa$& 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline +$11_6$ & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline +$*13_6$& 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline +\end{tabular} + +\footnotesize \noindent $\kappa = 1$, and a non-residue (mod $p$). + +\noindent $*$For $p=3$ these groups are isomorphic in Class II. +\end{center} + +A detailed analysis of congruences (32) for several cases is given below +as a general illustration of the methods used. + +\medskip +\begin{equation*} A_3 B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +\beta'xz' \equiv 0 \pmod{p}, \tag{II} \\ +a'(yz' - y'z) \equiv kx \pmod{p}, \tag{III} \\ +\beta v' \equiv 0, \qquad \beta z' \equiv 0, \qquad \beta y' \equiv + \beta'xz'' \pmod{p}, \tag*{(IV),(V),(VI)} \\ +a'(yz'' - y''z) + a'\beta'x\tbinom{z''}{2} \equiv +%% +\alpha x + \beta x' + a'\beta y'z \pmod{p}, \tag{VII} \\ +a'(y'z'' - y''z') \equiv ax \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v' \equiv 0 \pmod{p}, \tag{XI} \\ +\gamma z'' + \delta z' \equiv 0 \pmod{p}, \tag{XII} \\ +%% +\gamma y'' + \delta y' \equiv \beta'xz'' \pmod{p}, \tag{XIII} \\ +\begin{split} + a'(yz''' - y'''z) + a'\beta'x\tbinom{z'''}{2} \equiv \epsilon x &+ + \gamma x'' + \delta x + a'\delta y'z \\ &+ a'\gamma y''z + + a'\tbinom{\gamma}{2}y''z'' \pmod{p}, +\end{split} \tag{XIV} \\ +cv'' \equiv 0, \qquad cz'' \equiv 0, \qquad cy'' \equiv 0 \pmod{p}, + \tag*{(XV),(XVI),(XVII)} \\ +%% +a'(y'z''' - y'''z') \equiv ex \pmod{p}, \tag{XVIII} \\ +gv' \equiv 0, \qquad gz' \equiv 0, \qquad gy' \equiv 0 \pmod{p}, + \tag*{(XIX),(XX),(XXI)} \\ +a'(y''z''' - y'''z'') \equiv jx \pmod{p}, \tag{XXII} +\end{gather*} + +From (II) $z' \equiv 0 \pmod{p}$. + +The conditions of isomorphism give +\begin{equation*} + \Delta \equiv \left| \begin{matrix} + v' & v'' & v''' \\ + y' & y'' & y''' \\ + z' & z'' & z''' \\ \end{matrix}\right| \not\equiv 0 \pmod{p}. +\end{equation*} + +Multiply (IV), (V), (VI) by $\gamma$ and reduce by (XII), $\beta\gamma v' +\equiv 0$, $\beta\gamma z' \equiv 0$, $\beta\gamma y' \equiv 0 \pmod{p}$. Since +$\Delta \not\equiv 0 \pmod{p}$, one at least of the quantities, $v'$, $z'$ +or $y'$ is $\not\equiv 0 \pmod{p}$ and $\beta\gamma \equiv 0 \pmod{p}$. + +From (XV), (XVI) and (XVII) $c \equiv 0 \pmod{p}$, and from (XIX), (XX) and +(XXI) $g \equiv 0 \pmod{p}$. + +From (IV), (V), (VI) and (X) if $a \equiv 0$, then $\beta \equiv 0$ +and if $a \not\equiv 0$, then $\beta \not\equiv 0 \pmod{p}$. + +At least one of the three quantities $\beta$, $\gamma$ or $\delta$ is +$\not\equiv 0 \pmod{p}$ and one, at least, of $a$, $e$ or $j$ is $\not +\equiv 0 \pmod{p}$. + +\smallskip +$A_3$: Since $z''' \equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0$. +Elimination between (III), (X), (XIV) and (XXII) gives $a\epsilon - kj +\equiv 0 \pmod{p}$. Elimination between (VI) and (X) gives +$a'\beta'{z''}^2 \equiv a\beta \pmod{p}$ and $a\beta$ is a quadratic +residue or non-residue according as $a'\beta'$ is or is not, and there are +two types for this case. + +\smallskip +$A_4$: Since $y'$ and $z''$ are $\not\equiv 0 \pmod{p}$, $e \not\equiv 0 +\pmod{p}$. Elimination between (VI), (X), (XIII) and (XVIII) gives $a\delta +- \beta e \equiv 0 \pmod{p}$. + +This is a special form of (24). + +Elimination between (III), (VII), (X), (XIII), (XIV), (XVIII) and (XXII) +gives +\begin{equation*} +2jk + a\delta(a - e) + 2(a\epsilon - \alpha e) \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{24}$: Since from (XI), (XII) and (XIII) $y''$ and $z''' \not\equiv 0 +\pmod{p}$, and $z'' \equiv v'' \equiv 0 \pmod{p}$, (xxii) gives $j \not +\equiv 0 \pmod{p}$. + +Elimination between (III), (X), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{25}$: (XI), (XII) and (XIII) give $v' \equiv z' \equiv 0$ and $y', z''' +\not\equiv 0 \pmod{p}$ and this with (XVIII) gives $e \not\equiv 0$. + +Elimination between (III), (VII), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{28}$: Since $a \equiv 0$ then $e$ or $j \not\equiv 0 \pmod{p}$. + +Elimination between (III), (VII), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +Multiply (XIII) by $a'z'''$ and reduce +\begin{equation*} +\delta e + \gamma j \equiv a'\beta'{z'''}^2 \not\equiv 0 \pmod{p}. +\end{equation*} + +\medskip +\begin{equation*} A_{11} B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +\beta'xz' \equiv 0 \pmod{p}, \tag{II} \\ +kx \equiv 0 \pmod{p}, \tag{III} \\ +\beta v' \equiv \beta z' \equiv 0, \quad \beta y' \equiv \beta'xz'', + \tag*{(IV),(V),(VI)} \\ +\alpha x + \beta x' \equiv 0 \pmod{p}, \tag{VII} \\ +%% +ax \equiv 0 \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v \equiv 0 \pmod{p}, \tag{XI} \\ +\gamma z'' + \delta z \equiv 0 \pmod{p}, \tag{XII} \\ +\gamma y'' + \delta y \equiv \beta'xz''' \pmod{p}, \tag{XIII} \\ +%% +\epsilon x + \gamma x'' + \delta x' \equiv 0 \pmod{p}, \tag{XIV} \\ +cv'' \equiv cz'' \equiv cy'' \equiv 0 \pmod{p}, \tag*{(XV),(XVI),(XVII)} \\ +ex \equiv 0 \pmod{p}, \tag{XVIII} \\ +gv' \equiv gz' \equiv gy' \equiv 0 \pmod{p}, \tag*{(XIX),(XX),(XXI)} \\ +jx \equiv 0 \pmod{p}, \tag{XXII} +\end{gather*} + +(II) gives $z' = 0$, (III) gives $k \equiv 0$, (X) gives $a \equiv 0$, +(XV), (XVI), (XVII) give $c \equiv 0 (\Delta \not\equiv 0)$, (XVIII) gives +$e \equiv 0$, (XIX), (XX), (XXI) give $g \equiv 0$, (XXII) gives $j \equiv +0$. One of the two quantities $z''$ or $z''' \not\equiv 0 \pmod{p}$, +and by (VI) and (XIII) one of the three quantities $\beta$, $\gamma$ or +$\delta$ is $\not\equiv 0$. + +\smallskip +$A_{11}$: (XIV) gives $\epsilon \equiv 0 \pmod{p}$. Multiplying (IV), (V), +(VI) by $\gamma$ gives, by (XII), $\beta\gamma v' \equiv \beta\gamma z' +\equiv \beta\gamma y' \equiv 0 \pmod{p}$, and $\beta\gamma \equiv 0 +\pmod{p}$. + +\smallskip +$A_{14}$: Elimination between (VII) and (XIV) gives $\alpha\delta - +\beta\epsilon \equiv 0 \pmod{p}$. + +\smallskip +$A_{24}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv 0$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{25}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{28}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv$ +or $\not\equiv 0 \pmod{p}$. + +\newpage +\begin{equation*} A_{19} B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +c'(yv' - y'v) \equiv 0 \pmod{p}, \tag{I} \\ +kx \equiv 0 \pmod{p}, \tag{III} \\ +\beta v \equiv 0, \quad \beta z \equiv c'(yv'' - y''v), \quad \beta y' +\equiv 0 \pmod{p}, \tag*{(IV),(V),(VI)} \\ +%% +\alpha x + \beta x' \equiv 0 \pmod{p}, \tag{VII} \\ +c'(y'v'' - y''v') \equiv 0 \pmod{p}, \tag{VIII} \\ +ax \equiv 0 \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v' \equiv 0 \pmod{p}, \tag{XI} \\ +%% +\gamma z'' + \delta z' + c'\gamma\delta y''v + c'\tbinom{\delta}{2}v'y' + +c'\tbinom{\gamma}{2}v''y'' \equiv c'(yv''' - y'''v) \pmod {p}, \tag{XII} \\ +\gamma y'' + \delta y' \equiv 0 \pmod{p}, \tag{XIII} \\ +\epsilon x + \gamma x'' + \delta x' \equiv 0 \pmod{p}, \tag{XIV} \\ +%% +cv'' \equiv 0, \quad cz'' \equiv c'(y'v''' - y'''v'), \quad cy'' \equiv 0 + \pmod{p}, \tag*{(XV),(XVI),(XVII)} \\ +ex + cx'' \equiv 0 \pmod{p}, \tag{XVIII} \\ +gv' \equiv 0, \quad gz' \equiv c'(y''v''' - y'''v''), \quad gy' \equiv 0 + \pmod{p}, \tag*{(XIX),(XX),(XXI)} \\ +jx + gx' \equiv 0 \pmod{p}. \tag{XXII} \\ +\end{gather*} + +(III) gives $k\equiv 0$, (X) gives $a \equiv 0$. + +Since $dv[(y'v''' - y'''v'), (y''v''' - y'''v''), p] = 1$ then $dv[c, g, p] += 1$. + +If $c \not\equiv 0$, $v'' \equiv y'' \equiv 0 \pmod{p}$ and therefore $g +\equiv 0 \pmod{p}$ and if $g \not\equiv 0$, then $c \equiv 0 \pmod{p}$. + +\smallskip +$A_{12}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0 \pmod{p}$. + +\smallskip +$A_{15}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{16}$: (XVIII) gives $e \equiv 0$. Elimination between (XIV) and (XXII) +gives $\epsilon g - \delta j \equiv 0 \pmod{p}$, between (VII) and (XIV) +gives $\alpha\delta - \beta\epsilon \equiv 0$. + +\smallskip +$A_{18}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{19}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) gives $\epsilon +\equiv 0$, (XXII) gives $j \equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{20}$: (VII) gives $\alpha \equiv 0$, (XXII) gives $j \equiv 0$. +Elimination between (XIV) and (XVIII) gives $\epsilon c - e\gamma \equiv 0 +\pmod{p}$. + +\smallskip +$A_{21}$: (VII) gives $\alpha\equiv 0$, (XIV) gives $\epsilon \equiv 0$ +or $\not\equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0,$ or +$\not\equiv 0$, and (XXII) gives $j \equiv 0 \pmod{p}$. + +\smallskip +$A_{22}$: (VII) gives $\alpha \equiv 0$, (XIV) gives $\epsilon \equiv 0$ +or $\not\equiv 0$, (XVIII) gives $epsilon \equiv 0$ or $\not\equiv 0$, +(XXII) gives $j \equiv 0 \pmod{p}$. + +\smallskip +$A_{23}$: (VII) gives $\alpha \equiv 0$, (XIV) gives $\epsilon \equiv 0$, +(XVIII) gives $\epsilon \equiv 0$, (XXII) gives $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{26}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{27}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. Elimination between (XIV) and (XXII) gives +$\epsilon g - \delta j \equiv 0 \pmod{p}$. + +\smallskip +$A_{29}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + + + +\newpage +\small +\pagenumbering{gobble} +\begin{verbatim} + + + +End of the Project Gutenberg EBook of Groups of Order p^m Which Contain +Cyclic Subgroups of Order p^(m-3), by Lewis Irving Neikirk + +*** END OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M *** + +This file should be named 9930-t.tex or 9930-t.zip + +Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, +John Hagerson, and the Online Distributed Proofreading Team. + +Project Gutenberg eBooks are often created from several printed +editions, all of which are confirmed as Public Domain in the US +unless a copyright notice is included. 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If you are not located in the United States, you'll have +to check the laws of the country where you are located before using this ebook. + + + +Title: Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3) + +Author: Lewis Irving Neikirk + +Release Date: April 24, 2015 [EBook #9930] + +Language: English + +Character set encoding: ASCII + +*** START OF THIS PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M WHICH *** +\end{PGtext} +\end{minipage} +\end{center} +\clearpage + +%%%% Credits and transcriber's note %%%% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Cornell University, Joshua Hutchinson, Lee +Chew-Hung, John Hagerson, and the Online Distributed +Proofreading Team +\end{PGtext} +\end{minipage} +\vfill +\TranscribersNote +\end{center} +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% +\cleardoublepage +\iffalse %%%%% Start of original header %%%% +\documentclass[oneside]{article} +\usepackage{enumerate} +\usepackage[leqno]{amsmath} +\allowdisplaybreaks[1] +\begin{document} + +\thispagestyle{empty} +\small +\begin{verbatim} + +The Project Gutenberg EBook of Groups of Order p^m Which Contain Cyclic +Subgroups of Order p^(m-3), by Lewis Irving Neikirk + +Copyright laws are changing all over the world. 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You can also find out about how to make a +donation to Project Gutenberg, and how to get involved. + + +**Welcome To The World of Free Plain Vanilla Electronic Texts** + +**eBooks Readable By Both Humans and By Computers, Since 1971** + +*****These eBooks Were Prepared By Thousands of Volunteers!***** + + +Title: Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3) + +Author: Lewis Irving Neikirk + +Release Date: February, 2006 [EBook #9930] +[Yes, we are more than one year ahead of schedule] +[This file was first posted on November 1, 2003] + +Edition: 10 + +Language: English + +Character set encoding: US-ASCII + +*** START OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M *** + + +Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, +John Hagerson, and the Online Distributed Proofreading Team. + +\end{verbatim} +\normalsize +\newpage + +\fi +%%%%% End of original header %%%% + +\begin{center} +\noindent \Large GROUPS OF ORDER $p^m$ WHICH CONTAIN CYCLIC +SUBGROUPS OF ORDER $p^{m-3}$ + +\bigskip +\normalsize\textsc{by} +\bigskip + +\large LEWIS IRVING NEIKIRK + +\footnotesize\textsc{sometime harrison research fellow in +mathematics} + +\bigskip +\large 1905 +\end{center} + +\newpage +\begin{center} +\large \textbf{INTRODUCTORY NOTE.} +\end{center} +\normalsize + +This monograph was begun in 1902-3. Class I, Class II, Part I, and +the self-conjugate groups of Class III, which contain all the groups with +independent generators, formed the thesis which I presented to the Faculty +of Philosophy of the University of Pennsylvania in June, 1903, in partial +fulfillment of the requirements for the degree of Doctor of Philosophy. + +The entire paper was rewritten and the other groups added while the +author was Research Fellow in Mathematics at the University. + +I wish to express here my appreciation of the opportunity for scientific +research afforded by the Fellowships on the George Leib Harrison Foundation +at the University of Pennsylvania. + +I also wish to express my gratitude to Professor George H.\ +Hallett for his kind assistance and advice in the preparation of +this paper, and especially to express my indebtedness to Professor +Edwin S.\ Crawley for his support and encouragement, without which +this paper would have been impossible. + +\begin{flushright} +\textsc{Lewis I.\ Neikirk.} +\end{flushright} + +\footnotesize \textsc{ University Of Pennsylvania,} \textit{May, +1905.} \normalsize + +\MainMatter + +\begin{center} +\large GROUPS OF ORDER $p^m$, WHICH CONTAIN CYCLIC SUBGROUPS OF +ORDER $p^{(m-3)}$\footnote{Presented to the American Mathematical +Society April 25, 1903.} + +\bigskip \normalsize \textsc{by} + +\bigskip \textsc{lewis irving neikirk} + +\bigskip\textit{Introduction.} +\end{center} + +The groups of order $p^m$, which contain self-conjugate cyclic +subgroups of orders $p^{m-1}$, and $p^{m-2}$ respectively, have +been determined by \textsc{Burnside},\footnote{\textit{Theory of +Groups of a Finite Order}, pp.\ 75-81.} and the number of groups of +order $p^m$, which contain cyclic non-self-conjugate subgroups of +order $p^{m-2}$ has been given by +\textsc{Miller}.\footnote{Transactions, vol.\ 2 (1901), p.\ 259, and +vol.\ 3 (1902), p.\ 383.} + +Although in the present state of the theory, the actual tabulation +of all groups of order $p^m$ is impracticable, it is of importance +to carry the tabulation as far as may be possible. In this paper +\textit{all groups of order} $p^m$ ($p$ being an odd prime) +\textit{which contain cyclic subgroups of order $p^{m-3}$ and none +of higher order} are determined. The method of treatment used is +entirely abstract in character and, in virtue of its nature, it is +possible in each case to give explicitly the generational +equations of these groups. They are divided into three classes, +and it will be shown that these classes correspond to the three +partitions: $(m-3,\, 3)$, $(m-3,\, 2,\, 1)$ and $(m-3,\, 1,\, 1,\, 1)$, of +$m$. + +We denote by $G$ an abstract group $G$ of order $p^m$ containing +operators of order $p^{m-3}$ and no operator of order greater than +$p^{m-3}$. Let $P$ denote one of these operators of $G$ of order +$p^{m-3}$. The $p^3$ power of every operator in $G$ is contained +in the cyclic subgroup $\{P\}$, otherwise $G$ would be of order +greater than $p^m$. The complete division into classes is effected +by the following assumptions: +\begin{enumerate}[I.] +\item There is in $G$ at least one operator $Q_1$, such that +$Q{}_1^{p^2}$ is not contained in $\{P\}$. +\item The $p^2$ power of every operator in $G$ is contained in +$\{P\}$, and there is at least one operator $Q_1$, such that +$Q{}_1^p$ is not contained in $\{P\}$. +\item The $p$th power of every operator in $G$ is +contained in $\{P\}$. +\end{enumerate} + +\newpage +The number of groups for Class I, Class II, and Class III, +together with the total number, are given in the table below: +\bigskip + +\begin{tabular}{|c|c|c|c|c|c|c|c|} +\hline + & I & II$_1$ & II$_2$ & II$_3$ & II & III & Total \\ \hline +$p>3$ & & & & & & & \\ +$m>8$ & 9 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $64+5p$ \\ \hline +$p>3$ & & & & & & & \\ +$m=8$ & 8 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $63+5p$ \\ \hline +$p>3$ & & & & & & & \\ +$m=7$ & 6 & $20+p$ & $6+2p$ & $6+2p$ & $32+5p$ & 23 & $61+5p$ \\ \hline +$p=3$ & & & & & & & \\ +$m>8$ & 9 & 23 & 12 & 12 & 47 & 16 & 72 \\ \hline +$p=3$ & & & & & & & \\ +$m=8$ & 8 & 23 & 12 & 12 & 47 & 16 & 71 \\ \hline +$p=3$ & & & & & & & \\ +$m=7$ & 6 & 23 & 12 & 12 & 47 & 16 & 69 \\ \hline +\end{tabular} + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} I.\normalsize +\end{center} + +1. \textit{General notations and relations.}---The group $G$ is +generated by the two operators $P$ and $Q_1$. For brevity we +set\footnote{With J.~W.\ \textsc{Young}, \textit{On a certain +group of isomorphisms}, American Journal of Mathematics, vol.\ 25 +(1903), p.\ 206.} +\begin{equation*} +Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots = [a,\, b,\, c,\, d,\, \cdots]. +\end{equation*} + +Then the operators of $G$ are given each uniquely in the form +\begin{equation*} +[y,\, x] \quad \left( \begin{aligned}y &= 0,\, 1,\, 2,\, \cdots,\, p^3-1 \\ + x &= 0,\, 1,\, 2,\, \cdots,\, p^{m-3}-1 + \end{aligned} \right) . +\end{equation*} + +We have the relation +\begin{equation} +Q{}_1^{p^3} = P^{hp^3}. %% 1 +\end{equation} + +\noindent There is in $G$, a subgroup $H_1$ of order $p^{m-2}$, which +contains $\{P\}$ self-con\-ju\-gate\-ly.\footnote{\textsc{Burnside}: +\textit{Theory of Groups}, Art.\ 54, p.\ 64.} The subgroup $H_1$ is +generated by $P$ and some operator $Q{}_1^y P^x$ of $G$; it then +contains $Q{}_1^y$ and is therefore generated by $P$ and +$Q{}_1^{p^2}$; it is also self-conjugate in $H_2 = \{Q{}_1^p, P\}$ +of order $p^{m-1}$, and $H_2$ is self-conjugate in $G$. + +From these considerations we have the +equations\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{align} +Q{}_1^{-p^2}\,P\,Q{}_1^{p^2} &= P^{1+kp^{m-4}}, \\ %% 2 +Q{}_1^{-p}\,P\,Q{}_1^p &= Q{}_1^{\beta p^2}\,P^{\alpha_1}, \\ %% 3 +Q{}_1^{-1}\,P\,Q_1 &= Q{}_1^{bp}\,P^{a_1}. %% 4 +\end{align} + +\medskip +2. \textit{Determination of $H_1$. Derivation of a formula for +$[yp^2, x]^s$.}---From (2), by repeated multiplication we obtain +\begin{gather*} +[-p^2,\, x,\, p^2] = [0,\, x(1 + kp^{m-4})]; \\ +\intertext{and by a continued use of this equation we have} +[-yp^2,\, x,\, yp^2] = [0,\, x(1 + kp^{m-4})^y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4) +\end{gather*} +\noindent and from this last equation, +\begin{equation} +[yp^2,\, x]^s = \bigl[syp^2,\, x\{s + k \tbinom{s}{2}yp^{m-4}\}\bigr]. %% 5 +\end{equation} + +\medskip +3. \textit{Determination of $H_2$. Derivation of a formula for +$[yp,\, x]^s$.}---It follows from (3) and (5) that +\begin{equation*} +[-p^2,\, 1,\, p^2] = \left[\beta\frac{\alpha_1^p-1}{\alpha_1-1}p^2,\, + \alpha_1^p\left \{ 1+\frac{\beta k}{2} + \frac{\alpha_1^p-1}{\alpha_1-1} p^{m-4}\right \} \right] \quad (m > 4). +\end{equation*} +\noindent Hence, by (2), +\begin{gather*} +\beta\frac{\alpha_1^p - 1}{\alpha_1 - 1}p^2 \equiv 0 \pmod{p^3}, \\ +\alpha{}_1^p \left \{ 1 + \frac{\beta k}{2} + \frac{\alpha{}_1^p-1}{\alpha_1 - 1} p^{m-4} \right \} + + \beta\frac{\alpha{}_1^p - 1}{\alpha_1 - 1}hp^2 + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}. \\ +\intertext{From these congruences, we have for $m > 6$} +\alpha{}_1^p \equiv 1 \pmod{p^3}, \qquad \alpha_1 \equiv 1 \pmod{p^2}, \\ +\intertext{and obtain, by setting} +\alpha_1 = 1 + \alpha_2 p^2, \\ +\intertext{the congruence} +\frac{(1 + \alpha_2 p^2)^p - 1}{\alpha_2 p^3}(\alpha_2 + h\beta)p^3 + \equiv kp^{m-4} \pmod{p^{m-3}}; \\ +\intertext{and so} +(\alpha_2 + h\beta)p^3 \equiv 0 \pmod{p^{m-4}}, \\ +\intertext{since} +\frac{(1+\alpha_2 p^2)^p-1}{\alpha_2 p^3} \equiv 1 \pmod{p^2}. +\end{gather*} +\noindent From the last congruences +\begin{gather} +(\alpha_2 + h\beta)p^3 \equiv kp^{m-4} \pmod{p^{m-3}}. \\ %% 6 +\intertext{Equation (3) is now replaced by} +Q{}_1^{-p}\,P\, Q{}_1^{-p} = Q{}_1^{\beta p^2} P^{1 + \alpha_2 p^2}. %% 7 +\end{gather} +\noindent From (7), (5), and (6) +\begin{equation*} +[-yp,\, x,\, yp] = \left[\beta xyp^2,\, x\{1 + \alpha_2 yp^2\} + + \beta k \tbinom{x}{2}yp^{m-4}\right]. +\end{equation*} +\noindent A continued use of this equation gives +\begin{multline} +[yp,\, x]^s = [syp + \beta \tbinom{s}{2}xyp^2, \\ + xs + \tbinom{s}{2} \{\alpha_2 xyp^2 + \beta k\tbinom{x}{2}yp^{m-4}\} + + \beta k\tbinom{s}{3}x^2yp^{m-4}]. %% 8 +\end{multline} + +\medskip +4. \textit{Determination of $G$.}---From (4) and (8), +\begin{gather*} +[-p,\, 1,\, p] = [Np,\, a{}_1^p + Mp^2]. \\ +\intertext{From the above equation and (7),} +a{}_1^p \equiv 1 \pmod{p^2}, \qquad a_1 \equiv 1 \pmod{p}. +\end{gather*} + +Set $a_1 = 1 + a_2 p$ and equation (4) becomes +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q{}_1^{bp}\, P^{1 + a_2 p}. %% 9 +\end{equation} +\noindent From (9), (8) and (6) +\begin{gather*} +[-p^2,\, 1,\, p^2] = \left[\frac{(1 + a_2 p)^{p^2}-1}{a_2 p}bp, + (1 + a_2 p)^{p^2}\right], \\ +\intertext{and from (1) and (2)} +\frac{(1 + a_2 p)^{p^2} - 1}{a_2 p}bp \equiv 0 \pmod{p^3}, \\ +(1 + a_2 p)^{p^2} + bh\frac{(1 + a_2 p)^{p^2} - 1}{a_2 p}p + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}. +\end{gather*} +\noindent By a reduction similar to that used before, +\begin{equation} +(a_2 + bh)p^3 \equiv kp^{m-4} \pmod{p^{m-3}}. %% 10 +\end{equation} + +The groups in this class are completely defined by (9), (1) and (10). + +These defining relations may be presented in simpler form by a suitable +choice of the second generator $Q_1$. From (9), (6), (8) and (10) +\begin{gather*} +[1,\, x]^{p^3} = [p^3,\, xp^3] = [0,\, (x + h)p^3] \quad (m > 6), \\ +\intertext{and, if $x$ be so chosen that} +x + h \equiv 0 \pmod{p^{m-6}}, \\ +\intertext{$Q_1\, P^x$ is an operator of order $p^3$ whose $p^2$ +power is not contained in $\{P\}$. Let $Q_1\, P^x = Q$. The group +$G$ is generated by $Q$ and $P$, where} +Q^{p^3} = 1, \quad P^{p^{m-3}} = 1. \\ +\intertext{Placing $h = 0$ in (6) and (10) we find} +\alpha_2 p^3 \equiv a_2 p^3 \equiv k p^{m-4} \pmod{p^{m-3}}. +\end{gather*} +\noindent Let $\alpha_2 = \alpha p^{m-7}$, and $a_2 = ap^{m-7}$. +Equations (7) and (9) are now replaced by +\begin{equation} +\left. + \begin{aligned} + Q^{-p}\, P\, Q^p &= Q^{\beta p^2} P^{1 + \alpha p^{m-5}},\\ + Q^{-1}\, P\, Q &= Q^{bp} P^{1 + ap^{m-6}}. + \end{aligned} +\right. %% 11 +\end{equation} + +As a direct result of the foregoing relations, the groups in this +class correspond to the partition $(m-3,\, 3)$. From (11) we +find\footnote{For $m = 8$ it is necessary to add +$a^2\binom{y}{2}p^4$ to the exponent of $P$ and for $m = 7$ the +terms $a(a + \frac{abp}{2})\binom{y}{2}p^2 + a^3\binom{y}{3}p^3$ +to the exponent of $P$, and the term $ab\binom{y}{2}p^2$ to the +exponent of $Q$. The extra term $27ab^2 k\binom{y}{3}$ is to be +added to the exponent of $P$ for $m = 7$ and $p = 3$.} +\begin{equation*} +[-y,\, 1,\, y] = [byp,\, 1 + ayp^{m-6}] \qquad (m > 8). +\end{equation*} + +It is important to notice that by placing $y = p$ and $p^2$ in the +preceding equation we find that\footnote{For $m = 7,\, +ap^2-\frac{a^2p^3}{2} \equiv ap^2 \pmod{p^4},\, ap^3 \equiv kp^3 +\pmod{p^4}$. For $m = 7$ and $p = 3$ the first of the above +congruences has the extra terms $27(a^3 + ab\beta k)$ on the left +side.} +\begin{equation*} +b \equiv \beta \pmod{p}, \qquad a \equiv \alpha \equiv k \pmod{p^3} + \qquad (m > 7). +\end{equation*} + +A combination of the last equation with (8) yields\footnote{For $m += 8$ it is necessary to add the term $a\binom{y}{2}xp^4$ to the +exponent of $P$, and for $m = 7$ the terms $x\{a(a + +\frac{abp}{2})\binom{y}{2}p^2 + a^3\binom{y}{3}p^3\}$ to the +exponent of $P$, with the extra term $27ab^2 k\binom{y}{3}x$ for +$p = 3$, and the term $ab\binom{y}{2}xp^2$ to the exponent of +$Q$.} +\begin{multline} +[-y,\, x,\, y] = [bxyp + b^2\tbinom{x}{2}yp^2, \\ + x(1 + ayp^{m-6}) + ab\tbinom{x}{2}yp^{m-5} + + ab^2\tbinom{x}{3}yp^{m-4}] \qquad (m > 8). %% 12 +\end{multline} + +\newpage +From (12) we get\footnote{For $m = 8$ it is necessary to add the +term $\frac{1}{2} axy \binom{s}{2}[\frac{1}{3}y(2s - 1) - 1]p^4$ +to the exponent of $P$, and for $m=7$ the terms +\begin{multline*} + x \Bigl\{ \frac{a}{2} \bigl( a + \frac{ab}{2} p \bigr) + \bigl(\frac{2s-1}{3} y - 1 \bigr) \tbinom{s}{2}yp^2 + + \frac{a^{3}}{3!} \bigl(\tbinom{s}{2}y^2 - (2s - 1)y + 2 \bigr) yp^3 \\ + + \frac{a^2 bxy^2}{2} \tbinom{s}{3} \frac{3s-1}{2}p^3 + \frac{a^2 b}{2} + \bigl( \frac{s(s - 1)^2 (s - 4)}{4!}y - \tbinom{s}{3} \bigr) yp^3 \Bigr\} +\end{multline*} +\noindent with the extra terms +\begin{equation*} + 27abxy \Bigl\{ \frac{bk}{3!}\bigl[\tbinom{s}{2}y^2 - (2s - 1)y + 2\bigr] \tbinom{s}{3} + + x(b^2 k + a^2)(2y^2 + 1)\tbinom{s}{3} \Bigr\}, +\end{equation*} +\noindent for $p=3$, to the exponent of $P$, and the terms +$\frac{ab}{2} \bigl\{ 2s - \frac{1}{3}y - 1 \bigr\} \tbinom{s}{2}xyp^2$ +to the exponent of $Q$.} %% END OF FOOTNOTE +\begin{multline} +[y,\, x]^s = \bigl[ys + by\bigl\{(x +b\tbinom{x}{2}p)\tbinom{s}{2} + x\tbinom{s}{3}p\bigr\}p, \\ + xs + ay\bigl\{(x+b\tbinom{x}{2}p + b^2\tbinom{x}{3}p^2)\tbinom{s}{2} \\ + + (bx^2 p + 2b^2 x\tbinom{x}{2}p^2)\tbinom{s}{3} + bx^2\tbinom{s}{4}p^2\bigr\}p^{m-6}\bigr] + \qquad (m > 8). %% 13 +\end{multline} + +\medskip +5. \textit{Transformation of the Groups.}---The general group $G$ +of Class I is specified, in accordance with the relations (2) (11) +by two integers $a$, $b$ which (see (11)) are to be taken mod +$p^3$, mod $p^2$, respectively. Accordingly setting +\begin{gather*} +a = a_1 p^\lambda, \quad b = b_1 p^\mu, +\intertext{where} +dv[a_1,\, p] = 1, \quad dv[b_1,\, p] = 1 \qquad (\lambda = 0,\, 1,\, 2,\, 3;\; \mu = 0,\, 1,\, 2), +\intertext{we have for the group $G = G(a,\, b) = G(a,\, b)(P,\, Q)$ the +generational determination:} +G(a,\, b):\; \left \{ + \begin{gathered} + Q^{-1}\, P\, Q = Q^{b_1 p^{\mu + 1}} P^{1 + a_1 p^{m + \lambda - 6}} \\ + Q^{p^3} = 1, \quad P^{p^{m-3}} = 1. + \end{gathered} \right. +\end{gather*} + +Not all of these groups however are distinct. Suppose that +\begin{gather*} +G(a,\, b)(P,\, Q) \sim G(a',\, b')(P',\, Q'), +\intertext{by the correspondence} +C = \left[\begin{array}{cc} + Q, & P \\ + Q'_1, & P'_1 \\ + \end{array} \right], +\intertext{where} +Q'_1 = Q'^{y'} P'^{x'p^{m-6}}, \qquad \hbox{ and } \qquad P'_1 = Q'^y P'^x, +\end{gather*} +\noindent with $y'$ and $x$ prime to $p$. + +Since +\begin{gather*} +Q^{-1}\, P\, Q = Q^{bp} P^{1 + ap^{m-6}}, \\ +\intertext{then} +{Q'}_1^{-1}\, P'_1\, Q'_1 = {Q'}_1^{bp} {P'}_1^{1 + ap^{m-6}}, \\ +\intertext{or in terms of $Q'$, and $P'$} +\begin{aligned} + \bigl[y + b'xy'p &+ b'^2\tbinom{x}{2}y'p^2, x(1 + a'y'p^{m-6}) + a'b'\tbinom{x}{2}y'p^{m-5} \\ + &+ a'b'^2\tbinom{x}{3}y'p^{m-4}\bigr] = [y + by'p, x + (ax + bx'p)p^{m-6}] \qquad (m > 8) +\end{aligned} +\end{gather*} +\noindent and +\begin{gather} +by' \equiv b'xy' + b'^2\tbinom{x}{2}y'p \pmod{p^2}, \\ %% 14 +ax + bx'p \equiv a'y'x + a'b'\tbinom{x}{2}y'p + a'b'^2\tbinom{x}{3}y'p^2 \pmod{p^3}. %% 15 +\end{gather} +\noindent The necessary and sufficient condition for the simple +isomorphism of these two groups $G(a,\, b)$ and $G(a',\, b')$ is, that +the above congruences shall be consistent and admit of solution +for $x$, $y$, $x'$ and $y'$. The congruences may be written +\begin{gather*} +b_1 p^\mu \equiv b'_1 xp^{\mu'} + {b'}_1^2\tbinom{x}{2}p^{2\mu' + 1} \pmod{p^2}, \\ +\begin{aligned} + a_1 xp^{\lambda} + b_1 x'p^{\mu + 1} &\equiv \\ + y'\{a'_1 xp^{\lambda'} &+ a'_1 b'_1\tbinom{x}{2}p^{\lambda'+\mu'+1} + + a'_1 {b'}_1^2\tbinom{x}{3}p^{\lambda'+2\mu'+2}\} \pmod{p^3}. +\end{aligned} +\end{gather*} +\noindent Since $dv[x,\, p] = 1$ the first congruence gives $\mu = +\mu'$ and $x$ may always be so chosen that $b_1 = 1$. + +We may choose $y'$ in the second congruence so that $\lambda = +\lambda'$ and $a_1 = 1$ except for the cases $\lambda' \ge \mu + 1 += \mu' + 1$ when we will so choose $x'$ that $\lambda = 3$. + +The type groups of Class I for $m > 8$\footnote{For $m = 8$ the +additional term $ayp$ appears on the left side of the congruence +(14) and $G(1,\, p^2)$ and $G(1,\, p)$ become simply isomorphic. The +extra terms appearing in congruence (15) do not effect the result. +For $m = 7$ the additional term $ay$ appears on the left side of +(14) and $G(1,\, 1)$, $G(1,\, p)$, and $G(l,\, p^2)$ become simply +isomorphic, also $G(p,\, p)$ and $G(p,\, p^2)$.} are then given by +\begin{multline} +G(p^\lambda,\, p^\mu):\; Q^{-1}\, P\, Q = Q^{p^{1+\mu}} + P^{1+p^{m-6+\lambda}},\, Q^{p^3} = 1,\, P^{p^{m-3}} = 1 \\ +\left( + \begin{aligned} + \mu = 0,\, 1,\, 2;\;& \lambda = 0,\, 1,\, 2;\; \lambda \ge \mu; \\ + \mu = 0,\, 1,\, 2;\;& \lambda = 3 + \end{aligned} \right) +\tag{I}. +\end{multline} + +Of the above groups $G(p^\lambda,\, p^\mu)$ the groups for $\mu = 2$ have +the cyclic subgroup $\{P\}$ self-conjugate, while the group $G(p^3,\, p^2)$ +is the abelian group of type $(\mbox{$m-3$},\, 3)$. + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} II. \normalsize +\end{center} +\setcounter{equation}{0} +1. \textit{General relations.} + +There is in $G$ an operator $Q_1$ such that $Q{}_1^{p^2}$ is contained in +$\{P\}$ while $Q{}_1^p$ is not. +\begin{equation} +Q{}_1^{p^2} = P^{hp^2}. %% 1 +\end{equation} + +The operators $Q_1$ and $P$ either generate a subgroup $H_2$ of order +$p^{m-1}$, or the entire group $G$. + +\bigskip +\begin{center} +\large\textit{Section} 1. \normalsize +\end{center} + +2. \textit{Groups with independent generators.} + +Consider the first possibility in the above paragraph. There is in +$H_2$, a subgroup $H_1$ of order $p^{m-2}$, which contains $\{P\}$ +self-conjugately.\footnote{\textsc{Burnside}, \textit{Theory of +Groups}, Art.\ 54, p.\ 64.} $H_1$ is generated by $Q{}_1^p$ and $P$. +$H_2$ contains $H_1$ self-conjugately and is itself self-conjugate +in $G$. + +From these considerations\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{align} +Q{}_1^{-p}\, P\, Q{}_1^p &= P^{1 + kp^{m-4}}, \\ %% 2 +Q{}_1^{-1}\, P\, Q &= Q{}_1^{\beta p} P^{\alpha_1}. %% 3 +\end{align} + +\medskip +3. \textit{Determination of $H_1$ and $H_2$.} + +From (2) we obtain +\begin{equation} +[yp,\, x]^s = \bigl[syp,\, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\bigr] \quad (m > 4), %% 4 +\end{equation} +\noindent and from (3) and (4) +\begin{equation*} +[-p,\, 1,\, p] = \left[\frac{\alpha{}_1^p - 1}{\alpha_1 - 1}\beta p,\, + \alpha{}_1^p\left\{1 + \frac{\beta k}{2} + \frac{\alpha{}_1^p-1}{\alpha_1-1} p^{m-4} \right\} \right]. +\end{equation*} + +A comparison of the above equation with (2) shows that +\begin{gather*} +\frac{\alpha{}_1^p - 1}{\alpha_1 - 1} \beta p \equiv 0 \pmod{p^2}, \\ +\alpha{}_1^p \left\{1 + \frac{\beta k}{2}\frac{\alpha{}_1^p-1}{a_1-1} + p^{m-4} \right\} + \frac{\alpha{}_1^p - 1}{\alpha_1 - 1} \beta hp + \equiv 1 + kp^{m-4} \pmod{p^{m-3}}, +\end{gather*} +\noindent and in turn +\begin{equation*} +\alpha{}_1^p \equiv 1 \pmod{p^2}, \qquad \alpha_1 \equiv 1 \pmod{p} \qquad (m > 5). +\end{equation*} + +Placing $\alpha_1 = 1 + \alpha_2 p$ in the second congruence, we obtain +as in Class I +\begin{equation} +(\alpha_2 + \beta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}} \qquad (m > 5). %% 5 +\end{equation} + +Equation (3) now becomes +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q^\beta P^{1 + \alpha_2 p}. %% 6 +\end{equation} +\noindent The generational equations of $H_2$ will be simplified +by using an operator of order $p^2$ in place of $Q_1$. + +From (5), (6) and (4) +\begin{gather*} +[y,\, x]^s = [sy + U_s p,\, sx + W_s p] +\intertext{in which} +\begin{aligned} +U_s &= \beta \tbinom{s}{2}xy, \\ +W_s &= \alpha_2 \tbinom{s}{2}xy + \Bigl\{ \beta k \bigl[\tbinom{s}{2}\tbinom{x}{2} + + \tbinom{s}{3}x^2 y\bigr] \\ + & \qquad \qquad \qquad + \frac{1}{2}\alpha k\bigl[\frac{1}{3!}s(s - 1)(2s - 1)y^2 + - \tbinom{s}{2}y\bigr]x \Bigr\} p^{m-5}. +\end{aligned} +\end{gather*} + +Placing $s = p^2$ and $y = 1$ in the above +\begin{gather*} +[Q_1\, P^x]^{p^2} = Q{}_1^{p^2} P^{xp^2} = P^{(x+h)p^2}. +\intertext{If $x$ be so chosen that} +(x + h) \equiv 0 \pmod{p^{m-5}} \qquad (m > 5) +\end{gather*} +\noindent $Q_1 P^x$ will be the required $Q$ of order $p^2$. + +Placing $h = 0$ in congruence (5) we find +\begin{equation*} +\alpha_2 p^2 \equiv kp^{m-4} \pmod{p^{m-3}}. +\end{equation*} + +Let $\alpha_2 = \alpha p^{m-6}$. $H_2$ is then generated by +\begin{equation*} +Q^{p^2} = 1, \quad P^{p^{m-3}} = 1. +\end{equation*} +\begin{equation} +Q^{-1}\, P\, Q = Q^{\beta p} P^{1 + \alpha p^{m-5}}. %% 7 +\end{equation} + +Two of the preceding formul\ae\ now become +\begin{gather} +[-y,\, x,\, y] = \bigl[\beta xyp,\, x(1 + \alpha yp^{m-5}) + \beta k\tbinom{x}{2}yp^{m-4}\bigr], \\ %% 8 +[y,\, x]^s = [sy + U_s p,\, xs + W_s p^{m-5}], %% 9 +\end{gather} +\noindent where +\begin{equation*} +U_s = \beta \tbinom{s}{2}xy +\end{equation*} +\noindent and\footnote{For $m = 6$ it is necessary to add the terms +$\frac{ak}{2} \left \{ \frac{s(s - 1)(2s - 1)}{3!}y^2 - \tbinom{s}{2}y \right \}p$ +to $W_s$.} +\begin{equation*} +W_s = \alpha \tbinom{s}{2}xy + \beta k\bigl\{\tbinom{s}{2}\tbinom{x}{2} + + \tbinom{s}{3}x^2\bigr\}yp \quad (m > 6). +\end{equation*} + +\medskip +4. \textit{Determination of $G$.} + +Let $R_1$ be an operation of $G$ not in $H_2$. $R{}_1^p$ is in $H_2$. Let +\begin{equation} +R{}_1^p = Q^{\lambda p} P^{\mu p}. %% 10 +\end{equation} %% 10 + +Denoting $R{}_1^a\, Q^b\, P^c\, R{}_1^d\, Q^e\, P^f \cdots$ by the symbol $[a,\, b,\, c,\, +d,\, e,\, f,\, \cdots]$, all the operations of $G$ are contained in the set $[z,\, +y,\, x]$; $z = 0,\, 1,\, 2,\, \cdots,\, p - 1$; $y = 0,\, 1,\, 2,\, \cdots,\, p^2 - 1$; $x = 0,\, +1,\, 2,\, \cdots,\, p^{m-3} - 1$. + +The subgroup $H_2$ is self-conjugate in $G$. From +this\footnote{\textsc{Burnside}, \textit{Theory of Groups}, Art.\ +24, p.\ 27.} +\begin{gather} +R{}_1^{-1}\, P\, R_1 = Q^{b_1} P^{a_1}, \\ %% 11 +R{}_1^{-1}\, Q\, R_1 = Q^{d_1} P^{c_1 p^{m-5}}. %% 12 +\end{gather} +\noindent In order to ascertain the forms of the constants in (11) +and (12) we obtain from (12), (11), and (9) +\begin{gather*} +[-p,\, 1,\, 0,\, p] = [0,\, d{}_1^p + Mp,\, Np^{m-5}]. +\intertext{By (10) and (8)} +R{}_1^p\, Q\, R{}_1^p = P^{-\mu p}\, Q\, P^{\mu p} = Q\, P^{-a\mu p^{m-4}}. +\intertext{From these equations we obtain} +d{}_1^p \equiv 1 \pmod p \quad \hbox{ and } \quad d_1 \equiv 1 \pmod p . +\end{gather*} +\noindent Let $d_1 = 1 + dp$. Equation (12) is replaced by +\begin{equation} +R{}_1^{-1}\, Q\, R_1 = Q^{1+dp} P^{e_1 p^{m-5}}. %% 13 +\end{equation} +\noindent From (11), (13) and (9) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = \left[\frac{a{}_1^p - 1}{a_1 - 1}b_1 + Kp,\, a{}_1^p + b_1 Lp^{m-5}\right] +\intertext{in which} +K = a_1 b_1 \beta \sum_1^{p-1}\tbinom{a{}_1^y}{2}. +\intertext{By (10) and (8)} +R{}_1^{-p}\, P\, R{}_1^p = Q^{-\lambda p} P\, Q^{\lambda p} = P^{1 + a \lambda p^{m-4}}, +\intertext{and from the last two equations} +a{}_1^p \equiv 1 \pmod{p^{m-5}} +\intertext{and} +a_1 \equiv 1 \pmod{p^{m-6}} \quad (m > 6); \qquad a_1 \equiv 1 \pmod{p} \quad (m = 6). +\end{gather*} + +Placing $a_1 = 1 + a_2 p^{m-6} \quad (m > 6)$; \qquad $a_1 = 1 + a_2 p \quad (m=6)$. +\begin{equation*} +K \equiv 0 \pmod{p}, +\end{equation*} +\noindent and\footnote{$K$ has an extra term for $m = 6$ and $p = +3$, which reduces to $3b_1 c_1$. This does not affect the +reasoning except for $c_1 = 2$. In this case change $P^2$ to $P$ +and $c_1$ becomes $1$.} +\begin{gather*} +\frac{a{}_1^p - 1}{a_1 - 1}b_1 \equiv b_1 p \equiv 0 \pmod{p^2}, + \qquad b_1 \equiv 0 \pmod p. +\intertext{Let $b_1 = bp$ and we find} +a{}_1^p \equiv 1 \pmod{p^{m-4}}, \qquad a_1 \equiv 1 \pmod{p^{m-5}}. +\end{gather*} + +Let $a_1 = 1 + a_3 p^{m-5}$ and equation (11) is replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q^{bp} P^{1 + a_3 p^{m-5}}. %% 14 +\end{equation} +\noindent The preceding relations will be simplified by taking for +$R_1$ an operator of order $p$. This will be effected by two +transformations. + +From (14), (9) and (13)\footnote{The extra terms appearing in the +exponent of $P$ for $m=6$ do not alter the result.} +\begin{gather*} +[1,\, y]^p = \Bigl[p,\, yp,\, \frac{-c_1 y}{2} p^{m-4}\Bigr] + = \Bigl[0,\, (\lambda + y)p,\, \mu p - \frac{c_1 y}{2} p^{m-4}\Bigr], +\intertext{and if $y$ be so chosen that} +\lambda + y \equiv 0 \pmod{p}, +\end{gather*} +\noindent $R_2 = R_1\, Q^y$ is an operator such that $R{}_2^p$ is in +$\{P\}$. + +Let +\begin{gather*} +R{}_2^p = P^{lp}. +\intertext{Using $R_2$ in the place of $R_1$, from (15), (9) and (14)} +[1,\, 0,\, x]^p = \Bigl[p,\, 0,\, xp + \frac{ax}{2} p^{m-4}\Bigr] = + \Bigl[0,\, 0,\, (x + l)p + \frac{ax}{2} p^{m-4}\Bigr], +\intertext{and if $x$ be so chosen that} +x + l + \frac{ax}{2} p^{m-5} \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent then $R = R_2 P^x$ is the required operator of order $p$. + +$R^p = 1$ is permutable with both $Q$ and $P$. Preceding equations now +assume the final forms +\begin{align} +Q^{-1}\, P\, Q & = Q^{\beta p} P^{1 + ap^{m-5}}, \\ %% 15 +R^{-1}\, P\, R & = Q^{bp} P^{1 + ap^{m-4}}, \\ %% 16 +R^{-1}\, Q\, R & = Q^{1 + dp} P^{cp^{m-4}}, %% 17 +\end{align} +with $R^p = 1$, $Q^{p^2} = 1$, $P^{p^{m-3}} = 1$. + +The following derived equations are necessary\footnote{For $m=6$ +the term $a^2 \tbinom{x}{2} xp^2$ must be added to the exponent of +$P$ in (18).} +\begin{align} +[0,\, -y,\, x,\, 0,\, y] &= \bigl[0,\, \beta xyp,\, x(1 + \alpha yp^{m-5}) + \alpha + \beta \tbinom{x}{2}yp^{m-4}\bigr], \\ %% 18 +[-y,\, 0,\, x,\, -y] &= \bigl[0,\, bxyp,\, x(1 + ayp^{m-4}) + + ab\tbinom{x}{2} yp^{m-4}\bigr], \\ %% 19 +[-y,\, x,\, 0,\, y] &= [0,\, x(1 + dyp),\, cxyp^{m-4}]. %% 20 +\end{align} + +From a consideration of (18), (19) and (20) we arrive at the +expression for a power of a general operator of $G$. +\begin{equation} +[z,\, y,\, x]^s = [sz,\, sy + U_s p,\, sx + V_s p^{m-5}], %% 21 +\end{equation} +\noindent where\footnote{When $m = 6$ the following terms are to +be added to $V_s$: $\frac{a^2 x}{2} \left\{\frac{s(s - 1)(2s - 1)}{3!}y^2 + - \tbinom{s}{2}y\right\}p.$} +\begin{align*} +U_s &= \tbinom{s}{2} \{bxz +\beta xy + dyz \}, \\ +V_s &= \tbinom{s}{2} \Bigl\{\alpha xy + \bigl[axz + \alpha \beta \tbinom{x}{2}y + + cyz + ab\tbinom{x}{2}z\bigr]p\Bigr\} \\ + & \qquad \qquad \qquad + \alpha\tbinom{s}{3} \{bxz + \beta xy + dyz \} xp. +\end{align*} + +\medskip +5. \textit{Transformation of the groups.} All groups of this +section are given by equations (15), (16), and (17) with $a,\, b,\, +\beta,\, c,\, d = 0,\, 1,\, 2,\, \cdots ,\, p - 1$, and $\alpha = 0,\, 1,\, 2,\, +\cdots ,\, p^2 - 1$, independently. Not all these groups, however, +are distinct. Suppose that $G$ and $G'$ of the above set are +simply isomorphic and that the correspondence is given by +\begin{equation*} +C = \left[ + \begin{matrix} + R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ + \end{matrix} +\right], +\end{equation*} +\noindent in which +\begin{align*} +R'_1 &= R'^{z''} Q'^{y''p} P'^{x''p^{m-4}}, \\ +Q'_1 &= R'^{z'} Q'^{y'} P'^{x'p^{m-5}}, \\ +P'_1 &= R'^z Q'^y P'^x, +\end{align*} +\noindent where $x$, $y'$ and $z''$ \textit{are prime} to $p$. + +The operators $R'_1$, $Q'_1$, and $P'_1$ must be independent since +$R$, $Q$, and $P$ are, and that this is true is easily verified. +The lowest power of $Q'_1$ in $\{P'_1\}$ is $Q'{}_1^{p^2} = 1$ and +the lowest power of $R'_1$ in $\{Q'_1, P'_1\}$ is $R'{}_1^p = 1$. +Let $Q'{}_1^{s'} = P'{}_1^{sp^{m-5}}$. + +This in terms of $R'$, $Q'$, and $P'$ is +\begin{gather*} +\Bigl[s'z',\, y'\bigl\{s' + d'\tbinom{s'}{2}z'p\bigr\},\, s'x'p^{m-5} + +c'\tbinom{s'}{2}y'z'p^{m-4}\Bigr] = [0,\, 0,\, sxp^{m-5}]. \\ +\intertext{From this equation $s'$ is determined by} +s'z' \equiv 0 \pmod{p} \\ +y'\{s' + d'\tbinom{s}{2}z'p\} \equiv 0 \pmod{p^2}, +\intertext{which give} +s'y' \equiv 0 \pmod{p^2}. +\intertext{Since $y'$ is prime to $p$} +s' \equiv 0 \pmod{p^2} +\end{gather*} +\noindent and the lowest power of $Q'_1$ contained in $\{P'_1\}$ +is $Q'{}_1^{p^2} = 1$. + +Denoting by ${R'}_1^{s''}$ the lowest power of $R'_1$ contained in +$\{Q'_1, P'_1\}$. +\begin{equation*} +{R'}_1^{s''} = {Q'}_1^{s'p} {P'}_1^{sp^{m-4}}. +\end{equation*} + +This becomes in terms of $R'$, $Q'$, and $P'$ +\begin{gather*} +[s''z'',\, s''y''p,\, s''x''p^{m-4}] = [0,\, s'y'p,\, \{s'x' + sx\}p^{m-4}]. +\intertext{$s''$ is now determined by} +s''z'' \equiv 0 \pmod{p} +\intertext{and since $z''$ is prime to $p$} +s'' \equiv 0 \pmod{p}. +\end{gather*} +\noindent The lowest power of $R'_1$ contained in $\{Q'_1, P'\}$ is therefore +${R'}_1^p = 1$. + +Since $R$, $Q$, and $P$ satisfy equations (15), (16), and (17) $R'_1$, +$Q'_1$, and $P'_1$ also satisfy them. Substituting in these equations the +values of $R'_1$, $Q'_1$, and $P'_1$ and reducing we have in terms of +$R'$, $Q'$, and $P'$ +\begin{gather} +[z,\, y + \theta_1 p,\, x + \phi_1 p^{m-5}] = + [z,\, y + \beta y'p,\, x(1 + \alpha p^{m-5}) + \beta xp^{m-4}], \\ %% 22 +[z,\, y + \theta_2 p,\, x + \phi_2 p^{m-4}] = + [z,\, y + by'p,\, x(1 + ap^{m-4}) + bx'p^{m-4}], \\ %% 23 +[z',\, y' + \theta_3 p,\, (x' + \phi_3 p)p^{m-5}] = + [z',\, y'(1 + dp),\, x(1 + dp)p^{m-5} + cxp^{m-4}], %% 24 +\end{gather} +\noindent in which +\begin{align*} +\theta_1 &= d'(yz' - y'z) + x(b'z' + \beta'y'), \\ +\theta_2 &= d'yz'' + b'xz'', \\ +\theta_3 &= d'y'z'', \\ +\phi_1 &= \alpha'xy' + \bigl\{\alpha'(\beta'y' + b'z')\tbinom{x}{2} + + a'xz + c'(yz'-y'z)\bigr\}p, \\ +\phi_2 &= \alpha'xy'' + a'xz'' + \alpha'b'\tbinom{x}{2}z'' + c'yz'', \\ +\phi_3 &= c'yz''. +\end{align*} + +A comparison of the members of the above equations give six congruences +between the primed and unprimed constants and the nine indeterminates. + +\begin{align*} +\theta_1 &\equiv \beta y' \pmod{p}, \tag{I} \\ +\phi_1 &\equiv \alpha x + \beta x'p \pmod{p^2}, \tag{II} \\ +\theta_2 &\equiv by' \pmod{p}, \tag{III} \\ +\phi_2 &\equiv ax + bx' \pmod{p}, \tag{IV} \\ +\theta_3 &\equiv dy' \pmod{p}, \tag{V} \\ +\phi_3 &\equiv cx + dx' \pmod{p}. \tag{VI} +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of +the two groups $G$ and $G'$ is, \textit{that the above congruences shall be +consistent and admit of solution for the nine indeterminates, with the +condition that $x$, $y'$ and $z''$ be prime to $p$.} + +For convenience in the discussion of these congruences, the groups are +divided into six sets, and each set is subdivided into 16 cases. + +The group $G'$ is taken from the simplest case, and we associate with +this case all cases, which contain a group $G$, simply isomorphic with +$G'$. Then a single group $G$, in the selected case, simply isomorphic +with $G'$, is chosen as a type. + +$G'$ is then taken from the simplest of the remaining cases and we proceed +as above until all the cases are exhausted. + +Let $\kappa = \kappa_1 p^{\kappa_2}$, and $dv_1[\kappa_1 ,\, p] = 1$ +$(\kappa = a,\, b,\, \alpha ,\, \beta ,\, c,$ and $d)$. + +The six sets are given in the table below. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c||c|c|c|} +\hline + &$\alpha_2$&$d_2$& &$\alpha_2$&$d_2$ \\ \hline +$A$& 0 & 0 &$D$& 2 & 0 \\ \hline +$B$& 0 & 1 &$E$& 1 & 1 \\ \hline +$C$& 1 & 0 &$F$& 2 & 1 \\ \hline +\end{tabular} +\end{center} + +\medskip +The subdivision into cases and the results are given in Table II. + +\begin{center} +\large II. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} +\hline + &$a_2$&$b_2$&$\beta_2$&$c_2$& $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\ \hline + 1& 1 & 1 & 1 & 1 & & & & & & \\ \hline + 2& 0 & 1 & 1 & 1 &$A_1$& $B_1$ & &$C_2$& & $E_2$ \\ \hline + 3& 1 & 0 & 1 & 1 &$A_1$& &$C_1$&$D_1$& & \\ \hline + 4& 1 & 1 & 0 & 1 &$A_1$& &$C_1$&$D_1$& & $E_4$ \\ \hline + 5& 1 & 1 & 1 & 0 &$A_1$& &$C_1$&$D_1$& & $E_5$ \\ \hline + 6& 0 & 0 & 1 & 1 &$A_1$& $B_3$ &$C_2$&$C_2$& $E_3$ & $F_3$ \\ \hline + 7& 0 & 1 & 0 & 1 &$A_1$& $B_4$ &$C_2$&$C_2$& & $E_7$ \\ \hline + 8& 0 & 1 & 1 & 0 &$A_1$& $B_5$ &$C_2$&$C_2$& $E_5$ & $E_5$ \\ \hline + 9& 1 & 0 & 0 & 1 &$A_1$& $B_3$ &$C_1$&$D_1$& $E_3$ & $F_3$ \\ \hline +10& 1 & 0 & 1 & 0 &$A_1$& &$C_2$&$C_2$& &$E_{10}$ \\ \hline +11& 1 & 1 & 0 & 0 &$A_1$& & * &$C_1$& &$E_{11}$ \\ \hline +12& 0 & 0 & 0 & 1 &$A_1$& $B_3$ &$C_2$&$C_2$& * & $E_3$ \\ \hline +13& 0 & 0 & 1 & 0 &$A_1$&$B_{10}$& * & * &$E_{10}$&$E_{10}$ \\ \hline +14& 0 & 1 & 0 & 0 &$A_1$&$B_{11}$&$C_2$&$C_2$&$E_{11}$&$E_{11}$ \\ \hline +15& 1 & 0 & 0 & 0 &$A_1$&$B_{10}$&$C_2$&$C_2$&$E_{10}$&$E_{10}$ \\ \hline +16& 0 & 0 & 0 & 0 &$A_1$&$B_{10}$& * & * &$E_{10}$&$E_{10}$ \\ \hline +\end{tabular} + +\footnotesize The groups marked (*) divide into two or three parts. \normalsize +\end{center} + +\medskip +Let $ad - bc = \theta_1 p^{\theta_2}$, $\alpha_1 d - \beta c = +\phi_1 p^{\phi_2}$ and $\alpha_1 b - a\beta = \chi_1 p^{\chi_2}$ with +$\theta_1$, $\phi_1$, and $\chi_1$ prime to $p$. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c||c|c|c|c|c|} +\hline + * &$\theta_2$&$\phi_2$&$\chi_2$& & * &$\theta_2$&$\phi_2$&$\chi_2$& \\ \hline +$C_{11}$& & 1 & &$D_1$&$D_{13}$& 1 & & &$D_1$ \\ \hline +$C_{11}$& & 0 & &$C_1$&$D_{13}$& 0 & & &$C_2$ \\ \hline +$C_{13}$& 1 & & &$C_1$&$D_{16}$& 1 & & &$C_1$ \\ \hline +$C_{13}$& 0 & & &$C_2$&$D_{16}$& 0 & & &$C_2$ \\ \hline +$C_{16}$& 1 & 1 & &$D_1$&$E_{12}$& & & 1 &$F_3$ \\ \hline +$C_{16}$& 1 & 0 & &$C_1$&$E_{12}$& & & 0 &$E_3$ \\ \hline +$C_{16}$& 0 & & &$C_2$& & & & & \\ \hline +\end{tabular} + +\newpage +6. \textit{Types.} +\end{center} + +The type groups are given by equations (15), (16) and (17) with the +values of the constants given in Table IV. + +\begin{center} +\large IV. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c||c|c|c|c|c|c|c|} +\hline + & $a$ &$b$&$\alpha$&$\beta$& $c$ &$d$& &$a$&$b$&$\alpha$&$\beta$& $c$ &$d$ \\ \hline +$A_1$ & 0 & 0 & 1 & 0 & 0 & 1 &$E_1$ & 0 & 0 & $p$ & 0 & 0 & 0 \\ \hline +$B_1$ & 0 & 0 & 1 & 0 & 0 & 0 &$E_2$ & 1 & 0 & $p$ & 0 & 0 & 0 \\ \hline +$B_3$ & 0 & 1 & 1 & 0 & 0 & 0 &$E_3$ & 0 & 1 & $p$ & 0 & 0 & 0 \\ \hline +$B_4$ & 0 & 0 & 1 & 1 & 0 & 0 &$E_4$ & 0 & 0 & $p$ & 1 & 0 & 0 \\ \hline +$B_5$ & 0 & 0 & 1 & 0 & 1 & 0 &$E_5$ & 0 & 0 & $p$ & 0 & 1 & 0 \\ \hline +$B_{10}$& 0 & 1 & 1 & 0 &$\kappa$& 0 &$E_7$ & 1 & 0 & $p$ & 1 & 0 & 0 \\ \hline +$B_{11}$& 0 & 0 & 1 & 1 & 1 & 0 &$E_{10}$& 0 & 1 & $p$ & 0 &$\kappa$& 0 \\ \hline +$C_1$ & 0 & 0 & $p$ & 0 & 0 & 1 &$E_{11}$& 0 & 0 & $p$ & 1 & 1 & 0 \\ \hline +$C_2$ &$\omega$& 0 & $p$ & 0 & 0 & 1 &$F_1$ & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$D_1$ & 0 & 0 & 0 & 0 & 0 & 1 &$F_3$ & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +\end{tabular} + +\footnotesize +\begin{align*} +\kappa &= 1, \hbox{ and a non-residue } \pmod{p}, \\ +\omega &= 1, 2, \cdots, p - 1. +\end{align*} +\normalsize +\end{center} + +\medskip +The congruences for three of these cases are completely analyzed as +illustrations of the methods used. + +\medskip +\begin{equation*} B_{10}. \end{equation*} + +The congruences for this case have the special forms. +\begin{gather*} +b'xz' \equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha'y' \equiv \alpha \pmod{p}, \tag{II} \\ +b'xz'' \equiv by' \pmod{p}, \tag{III} \\ +\alpha'xy'' + \alpha'b'\tbinom{x}{2}z'' + c'yz'' \equiv ax + bx' \pmod{p}, \tag{IV} \\ +d \equiv 0 \pmod{p}, \tag{V} \\ +c'y'z'' \equiv cx \pmod{p}. \tag{VI} +\end{gather*} + +Since $z'$ is unrestricted (I) gives $\beta \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +From (II) since $y' \not\equiv 0, \alpha \not\equiv 0 \pmod{p}$. + +From (III) since $x, y', z'' \not\equiv 0$, $b \not\equiv 0 \pmod{p}$. + +In (IV) $b \not\equiv 0$ and $x'$ is contained in this congruence alone, +and, therefore, $a$ may be taken $\equiv 0$ or $\not\equiv 0 \pmod{p}$. + +(V) gives $d \equiv 0 \pmod{p}$ and (VI), $c \not\equiv 0 \pmod{p}$. + +Elimination of $y'$ between (III) and (VI) gives +\begin{equation*} +b'c'z''^{2} \equiv bc \pmod{p} +\end{equation*} +\noindent so that $bc$ is a quadratic residue or non-residue (mod $p$) +according as $b'c'$ is a residue or non-residue. + +The types are given by placing $a = 0$, $b = 1$, $\alpha = 1$, $\beta = 0$, +$c = \kappa$, and $d = 0$ where $\kappa$ has the two values, 1 and a +representative non-residue of $p$. + +\medskip +\begin{equation*} C_2. \end{equation*} + +The congruences for this case are +\begin{gather*} +d'(yz' - y'z) \equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha'_1 xy' + a'xz' \equiv \alpha_1 x + \beta x' \pmod{p}, \tag{II} \\ +d'yz'' \equiv by' \pmod{p}, \tag{III} \\ +a'xz'' \equiv ax + bx' \pmod{p}, \tag{IV} \\ +d'z'' \equiv d \pmod{p}, \tag{V} \\ +cx + dx' \equiv 0 \pmod{p}. \tag{VI} +\end{gather*} + +Since $z$ appears in (I) alone, $\beta$ can be either $\equiv 0$ or +$\not\equiv 0 \pmod{p}$. (II) is linear in $z'$ and, therefore, $\alpha +\equiv 0$ or $\not\equiv 0 \pmod{p}$, (III) is linear in $y$ and, +therefore, $b \equiv 0$ or $\not\equiv 0$. + +Elimination of $x'$ and $z''$ between (IV), (V), and (VI) gives +\begin{equation*} +a'd^2 \equiv d'(ad - bc) \pmod{p}. +\end{equation*} +\noindent Since $z''$ is prime to $p$, (V) gives $d \not\equiv 0 \pmod{p}$, so that +$ad - bc \not\equiv 0 \pmod{p}$. We may place $b = 0$, $\alpha = p$, +$\beta = 0$, $c = 0$, $d = 1$, then $a$ will take the values $1, 2, 3, \cdots, +p - 1$ giving $p - 1$ types. + +\medskip +\begin{equation*} D_1. \end{equation*} + +The congruences for this case are +\begin{align*} +d'(yz' - y'z) &\equiv \beta y' \pmod{p}, \tag{I} \\ +\alpha_1 x + \beta x' &\equiv 0 \pmod{p}, \tag{II} \\ +d'yz'' &\equiv by' \pmod{p}, \tag{III} \\ +ax + bx' &\equiv 0 \pmod{p}, \tag{IV} \\ +d'z'' &\equiv d \pmod{p}, \tag{V} \\ +cx + dx' &\equiv 0 \pmod{p}. \tag{VI} +\end{align*} +\noindent $z$ is contained in (I) alone, and therefore $\beta \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +(III) is linear in $y$, and $b \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +(V) gives $d \not\equiv 0 \pmod{p}$. + +Elimination of $x'$ between (II) and (VI) gives $\alpha_1 d - \beta c +\equiv 0 \pmod{p}$, and between (IV) and (VI) gives $ad - bc \equiv 0 +\pmod{p}$. The type group is derived by placing $a = 0$, $b = 0$, $\alpha = 0$, +$\beta = 0$, $c = 0$ and $d = 1$. + +\bigskip +\begin{center} +\large\textit{Section} 2. \normalsize +\end{center} +\setcounter{equation}{0} + +1. \textit{Groups with dependent generators.} In this section, $G$ is generated +by $Q_1$ and $P$ where +\begin{equation} +Q{}_1^{p^2} = P^{hp^2}. %% 1 +\end{equation} +\noindent There is in $G$, a subgroup $H_1$, of order $p^{m-2}$, which contains $\{P\}$ +self-con\-ju\-gate\-ly.\footnote{\textsc{Burnside}, \textit{Theory of Groups}, +Art.\ 54, p.\ 64.} $H_1$ either contains, or does not contain $Q{}_1^p$. We will +consider the second possibility in the present section, reserving the first for +the next section. + +\medskip +2. \textit{Determination of $H_1$.} $H_1$ is generated by $P$ and some other operator +$R_1$ of $G$. $R{}_1^p$ is contained in $\{P\}$. Let +\begin{equation} +R{}_1^p = P^{lp}. %% 2 +\end{equation} +\noindent Since $\{P\}$ is self-conjugate in $H_1$,\footnote{\textsc{Burnside}, +\textit{Theory of Groups}, Art.\ 56, p.\ 66.} +\begin{equation} +R{}_1^{-1}\, P\, R_1 = P^{1 + kp^{m-4}} %% 3 +\end{equation} +\noindent Denoting $R{}_1^a\, P^b\, R{}_1^c\, P^d \cdots$ by the symbol $[a,\, b,\, c,\, d,\, +\cdots]$ we derive from (3) +\begin{gather} +[-y,\, x,\, y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4), %% 4 +\intertext{and} +[y,\, x]^s = \Bigl[sy,\, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\Bigr] %% 5 +\end{gather} +\noindent Placing $s = p$ and $y = 1$ in (5) we have, from (2) +\begin{gather*} +[R_1\, P^x]^p = R{}_1^p P^{xp} = P^{(l + x)p}. +\intertext{Choosing $x$ so that} +x + l \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent $R = R_1 P^x$ is an operator of order $p$, which will be used in the place +of $R_1$, and $H = \{R, P\}$ with $R^p = 1$. + +\medskip +3. \textit{Determination of $H_2$.} We will now use the symbol $[a,\, b,\, c,\, d,\, e,\, f,\, +\cdots]$ to denote $Q{}_1^a\, R^b\, P^c\, Q{}_1^d\, R^e\, P^f \cdots$. + +$H_1$ and $Q_1$ generate $G$ and all the operations of $G$ are given by +$[x,\, y,\, z]$ ($z = 0,\, 1,\, 2,\, \cdots,\, p^2 - 1$; $y = 0,\, 1,\, 2,\, \cdots,\, p - 1$; +$x = 0,\, 1,\, 2,\, \cdots,\, p^{m-3} - 1$), since these are $p^m$ in number and +are all distinct. There is in $G$ a subgroup $H_2$ of order $p^{m-1}$ +which contains $H_1$ self-conjugately. $H_2$ is generated by $H_1$ and +some operator $[z,\, y,\, x]$ of $G$. $Q{}_1^z$ is then in $H_2$ and $H_2$ +is the subgroup $\{Q{}_1^p, H_1\}$. Hence, +\begin{gather} +Q{}_1^{-p}\, P\, Q{}_1^p = R^\beta P^{\alpha_1}, \\ %% 6 +Q{}_1^{-p}\, P\, Q{}_1^p = R^{b_1} P^{ap^{m-4}}. %% 7 +\end{gather} +\noindent To determine $\alpha_1$ and $\beta$ we find from (6), (5) and (7) +\begin{multline*} +[-p^2,\, 0,\, 1,\, p^2] = \biggl[ 0,\, \frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1} + \beta,\, \alpha{}_1^p\Bigl\{1 + \frac{\beta k}{2}\frac{\alpha{}_1^p - 1} + {\alpha_1 - 1}p^{m-4} \Bigr\} \\ + + a\beta\Bigl\{ p\frac{\alpha{}_1^{p-1}}{\alpha_1 - b_1} - + \frac{\alpha{}_1^p - b{}_1^p}{(\alpha_1 - b_1)^2} + \Bigr\}p^{m-4} \biggr]. +\end{multline*} +\noindent By (1) +\begin{gather*} +Q{}_1^{-p^2}\, P\, Q{}_1^{p^2} = P, +\intertext{and, therefore,} +\frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta \equiv 0 \pmod{p}, \\ + \alpha{}_1^p \equiv 1 \pmod{p^{m-4}}, \qquad \hbox{ and } \qquad + \alpha_1 \equiv 1 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} + +Let $\alpha_1 = 1 + \alpha_2 p^{m-5}$ and equation (6) is replaced by +\begin{equation} +Q{}_1^{-p}\, P\, Q{}_1^p = R^\beta P^{1 + \alpha_2 p^{m-5}}. %% 8 +\end{equation} + +To find $a$ and $b_1$ we obtain from (7), (8) and (5) +\begin{gather*} +[-p^2,\, 1,\, 0,\, p^2] = \Bigl[0,\, b{}_1^p,\, a\frac{b{}_1^p - 1}{b_1 - 1}p^{m-4} \Bigr]. +\intertext{By (1) and (4)} +Q{}_1^{-p^2}\, R\, Q{}_1^{p^2} = P^{-lp^2} R\, P^{lp^2} = R, +\intertext{and, hence,} +b{}_1^p \equiv 1 \pmod{p}, \qquad a\frac{b{}_1^p - 1}{b_1 - 1} \equiv 0 \pmod{p}, +\end{gather*} +\noindent therefore $b_1 = 1$. + +Substituting $b_1 = 1$ and $\alpha_1 = 1 + \alpha_2 p^{m-5}$ in the +congruence determining $\alpha_1$ we obtain $(1 + \alpha_2 p^{m-5})^p +\equiv 1 \pmod{p^{m-3}}$, which gives $\alpha_2 \equiv 0 \pmod{p}$. + +Let $\alpha_2 = \alpha p$ and equations (8) and (7) are now replaced by +\begin{align} +Q{}_1^p\, P\, Q{}_1^p &= R^\beta P^{1 + \alpha p^{m-4}}, \\ %% 9 +Q{}_1^{-p}\, R\, Q{}_1^p &= RP^{ap^{m-4}}. %% 10 +\end{align} + +From these we derive +\begin{align} +[-yp,\, 0,\, x,\, yp] &= \Bigl[0,\, \beta xy,\, x + \bigl\{\alpha xy + + a\beta x\tbinom{y}{2} + \beta k\tbinom{x}{2}y\bigr\}p^{m-4}\Bigr], \\ %% 11 +[-yp,\, x,\, 0,\, yp] &= [0,\, x,\, axyp^{m-4}]. %% 12 +\end{align} + +A continued use of (4), (11), and (12) yields +\begin{equation} +[zp,\, y,\, x]^s = [szp,\, sy + U_s,\, sx + V_sp^{m-4}] %% 13 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta\tbinom{s}{2}xz, \\ +V_s &= \tbinom{s}{2}\Bigl\{\alpha xz + \beta k\tbinom{s}{2}z + kxy + + ayz\Bigr\} + \beta k\tbinom{s}{3}x^2 z \\ + & \qquad \qquad \qquad + \frac{1}{2}a\beta\Bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 + - \tbinom{s}{2}z\Bigr\}. +\end{align*} + +\medskip +4. \textit{Determination of $G$.} + +Since $H_2$ is self-conjugate in $G_1$ we have +\begin{align} +Q{}_1^{-1}\, P\, Q_1 &= Q{}_1^{\gamma p} R^\delta P^{\epsilon_1}, \\ %% 14 +Q{}_1^{-1}\, R\, Q_1 &= Q{}_1^{cp} R^d P^{ep^{m-4}}. %% 15 +\end{align} + +From (14), (15) and (13) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = [\lambda p,\, \mu,\, \epsilon{}_1^p + vp^{m-4}] +\intertext{and by (9) and (1)} +\lambda p \equiv 0 \pmod{p^2}, \qquad +\epsilon{}_1^p + \nu p^{m-4} + \lambda hp \equiv 1 + \alpha p^{m-4} + \pmod{p^{m-3}}, +\intertext{from which} +\epsilon{}_1^p \equiv 1 \pmod{p^2}, \quad \hbox{ and } \quad \epsilon_1 \equiv 1 \pmod{p} + \qquad (m > 5). +\end{gather*} + +Let $\epsilon_1 = 1 + \epsilon_2 p$ and equation (14) is replaced by +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = Q{}_1^{\gamma p} R^\delta P^{1 + \epsilon_2 p}. %% 16 +\end{equation} + +From (15), (16), and (13) +\begin{gather*} +[-p,\, 1,\, 0,\, p] = \left[c\frac{d^p - 1}{d - 1}p,\, d^p,\, Kp^{m-4} \right] +\intertext{where} +K = \frac{d^p - 1}{d - 1}e + \sum_{1}^{p-1} acd\frac{d^n(d^n - 1)}{2}. +\end{gather*} + +By (10) +\begin{gather*} +d^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad d = 1 +\intertext{and by (1)} +chp^2 \equiv ap^{m-4} \pmod{p^{m-3}}. +\end{gather*} + +Equation (15) is now replaced by +\begin{equation} +Q{}_1^{-1}\, R\, Q_1 = Q{}_1^{cp} R P^{ep^{m-4}}. %% 17 +\end{equation} + +A combination of (17), (16) and (13) gives +\begin{equation*} +[-p,\, 0,\, 1,\, p] = \Bigl[\bigl\{\gamma\frac{(1 + \epsilon_2 p)^p - 1} +{\epsilon_2 p^2} + c\delta\frac{p - 1}{2} \bigr\}p^2,\, 0,\, (1 + +\epsilon_2 p)^p \Bigr]. +\end{equation*} + +By (9) +\begin{equation*} +\Bigl\{\gamma\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} + c\delta +\frac{p - 1}{2}\Bigr\}hp^2 + (1 + \epsilon_2 p)^p \equiv 1 + \alpha p^{m-4} +\pmod{p^{m-3}}, +\end{equation*} +\noindent $\beta \equiv 0 \pmod{p}.$ + +A reduction of the first congruence gives +\begin{gather*} +\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon p^2}\bigl\{\epsilon_2 + \gamma h\bigr\}p^2 + \equiv \Bigl\{\alpha - a\delta\frac{p - 1}{2}\Bigr\}p^{m-4} \pmod{p^{m-3}} +\intertext{and, since} +\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} \equiv 1 \pmod{p}, \qquad +(\epsilon_2 + \gamma h)p^2 \equiv 0 \pmod{p^{m-4}} +\end{gather*} +\noindent and +\begin{equation} +(\epsilon_2 + \gamma h)p^2 \equiv \bigl(\alpha + \frac{a\delta}{2}\bigr)p^{m-4} + \pmod{p^{m-3}}. %% 18 +\end{equation} + +From (17), (16), (13) and (18) +\begin{align} +[-y,\, x,\, 0,\, y] &= \Bigl[cxyp,\, x,\, \bigl\{exy + ac\tbinom{x}{2}y\bigr\}p^{m-4}\Bigr], \\ %% 19 +[-y,\, 0,\, x,\, y] &= \Bigl[x\bigl\{\gamma y + c\delta\tbinom{y}{2}\bigr\}p,\, + \delta xy,\, x(1 + \epsilon_2 yp) + \theta p^{m-4}\Bigr] %% 20 +\end{align} +\noindent where +\begin{multline*} +\theta = \Bigl\{e\delta x + a\delta\gamma x + \epsilon_2 \left(\alpha + + \frac{a\delta}{2}\right)x\Bigr\}\tbinom{y}{2} \\ + + \frac{1}{2}ac \Bigl\{\frac{1}{3!}y(y-1)(2y-1)\delta^2 + - \tbinom{y}{2}\delta\Bigr\} \\ + + \bigl\{\alpha\gamma y + \delta ky + a\delta xy^2 + + (ac\delta^2 y + ac\delta) \tbinom{y}{2}\bigr\} \tbinom{x}{2}. +\end{multline*} + +From (19), (20), (4) and (18) +\begin{equation*} +\{Q_1\, P^x\}^{p^2} = Q{}_1^{p^2} P^{xp^2} = P^{(h+x)p^2}. +\end{equation*} + +If $x$ be so chosen that +\begin{equation*} +h + x \equiv 0 \pmod{p^{m-5}} +\end{equation*} +\noindent $Q = Q_1\, P^x$ is an operator of order $p^2$ which will be used in place +$Q_1$ and $Q^{p^2} = 1$. + +Placing $h = 0$ in (18) we get +\begin{equation*} +\epsilon_2 p^2 \equiv 0 \pmod{p^{m-4}}. +\end{equation*} + +Let $\epsilon_2 = \epsilon p^{m-6}$ and equation (16) is replaced by +\begin{equation} +Q^{-1}\, P\, Q = Q^{\gamma p} R^\delta P^{1 + \epsilon p^{m-5}} %% 21 +\end{equation} + +The congruence +\begin{gather*} +ap^{m-4} \equiv chp^2 \pmod{p^{m-3}} +\intertext{becomes} +ap^{m-4} \equiv 0 \pmod{p^{m-3}}, \qquad \hbox{ and } \qquad a \equiv 0 \pmod{p}. +\end{gather*} +\noindent Equations (19) and (20) are replaced by +\begin{align} +[-y,\, x,\, 0,\, y] &= [cxyp,\, x,\, exyp^{m-4}] \\ %% 22 +[-y,\, 0,\, x,\, y] &= \Bigl[ \bigl\{\gamma y + c\delta\tbinom{y}{2} \bigr\}xp,\, + \delta xy,\, x(1 + \epsilon yp^{m-5}) + \theta p^{m-4} \Bigr] %% 23 +\end{align} +\noindent where +\begin{equation*} +\theta = e\delta x\tbinom{y}{2} + \bigl\{\alpha\gamma y + \delta ky + + \alpha c\delta\tbinom{y}{2}\bigr\}\tbinom{x}{2}. +\end{equation*} + +A formula for any power of an operation of $G$ is derived from (4), +(22) and (23) +\begin{equation} +[z,\, y,\, x]^s = [sz + U_s p,\, sy + V_s,\, sx + W_s p^{m-5}] %% 24 +\end{equation} +\noindent where +\begin{align*} +U_s &= \tbinom{s}{2}\bigl\{\gamma xz + cyz\bigr\} + \frac{1}{2}c\delta x + \Bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\Bigr\}, \\ +V_s &= \delta \tbinom{s}{2}xz, \\ +W_s &= \tbinom{s}{2} \Bigl\{\epsilon xz + \bigl[(a\gamma + \delta k)\tbinom{x}{2}z + eyz + kxy\bigr]p\Bigr\} \\ + & \qquad \qquad + \tbinom{s}{3}\bigl\{\epsilon \gamma x + \epsilon y + \delta kx \bigr\}xzp + + \frac{1}{2}c \delta \epsilon \bigl\{\frac{1}{2}(s - 1)z^2 - z\bigr\} \tbinom{s}{3}xp \\ + & \qquad \qquad + \frac{1}{2}\bigl\{\delta ex + \alpha c \delta \tbinom{x}{2}\bigr\} + \bigl\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\bigr\}p. +\end{align*} + +\medskip +5. \textit{Transformations of the groups.} Placing $y = 1$ and $x = -1$ in (22) +we obtain (17) in the form +\begin{equation*} +R^{-1}\, Q\, R = Q^{1-cp} P^{-ep^{m-4}}. +\end{equation*} +\noindent A comparison of the generational equations of the present section with +those of Section 1, shows that groups, in which $\delta \equiv 0 \pmod{p}$, +are simply isomorphic with those in Section 1, so we need consider only +those cases in which $\delta \not\equiv 0 \pmod{p}$. + +All groups of this section are given by +\begin{equation*} +G: \left\{ \begin{aligned} + R^{-1}\, P\, R &= P^{1 + kp^{m-4}}, \\ + Q^{-1}\, P\, Q &= Q^{\gamma p} R^\delta P^{1 + \epsilon p^{m-5}}, \\ + Q^{-1}\, R\, Q &= Q^{cp} R\, P^{\epsilon p^{m-4}}. \\ \end{aligned} +\right. \tag*{(25), (26), (27)} +\end{equation*} +\noindent $R^p = 1$, $Q^{p^2} = 1$, and $P^{p^{m-3}} = 1$, $(k,\, \gamma,\, c,\, e = 0,\, 1,\, +2,\, \cdots ,\, p - 1$; $\delta = 1,\, 2,\, \cdots,\, p - 1$; $\epsilon = 0,\, 1,\, 2,\, \cdots,\, +p^2 - 1)$. + +Not all these groups, however, are distinct. Suppose that $G$ and $G'$ of +the above set are simply isomorphic and that the correspondence is given by +\begin{equation*} +C = \left[\begin{matrix}R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right] . +\end{equation*} +\noindent Since $R^p = 1$, $Q^{p^2} = 1$, and $P^{p^{m-3}} = 1$, $R'{}_1^p = 1$, +$Q'{}_1^{p^2} = 1$ and $P'{}_1^{p^{m-3}}$. + +The forms of these operators are then +\begin{align*} +P'_1 &= Q'^z R'^y P'^x, \\ +R'_1 &= Q'^{z'p} R'^{y'} P'^{x'p^{m-4}}, \\ +Q'_1 &= Q'^{z''} R'^{y''}P'^{x''p^{m-5}}, \ +\end{align*} +\noindent where $dv[x,\, p] = 1$. + +Since $R$ is not contained in $\{P\}$, and $Q^p$ is not contained in +$\{R, P\}$ $R'_1$ is not contained in $\{P'_1\}$, and $Q'{}_1^p$ is not contained in +$\{R'_1, P'_1\}$. + +Let +\begin{gather*} +{R'}_1^{s'} = {P'}_1^{sp^{m-4}}. +\intertext{This becomes in terms of $Q'$, $R'$ and $P'$} +[s'z'p,\, s'y',\, s'x'p^{m-4}] = [0,\, 0,\, sxp^{m-4}], +\intertext{and} +s'y' \equiv 0 \pmod{p}, \qquad s'z' \equiv 0 \pmod{p}. +\end{gather*} +\noindent Either $y'$ or $z'$ is prime to $p$ or $s'$ may be taken $= 1$. + +Let +\begin{gather*} +{Q'}_1^{s''p} = {R'}_1^{s'} P'{}_1^{sp^{m-4}}, +\intertext{and in terms of $Q'$, $R'$ and $P'$} +[s''z''p,\, 0,\, s''x''p^{m-4}] = [s'z'p,\, s'y',\, (s'x' + sx)p^{m-4}], +\intertext{from which} +s''z'' \equiv s'z' \pmod{p}, \qquad \hbox{ and } \qquad s'y' \equiv 0 \pmod{p}. +\intertext{Eliminating $s'$ we find} +s''y'z'' \equiv 0 \pmod{p}, +\end{gather*} +\noindent $dv[y'z'',\, p] = 1$ or $s''$ may be taken $= 1$. We have then $z''$, $y'$ +and $x$ \textit{prime to} $p$. + +Since $R$, $Q$ and $P$ satisfy equations (25), (26) and (27) $R'_1$, $Q'_1$ +and $P'_1$ do also. These become in terms of $R'$, $Q'$ and $P'$. +\begin{align*} +[z + \Phi'_1 p,\, y,\, x + \Theta'_1 p^{m-4}] &= [z,\, y,\, x(1 + kp^{m-4})], \\ +[z + \Phi'_2 p,\, y + \delta'xz'',\, x + \Theta'_2 p^{m-5}] &= [z + \Phi_2 p,\, + y + \delta y',\, x + \Theta_2 p^{m-5}], \\ +[(z' + \Phi'_3)p,\, y',\, \Theta'_3 p^{m-4}] &= [(z' + \Phi_3)p,\, y,\, \Theta'_3 p^{m-4}], +\end{align*} +\noindent where +\begin{align*} +\Phi'_1 &= -c'yz', \quad \Theta'_1 = \epsilon'xz' + k'xy' - e'y'z, \\ +\Phi'_2 &= \Bigl\{\gamma'z'' + c'\delta'\tbinom{z}{2}\Bigr\}x + c'(yz'' - y''z), \\ +\Theta'_2 &= \epsilon'xz'' + \Bigl\{\tbinom{x}{2}\bigl[\alpha'\gamma'z'' + + \alpha'c'\delta'\tbinom{z''}{2} + \delta'k'z''\bigr] \\ + & \qquad \qquad \qquad + \delta'e'x \tbinom{z''}{2} + e'(yz'' - y''z) + k'xy''\Bigr\}p, \\ +\Phi_2 &= \gamma z'' + \delta z' + c'\delta y'z, \quad \Theta_2 \equiv + \epsilon x + (\gamma x'' + \delta x + e'\delta y'z)p, \\ +\Phi'_3 &= c'y'z'', \quad \Theta'_3 = e'y'z'', \quad \Phi_3 = cz'', \quad + \Theta_3 = ex + cx''. +\end{align*} + +A comparison of the members of these equations give seven congruences +\begin{align*} +\Phi'_1 &\equiv 0 \pmod{p}, \tag{I} \\ +\Theta'_1 &\equiv kx \pmod{p}, \tag{II} \\ +\Phi'_2 &\equiv \Phi_2 \pmod{p}, \tag{III} \\ +\delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ +\Theta'_2 &\equiv \Theta_2 \pmod{p^2}, \tag{V} \\ +\Phi_3' &\equiv cz'' \pmod{p}, \tag{VI} \\ +\Theta'_3 &\equiv \Theta_3 \pmod{p}. \tag{VII} +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of $G$ +and $G'$ is, \textit{that these congruences be consistent and admit of solution +for the nine indeterminants with $x$, $y'$, and $z''$ prime to $p$}. + +Let $\kappa = \kappa_1 p^{\kappa_2},\, dv[\kappa_1,\, p] = 1\; (\kappa = k,\, +\delta,\, \gamma,\, \epsilon,\, c,\, e)$. + +The groups are divided into three parts and each part is subdivided into +16 cases. + +The methods used in discussing the congruences are the same as those +used in Section 1. + +\medskip +6. \textit{Reduction to types.} The three parts are given by + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|} \hline + & $\epsilon_2$ & $\delta_2$ \\ \hline + $A$ & 0 & 0 \\ \hline + $B$ & 1 & 0 \\ \hline + $C$ & 2 & 0 \\ \hline +\end{tabular} +\end{center} + +The subdivision into cases and the results of the discussion of the +congruences are given in Table II. + +\medskip +\begin{center} +\large II. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline + &$k_2$&$\gamma_2$&$c_2$&$e_2$& $A$ & $B$ & $C$ \\ \hline + 1 & 1 & 1 & 1 & 1 & & & $B_1$ \\ \hline + 2 & 0 & 1 & 1 & 1 & & & $B_2$ \\ \hline + 3 & 1 & 0 & 1 & 1 & $A_2$ & $B_1$ & $B_1$ \\ \hline + 4 & 1 & 1 & 0 & 1 & & & $B_4$ \\ \hline + 5 & 1 & 1 & 1 & 0 & & & $B_5$ \\ \hline + 6 & 0 & 0 & 1 & 1 & * & $B_2$ & $B_2$ \\ \hline + 7 & 0 & 1 & 0 & 1 & $A_4$ & & $B_7$ \\ \hline + 8 & 0 & 1 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 9 & 1 & 0 & 0 & 1 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 10 & 1 & 0 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 11 & 1 & 1 & 0 & 0 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 12 & 0 & 0 & 0 & 1 & $A_4$ & $B_7$ & $B_7$ \\ \hline + 13 & 0 & 0 & 1 & 0 & $A_5$ & $B_5$ & $B_5$ \\ \hline + 14 & 0 & 1 & 0 & 0 & $A_4$ & $B_7$ & $B_7$ \\ \hline + 15 & 1 & 0 & 0 & 0 & $A_4$ & $B_4$ & $B_4$ \\ \hline + 16 & 0 & 0 & 0 & 0 & $A_4$ & $B_7$ & $B_7$ \\ \hline +\end{tabular} +\end{center} + +$A_6$ divides into two parts. + +The groups of $A_6$ in which $\delta k + \epsilon\gamma \equiv 0 \pmod{p}$ +are simply isomorphic with the groups of $A_1$ and those in which $\delta +k + \epsilon\gamma \not\equiv 0 \pmod{p}$ are simply isomorphic with the +groups of $A_2$. The types are given by equations (25), (26) and (27) where +the constants have the values given in Table III. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|} \hline + & $k$ & $\delta$ & $\gamma$ & $\epsilon$ & $c$ & $e$ \\ \hline + $A_1$ & 0 & 1 & 0 & 1 & 0 & 0 \\ \hline + $A_2$ & 1 & 1 & 0 & 1 & 0 & 0 \\ \hline + $A_4$ & 0 & 1 & 0 & 1 & 1 & 0 \\ \hline + $A_5$ & 0 & 1 & 0 & 1 & 0 & $\omega$ \\ \hline + $B_1$ & 0 & 1 & 0 & $p$ & 0 & 0 \\ \hline + $B_2$ & 1 & 1 & 0 & $p$ & 0 & 0 \\ \hline + $B_4$ & 0 & 1 & 0 & $p$ & 1 & 0 \\ \hline + $B_5$ & 0 & 1 & 0 & $p$ & 0 & $\kappa$ \\ \hline + $B_7$ & 1 & 1 & 0 & $p$ & $\omega$ & 0 \\ \hline +\end{tabular} + +\footnotesize +\begin{align*} +\kappa &= 1, \hbox { and a non-residue } \pmod{p}, \\ +\omega &= 1, 2, \cdots, p - 1. +\end{align*} +\normalsize +\end{center} + +A detailed analysis of several cases is given below, as a general +illustration of the methods used. + +\medskip +\begin{equation*} A_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{align*} + \epsilon'xz' &\equiv kx \pmod{p}, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ + \epsilon'xz'' &\equiv \epsilon x \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + ex &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} +\noindent Congruence (IV) gives $\delta \not\equiv 0 \pmod{p}$, from (II) $k$ can be +$\equiv 0$ or $\not\equiv 0 \pmod{p}$, (III) gives $\gamma \equiv 0$ +or $\not\equiv 0$, (V) gives $\epsilon \not\equiv 0$, (VI) and (VII) +give $c \equiv e \equiv 0 \pmod{p}$. Elimination of $x$, $z'$ and $z''$ +between (II), (III) and (V) gives $\delta k + \gamma\epsilon \equiv 0 +\pmod{p}$. If $k \equiv 0$, then $\gamma \equiv 0 \pmod{p}$ and +if $k \not\equiv 0$, then $\gamma \not\equiv 0 \pmod{p}$. + +\medskip +\begin{equation*} A_2. \end{equation*} + +The congruences for this case are +\begin{align*} + \epsilon'xz' + k'xy' &\equiv kx \pmod{p}, \tag{II} \\ +\gamma x'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ + \epsilon'xz'' &\equiv \epsilon x \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + ex &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} +\noindent Congruence (III) gives $\gamma \equiv 0$ or $\not\equiv 0$, (IV) gives +$\delta \not\equiv 0$, (V) $\epsilon \not\equiv 0$, (VI) and (VII) give $c +\equiv e \equiv 0 \pmod{p}$. Elimination of $x$, $z'$, and $z''$ between +(II), (III) and (V) gives +\begin{gather*} +\delta k + \gamma\epsilon \equiv k'\delta y' \pmod{p} +\intertext{from which} +\delta k + \gamma\epsilon \not\equiv 0 \pmod{p}. +\end{gather*} +\noindent If $k \equiv 0$, then $\gamma\not\equiv 0$, and if $\gamma \equiv 0$ then +$k \not\equiv 0 \pmod {p}$. + +Both $\gamma$ and $k$ can be $\not\equiv 0 \pmod{p}$ provided the above +condition is fulfilled. + +\medskip +\begin{equation*} A_5. \end{equation*} + +The congruences for this case are +\begin{align*} + \epsilon'xz'-e'y'z &\equiv kx \pmod p, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod p, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod p, \tag{IV} \\ + \epsilon'xz'' &\equiv ex \pmod p, \tag{V} \\ + cz'' &\equiv 0 \pmod p, \tag{VI} \\ + e'y'z'' &\equiv ex \pmod p. \tag{VII} +\end{align*} +\noindent (II) and (III) are linear in $z$ and $z'$ so $k$ and $\gamma$ are $\equiv +\hbox{ or } \not\equiv 0 \pmod{p}$ independently, (IV) gives $\delta \not +\equiv 0$, (V) give $\epsilon \not\equiv 0$, (VI) $c \equiv 0$, and (VII) +$e \not\equiv 0$. + +Elimination between (IV), (V), and (VII) gives +\begin{equation*} +\delta'e'\epsilon^2 \equiv \delta e \epsilon'^2 \pmod{p}. +\end{equation*} + +The types are derived by placing $\epsilon = \delta = 1$, and $e = 1, 2, +\cdots, p - 1$. + +\begin{equation*} B_5. \end{equation*} + +The congruences for this case are +\begin{align*} + -e'y'z &\equiv kx \pmod{p}, \tag{II} \\ +\gamma z'' + \delta z' &\equiv 0 \pmod{p}, \tag{III} \\ + \delta'xz'' &\equiv \delta y' \pmod{p}, \tag{IV} \\ +\epsilon'_1 xz'' + \delta'e'x\tbinom{z''}{2} + e'yz'' + &\equiv e_1 x + \gamma x'' + \delta x' + \pmod{p}, \tag{V} \\ + cz'' &\equiv 0 \pmod{p}, \tag{VI} \\ + e'y'z'' &\equiv ex \pmod{p}. \tag{VII} +\end{align*} +\noindent (II), and (III) being linear in $z$ and $z'$ give $k \equiv 0 \hbox{ or } +\not\equiv 0$, and $\gamma \equiv 0 \hbox{ or } \not\equiv 0 \pmod{p}$, +(IV) gives $\delta \not\equiv 0$, (V) being linear in $x'$ gives +$\epsilon_1 \equiv 0 \hbox{ or } \not\equiv 0 \pmod{p}$, (VI) gives $c +\equiv 0$ and (VII) $e \not\equiv 0$. + +Elimination of $x$ and $y'$ from (IV) and (VII) gives +\begin{equation*} +\delta'e'z''^2 \equiv \delta e \pmod{p}. +\end{equation*} + +$\delta e$ is a quadratic residue or non-residue $\pmod{p}$ according as +$\delta'e'$ is a residue or non-residue. + +The two types are given by placing $\delta = 1$, and $e = 1$ and a +non-residue $\pmod{p}$. + +\bigskip +\begin{center} +\textit{Section} 3. +\end{center} +\setcounter{equation}{0} + +1. \textit{Groups with dependent generators continued.} As in Section 2, $G$ +is here generated by $Q_1$ and $P$, where +\begin{equation*} +Q{}_1^{p^2} = P^{hp^2}. +\end{equation*} +\noindent $Q{}_1^p$ is contained in the subgroup $H_1$ of order $p^{m-2}$, $H_1 += \{Q{}_1^p, P\}$. + +\medskip +2. \textit{Determination of $H_1$.} Since $\{P\}$ is self-conjugate in $H_1$ +\begin{equation} +Q{}_1^{-p}\, P\, Q{}_1^p = P^{1 + kp^{m-4}}. %% 1 +\end{equation} +\noindent Denoting $Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots$ by the symbol $[a, b, c, d, +\cdots]$, we have from (1) +\begin{equation} +[-yp,\, x,\, yp] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4). %% 2 +\end{equation} +\noindent Repeated multiplication with (2) gives +\begin{equation} +[yp, x]^s = \Bigl[syp, x\bigl\{s + k\tbinom{s}{2}yp^{m-4}\bigr\}\Bigr]. %% 3 +\end{equation} + +\medskip +3. \textit{Determination of $H_2$.} There is a subgroup $H_2$ of order $p^{m-1}$ +which contains $H_1$ self-conjugately.\footnote{\textsc{Burnside}, \textit{Theory of +Groups}, Art.\ 54, p.\ 64.} $H_2$ is generated by $H_1$ and some operator +$R_1$ of $G$. $R{}_1^p$ is contained in $H_1$, in fact in $\{P\}$, +since if $R{}_1^{p^2}$ is the first power of $R_1$ in $\{P\}$, then $H_2 += \{R_1, P\}$, which case was treated in Section 1. +\begin{equation} +R{}_1^p = P^{lp}. %% 4 +\end{equation} + +Since $H_1$ is self-conjugate in $H_2$ +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q{}_1^{\beta p} P^{\alpha_1}, \\ +R{}_1^{-1}\, Q^p\, R_1 &= Q{}_1^{bp} P^{\alpha_1 p}. +\end{align} + +Using the symbol $[a, b, c, d, e, f, \cdots]$ to denote $R{}_1^a\, Q{}_1^b\, +P^c\, R{}_1^d\, Q{}_1^e\, P^f \cdots$, we have from (5), (6) and (3) +\begin{equation} +[-p,\, 0,\, 1,\, p] = [0,\, \beta Np,\, \alpha{}_1^p + Mp], %% 7 +\end{equation} +\noindent and by (4) +\begin{equation*} +\alpha{}_1^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad \alpha_1 \equiv 1 \pmod{p}. +\end{equation*} + +Let $\alpha_1 = 1 + \alpha_2 p$ and (5) is now replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q{}_1^{\beta p} P^{1 + \alpha_2 p}. +\end{equation} + +From (6), (8) and (3) +\begin{gather*} +[-p,\, p,\, 0,\, p] = \bigl[0,\, b^p p,\, a_1 \frac{b^p - 1}{b - 1}p + a_1 Up^2\bigr], +\intertext{and by (4) and (2)} +R{}_1^{-p}\, Q{}_1^p\, R{}_1^p = Q{}_1^p +\end{gather*} +\noindent and therefore $b^p \equiv 1 \pmod{p}$, and $b = 1$. Placing +$b = 1$ in the above equation the exponent of $P$ takes the form +\begin{gather*} +a_1 p^2 (1 + U'p) = a_1 \frac{\left\{1 + (\alpha_2 + \beta h)p\right\}^p +- 1}{(\alpha_2 + \beta h)p^2}p^2 +\intertext{from which} +a_1 p^2 (1 + U'p) \equiv 0 \pmod{p^{m-3}} +\intertext{or} +a_1 \equiv 0 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} + +Let $a_1 = ap^{m-5}$ and (6) is replaced by +\begin{equation} +R{}_1^{-1}\, Q{}_1^p\, R_1 = Q{}_1^p\, P^{ap^{m-4}}. %% 9 +\end{equation} + +(7) now has the form +\begin{gather*} +[-p,\, 0,\, 1,\, p] = [0,\, \beta Np,\, (1 + \alpha_2 p)^p + Mp^2], +\intertext{where} +N = p \quad \hbox{ and } \quad M = \beta h \left\{ \frac{(1 + \alpha_2 p)^p - 1} + {\alpha_2 p^2} -1 \right\}, +\intertext{from which} +(1 + \alpha_2 p)^p + \frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2}\beta hp^2 + \equiv 1 \pmod{p^{m-3}} +\intertext{or} +\frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2}\{\alpha_2 + \beta h\}p^2 + \equiv 0 \pmod{p^{m-3}} +\intertext{and since} +\frac{(1 + \alpha_2 p)^p - 1}{\alpha_2 p^2} \equiv 1 \pmod{p} +\end{gather*} +\begin{equation} +(\alpha_2 + \beta h)p^2 \equiv 0 \pmod{p^{m-3}}. %% 10 +\end{equation} + +From (8), (9), (10) and (3) +\begin{align} +[-y,\, 0,\, x,\, y] &= [0,\, \beta xyp,\, x(1+\alpha_2 yp)+\theta p^{m-4}], \\ %% 11 +[-y,\, xp,\, 0,\, y] &= [0,\, xp,\, axyp^{m-4}], %% 12 +\end{align} +\noindent where +\begin{equation*} +\theta = a \beta x \tbinom{y}{2} + \beta k \tbinom{x}{2} y. +\end{equation*} + +By continued use of (11), (12), (2) and (10) +\begin{equation} +[z,\, yp,\, x]^s = [sz,\, (sy + U_s)p,\, xs+ V_s p], %% 13 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta \tbinom{s}{2} xz \\ +V_s &= \tbinom{s}{2} \Bigl\{ \alpha_2 xz + \bigl[ayz + kxy + \beta k\tbinom{x}{2}z \bigr] + p^{m-5} \Bigr\} \\ & \qquad \qquad +\Bigl\{\beta\tbinom{s}{3}x^2 z + \frac{1}{2}a\beta + \Bigl[\frac{1}{3!} s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z\Bigr]x \Bigr\}p^{m-5}. +\end{align*} + +Placing in this $y = 0$, $z = 1$ and $s = p$,\footnote{Terms of the form +$(Ax^2 + Bx)p^{m-4}$ in the exponent of $P$ for $p = 3$ and $m > 5$ do not +alter the result.} +\begin{gather*} +(R_1\, P^x)^p = R{}_1^p\, P^{xp} = P^{(x+l)p}, +\intertext{determine $x$ so that} +x + l \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +then $R = R_1 P^x$ is an operator of order $p$ which will be used in the +place of $R_1$, $R^p = 1$. + +\medskip +4. \textit{Determination of $G$.} Since $H_2$ is self-conjugate in $G$ +\begin{align} +Q{}_1^{-1}\, P\, Q_1 &= R^\gamma\, Q{}_1^{\delta p}\, P^{\epsilon_1}, \\ %% 14 +Q{}_1^{-1}\, R\, Q_1 &= R^c\, Q{}_1^{dp}\, P^{e_1 p}. %% 15 +\end{align} + +From (15) +\begin{gather*} +(R^c\, Q{}_1^{dp}\, P^{e_1 p})^p = 1, +\intertext{by (13)} +Q{}_1^{dp^2}\, P^{e_1 p^2} = P^{(e_1 + dh)p^2} = 1, +\end{gather*} +\noindent and +\begin{equation} +(e_1+ dh)p^2 \equiv 0 \pmod{p^{m-3}}. %% 16 +\end{equation} + +From (14), (15) and (13) +\begin{equation} +[0,\, -p,\, 1,\, 0,\, p] = [L,\, Mp,\, \epsilon_1^p + Np]. %% 17 +\end{equation} + +By (1) +\begin{equation*} +\epsilon{}_1^p \equiv 1 \pmod{p}, \qquad \hbox{ and } \qquad \epsilon_1 \equiv 1 \pmod{p}. +\end{equation*} + +Let $\epsilon_1 = 1 + \epsilon_2 p$ and (14) is replaced by +\begin{equation} +Q{}_1^{-1}\, P\, Q_1 = R^\gamma\, Q{}_1^{\delta p}\, P^{1 + \epsilon_2 p}. %% 18 +\end{equation} + +From (15), (18), and (13) +\begin{equation*} +[0,\, -p,\, 0,\, 1,\, p] = \left[c_p,\, \frac{c^p - 1}{c - 1}dp,\, Kp \right]. +\end{equation*} + +Placing $x = 1$ and $y =-1$ in (12) we have +\begin{equation} +[0,\, -p,\, 0,\, 1,\, p] = [1,\, 0,\, -ap^{m-4}], %% 19 +\end{equation} +and therefore $c^p \equiv 1 \pmod{p}$, and $c = 1$. (15) is now replaced by +\begin{equation} +Q{}_1^{-1}\, R\, Q_1 = R\, Q{}_1^{dp}\, P^{e_1 p}. %% 20 +\end{equation} + +Substituting $1 + \epsilon_2 p$ for $\epsilon_1$ and 1 for $c$ in (17) +gives, by (16) +\begin{gather*} +[0,\, -p,\, 1,\, p] = [0,\, Mp^2,\, (1 + \epsilon_2 p)^p + Np^2], +\intertext{where} +M = \gamma d \frac{p-1}{2} + \delta\frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} +\intertext{and} +N = \frac{e_1\gamma}{(\epsilon_2 + \delta h)p^2} \left\{\frac{[1 + (\epsilon_2 + +\delta h)p]^p - 1}{(\epsilon_2 + \delta h)p} - p \right\}. +\end{gather*} + +By (1) +\begin{gather*} +(1 + \epsilon_2 p)^p + (N + Mh)p^2 \equiv 1 + kp^{m-4} \pmod{p^{m-3}}, +\intertext{or reducing} +\psi(\epsilon_2 + \delta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}}, +\intertext{where} +\psi = \frac{(1 + \epsilon_2 p)^p - 1}{\epsilon_2 p^2} + N - + e_1 \gamma \frac{p-1}{2}. +\end{gather*} + +Since +\begin{equation*} +\psi = 1 \pmod{p}. +\end{equation*} +\begin{equation} +(\epsilon_2 + \delta h)p^2 \equiv kp^{m-4} \pmod{p^{m-3}}. %% 21 +\end{equation} + +From (18), (20), (13), (16) and (21) +\begin{align} +[0,\, -y,\, x,\, 0,\, y] &= [\gamma xy,\, \theta_1 p,\, x + \phi_1 p], \\ %% 22 +[0,\, -y,\, 0,\, x,\, y] &= [x,\, dxyp,\, \phi_2 p], %% 23 +\end{align} +\noindent where +\begin{align*} +\theta_1 &= d\gamma x\tbinom{y}{2} + \delta xy + \beta\gamma\tbinom{x}{2}y, \\ +\phi_1 &= \epsilon_2 xy + \alpha_2\gamma\tbinom{x}{2}y + e_1\gamma + \tbinom{y}{2}x + \bigl\{x\tbinom{y}{2}(\epsilon_2 k + \delta \gamma) \\ +& \qquad + \frac{1}{2}ad\left[\frac{1}{3!}y(y - 1)(2y - 1)\gamma^2 - + \frac{y}{2}\gamma\right]x + a\gamma^2 dx\frac{1}{3!}y(y + 1)(y - 1) \\ +& \qquad + e_1\gamma k\tbinom{y}{3}x + \frac{1}{2}a\beta\left[\frac{1}{3!}x(x - 1)(2x - 1)\gamma^2 y^2 + - \tbinom{x}{2}\gamma y\right] \\ +& \qquad \qquad + \tbinom{x}{2}(a + k)\left[dy\tbinom{y}{2} + + \delta y\right] + \beta\gamma \tbinom{x}{3}\bigr\}p^{m-5}, \\ +\phi_2 &= e_1 xy + \left\{e_1 k\tbinom{y}{2} + ad\tbinom{x}{2}y \right\}p^{m-5}. +\end{align*} + +Placing $x = 1$ and $y = p$ in (23) and by (16) +\begin{gather*} +Q{}_1^{-p}\, R\, Q{}_1^p = R, +\intertext{and by (19)} +a \equiv 0 \pmod{p}. +\end{gather*} + +A continued multiplication, with (11), (22), and (23), gives +\begin{gather*} +(Q_1\, P^x)^{p^2} = Q{}_1^{p^2}\, P^{xp^2} = P^{(x + l)p^2}. +\intertext{Let $x$ be so chosen that} +(x + l) \equiv 0 \pmod{p^{m-5}}, +\end{gather*} +\noindent then $Q = Q_1\, P^x$ is an operator of order $p^2$ which will be used in +place of $Q_1$, $Q^{p^2} = 1$ and +\begin{equation*} +h \equiv 0 \pmod{p^{m-5}}. +\end{equation*} + +From (21), (10) and (16) +\begin{equation*} +\epsilon_2 p^2 \equiv kp^{m-4}, \qquad \alpha_2 p^2 \equiv 0 \qquad \hbox{ and } + \qquad e_1 p^2 \equiv 0 \pmod{p^{m-3}}. +\end{equation*} +\noindent Let $\epsilon_2 = \epsilon p^{m-6}$, $\alpha_2 = \alpha p^{m-5}$ and +$e_1 = ep^{m-5}$. Then equations (18), (20) and (8) are replaced by +\begin{gather*} +G: \left\{ \begin{aligned} +Q^{-1}\, P\, Q &= R^\gamma\, Q^{\delta p}\, P^{1 + \epsilon p^{m-5}}, \\ +Q^{-1}\, R\, Q &= R\, Q^{dp}\, P^{ep^{m-4}}, \\ +R^{-1}\, P\, R &= Q^{\beta p}\, P^{1 + \alpha p^{m-4}}, \\ \end{aligned} \right. +\tag*{(24), (25), (26)} \\ +R^p = 1, \qquad Q^{p^2} = 1, \qquad P^{p^{m-3}} = 1. +\end{gather*} + +\setcounter{equation}{26} +(11), (22) and (23) are replaced by +\begin{align} + [-y,\, 0,\, x,\, y] &= [0,\, \beta xyp,\, x + \phi p^{m-4}], \\ %% 27 +[0,\, -y,\, x,\, 0,\, y] &= [\gamma xy,\, \theta_1 p,\, x + \phi_1 p^{m-5}], \\ %% 28 +[0,\, -y,\, 0,\, x,\, y] &= [x,\, dxyp,\, \phi_2 p^{m-4}], %% 29 +\end{align} +\noindent where +\begin{gather*} +\phi = \alpha xy + \beta k\tbinom{x}{2}y, \quad + \theta_1 = d\gamma\tbinom{y}{2}x + \delta xy + \beta\gamma\tbinom{x}{2}y, \\ +\phi_1 = exy + \left\{e\gamma x\tbinom{y}{2} + \tbinom{x}{2}\left(\alpha\gamma y + + d\gamma k\tbinom{y}{2} + \delta ky\right) + \beta\gamma y\tbinom{x}{3}\right\}p, \\ +\phi_2 = exy. +\end{gather*} +\noindent A formula for a general power of any operator of $G$ is derived from (27), +(28) and (29) +\begin{equation} +[0,\, z,\, 0,\, y,\, 0,\, z]^s = [0,\, sz + U_s p,\, 0,\, sy + V_s,\, 0,\, sx + W_s p^{m-5}], %% 30 +\end{equation} +\noindent where +\begin{align*} +U_s &= \tbinom{s}{2}\left\{\delta xz + dyz + \beta xy + \beta\gamma \tbinom{x}{2}z \right\} \\ + & \qquad + \frac{1}{2}dx \left\{ \frac{1}{3!}s(s - 1)(2s - 1)z^2 - \tbinom{s}{2}z \right\}x + + \beta\gamma\tbinom{s}{2}x^2 z, \\ +V_s &= \gamma\tbinom{s}{2}xz, \displaybreak \\ +W_s &= \tbinom{s}{2} \left\{\epsilon xz + \left[axy + eyz + (\beta ky + \alpha\beta + \gamma + \delta kz)\tbinom{x}{2}\right]p \right\} \\ + & \qquad + \tbinom{s}{3}\left\{\alpha\gamma x^2 z + dkxyz + \delta kx^2 z + + \beta kx^2 y + 2\beta\gamma k\tbinom{x}{2}xz\right\}p \\ + & \qquad + \beta yk\tbinom{s}{4}x^3 zp + \frac{1}{2}\left\{\frac{1}{3!}s(s - 1)(2s - 1)z^2 + - \frac{s}{2}z \right\} \left\{e\gamma x + d\gamma k\tbinom{x}{2}\right\}p \\ + & \qquad + \frac{1}{2}d\gamma k\left[\frac{1}{2}(s - 1)z^2 - z\right]\tbinom{s}{3}x^2. +\end{align*} +\noindent A comparison of the generational equations of the present section with those +of Sections 1 and 2, shows that, $\gamma \equiv 0 \pmod{p}$ gives groups +simply isomorphic with those of Section 1, while $\beta \equiv 0 \pmod{p}$, +groups simply isomorphic with those of Section 2 and we need consider only +the groups in which $\beta$ and $\gamma$ are prime to $p$. + +\medskip +5. \textit{Transformation of the groups.} All groups of this section are given by +equations (24), (25), and (26), where $\gamma, \beta = 1, 2, \cdots, p - 1$; +$\alpha, \delta, d, e = 0, 1, 2, \cdots, p - 1$; and $\epsilon = 0, 1, +2, \cdots, p^2 - 1$. + +Not all of these, however are distinct. Suppose that $G$ is simply +isomorphic with $G'$ and that the correspondence is given by +\begin{equation*} +C = \left[\begin{matrix}R, & Q, & P \\ + R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right]. +\end{equation*} +\noindent An inspection of (30) gives +\begin{align*} +R'_1 &= Q'^{z''p}\, R'^{y''}\, P'^{x''p^{m-4}}, \\ +Q'_1 &= Q'^{z'}\, R'^{y'}\, P'^{x'p^{m-5}}, \\ +P'_1 &= Q'^z\, R'^y\, P'^x, +\end{align*} +\noindent with $dv[x,\, p] = 1$. Since $Q^p$ is not in $\{P\}$, and $R$ is not in +$\{Q^p, P\}$, ${Q'}_1^p$ is not in $\{P'_1\}$ and $R'_1$ is not in +$\{{Q'}_1^p, P'_1\}$. Let +\begin{gather*} +{Q'}_1^{s'p} = {P'}_1^{sp^{m-4}}. +\intertext{This is in terms of $R'$, $Q'$, and $P'$,} +[0,\, s'z'p,\, s'x'p^{m-4}] = [0,\, 0,\, sxp^{m-4}]. +\intertext{From which} +s'z'p \equiv 0 \pmod{p^2}, +\intertext{and $z'$ must be prime to $p$, since otherwise $s' \hbox{ can } = 1$. Let} +{R'}_1^{s''} = {Q'}_1^{s'p}\, {P'}_1^{sp^{m-4}}, +\intertext{or in terms of $R'$, $Q'$, and $P'$,} +[s''y'',\, s''z''p,\, s''x''p^{m-4}] = [0,\, s'z'p,\, (sx + s'x')p^{m-4}] +\intertext{and} +s''z'' \equiv s'z' \pmod{p}, \qquad s''y'' \equiv 0 \pmod{p}, +\end{gather*} +\noindent and $y''$ is prime to $p$, since otherwise $s''$ can $= 1$. Since $R$, +$Q$, and $P$ satisfy equations (24), (25) and (26), $R'_1$, $Q'_1$, and +$P'_1$ must also satisfy them. These become when reduced in terms of $R'$, +$Q'$ and $P'$ +\begin{align*} +[0,&\, z + \theta'_1 p,\, 0,\, y + \gamma'xz',\, 0,\, x + \psi'_1 p^{m-5}] \\ + & \qquad \qquad \qquad \qquad \qquad = [0,\, z + \theta_1 p,\, 0,\, y + \gamma y'',\, 0,\, x + \psi_1 p^{m-5}], \\ +[0,&\, (z'' + \theta'_2)p,\, 0,\, y'',\, 0,\, (x'' + \psi_2)p^{m-4}] \\ + & \qquad \qquad \qquad \qquad \qquad = [0,\, (z'' + \theta_2)p,\, 0,\, y'',\, 0,\, (x'' + \psi_2)p^{m-4}], \\ +[0,&\, z + \theta'_3p,\, 0,\, y,\, 0,\, x + \psi'_3 p^{m-4}] + = [0,\, z + \theta_3p,\, 0,\, y,\, 0,\, x + \psi_3 p^{m-4}], +\end{align*} +\noindent where +\begin{align*} +\theta'_1 &= d'(yz' - y'z) + x\left\{d'\gamma'\tbinom{z'}{2} + \delta'z' + +\beta'y'\right\} + \beta'\gamma'\tbinom{x}{2}z', \\ +\theta_1 &= \gamma z'' + \delta z' + d'\gamma y''z, \\ +\psi'_1 &= \epsilon'xz' + \bigl\{e'\gamma'x\tbinom{z'}{2} + \tbinom{x}{2}\left[\alpha' + \gamma'z' + \gamma'\epsilon'd'k'\tbinom{z'}{2} + \delta'\epsilon k'z' + + \beta'k'y'\right] \\ + & \qquad + \beta'\gamma'\tbinom{x}{3}z' + e'(yz' - y'z) + \alpha'xy'\bigr\}p, \\ +\psi_1 &= \epsilon x + \{\delta x' + \gamma x'' + e'\gamma y''z\}p, \\ +\theta'_2 &= d'y''z', \qquad \theta_2 = dz', \qquad \psi'_2 = e'y''z, \qquad + \psi_2 = dx' + ex, \\ +\theta'_3 &= \beta'xy'' - d'y''z, \qquad \theta_3 = \beta z', \\ +\psi_3 &= \epsilon'xz'' - e'y''z + \alpha'xy'' + \beta'\epsilon'\tbinom{x}{2}y'', + \qquad \psi_3 = \alpha x + \beta x'. +\end{align*} + +A comparison of the two sides of these equations give seven congruences +\begin{align*} +\theta'_1 &\equiv \theta_1 \pmod{p}, \tag{I} \\ +\gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\psi'_1 &\equiv \psi_1 \pmod{p^2}, \tag{III} \\ +\theta'_2 &\equiv \theta_2 \pmod{p}, \tag{IV} \\ +\psi'_2 &\equiv \psi_2 \pmod{p}, \tag{V} \\ +\theta'_3 &\equiv \theta_3 \pmod{p}, \tag{VI} \\ +\psi'_3 &\equiv \psi_3 \pmod{p}. \tag{VII} +\end{align*} + +(VI) is linear in $z$ provided $d' \not\equiv 0 \pmod{p}$ and $z$ may be +so determined that $\beta \equiv 0 \pmod{p}$ and therefore all groups in +which $d' \not\equiv 0 \pmod{p}$ are simply isomorphic with groups in +Section 2. + +Consequently we need only consider groups in which $d' \equiv 0 \pmod{p}$. + +As before we take for $G'$ the simplest case and associate with it all +simply isomorphic groups $G$. We then take as $G'$ the simplest case left +and proceed as above. + +Let $\kappa = \kappa_1 p^{\kappa_2}$ where $dv[\kappa_1, p] = 1, (\kappa = +\alpha, \beta, \gamma, \delta, \epsilon, d, e)$. + +For convenience the groups are divided into three sets and each set is +subdivided into eight cases. + +The sets are given by +\begin{equation*} +\begin{matrix} +A: & \epsilon_2 = 0, & \beta_2 = 0, & \gamma_2 = 0, \\ +B: & \epsilon_2 = 1, & \beta_2 = 0, & \gamma_2 = 0, \\ +C: & \epsilon_2 = 2, & \beta_2 = 0, & \gamma_2 = 0. +\end{matrix} +\end{equation*} + +The subdivision into cases and results of the discussion are given in +Table I. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|c|c|c|c|c|c|c|} \hline + & $\delta_2$ & $e_2$ & $\alpha_2$ & $A$ & $B$ & $C$ \\ \hline + 1 & 1 & 1 & 1 & & & $B_1$ \\ \hline + 2 & 0 & 1 & 1 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 3 & 1 & 0 & 1 & & & $B_3$ \\ \hline + 4 & 1 & 1 & 0 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 5 & 0 & 0 & 1 & $A_3$ & $B_3$ & $B_3$ \\ \hline + 6 & 0 & 1 & 0 & $A_1$ & $B_1$ & $B_1$ \\ \hline + 7 & 1 & 0 & 0 & $A_3$ & $B_3$ & $B_3$ \\ \hline + 8 & 0 & 0 & 0 & $A_3$ & $B_3$ & $B_3$ \\ \hline +\end{tabular} +\end{center} + +\medskip +6. \textit{Reduction to types.} The types of this section are given by equations +(24), (25) and (26) with $\alpha = 0, \beta = 1, \lambda = 1$ or a +quadratic non-residue (mod $p$), $\delta \equiv 0; \epsilon = l, e = 0, 1, +2, \cdots, p - 1;$ and $\epsilon = p, e = 0, 1,$ or a +non-residue (mod $p$), $2p+6$ in all. + +The special forms of the congruences for these cases are given below. + +\medskip +\begin{equation*} A_1. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'xz' &\equiv \epsilon x \pmod{p}, \tag{III} \\ + dz' &\equiv 0 \pmod{p}, \tag{IV} \\ + ex &\equiv 0 \pmod{p}, \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\epsilon'xz'' + \beta'\epsilon'\tbinom{x}{2}y' + &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} +\end{align*} + +(I) is linear in $z''$ and $\delta \equiv 0$ or $\not\equiv 0$, (II) +gives $\gamma \not\equiv 0$, (III) $\epsilon \not\equiv 0$, (IV) and (V) +$d \equiv e \equiv 0$, (VI) $\beta \not\equiv 0$, (VII) is linear in $x'$ +and $\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +Elimination of $y''$ and $z'$ between (II) and (VI) gives +\begin{equation*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p} +\end{equation*} +\noindent and $\beta\gamma$ is a residue or non-residue $\pmod{p}$ according as +$\beta'\gamma'$ is a residue or non-residue. + +\medskip +\begin{equation*} A_3. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'z' &\equiv \epsilon \pmod{p}, \tag{III} \\ + d &\equiv 0 \pmod{p}, \tag{IV} \\ + e'y''z' &\equiv ex \pmod{p}, \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\epsilon'xz'' - e'y''z + \beta'\epsilon'\tbinom{x}{2}y' + &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} +\end{align*} + +(I) is linear in $z''$ and $\delta \equiv 0$ or $\not\equiv 0$. (II) +gives $\gamma \not\equiv 0$, (III) $\epsilon \not\equiv 0$, (V) $e \not +\equiv 0$ and (VI) $\beta \not\equiv 0$. (VII) is linear in $x'$ and +$\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +Elimination between (II) and (VI) gives +\begin{gather*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}, +\intertext{and between (II), (III), and (IV) gives} +\epsilon'^2 \gamma e \equiv \epsilon^2\gamma'e' \pmod{p}. +\end{gather*} + +$\beta\gamma$ is a residue, or non-residue, according as $\beta'\gamma'$ is +or is not, and if $\gamma$ and $\epsilon$ are fixed, $e$ must take the +$(p - 1)$ values $1, 2, \cdots, p - 1$. + +\medskip +\begin{equation*} B_1. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma z'' + \delta z' \pmod{p}, \tag{I} \\ +\gamma'xz' &\equiv \gamma y'' \pmod{p}, \tag{II} \\ +\epsilon'_1 xz' + \beta'xz'\tbinom{x}{3} + &\equiv \epsilon_1 x + \delta x' + \gamma x'' \pmod{p}, \tag{III} \\ + ex &\equiv 0 \pmod{p}, \tag{IV} \\ +\beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ +\alpha x + \beta x' &\equiv 0 \pmod{p}. \tag{VII} +\end{align*} + +(I) gives $\delta \equiv 0$ or $\not\equiv 0$, (II) $\gamma \not +\equiv 0$, (III) is linear in $x''$ and gives $\epsilon_1 \equiv 0$ +or $\not\equiv 0$, (V) $e = 0$, (VI) $\beta \not\equiv 0$ and +(VII) is linear in $x'$ and gives $\alpha \equiv 0$ or $\not\equiv 0$. + +Elimination between (II) and (VI) gives +\begin{equation*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}. +\end{equation*} + +\newpage +\begin{equation*} B_3. \end{equation*} +\begin{align*} +\beta'\gamma'\tbinom{x}{2}z' + \beta'xy' + &\equiv \gamma\beta'' + \delta z' \pmod{p}, \tag{I} \\ + \gamma'xz' &\equiv \gamma y' \pmod{p}, \tag{II} \\ + \epsilon'_1 xz' + e'\gamma'x\tbinom{z'}{2} + \beta'\gamma'\tbinom{x}{3} &+ e'(yz' - y'z) \\ + &\equiv \epsilon_1 x + \delta x' + \gamma x'' + e'\gamma zy'' \pmod{p}, + \tag{III} \\ + e'y''z' &\equiv ex \pmod{p}. \tag{V} \\ + \beta'xy'' &\equiv \beta z' \pmod{p}, \tag{VI} \\ + -e'y''z &\equiv \alpha x + \beta x' \pmod{p}. \tag{VII} \\ +\end{align*} +(I) gives $\delta\equiv 0$ or $\not\equiv 0$, (II) $\gamma \not +\equiv 0$, (III) is linear in $x''$ and gives $\epsilon_1 \equiv 0$ or +$\not\equiv 0$, (V) $e \not\equiv 0$, (VI) $\beta \not\equiv 0$, (VII) is +linear in $x'$ and gives $\alpha \equiv 0$ or $\not\equiv 0 \pmod{p}$. +Elimination of $y''$ and $z'$ between (II) and (VI) gives +\begin{gather*} +\beta'\gamma'x^2 \equiv \beta\gamma \pmod{p}, +\intertext{and between (V) and (VI) gives} +\beta'e'y''^2 \equiv \beta e \pmod{p} +\end{gather*} +\noindent and $\beta\gamma$ and $\beta e$ are residues or non-residues, independently, +according as $\beta'\gamma'$ and $\beta'e'$ are residues or non-residues. + +\bigskip \bigskip +\begin{center} +\Large\textit{Class} III.\normalsize +\end{center} +\setcounter{equation}{0} + +1. \textit{General relations.} In this class, the $p$th power of every operator of +$G$ is contained in $\{P\}$. There is in $G$ a subgroup $H_1$ of order +$p^{m-2}$, which contains $\{P\}$ self-conjugately.\footnote{\textsc{Burnside}, +\textit{Theory of Groups}, Art.\ 54, p.\ 64.} + +\medskip +2. \textit{Determination of $H_1$.} $H_1$ is generated by $P$ and some operator +$Q_1$ of $G$. +\begin{equation*} +Q{}_1^p = P^{hp}. +\end{equation*} +\noindent Denoting $Q{}_1^a\, P^b\, Q{}_1^c\, P^d \cdots$ by the symbol $[a, b, c, d, +\cdots]$, all operators of $H_1$ are included in the set $[y, x]; (y = 0, +1, 2, \cdots, p - 1, x = 0, 1, 2, \cdots, p^{m-3} - 1)$. + +Since $\{P\}$ is self-conjugate in $H_1$\footnote{\textit{Ibid.}, Art.\ 56, p.\ 66.} +\begin{gather} +Q{}_1^{-1}\, P\, Q_1 = P^{1 + kp^{m-4}}. %% 1 +\intertext{Hence} +[-y,\, x,\, y] = [0,\, x(1 + kyp^{m-4})] \qquad (m > 4). %% 2 +\intertext{and} +[y,\, x]^s = \left[sy, x\left\{s + ky \tbinom{s}{2}p^{m-4} \right\}\right]. %% 3 +\end{gather} +\noindent Placing $y = 1$ and $s = p$ in (3), we have, +\begin{gather*} +[Q_1\, P^x]^p = Q{}_1^p\, P^{xp} = P^{(x + h)p} +\intertext{and if $x$ be so chosen that} +(x + h) \equiv 0 \pmod{p^{m-4}}, +\end{gather*} +\noindent $Q = Q_1\, P^x$ will be an operator of order $p$ which will be used in place +of $Q_1$, $Q^p = 1$. + +\medskip +3. \textit{Determination of $H_2$.} There is in $G$ a subgroup $H_2$ of order +$p^{m-1}$, which contains $H_1$ self-conjugately. $H_2$ is generated by +$H_1$, and some operator $R_1$ of $G$. +\begin{equation*} +R{}_1^p = P^{lp}. +\end{equation*} + +We will now use the symbol $[a, b, c, d, e, f, \cdots]$ to denote $R{}_1^a$ $Q^b$ +$P^c$ $R{}_1^d$ $Q^e$ $P^f$ $\cdots$. + +The operations of $H_2$ are given by $[z, y, x];$ $(z, y = 0, 1, \cdots, p - 1$; +$x = 0, 1, \cdots, p^{m-3} - 1)$. Since $H_1$ is self-conjugate in $H_2$ +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q{}_1^\beta P^{\alpha_1}, \\ %% 4 +R{}_1^{-1}\, Q\, R_1 &= Q{}_1^{b_1} P^{\alpha p^{m-4}}. %% 5 +\end{align} + +From (4), (5) and (3) +\begin{gather*} +[-p,\, 0,\, 1,\, p] = \left[0,\, \frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta,\, + \alpha{}_1^p + \theta p^{m-4} \right] = [0,\, 0,\, 1], \\ +\intertext{where} +\theta = \frac{\alpha{}_1^p \beta k}{2}\frac{\alpha{}_1^p - 1}{\alpha_1-1} + + a\beta \left\{\frac{\alpha{}_1^p - 1}{\alpha_1 - b_1}p - + \frac{\alpha{}_1^p - b{}_1^p}{(\alpha_1 - b_1)^2}\right\}. +\end{gather*} +\noindent Hence +\begin{equation} +\frac{\alpha{}_1^p - b{}_1^p}{\alpha_1 - b_1}\beta \equiv 0 \pmod{p}, \qquad +\alpha{}_1^p + \theta p^{m-4} \equiv 1 \pmod{p^{m-3}}, %% 6 +\end{equation} +\noindent and $\alpha{}_1^p \equiv 1 \pmod{p^{m-4}}$, or $\alpha_1 \equiv 1 +\pmod{p^{m-5}} \qquad (m > 5)$, $\alpha_1 = 1 + \alpha_2 p^{m-5}.$ Equation (4) +is replaced by +\begin{equation} +R{}_1^{-1}\, P\, R_1 = Q^\beta P^{1 + \alpha_2 p^{m-5}}, +\end{equation} + +From (5), (7) and (3). +\begin{equation*} +[-p,\, 1,\, 0,\, p] = \left[0,\, b{}_1^p,\, a\frac{b{}_1^p - 1}{b - 1} p^{m-4} \right]. +\end{equation*} +\noindent Placing $x = lp$ and $y = 1$ in (2) we have $Q^{-1} P^{lp} Q = P^{lp}$, and +\begin{equation*} +b{}_1^p \equiv 1 \pmod{p}, \qquad a\frac{b{}_1^p - 1}{b_1 - 1} \equiv 0 +\pmod{p}. +\end{equation*} +\noindent Therefore, $b_1 = 1$. + +Substituting 1 for $b_1$ and $1 + \alpha_2 p^{m-5}$ for $\alpha_1$ in +congruence (6) we find +\begin{equation*} +(1 + \alpha_2 p^{m-5})^p \equiv 1 \pmod{p^{m-3}}, \qquad \hbox{ or } \qquad +\alpha_2 \equiv 0 \pmod{p}. +\end{equation*} + +Let $\alpha_2 = \alpha p$ and equations (7) and (5) are replaced by +\begin{align} +R{}_1^{-1}\, P\, R_1 &= Q^\beta P^{1 + \alpha p^{m-4}}, \\ %% 8 +R{}_1^{-1}\, Q\, R_1 &= Q P^{\alpha p^{m-4}}. %% 9 +\end{align} + +From (8), (9) and (3) +\begin{align} +[-y,\, 0,\, x,\, y] &= \Bigl[0,\, \beta xy,\, x + \bigl\{\alpha xy + a\beta x \tbinom{y}{2} + + \beta ky\tbinom{x}{2}\bigr\}p^{m-4}\Bigr], \\ %% 10 +[-y,\, x,\, 0,\, y] &= [0,\, x,\, axyp^{m-4}]. %% 11 +\end{align} + +From (2), (10), and (11) +\begin{equation} +[z,\, y,\, x]^s = [sz,\, sy + U_s,\, sx + V_s p^{m-4}], %% 12 +\end{equation} +\noindent where +\begin{align*} +U_s &= \beta \tbinom{s}{2}xz, \\ +V_s &= \tbinom{s}{2}\left\{\alpha xz + kxy + ayz + \beta k\tbinom{x}{2}z\right\} \\ + & \qquad \qquad + \beta k\tbinom{s}{3}x^2 z + \frac{1}{2}a\beta\tbinom{s}{2} \left\{\frac{1}{3!} + (2s - 1)z - 1\right\}xz. +\end{align*} + +Placing $z = 1$, $y = 0$, and $s = p$ in (12)\footnote{The terms of the form +$(Ax + Bx^2)p^{m-4}$ which appear in the exponent of $P$ for $p = 3$ +do not alter the conclusion for $m > 5$.} +\begin{equation*} +[R_1\, P^x]^p = R{}_1^p\, P^{xp} = P^{(x+l)p}. +\end{equation*} + +If $x$ be so chosen that +\begin{equation*} +x + l \equiv 0 \pmod{p^{m-4}} +\end{equation*} +\noindent then $R =R_1 P^x$ is an operator of order $p$ which will be used in +place of $R_1$, and $R^p = 1$. + +\medskip +4. \textit{Determination of $G$.} $G$ is generated by $H_2$ and some operation +$S_1$. +\begin{equation*} +S{}_1^p = P^{\lambda p}. +\end{equation*} + +Denoting $S{}_1^a\, R^b\, Q^c\, P^d \cdots$ by the symbol $[a, b, c, d, \cdots]$ +all the operators of $G$ are given by +\begin{equation*} +[v,\, z,\, y,\, x];\, (v,\, z,\, y = 0, 1, \cdots, p - 1; x = 0, 1, \cdots, p^{m-3} - 1). +\end{equation*} + +Since $H_2$ is self-conjugate in $G$ +\begin{align} +S{}_1^{-1}\, P\, S_1 &= R^\gamma Q^s P^{\epsilon_1}, \\ %% 13 +S{}_1^{-1}\, Q\, S_1 &= R^c Q^d P^{ep^{m-4}}, \\ %% 14 +S_1\, R\, S_1 &= R^f Q^g P^{jp^{m-4}}. %% 15 +\end{align} + +From (13), (14), (15), and (12) +\begin{gather*} +[-p,\, 0,\, 0,\, 1,\, p] = [0,\, L,\, M,\, \epsilon{}_1^p + Np^{m-4}] = [0,\, 0,\, 0,\, 1] +\intertext{and} +\epsilon{}_1^p \equiv 1 \pmod{p^{m-4}} \qquad \hbox{ or } \qquad + \epsilon_1 \equiv 1 \pmod{p^{m-5}} \quad (m > 5). +\end{gather*} +\noindent Let $\epsilon_1 = 1 + \epsilon_2 p^{m-5}$. Equation (13) is +now replaced by +\begin{equation} +S{}_1^{-1}\, P\, S_1 = R^\gamma Q^\delta P^{1 + \epsilon_2 p^{m-5}}. %% 16 +\end{equation} + +If $\lambda = 0 \pmod{p}$ and $\lambda = \lambda'p,$ +\begin{multline*} +[1,\, 0,\, 0,\, 1]^p = \left[p,\, 0,\, 0,\, p + \epsilon\tbinom{p}{2}p^{m-5} + Wp^{m-4}\right] \\ += [0,\, 0,\, 0,\, p + \lambda'p^2 + W'p^{m-4}] +\end{multline*} +\noindent and for $m>5$ $S_1 P$ is of order $p^{m-3}$. We will take this in place +of $S_1$ and assume $dv [\lambda, p] = 1$. +\begin{equation*} +S{}_1^{p^{m-3}} = 1. +\end{equation*} +\noindent There is in $G$ a subgroup $H'_1$ of order $p^{m-2}$ which contains +$\{S_1\}$ self-conjugately. $H'_1 = \{S_1,\, S^v_1\, R^z\, Q^y\, P^x\}$ and the +operator $T= R^z\, Q^y\, P^x$ is in $H'_1$. + +There are two cases for discussion. + +\smallskip +$1^\circ$. Where $x$ is prime to $p$. + +$T$ is an operator of $H_2$ of order $p^{m-3}$ and will be taken as $P$. +Then +\begin{gather*} +H'_1 = \{S_1, P\}. +\intertext{Equation (16) becomes} +S{}_1^{-1}\, P\, S_1 = P^{1 + \epsilon p^{m-4}}. +\end{gather*} +\noindent There is in $G$ a subgroup $H'_2$ of order $p^{m-1}$ which contains $H'_1$ +self-conjugately. +\begin{equation*} +H'_2 = \{H'_1,\, S{}_1^{v'}\, R^{z'}\, Q^{y'}\, P^{x'}\}. +\end{equation*} +\noindent $T' = R^{z'}Q^{y'}$ is in $H'_2$ and also in $H_2$ and is taken as $Q$, +since $\{P, T'\}$ is of order $p^{m-2}$. + +$H'_2 = \{H'_1, Q\} = \{S_1, H_1\}$ and in this case $c$ may be taken +$\equiv 0 \pmod{p}$. + +\smallskip +$2^\circ$. \textit{Where $x = x_1 p$.} $P^p$ is in $\{S_1\}$ since $\lambda$ is +prime to $p$. In the present case $R^z\, Q^y$ is in $H'_1$ and also in $H_2$. +If $z \not\equiv 0 \pmod{p}$ take $R^z\, Q^y$ as $R$; if $z \equiv 0 +\pmod{p}$ take it as $Q$. +\begin{gather*} +H'_1 = \{S_1, R\} \quad \hbox{ or } \quad \{S_1,Q\}, +\intertext{and} +R^{-1}\, S_1\, R = S{}_1^{1 + k'p^{m-4}} \qquad \hbox{ or } \qquad +Q^{-1}\, S_1\, Q = S{}_1^{1 + k''p^{m-4}}. +\intertext{On rearranging these take the forms} +S{}_1^{-1}\, R\, S_1 = R\,S{}_1^{np^{m-4}} = R\,P^{jp^{m-4}} \quad \hbox{ or } + \quad S_1^{-1}\, Q\, S_1 = Q\,S{}_1^{n'p^{m-4}} = Q\,P^{ep^{m-4}}, +\end{gather*} +\noindent and either $c$ or $g$ may be taken $\equiv 0 \pmod{p}$, +\begin{equation} +cg \equiv 0 \pmod{p}. %% 17 +\end{equation} +\noindent From (14), (15), (16), (12) and (17) +\begin{equation*} +[-p,\, 0,\, 1,\, 0,\, p] = \left[0,\, c\frac{d^p - f^p}{d - f},\, d^p,\, Wp^{m-4}\right]. +\end{equation*} + +Place $x = \lambda p$ and $y = 1$ in (12) +\begin{gather*} +Q^{-1}\, P^{\lambda p}\,Q = P^{\lambda p} \qquad \hbox{ or } \qquad + S{}_1^p\, Q\, S{}_1^p = Q, +\intertext{and} +d^p \equiv 1 \pmod{p}, \qquad d = 1. +\end{gather*} +\noindent Equation (14) is replaced by +\begin{equation} +S{}_1^{-1}\, Q\, S_1 = R^c\, Q\, P^{ep^{m-4}}. %% 18 +\end{equation} + +From (15), (18), (17), (16) and (12) +\begin{equation*} +[-p,\, 1,\, 0,\, 0,\, p] = \left[0,\, f^p,\, \frac{d^p - f^p}{d - f}g, W'p^{m-4}\right]. +\end{equation*} +\noindent Placing $x = \lambda p,\, y = 1$ in (10) +\begin{equation*} +R^{-1}\, P^{\lambda p}\, R = P^{\lambda p}, +\end{equation*} +\noindent and $f^p \equiv 1 \pmod{p},\, f = 1$. Equation (15) is replaced by +\begin{equation} +S{}_1^{-1}\, R\, S_1 = R\, Q^g\, P^{jp^{m-4}}. %% 19 +\end{equation} +\noindent From (16), (18), (19) and (12) +\begin{equation*} +S{}_1^{-p}\, P\, S{}_1^p = P^{1 + \epsilon_2 p^{m-4}} = P +\end{equation*} +\noindent and $\epsilon_2 \equiv 0 \pmod{p}$. Let $\epsilon_2 = \epsilon p$ and (16) +is replaced by +\begin{equation} +S{}_1^{-1}\, P\, S_1 = R^\gamma\, Q^\delta\, P^{1 + \epsilon p^{m-4}}. %% 20 +\end{equation} +\noindent Transforming both sides of (1), (8) and (9) by $S_1$ +\begin{align*} +S{}_1^{-1} Q^{-1} S_1 \cdot S{}_1^{-1} P S_1 \cdot S{}_1^{-1} Q S_1 &= + S{}_1^{-1} P^{1 + kp^{m-4}} S_1, \\ +S{}_1^{-1} R^{-1} S_1 \cdot S{}_1^{-1} P S_1 \cdot S{}_1^{-1} R S_1 &= + S{}_1^{-1} Q^\beta S_1 \cdot S{}_1^{-1} P^{1 + \alpha p^{m-4}} S_1, \\ +S{}_1^{-1} R^{-1} S_1 \cdot S{}_1^{-1} Q S_1 \cdot S{}_1^{-1} R S_1 &= + S{}_1^{-1} Q S_1 \cdot S{}_1^{-1} P^{ap^{m-4}} S_1. +\end{align*} +\noindent Reducing these by (18), (19), (20) and (12) and rearranging +\begin{align*} +\bigl[0,&\, \gamma,\, \delta + \beta c,\, 1 + \left \{ \epsilon + \alpha c + k + ac\delta + + a\beta\tbinom{c}{2} - a\gamma \right \} p^{m-4} \bigr] \\ + & \qquad \qquad \qquad = [0,\, \gamma,\, \delta, 1 + (\epsilon + k)p^{m-4}]. \\ +[0,&\, \gamma,\, \beta + \delta,\, 1 + \{kg + \epsilon + \alpha + a\delta - + a\gamma g\}p^{m-4}] \\ + & \qquad \qquad \qquad = \Bigl[0,\, \gamma + \beta c,\, \beta + \delta,\, 1 + + \bigl\{\epsilon + \alpha + \beta e + \alpha\tbinom{\beta}{2}c + + a\beta\gamma\bigr\}p^{m-4}\Bigr], \\ +[0,&\, c,\, 1,\, (e + a)p^{m-4}] = [0,\, c,\, 1,\, (e + a)p^{m-4}]. +\end{align*} + +The first gives +\begin{gather} +\beta c \equiv 0 \pmod{p}, \\ %% 21 +ac + ac\delta - a\gamma \equiv 0 \pmod{p}. %% 22 +\intertext{Multiplying this last by $g$} +ag\gamma \equiv 0 \pmod{p}. %% 23 +\intertext{From the second equation above} +gk + \alpha\delta \equiv \beta e + a\beta\gamma \pmod{p}. %% 24 +\intertext{Multiplying by $c$} +ac\delta \equiv 0 \pmod{p}. %% 25 +\end{gather} + +These relations among the constants \textit{must be satisfied} in order that our +equations should define a group. + +From (20), (19), (18) and (12) +\begin{align} +[-y,\, 0,\, 0,\, x,\, y] &= [0,\, \gamma xy + \chi_1 (x, y),\, \delta xy + \phi_1 + (x, y),\, x + \Theta_1 (x, y)p^{m-4}], \\ %% 26 +[-y,\, 0,\, x,\, 0,\, y] &= [0,\, cxy,\, x,\, \Theta_2 (x, y)p^{m-4}], \\ %% 27 +[-y,\, x,\, 0,\, 0,\, y] &= [0,\, x,\, gxy,\, \Theta_3 (x, y)p^{m-4}], %% 28 +\end{align} +\noindent where +\begin{align*} +\chi_1 (x,\, y) &= c\delta x\tbinom{y}{2}, \\ +\phi_1 (x,\, y) &= \gamma gx\tbinom{y}{2} + \beta\gamma\tbinom{x}{2}y, \\ +\Theta_1 (x,\, y) &= \epsilon xy + \tbinom{y}{2}\left[\gamma jx + e\delta x + + a\delta\gamma + (\alpha\gamma + k\delta)\tbinom{x}{2}\right] \\ + & \qquad \qquad + \tbinom{y}{3} [c\delta j + eg\gamma]x + + \tbinom{x}{2}[\alpha\gamma y + \delta ky + a\delta\gamma y^2] + + \beta\gamma k\tbinom{x}{3}y^2, \\ +\Theta_2 (x, y) &= exy + cjx\tbinom{y}{2} + ac\tbinom{x}{2}y, \\ +\Theta_3 (x, y) &= jxy + egx\tbinom{y}{2} + ag\tbinom{x}{2}y. +\end{align*} + +Let a general power of any operator be +\begin{equation} +[v,\, z,\, y,\, x]^s = [sv,\, sz + U_s,\, sy + V_s,\, sx + W_s p^{m-4}]. %% 29 +\end{equation} + +Multiplying both sides by $[v,\, z,\, y,\, x]$ and reducing by (2), (10), (11), +(26), (27) and (28), we find +\begin{align*} +U_{s+1} &\equiv U_s + (cy + \gamma x)sv + c \delta\tbinom{sv}{2}x \pmod{p}, \\ +V_{s+1} &\equiv V_s + (gz + \delta x)sv + \gamma g\tbinom{sv}{2}x + + \beta\gamma\tbinom{x}{2}sv + \beta (sz + U_s)x \pmod{p}, \\ +W_{s+1} &\equiv W_s + \Theta_1 (x, sv) + \left\{ey + jz + a \gamma xy + + ac\tbinom{y}{2} + ag\tbinom{z}{2} \right\}sv \\ + & \qquad \qquad + \left\{\alpha x + \beta k\tbinom{x}{2} + + ay + a\delta sx + \alpha gsvz\right\}sz + ksxy \\ + & \qquad \qquad + \tbinom{sv}{2}\{cjy + egz\} + U_s \left\{\alpha x + + \beta k\tbinom{x}{2} + ay + a(\delta x + gz)sv\right\} \\ + & \qquad \qquad + a\beta\tbinom{sz + Us}{2}x + kV_s x \pmod{p}. +\end{align*} + +From (29) +\begin{equation*} +U_1 \equiv 0, \qquad V_1 \equiv 0, \qquad W_1 \equiv 0 \pmod{p}. +\end{equation*} + +A continued use of the above congruences give +\begin{align*} +U_s &\equiv (cy + \gamma x)\tbinom{s}{2}v + \frac{1}{2} c\delta xv + \{\frac{1}{3} (2s - 1)v - 1\}\tbinom{s}{2} \pmod{p}, \\ +V_s &\equiv \{[gz + \delta x + \beta\gamma\tbinom{x}{2}v + \beta xz\} + \tbinom{s}{2} \\ & \qquad + \frac{1}{2} \gamma gxv\{\frac13 (2s - 1)v -1\} + \tbinom{s}{2} + \beta\gamma\tbinom{s}{3}x^2 v \pmod{p}, \displaybreak \\ +%% +W_s &\equiv \tbinom{s}{2} + \Bigl\{ + \epsilon xv + egv + (\alpha\gamma + \delta kv + \beta kz)\tbinom{s}{2} + + \beta\gamma\ k\tbinom{x}{3}v + ac\tbinom{y}{2}v \\ + & \qquad + jvz + ag\tbinom{z}{2}v + \alpha xz + kxy + a\gamma xyv + ayz + \Bigr\} ++ \tbinom{s}{3} + \Bigl\{ + \alpha cxyv \\ & \qquad + \alpha\gamma x^2 v + 2\beta\gamma k\tbinom{x}{2} xv + + gkxzv + \delta kx^2 v + + \beta kx^2 z + acvy^2 \\ & \qquad + a\gamma xvy + \Bigr\} ++ \beta k \gamma\tbinom{s}{4}x^3 v + \tbinom{s}{2}\frac{2s-1}{3} + \Bigl\{ + a\delta\gamma\tbinom{x}{2}v^2 + a\delta xzv \\ + & \qquad + agvz^2 + \Bigr\} ++ \frac{1}{2}v\tbinom{s}{2} + \Bigl\{ + \frac13(2s-1)v - 1 + \Bigr\} + \Bigl\{ + \gamma jx + e\delta x + a\delta\gamma x \\ + & \qquad + \alpha c \delta\tbinom{x}{2} + \gamma gk \tbinom{x}{2} + cjy + egz + \Bigr\} ++ \frac{1}{6}\tbinom{s}{2} + \Bigl\{ + \tbinom{s}{2}v^2 - (2s-1)v \\ & \qquad + 2 + \Bigr\} + \bigl\{ + c\delta jx + eg\gamma x + \bigr\}v ++ \frac{1}{2}\tbinom{s}{3} + \Bigl\{ + \frac{1}{2}(s-1)v-1 + \Bigr\} + \bigl\{ + \alpha c \delta \\ & \qquad + \gamma gk + \bigr\} x^2 v ++ \frac12 a\beta x \tbinom{s}{2} + \Bigl\{ + \frac{1}{3}(2s-1)z - 1 + \Bigr\} z \\ +& \qquad + \frac{1}{2} a\delta\gamma x^2 v\tbinom{s}{3} \frac{1}{2}(3s-1) \pmod{p} +\end{align*} + +Placing $v = 1,\, z = y = s = p$ in (29)\footnote{For $p = 3$ and +$c\delta \equiv \gamma g \equiv \beta \gamma +\equiv 0 \pmod{p}$ there are terms of the form $(A + Bx + Cx^2 + Dx^3) +p^{m-4}$ in the exponent of $P$. For $m > 5$ these do not vitiate our +conclusion. For $p = 3$ and $c\delta$, $\gamma g$, or $\beta\gamma$ prime +to $p$, $[S_1\, P^x]^p$ is not contained in $\{P\}$ and the groups defined +belong to Class II.} +\begin{gather*} +[S_1\, P^x]^p = S{}_1^p P^{xp} = P^{(\lambda + x)p} \qquad (p > 3). +\intertext{If $x$ be so chosen that} +x + \lambda \equiv 0 \pmod{p^{m-4}}. +\intertext{$S = S_1\, P^x$ is an operator of order $p$ and is taken in place +of $S_1$.} +S^p = 1. +\end{gather*} + +The substitution of $S$ for $S_1$ leaves congruence (17) invariant. + +\medskip +5. \textit{Transformation of the groups.} All groups of this class are given by + +\begin{equation} +G: \begin{cases} +Q^{-1} P\, Q = P^{1 + kp^{m-4}}, \\ +R^{-1} P\, R = Q^\beta\, P^{1 + \alpha p^{m-4}}, \\ +R^{-1} Q\, R = Q\, P^{ap^{m-4}}, \\ +S^{-1} P\, S = R^\gamma\, Q^\delta P^{1 + \epsilon p^{m-4}}, \\ +S^{-1} Q\, S = R^c\, Q\, P^{ep^{m-4}}, \\ +S^{-1} R\, S = R\, Q^g\, P^{jp^{m-4}}, \\ +\end{cases} %% 30 +\end{equation} +\noindent with +\begin{equation*} +P^{p^{m-3}} = 1, \quad Q^p = R^p = S^p = 1, +\end{equation*} +\noindent$(k,\, \beta,\, \alpha,\, a,\, \gamma,\, \delta,\, \epsilon,\, c,\, +e,\, g,\, j = 0,\, 1,\, 2,\, \cdots,\, p - 1)$. + +These constants are however subject to conditions (17), (21), (22), (23), +(24) and (25). Not all these groups are distinct. Suppose that $G$ and +$G'$ of the above set are simply isomorphic and that the correspondence is +given by + +\begin{equation*} +C = \left[ \begin{matrix}S, & R, & Q, & P \\ + S'_1, & R'_1, & Q'_1, & P'_1 \\ \end{matrix} \right]. +\end{equation*} + +Inspection of (29) gives +\begin{align*} +S'_1 &= S'^{v'''} R'^{z'''} Q'^{y'''} P'^{x'''p^{m-4}}, \\ +R'_1 &= S'^{v''} R'^{z''} Q'^{y''} P'^{x''p^{m-4}}, \\ +Q'_1 &= S'^{v'} R'^{z'} Q'^{y'} P'^{x'p^{m-4}}, \\ +P'_1 &= S'^v R'^z Q'^y P'^x, +\end{align*} +\noindent in which $x$ and one out of each of the sets $v'$, $z'$, $y'$, $x'$; $v''$, +$z''$, $y''$, $x''$; $v'''$, $z'''$, $y'''$, $x'''$ are prime to $p$. + +Since $S$, $R$, $Q$, and $P$ satisfy equations (30), $S'_1$, $R'_1$, $Q'_1$ +and $P'_1$ also satisfy them. Substituting these operators and reducing in +terms of $S'$, $R'$, $Q'$, and $P'$ we get the six equations +\begin{equation} +[V'_\kappa,\, Z'_\kappa,\, Y'_\kappa,\, X'_\kappa] = [V_\kappa,\, Z_\kappa,\, +Y_\kappa,\, X_\kappa] \qquad (\kappa = 1,\, 2,\, 3,\, 4,\, 5,\, 6), %% 31 +\end{equation} +\noindent which give the following twenty-four congruences +\begin{equation} + \begin{cases} + V'_\kappa \equiv V_\kappa \pmod{p}, \\ + Z'_\kappa \equiv Z_\kappa \pmod{p}, \\ + Y'_\kappa \equiv Y_\kappa \pmod{p}, \\ + X'_\kappa \equiv X_\kappa \pmod{p^{m-3}}, + \end{cases} %% 32 +\end{equation} +\noindent where +\begin{align*} +V'_1 &= v, \quad V_1 = v, \\ +Z'_1 &= Z + c'(yv' - y'v) + \gamma'xv' + c\delta x\tbinom{v'}{2}, \quad Z_1 = z, \\ +Y'_1 &= y + g'(zv' - z'v) + \delta'xv' + \gamma'g'x\tbinom{v}{2} + \beta'xz', \quad Y_1 = y, \\ +%% +X'_1 &= x + \Bigl\{\epsilon'xv' + (\gamma'j'x + e'\delta'x + a'\delta'\gamma'x) + \tbinom{v'}{2} + c'\delta'j'\tbinom{v'}{3} + (\alpha'\gamma'v' + \delta'k'v' \\ + &\quad+ a'\delta'\gamma'v^2 + \beta'k'z')\tbinom{x}{2} + j'(zv' - z'v) + + e'g'[z\tbinom{v'}{2} - z'\tbinom{v}{2}] \\ + &\quad+ a'g'[\tbinom{z}{2}v' + \tbinom{z'}{2}v - zz'v] + e'(yv' - y'v) + + c'j'[y\tbinom{v'}{2} - y'\tbinom{v}{2}] \\ + &\quad+ a'c'[\tbinom{y}{2}v' + v\tbinom{-y'}{2} - yy'v] + a'(yz' - y'z) + - a'\beta'xz'^2 + \alpha'xz' \\ + &\quad+ \alpha'\beta'x\tbinom{z'}{2} + a'\gamma'x(y - y')v' + k'xy'\Bigr\}p^{m-4}, \\ +%% +X_1 &= x + kxp^{m-4}, \\ +V'_2 &= v, \quad V_2 = v + \beta v', \\ +Z'_2 &= z + c'(yv'' - y''v) + \gamma'xv'' + e'\delta'\tbinom{v''}{2}, + \quad Z_2 = z + \beta z' + c'\beta y'v, \\ +Y'_2 &= y + g'(zv'' - z''v) + \delta'xv'' + \gamma'g'x\tbinom{v''}{2} + + \beta'\gamma'\tbinom{x}{2}v'' + \beta'xz'', \\ +%% +Y_2 &= y + \beta y' + g'\beta z'v, \\ +X'_2 &= x + \Bigl\{\Theta'_1(x, v'') + j'(zv'' - z''v) + e'g'[z\tbinom{v''}{2} - + z''\tbinom{v}{2}] + a'g'[\tbinom{x}{2}v'' \\ + &\quad+ \tbinom{-z''}{2}v - zz''v] + e'(yv'' - y''v) + c'j'[y\tbinom{v''}{2} + - y''\tbinom{v}{2}] + a'c'[\tbinom{y}{2}v'' \\ + &\quad+ \tbinom{-y''}{2}v - yy''v''] + a'g'(zv'' - z''v)z'' + a'(yz'' - y''z) + + a'\delta'v''z'' \\ + &\quad+ a'\gamma'(y - y'')v''x + \alpha'xz'' + a'\beta'x\tbinom{z''}{2} + + \beta'k'\tbinom{x}{2}z'' + k'xy''\Bigr\}p^{m-4}, \\ +%% +X_2 &= x + \Bigl\{\alpha x + \beta x' + a'\tbinom{\beta}{2}y'z' + e'\beta vy' + + (c'j'\beta + e'g'\beta z')\tbinom{v}{2} \\ + &+ a'c'\tbinom{\beta y'}{2}v + j'\beta vz' + a'g'\tbinom{\beta z'}{2} + + a'\beta(g'z'v + y')z\Bigr\}p^{m-4}, \\ +V'_3 &= v', \quad V_3 = v', \\ +Z'_3 &= z' + c'(y'v'' - y''v'), \quad Z_3 = z', \\ +Y'_3 &= y' + g'(z'v'' - z''v'), \quad Y_3 = y', \\ +%% +X'_3 &= \Bigl\{x' + j'(z'v'' - z''v') + e'g'[\tbinom{v''}{2}z' - + \tbinom{v'}{2}z''] + a'g'[\tbinom{z'}{2}v'' + \tbinom{-z''}{2}v' \\ + &\quad- z'z''v'] + e'(y'v'' - y''v') + c'j'[y'\tbinom{v''}{2} - y''\tbinom{v'}{2}] + + a'c'[\tbinom{y'}{2}v'' + \tbinom{-y''}{2}v' \\ + &\quad- y''y'v''] + a'(y'z'' - y''z')\Bigr\}p^{m-4}, \\ +X_4 &= (x' + a'x)p^{m-4}, \\ +%% +V'_4 &= v, \quad V_4 = v + \gamma v'' + \delta v', \\ +Z'_4 &= z + c'(yv''' - y'''v) + \gamma'xv''' + c'\delta'x\tbinom{v'''}{2}, \\ +Z_4 &= z + \gamma z'' + \delta z' + c'[\tbinom{\gamma}{2}v''y'' + + \tbinom{\delta}{2}v'y'] + c'(\gamma y'' + \delta y')v + c'\gamma\delta y''v, \\ +Y'_4 &= y + g'(zv''' - z'''v) + \delta' xv''' + \gamma' g'x\tbinom{v'''}{2} + + \beta'\gamma'\tbinom{x}{2}v''' + \beta'xz''', \\ +%% +Y_4 &= y + \gamma y'' + \delta y' + g'[\tbinom{\gamma}{2}v''z'' + + \tbinom{\delta}{2}v'z'] + g'(\gamma z'' + \delta z')v + g'\delta\gamma v'z'', \\ +X'_4 &= x + \Bigl\{\Theta'_1 (x, v''') + j'(zv''' - z'''v) + + e'g'[\tbinom{v'''}{2}z - \tbinom{v}{2}z'''] + a'g'\left[\tbinom{z}{2}v''' \right. \\ + &\quad \left. + \tbinom{-z'''}{2}v - zz'''v\right] + e'(yv''' - y'''v) + + c'j'[y\tbinom{v'''}{2} - y'''\tbinom{v}{2}] \\ + &\quad+ a'c'[\tbinom{y}{2}v''' + \tbinom{-y'''}{2}v - yy'''v'''] + + a'g'(v'''z - vz''')z''' \\ + &\quad+ a'(yz''' - y'''z) + a'\delta' xz'''v''' + a'\gamma' x(y - y''')v''' + \alpha' xz''' \\ + &\quad+ a'\beta' x\tbinom{z'''}{2} + \beta' k'z'''\tbinom{x}{2} + k'xy''' \Bigr\} p^{m-4}. \displaybreak \\ +%% +X_4 &= x + \Bigl\{\epsilon x + \delta x' + \gamma x'' + \tbinom{\gamma}{2} + [a'c'\tbinom{y''}{2}v'' + a'y''z'' + e'v''y'' + j'v''z'' \\ + &\quad+ a'g'\tbinom{z''}{2}v'' + (c'j'v''y'' + e'g'v''z'')(v + \delta v') + + a'(z + \delta z')v''z'' \\ + &\quad+ \frac{2\gamma - 1}{3}a'g'v''z''^2 + \frac{1}{2}[\frac{1}{3}(2\gamma - 1)v'' + - 1](c'j'y'' + e'g'z'')v''] \\ + &\quad+ \tbinom{\gamma}{3}a'c'v''y'' + \tbinom{\delta}{2}[a'c'\tbinom{y'}{2}v' + + a'y'z' + e'v'y' + j'v'z' \\ + &\quad+ a'g'\tbinom{z'}{2}v' + j'c'vv'y' + e'g'vv'z' + a'g'v'zz' + a'c'\gamma y'y''v' \\ + &\quad+ \frac{2\delta - 1}{3}a'g'v'z'^2 + \frac{1}{2}\{\frac{1}{3}(2\delta - 1)v' + - 1\}(c'j'y' + e'g'z')] \\ + &\quad+ \tbinom{\delta}{3}a'c'v'y'^2 + (v + \delta v')[j'\gamma z'' + + \tbinom{\gamma z''}{2}a'g' + e'\gamma y'' + \tbinom{\gamma y''}{2}a'c' \\ + &\quad+ a'g'(z + \delta z')] + \tbinom{v + \delta v'}{2}[e'g'\gamma z'' + + c'j'\gamma y''] + \delta[(e'g'z' \\ + &\quad+ c'j'y')\tbinom{v}{2} + e'vy' + j'z' + a'zy + a'g'vzz' + a'\gamma z'y'' + + a'c'\gamma vy'y''] \\ + &\quad+ a'g'\tbinom{\delta z'}{2}v + a'c'\tbinom{\delta y'}{2} v + + a'\gamma zy''\Bigr\}p^{m-4}, \\ +%% +V'_5 &= v', \quad V_5 = v' + cv'', \\ +Z'_5 &= z' + c'(y'v''' - y'''v'), \quad Z_5 = z' + cz'' + c'cy''v, \\ +Y'_5 &= y' + g'(z'v''' - z'''v'), \quad Y_5 = y' + cy'' + g'cv'z'', \\ +X'_5 &= \Bigl\{x' + j'(z'v''' - z'''v') + e'g'[\tbinom{v'''}{2}z' + - \tbinom{v'}{2}z'''] + a'g'[\tbinom{z'}{2}v''' \\ + &\quad+ \tbinom{-z'''}{2}v' - z'z'''v'] + c'(y'v''' - y'''v') + + c'j'[y'\tbinom{\delta'''}{2} - y'''\tbinom{v'}{2}] \\ + &\quad+ a'c'[\tbinom{y'}{2}v''' + \tbinom{-y'''}{2}v' - y'y'''v'''] + + a'(y'z''' - y'''z')\Bigr\}p^{m-4}, \\ +%% +X_5 &= \Bigl\{x' + ex + cx'' + a'\tbinom{c}{2}y''z'' + j'cv'z'' + + (e'g'cz'' + c'cj'y'')\tbinom{v'}{2} + e'cy''v \\ + &\quad+ a'cy''z' + a'g'z'v' + a'g'\tbinom{cz''}{2} + a'c'\tbinom{cy''}{2}\Bigr\}p^{m-4}, \\ +V'_6 &= v'', \quad V_6 = v'' + gv', \\ +Z'_6 &= z'' + c'(y''v''' - y'''v''), \quad Z_6 = z'' + gz', \\ +Y'_6 &= y'' + g'(z''v''' - z'''v''), \quad Y_6 = y'' + gy', \\ +%% +X'_6 &= \Bigl\{x'' + j'(z''v''' - z'''v'') + e'g'[\tbinom{v'''}{2}z'' + - \tbinom{v''}{2}z'''] + a'g'\left[\tbinom{z''}{2}v''' \right. \\ + &\quad+ \left. \tbinom{-z'''}{2}v'' - z''z'''v''\right] + e'(y''v''' - y'''v'') + + c'j'[y''\tbinom{v'''}{2} - y'''\tbinom{v''}{2}] \\ + &\quad+ a'c'[\tbinom{y''}{2}v''' + \tbinom{-y'''}{2}v'' - y''y'''v'''] + + a'(y''z''' - y'''z'')\Bigr\}p^{m-4}, \\ +X_6 &= \{x'' + jx + gx' + a'gy''z'\}p^{m-4}. +\end{align*} + +The necessary and sufficient condition for the simple isomorphism of the +two groups $G$ and $G'$ is \textit{that congruences (32) shall be consistent and +admit of solution} subject to conditions derived below. + +\medskip +6. \textit{Conditions of transformation.} Since $Q$ is not contained in $\{P\}$, +$R$ is not contained in $\{Q, P\}$, and $S$ is not contained in $\{R, Q, +P\}$, then $Q'_1$ is not contained in $\{P'_1\}$, $R'_1$ is not contained +in $\{Q'_1, P'_1\}$, and $S'_1$ is not contained in $\{R'_1, Q'_1, P'_1\}$. + +Let +\begin{equation*} +{Q'}_1^{s'} = {P'}_1^{sp^{m-4}}. +\end{equation*} + +This equation becomes in terms of $S'$, $R'$, $Q'$ and $P'$ +\begin{gather*} +[s'v',\, s'z' + c'\tbinom{s'}{2}v'y',\, s'y' + g'\tbinom{s'}{2}v'z',\, Dp^{m-4}] + = [0,\, 0,\, 0,\, sxp^{m-4}], +\intertext{and} +s'v' \equiv s'z' \equiv s'y' \equiv 0 \pmod{p}. +\end{gather*} + +At least one of the three quantities $v'$, $z'$ or $y'$ is prime to $p$, +since otherwise $s'$ may be taken $= 1$. + +Let +\begin{equation*} +{R'}_1^{s''} = {Q'}_1^{s'} {P'}_1^{sp^{m-4}}, +\end{equation*} +\noindent or in terms of $S'$, $R'$, $Q'$ and $P'$ +\begin{multline*} +[s''v'',\, s''z'' + c'\tbinom{s''}{2}v''y'',\, s''y'' + g'\tbinom{s''}{2}v'' + z'',\, Ep^{m-4}] \\ = [s'v',\, s'z' + c'\tbinom{s'}{2}v'y',\, s'y' + + g'\tbinom{s'}{2}v'z',\, E_1 p^{m-4}], +\end{multline*} +\noindent and +\begin{align*} +s''v'' &\equiv s'v' \pmod{p}, \\ +s''z'' + c'\tbinom{s''}{2}v''y'' &\equiv s'z' + c'\tbinom{s'}{2}v'y' \pmod{p}, \\ +s''y'' + g'\tbinom{s''}{2}v''z'' &\equiv s'y' + g'\tbinom{s'}{2}v'z' \pmod{p}. +\end{align*} + +Since $c'g' \equiv 0 \pmod{p}$, suppose $g' \equiv 0 \pmod{p}$. Elimination +of $s'$ between the last two give by means of the congruence $Z'_3 \equiv +Z_3 \pmod{p}$, +\begin{equation*} +s''\{2(y'z'' - y''z') + c'y'y''(v' - v'')\} \equiv 0 \pmod{p}, +\end{equation*} +\noindent between the first two +\begin{equation*} +s''\{2(v'z'' - v''z') + c'v'v''(y' - y'')\} \equiv 0 \pmod{p}, +\end{equation*} +\noindent and between the first and last +\begin{equation*} +s''(y'v'' - y''v') \equiv 0 \pmod{p}. +\end{equation*} + +At least one of the three above coefficients of $s''$ is prime to $p$, +since otherwise $s''$ may be taken $= 1$. + +Let +\begin{equation*} +{S'}_1^{s'''} = {R'}_1^{s''} {Q'}_1^{s'} {P'}_1^{sp^{m-4}} +\end{equation*} +\noindent or, in terms of $S'$, $R'$, $Q'$, and $P'$ +\begin{multline*} +[s'''v''', s'''z''' + c'\tbinom{s'''}{2}v'''y''', s'''y''' + +g'\tbinom{s'''}{2}v'''z''', E_2 p^{m-4}] \\ = [s''v'' + s'v', s''z'' + s'z' + +c'\{\tbinom{s''}{2}v''y'' + \tbinom{s'}{2}v'y' + s's''y''v'\}, \\ s''y'' + +s'y' + g'\{\tbinom{s''}{2}v''z'' + \tbinom{s'}{2}v'z' + s's''v'z''\}, E_3 p^{m-4}] +\end{multline*} +\noindent and +\begin{align*} +s'''v''' & \equiv s''v'' + s'v' \pmod{p}, \\ +s'''z''' & + c'\tbinom{s'''}{2}v'''y''' \\ + & \qquad \equiv s''z'' + s'z' + c'\{\tbinom{s''}{2}v''y'' + \tbinom{s'}{2}v'y' + + s's''y''v'\} \pmod{p}, \\ +s'''y''' & + g'\tbinom{s'''}{2}v'''z''' \\ + & \qquad \equiv s''y'' + s'y' + g'\{\tbinom{s''}{2}v''z'' + \tbinom{s'}{2}v'z' + + s's''z''v'\} \pmod{p}. +\end{align*} + +If $g' \equiv 0$ and $c' \not\equiv 0 \pmod{p}$ the congruence $Z'_3 +\equiv Z_3 \pmod{p}$ gives +\begin{equation*} +(y'v'' - y''v') \equiv 0 \pmod{p}. +\end{equation*} + +Elimination in this case of $s''$ between the first and last congruences +gives +\begin{equation*} +s'''(y''v''' - y'''v'') \equiv 0 \pmod{p}. +\end{equation*} + +Elimination of $s''$ between the first and second, and between the second +and third, followed by elimination of $s'$ between the two results, gives +\begin{equation*} +s'''\left(z''^2 - c'y''z''v' + \frac{c'^2}{4}y''v''\right) +(y'v''' - y'''v') \equiv 0 \pmod{p}. +\end{equation*} + +Either $(y''v''' - y'''v'')$, or $(y'v''' - y'''v')$ is prime to $p$, +since otherwise $s'''$ may be taken $= 1$. + +A similar set of conditions holds for $c' \equiv 0$ and $g' \not\equiv 0 +\pmod{p}$. + +When $c' \equiv g' \equiv 0 \pmod{p}$ elimination of $s'$ and $s''$ between +the three congruences gives +\begin{equation*} +s'''\Delta \equiv s''' \left|\begin{matrix} + v' & v'' & v''' \\ + y' & y'' & y''' \\ + z' & z'' & z''' \\ \end{matrix}\right| +\equiv 0 \pmod{p} +\end{equation*} +\noindent and $\Delta$ is prime to $p$, since otherwise $s'''$ +may be taken $= 1$. + +\newpage +7. \textit{Reduction to types.} In the discussion of congruences (32), the +group $G'$ is taken from the simplest case and we associate with it all +simply isomorphic groups $G$. + +\begin{center} +\large I. \normalsize + +\smallskip +\begin{tabular}{|r|c|c|c|c|c|c||r|c|c|c|c|c|} +\multicolumn{7}{c}{A.}&\multicolumn{6}{c}{B.} \\ \hline + &$a_2$&$\beta_2$&$c_2$&$g_2$&$\gamma_2$&$\delta_2$& &$k_2$&$\alpha_2$&$\epsilon_2$&$e_2$&$j_2$ \\ \hline +\textbf{ 1}& 1 & 1 & 1 & 1 & 1 & 1 &\textbf{ 1}& 1 & 1 & 1 & 1 & 1 \\ \hline +\textbf{ 2}& 0 & 1 & 1 & 1 & 1 & 1 &\textbf{ 2}& 0 & 1 & 1 & 1 & 1 \\ \hline +\textbf{ 3}& 0 & 0 & 1 & 1 & 1 & 1 &\textbf{ 3}& 1 & 0 & 1 & 1 & 1 \\ \hline +\textbf{ 4}& 0 & 0 & 1 & 1 & 1 & 0 &\textbf{ 4}& 1 & 1 & 0 & 1 & 1 \\ \hline +\textbf{ 5}& 0 & 0 & 1 & 0 & 1 & 1 &\textbf{ 5}& 1 & 1 & 1 & 0 & 1 \\ \hline +\textbf{ 6}& 0 & 0 & 1 & 0 & 1 & 0 &\textbf{ 6}& 1 & 1 & 1 & 1 & 0 \\ \hline +\textbf{ 7}& 0 & 1 & 0 & 1 & 1 & 1 &\textbf{ 7}& 0 & 0 & 1 & 1 & 1 \\ \hline +\textbf{ 8}& 0 & 1 & 0 & 1 & 0 & 1 &\textbf{ 8}& 0 & 1 & 0 & 1 & 1 \\ \hline +\textbf{ 9}& 0 & 1 & 1 & 0 & 1 & 1 &\textbf{ 9}& 0 & 1 & 1 & 0 & 1 \\ \hline +\textbf{10}& 0 & 1 & 1 & 0 & 1 & 0 &\textbf{10}& 0 & 1 & 1 & 1 & 0 \\ \hline +\textbf{11}& 1 & 0 & 1 & 1 & 1 & 1 &\textbf{11}& 1 & 0 & 0 & 1 & 1 \\ \hline +\textbf{12}& 1 & 0 & 1 & 0 & 1 & 1 &\textbf{12}& 1 & 0 & 1 & 0 & 1 \\ \hline +\textbf{13}& 1 & 0 & 1 & 1 & 0 & 1 &\textbf{13}& 1 & 0 & 1 & 1 & 0 \\ \hline +\textbf{14}& 1 & 0 & 1 & 1 & 1 & 0 &\textbf{14}& 1 & 1 & 0 & 0 & 1 \\ \hline +\textbf{15}& 1 & 0 & 1 & 0 & 0 & 1 &\textbf{15}& 1 & 1 & 0 & 1 & 0 \\ \hline +\textbf{16}& 1 & 0 & 1 & 0 & 1 & 0 &\textbf{16}& 1 & 1 & 1 & 0 & 0 \\ \hline +\textbf{17}& 1 & 0 & 1 & 1 & 0 & 0 &\textbf{17}& 0 & 0 & 0 & 1 & 1 \\ \hline +\textbf{18}& 1 & 0 & 1 & 0 & 0 & 0 &\textbf{18}& 0 & 0 & 1 & 0 & 1 \\ \hline +\textbf{19}& 1 & 1 & 0 & 1 & 1 & 1 &\textbf{19}& 0 & 0 & 1 & 1 & 0 \\ \hline +\textbf{20}& 1 & 1 & 0 & 1 & 0 & 1 &\textbf{20}& 0 & 1 & 0 & 0 & 1 \\ \hline +\textbf{21}& 1 & 1 & 0 & 1 & 1 & 0 &\textbf{21}& 0 & 1 & 0 & 1 & 0 \\ \hline +\textbf{22}& 1 & 1 & 0 & 1 & 0 & 0 &\textbf{22}& 0 & 1 & 1 & 0 & 0 \\ \hline +\textbf{23}& 1 & 1 & 1 & 0 & 1 & 1 &\textbf{23}& 1 & 0 & 0 & 0 & 1 \\ \hline +\textbf{24}& 1 & 1 & 1 & 1 & 0 & 1 &\textbf{24}& 1 & 0 & 0 & 1 & 0 \\ \hline +\textbf{25}& 1 & 1 & 1 & 1 & 1 & 0 &\textbf{25}& 1 & 0 & 1 & 0 & 0 \\ \hline +\textbf{26}& 1 & 1 & 1 & 0 & 0 & 1 &\textbf{26}& 1 & 1 & 0 & 0 & 0 \\ \hline +\textbf{27}& 1 & 1 & 1 & 0 & 1 & 0 &\textbf{27}& 0 & 0 & 0 & 0 & 1 \\ \hline +\textbf{28}& 1 & 1 & 1 & 1 & 0 & 0 &\textbf{28}& 0 & 0 & 0 & 1 & 0 \\ \hline +\textbf{29}& 1 & 1 & 1 & 0 & 0 & 0 &\textbf{29}& 0 & 0 & 1 & 0 & 0 \\ \hline + & & & & & & &\textbf{30}& 0 & 1 & 0 & 0 & 0 \\ \hline + & & & & & & &\textbf{31}& 1 & 0 & 0 & 0 & 0 \\ \hline + & & & & & & &\textbf{32}& 0 & 0 & 0 & 0 & 0 \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &1 &2 &3 &4&5 &6 &7 &8 &9 &10 \\ \hline +\textbf{1} &$\times$&$\times$&$\times$& &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{2} &$\times$&$2_1$ &$3_1$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{3} &$1_2$ &$2_1$ &$3_1$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{4} &$1_2$ &$\times$&$\times$& &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{5} &$2_1$ &$2_1$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{6} &$2_1$ &$2_1$ &$3_1$ & &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{7} &$1_2$ &$2_1$ &$3_1$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{8} &$1_2$ &$2_4$ &$3_4$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{9} &$2_1$ &$2_1$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{10}&$2_4$ &$2_4$ &$3_4$ & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{11}&$1_2$ &$2_4$ &$3_4$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{12}&$2_4$ &$2_4$ & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{13}&$2_1$ &$2_1$ &$3_1$ & &$19_6$& & &$19_6$&$19_6$& \\ \hline +\textbf{14}&$2_1$ &$2_4$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{15}&$2_1$ &$2_4$ &$3_4$ & &$19_6$& &$19_6$& &$19_6$& \\ \hline +\textbf{16}&$2_1$ &$2_1$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{17}&$1_2$ &$2_4$ &$3_4$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{18}&$2_4$ &$2_4$ & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{19}&$2_4$ &$2_4$ &$3_4$ & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{20}&$2_1$ &$2_4$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{21}&$2_4$ &* &* & & &$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{22}&$2_4$ &$2_4$ & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{23}&$2_4$ &* & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{24}&$2_1$ &$2_4$ &$3_4$ & & &$19_6$& & &$19_6$&$19_6$ \\ \hline +\textbf{25}&$2_4$ &$2_4$ & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{26}&$2_1$ &$2_4$ & &*& &$19_6$&$19_6$& &$19_6$& \\ \hline +\textbf{27}&$2_4$ &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{28}&$2_4$ &* &* & & &$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{29}&* &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\textbf{30}&$2_4$ &* & &*&$19_6$&$19_6$&$19_6$& & &$19_6$ \\ \hline +\textbf{31}&$2_4$ &* & &*& &$19_6$& &$19_6$&$19_6$& \\ \hline +\textbf{32}&* &* & &*&$19_6$&$19_6$& &$19_6$& &$19_6$ \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize (continued) + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &11 &12 &13 &14 &15 &16 &17 &18 &19 \\ \hline +\textbf{1} &$\times$ &$19_1$&$\times$ &$11_1$ &$19_1$&$19_1$&$13_1$ &$19_1$&$\times$ \\ \hline +\textbf{2} &$25_2$ & &$\times$ &$25_2$ & & &$13_2$ & &$\times$ \\ \hline +\textbf{3} &$11_1$ &$19_2$&$13_1$ &$24_2$ &$21_2$&$19_2$&$13_1$ &$21_2$& \\ \hline +\textbf{4} &$24_2$ &$19_2$&$13_1$ &$24_2$ &$19_1$&$19_2$&$13_1$ &$19_1$&$19_2$ \\ \hline +\textbf{5} & & & & & & & & &$19_1$ \\ \hline +\textbf{6} &$\times$ &$19_2$&$\times$ &$11_6$ &$21_2$&$19_2$&$13_6$ &$21_2$&$\times$ \\ \hline +\textbf{7} &$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & & \\ \hline +\textbf{8} &$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & &$19_2$ \\ \hline +\textbf{9} & &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_2$ \\ \hline +\textbf{10}&$25_{10}$& &$\times$ &$25_{10}$& & &$13_{10}$& &$19_6$ \\ \hline +\textbf{11}&$24_2$ &$19_2$&$13_1$ &* &$21_2$&$19_2$&$13_1$ &$21_2$& \\ \hline +\textbf{12}& & & & & & & & & \\ \hline +\textbf{13}&$11_6$ &* &$13_6$ &$11_6$ &* &$19_2$&$13_6$ &* & \\ \hline +\textbf{14}& & & & & & & & &$19_2$ \\ \hline +\textbf{15}&$11_6$ &$19_2$&$13_6$ &$11_6$ &$21_2$&$19_2$&$13_6$ &$21_2$&$19_6$ \\ \hline +\textbf{16}& & & & & & & & &$19_6$ \\ \hline +\textbf{17}&$25_2$ & &$13_2$ &$25_2$ & & &$13_2$ & & \\ \hline +\textbf{18}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{19}&$25_{10}$& &$13_{10}$&$25_{10}$& & &$13_{10}$& & \\ \hline +\textbf{20}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_2$ \\ \hline +\textbf{21}&$25_{10}$& &$13_{10}$&$25_{10}$& & &$13_{10}$& &$19_6$ \\ \hline +\textbf{22}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_6$ \\ \hline +\textbf{23}& & & & & & & & & \\ \hline +\textbf{24}& &$11_6$&$19_2$ &$13_6$ &$11_6$&* &* &$13_6$&* \\ \hline +\textbf{25}& & & & & & & & & \\ \hline +\textbf{26}& & & & & & & & &$19_6$ \\ \hline +\textbf{27}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{28}&$25_{10}$& &$13_{10}$&$25_10$ & & &$13_{10}$& & \\ \hline +\textbf{29}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\textbf{30}& &$19_6$& & &$21_6$&$19_6$& &$21_6$&$19_6$ \\ \hline +\textbf{31}& & & & & & & & & \\ \hline +\textbf{32}& &$19_6$& & &$21_6$&$19_6$& &$21_6$& \\ \hline +\end{tabular} + +\newpage +\large II. \normalsize (concluded) + +\smallskip +A. + +B.\ +\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline + &20 &21 &22 &23 &24 &25 &26 &27 &28 &29 \\ \hline +\textbf{1} &$19_1$&$19_1$ &$19_1$&$19_1$&$11_1$ &$11_1$ &$19_1$ &$19_1$&$11_1$&$19_1$ \\ \hline +\textbf{2} &$19_2$&$\times$&$21_2$& &$\times$ &$\times$ & & &$25_2$& \\ \hline +\textbf{3} & & & &$19_2$&$25_2$ &$24_2$ &$21_2$ &$19_2$&$25_2$&$21_2$ \\ \hline +\textbf{4} &$19_2$&$19_1$ &$19_1$&$19_2$&$11_1$ &$11_1$ &$19_1$ &$19_2$&$11_1$&$19_1$ \\ \hline +\textbf{5} &$19_2$&$19_1$ &$19_1$&$19_6$&$11_6$ &$3_1$ &$21_6$ &$19_6$&$3_1$ &$21_6$ \\ \hline +\textbf{6} &$19_6$&$\times$&$21_6$&$19_1$&$3_1$ &$11_6$ &$19_1$ &$19_2$&$3_1$ &$19_1$ \\ \hline +\textbf{7} & & & & &$25_2$ &$25_2$ & & &* & \\ \hline +\textbf{8} &$19_2$&$21_2$ &$21_2$& &$24_2$ &$25_2$ & & &$25_2$& \\ \hline +\textbf{9} &$19_2$&$21_2$ &$21_2$& &$11_6$ &$3_1$ & & &$3_1$ & \\ \hline +\textbf{10}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$\times$ & & &$3_4$ & \\ \hline +\textbf{11}& & & &$19_2$&$25_2$ &$24_2$ &$21_2$ &$19_2$&$25_2$&$21_2$ \\ \hline +\textbf{12}& & & &$19_6$&$25_{10}$ &$3_4$ &$21_6$ &$19_6$&$3_4$ &$21_6$ \\ \hline +\textbf{13}& & & &$19_2$&$3_1$ &$11_6$ &$21_2$ &$19_2$&$3_1$ &$21_2$ \\ \hline +\textbf{14}&* &$19_1$ &$19_1$&$19_6$&$11_6$ &$3_1$ &$21_6$ &$19_6$&$3_1$ &$21_6$ \\ \hline +\textbf{15}&$19_6$&$21_6$ &$21_6$&$19_2$&$3_1$ &$11_6$ &$19_1$ &* &$3_1$ &$19_1$ \\ \hline +\textbf{16}&$19_6$&$21_6$ &$21_6$&$19_6$&$3_1$ &$3_1$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{17}& & & & &$25_2$ &$25_2$ & & &* & \\ \hline +\textbf{18}& & & & &$25_{10}$ &$3_4$ & & &$3_4$ & \\ \hline +\textbf{19}& & & & &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{20}&$19_2$&$21_2$ &$21_2$& &$11_6$ &$3_1$ & & &$3_1$ & \\ \hline +\textbf{21}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{22}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$3_4$ & & &* & \\ \hline +\textbf{23}& & & &$19_6$&$25_{10}$ &$3_4$ &$21_6$ &$19_6$&$3_4$ &$21_6$ \\ \hline +\textbf{24}& & & &$19_2$&$3_1$ &$11_6$ &$21_2$ &$19_2$&$3_1$ &$21_2$ \\ \hline +\textbf{25}& &$21_6$ &$21_6$&$19_6$&$3_4$ &$3_4$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{26}&$19_6$&$21_6$ &$21_6$&$19_6$&$3_1$ &$3_1$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{27}& & & & &$25_{10}$ &$3_4$ & & &$3_4$ & \\ \hline +\textbf{28}& & & & &$3_4$ &$25_{10}$& & &$3_4$ & \\ \hline +\textbf{29}& & & & &* &* & & &* & \\ \hline +\textbf{30}&$19_6$&$21_6$ &$21_6$& &$3_4$ &$3_4$ & & &* & \\ \hline +\textbf{31}& & & &$19_6$&$3_4$ &$3_4$ &$21_6$ &$19_6$&* &$21_6$ \\ \hline +\textbf{32}& & & & &* &* & & &* & \\ \hline +\end{tabular} +\end{center} + +For convenience the groups are divided into cases. + +The double Table I gives all cases consistent with congruences (17), (21), +(23) and (25). The results of the discussion are given in Table II. The +cases in Table II left blank are inconsistent with congruences (22) and +(24), and therefore have no groups corresponding to them. + +Let $\kappa = \kappa_1 p^{k_2}$ where $dv[\kappa_1,\, p] = 1\; (\kappa = a,\, +\beta,\, c,\, g,\, \gamma,\, d,\, k,\, \alpha,\, \epsilon,\, e,\, j)$. + +In explanation of Table II the groups in cases marked +$\boxed{r_s}$ are simply isomorphic with groups in $A_r B_s$. + +The group $G'$ is taken from the cases marked +$\boxed{\times}$. The types are also selected from these cases. + +The cases marked $\boxed{*}$ divide into two or more parts. Let +\begin{align*} +a\epsilon - \alpha e + jk &= I_1, & a\epsilon - jk &= I_2, \\ +a\delta(a - e) + 2I_1 &= I_3, & \alpha g - \beta j &= I_4, \\ +\alpha\delta - \beta\epsilon &= I_5, & \epsilon g - \delta j &= I_6, \\ +c\epsilon - e\gamma &= I_7, & \alpha e - jk &= I_8, \\ +\delta e + \gamma j &= I_9, & \alpha\gamma + \delta k &= I_{10}. +\end{align*} + +The parts into which these groups divide, and the cases with which they are +simply isomorphic, are given in Table III. + +\begin{center} +\large III. \normalsize + +\smallskip +\begin{tabular}{|l|l|c|l|c|} \hline +$A_{1,2}B^*$ &$dv[I_1,p]=p$ &$2_1$ &$dv[I_1,p]=1$ &$2_4$ \\ \hline +$A_{3}B^*$ &$dv[I_2,p]=p$ &$3_1$ &$dv[I_2,p]=1$ &$3_4$ \\ \hline +$A_{4}B^*$ &$dv[I_3,p]=p$ &$3_1$ &$dv[I_3,p]=1$ &$3_4$ \\ \hline +$A_{12}B_{13}$ &$dv[I_4,p]=p$ &$19_1$ &$dv[I_4,p]=1$ &$19_2$ \\ \hline +$A_{14}B_{11}$ &$dv[I_5,p]=p$ &$11_1$ &$dv[I_5,p]=1$ &$24_2$ \\ \hline +$A_{15,18}B^*$ &$dv[I_4,p]=p$ &$19_1$ &$dv[I_4,p]=1$ &$21_2$ \\ \hline +$A_{16}B_{24}$ &$dv[I_6,I_5,p]=p$ &$19_1$ &$dv[I_6,I_5,p]=1$ &$19_2$ \\ \hline +$A_{20}B_{14}$ &$dv[I_7,p]=p$ &$19_1$ &$dv[I_7,p]=1$ &$19_2$ \\ \hline +$A_{24,25}B^*$ &$dv[I_8,p]=p$ &$3_1$ &$dv[I_8,p]=1$ &$3_4$ \\ \hline +$A_{27}B_{15}$ &$dv[I_6,p]=p$ &$19_1$ &$dv[I_6,p]=1$ &$19_2$ \\ \hline +$A_{29}B_{7,17}$ &$dv[I_{10},p]=p$ &$24_2$ &$dv[I_{10},p]=1$ &$25_2$ \\ \hline +$A_{29}B_{16,26}$ &$dv[I_9,p]=p$ &$11_6$ &$dv[I_9,p]=1$ &$3_1$ \\ \hline +$A_{29}B_{22,25,30,31}$&$dv[I_9,p]=p$ &$25_{10}$&$dv[I_9,p]=1$ &$3_4$ \\ \hline +$A_{29}B_{29,32}$ &$dv[I_8,I_9,p]=p$ &$11_6$ &$[I_8,p]=p,[I_9,p]=1$&$3_1$ \\ \hline +$A_{29}B_{29,32}$ &$[I_8,p]=1,[I_9,p]=p$&$25_{10}$&$[I_8,p]=1,[I_9,p]=1$&$3_4$ \\ \hline +\end{tabular} +\end{center} + +\newpage +8. \textit{Types.} The types for this class are given by equations (30) where the +constants have the values given in Table IV. + +\begin{center} +\large IV. \normalsize + +\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline + & $a$ &$\beta$&$c$&$g$&$\gamma$&$\delta$&$k$&$\alpha$&$\epsilon$&$e$&$j$ \\ \hline +$1_1$ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$2_1$ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$3_1$ &$\kappa$& 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$11_1$ & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$*13_1$& 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$19_1$ & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline +$1_2$ & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$*13_2$& 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$19_2$ & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$*21_2$& 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ \hline +$24_2$ & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ \hline +$25_2$ & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ \hline +$2_4$ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline +$3_4$ &$\kappa$& 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \hline +$11_6$ & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline +$*13_6$& 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline +\end{tabular} + +\footnotesize \noindent $\kappa = 1$, and a non-residue (mod $p$). + +\noindent $*$For $p=3$ these groups are isomorphic in Class II. +\end{center} + +A detailed analysis of congruences (32) for several cases is given below +as a general illustration of the methods used. + +\medskip +\begin{equation*} A_3 B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +\beta'xz' \equiv 0 \pmod{p}, \tag{II} \\ +a'(yz' - y'z) \equiv kx \pmod{p}, \tag{III} \\ +\beta v' \equiv 0, \qquad \beta z' \equiv 0, \qquad \beta y' \equiv + \beta'xz'' \pmod{p}, \tag*{(IV),(V),(VI)} \\ +a'(yz'' - y''z) + a'\beta'x\tbinom{z''}{2} \equiv +%% +\alpha x + \beta x' + a'\beta y'z \pmod{p}, \tag{VII} \\ +a'(y'z'' - y''z') \equiv ax \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v' \equiv 0 \pmod{p}, \tag{XI} \\ +\gamma z'' + \delta z' \equiv 0 \pmod{p}, \tag{XII} \\ +%% +\gamma y'' + \delta y' \equiv \beta'xz'' \pmod{p}, \tag{XIII} \\ +\begin{split} + a'(yz''' - y'''z) + a'\beta'x\tbinom{z'''}{2} \equiv \epsilon x &+ + \gamma x'' + \delta x + a'\delta y'z \\ &+ a'\gamma y''z + + a'\tbinom{\gamma}{2}y''z'' \pmod{p}, +\end{split} \tag{XIV} \\ +cv'' \equiv 0, \qquad cz'' \equiv 0, \qquad cy'' \equiv 0 \pmod{p}, + \tag*{(XV),(XVI),(XVII)} \\ +%% +a'(y'z''' - y'''z') \equiv ex \pmod{p}, \tag{XVIII} \\ +gv' \equiv 0, \qquad gz' \equiv 0, \qquad gy' \equiv 0 \pmod{p}, + \tag*{(XIX),(XX),(XXI)} \\ +a'(y''z''' - y'''z'') \equiv jx \pmod{p}, \tag{XXII} +\end{gather*} + +From (II) $z' \equiv 0 \pmod{p}$. + +The conditions of isomorphism give +\begin{equation*} + \Delta \equiv \left| \begin{matrix} + v' & v'' & v''' \\ + y' & y'' & y''' \\ + z' & z'' & z''' \\ \end{matrix}\right| \not\equiv 0 \pmod{p}. +\end{equation*} + +Multiply (IV), (V), (VI) by $\gamma$ and reduce by (XII), $\beta\gamma v' +\equiv 0$, $\beta\gamma z' \equiv 0$, $\beta\gamma y' \equiv 0 \pmod{p}$. Since +$\Delta \not\equiv 0 \pmod{p}$, one at least of the quantities, $v'$, $z'$ +or $y'$ is $\not\equiv 0 \pmod{p}$ and $\beta\gamma \equiv 0 \pmod{p}$. + +From (XV), (XVI) and (XVII) $c \equiv 0 \pmod{p}$, and from (XIX), (XX) and +(XXI) $g \equiv 0 \pmod{p}$. + +From (IV), (V), (VI) and (X) if $a \equiv 0$, then $\beta \equiv 0$ +and if $a \not\equiv 0$, then $\beta \not\equiv 0 \pmod{p}$. + +At least one of the three quantities $\beta$, $\gamma$ or $\delta$ is +$\not\equiv 0 \pmod{p}$ and one, at least, of $a$, $e$ or $j$ is $\not +\equiv 0 \pmod{p}$. + +\smallskip +$A_3$: Since $z''' \equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0$. +Elimination between (III), (X), (XIV) and (XXII) gives $a\epsilon - kj +\equiv 0 \pmod{p}$. Elimination between (VI) and (X) gives +$a'\beta'{z''}^2 \equiv a\beta \pmod{p}$ and $a\beta$ is a quadratic +residue or non-residue according as $a'\beta'$ is or is not, and there are +two types for this case. + +\smallskip +$A_4$: Since $y'$ and $z''$ are $\not\equiv 0 \pmod{p}$, $e \not\equiv 0 +\pmod{p}$. Elimination between (VI), (X), (XIII) and (XVIII) gives $a\delta +- \beta e \equiv 0 \pmod{p}$. + +This is a special form of (24). + +Elimination between (III), (VII), (X), (XIII), (XIV), (XVIII) and (XXII) +gives +\begin{equation*} +2jk + a\delta(a - e) + 2(a\epsilon - \alpha e) \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{24}$: Since from (XI), (XII) and (XIII) $y''$ and $z''' \not\equiv 0 +\pmod{p}$, and $z'' \equiv v'' \equiv 0 \pmod{p}$, (xxii) gives $j \not +\equiv 0 \pmod{p}$. + +Elimination between (III), (X), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{25}$: (XI), (XII) and (XIII) give $v' \equiv z' \equiv 0$ and $y', z''' +\not\equiv 0 \pmod{p}$ and this with (XVIII) gives $e \not\equiv 0$. + +Elimination between (III), (VII), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +\smallskip +$A_{28}$: Since $a \equiv 0$ then $e$ or $j \not\equiv 0 \pmod{p}$. + +Elimination between (III), (VII), (XVIII) and (XXII) gives +\begin{equation*} +\alpha e - jk \equiv 0 \pmod{p}. +\end{equation*} + +Multiply (XIII) by $a'z'''$ and reduce +\begin{equation*} +\delta e + \gamma j \equiv a'\beta'{z'''}^2 \not\equiv 0 \pmod{p}. +\end{equation*} + +\medskip +\begin{equation*} A_{11} B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +\beta'xz' \equiv 0 \pmod{p}, \tag{II} \\ +kx \equiv 0 \pmod{p}, \tag{III} \\ +\beta v' \equiv \beta z' \equiv 0, \quad \beta y' \equiv \beta'xz'', + \tag*{(IV),(V),(VI)} \\ +\alpha x + \beta x' \equiv 0 \pmod{p}, \tag{VII} \\ +%% +ax \equiv 0 \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v \equiv 0 \pmod{p}, \tag{XI} \\ +\gamma z'' + \delta z \equiv 0 \pmod{p}, \tag{XII} \\ +\gamma y'' + \delta y \equiv \beta'xz''' \pmod{p}, \tag{XIII} \\ +%% +\epsilon x + \gamma x'' + \delta x' \equiv 0 \pmod{p}, \tag{XIV} \\ +cv'' \equiv cz'' \equiv cy'' \equiv 0 \pmod{p}, \tag*{(XV),(XVI),(XVII)} \\ +ex \equiv 0 \pmod{p}, \tag{XVIII} \\ +gv' \equiv gz' \equiv gy' \equiv 0 \pmod{p}, \tag*{(XIX),(XX),(XXI)} \\ +jx \equiv 0 \pmod{p}, \tag{XXII} +\end{gather*} + +(II) gives $z' = 0$, (III) gives $k \equiv 0$, (X) gives $a \equiv 0$, +(XV), (XVI), (XVII) give $c \equiv 0 (\Delta \not\equiv 0)$, (XVIII) gives +$e \equiv 0$, (XIX), (XX), (XXI) give $g \equiv 0$, (XXII) gives $j \equiv +0$. One of the two quantities $z''$ or $z''' \not\equiv 0 \pmod{p}$, +and by (VI) and (XIII) one of the three quantities $\beta$, $\gamma$ or +$\delta$ is $\not\equiv 0$. + +\smallskip +$A_{11}$: (XIV) gives $\epsilon \equiv 0 \pmod{p}$. Multiplying (IV), (V), +(VI) by $\gamma$ gives, by (XII), $\beta\gamma v' \equiv \beta\gamma z' +\equiv \beta\gamma y' \equiv 0 \pmod{p}$, and $\beta\gamma \equiv 0 +\pmod{p}$. + +\smallskip +$A_{14}$: Elimination between (VII) and (XIV) gives $\alpha\delta - +\beta\epsilon \equiv 0 \pmod{p}$. + +\smallskip +$A_{24}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv 0$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{25}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{28}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) $\epsilon \equiv$ +or $\not\equiv 0 \pmod{p}$. + +\newpage +\begin{equation*} A_{19} B_1. \end{equation*} + +The special forms of the congruences for this case are +\begin{gather*} +c'(yv' - y'v) \equiv 0 \pmod{p}, \tag{I} \\ +kx \equiv 0 \pmod{p}, \tag{III} \\ +\beta v \equiv 0, \quad \beta z \equiv c'(yv'' - y''v), \quad \beta y' +\equiv 0 \pmod{p}, \tag*{(IV),(V),(VI)} \\ +%% +\alpha x + \beta x' \equiv 0 \pmod{p}, \tag{VII} \\ +c'(y'v'' - y''v') \equiv 0 \pmod{p}, \tag{VIII} \\ +ax \equiv 0 \pmod{p}, \tag{X} \\ +\gamma v'' + \delta v' \equiv 0 \pmod{p}, \tag{XI} \\ +%% +\gamma z'' + \delta z' + c'\gamma\delta y''v + c'\tbinom{\delta}{2}v'y' + +c'\tbinom{\gamma}{2}v''y'' \equiv c'(yv''' - y'''v) \pmod {p}, \tag{XII} \\ +\gamma y'' + \delta y' \equiv 0 \pmod{p}, \tag{XIII} \\ +\epsilon x + \gamma x'' + \delta x' \equiv 0 \pmod{p}, \tag{XIV} \\ +%% +cv'' \equiv 0, \quad cz'' \equiv c'(y'v''' - y'''v'), \quad cy'' \equiv 0 + \pmod{p}, \tag*{(XV),(XVI),(XVII)} \\ +ex + cx'' \equiv 0 \pmod{p}, \tag{XVIII} \\ +gv' \equiv 0, \quad gz' \equiv c'(y''v''' - y'''v''), \quad gy' \equiv 0 + \pmod{p}, \tag*{(XIX),(XX),(XXI)} \\ +jx + gx' \equiv 0 \pmod{p}. \tag{XXII} \\ +\end{gather*} + +(III) gives $k\equiv 0$, (X) gives $a \equiv 0$. + +Since $dv[(y'v''' - y'''v'), (y''v''' - y'''v''), p] = 1$ then $dv[c, g, p] += 1$. + +If $c \not\equiv 0$, $v'' \equiv y'' \equiv 0 \pmod{p}$ and therefore $g +\equiv 0 \pmod{p}$ and if $g \not\equiv 0$, then $c \equiv 0 \pmod{p}$. + +\smallskip +$A_{12}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0 \pmod{p}$. + +\smallskip +$A_{15}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{16}$: (XVIII) gives $e \equiv 0$. Elimination between (XIV) and (XXII) +gives $\epsilon g - \delta j \equiv 0 \pmod{p}$, between (VII) and (XIV) +gives $\alpha\delta - \beta\epsilon \equiv 0$. + +\smallskip +$A_{18}$: (XVIII) gives $e \equiv 0 \pmod{p}$. Elimination between (VII) +and (XXII) gives $\alpha g - \beta j \equiv 0 \pmod{p}$, (XIV) gives +$\epsilon \equiv 0$ or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{19}$: (VII) gives $\alpha \equiv 0 \pmod{p}$, (XIV) gives $\epsilon +\equiv 0$, (XXII) gives $j \equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0$ +or $\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{20}$: (VII) gives $\alpha \equiv 0$, (XXII) gives $j \equiv 0$. +Elimination between (XIV) and (XVIII) gives $\epsilon c - e\gamma \equiv 0 +\pmod{p}$. + +\smallskip +$A_{21}$: (VII) gives $\alpha\equiv 0$, (XIV) gives $\epsilon \equiv 0$ +or $\not\equiv 0 \pmod{p}$, (XVIII) gives $e \equiv 0,$ or +$\not\equiv 0$, and (XXII) gives $j \equiv 0 \pmod{p}$. + +\smallskip +$A_{22}$: (VII) gives $\alpha \equiv 0$, (XIV) gives $\epsilon \equiv 0$ +or $\not\equiv 0$, (XVIII) gives $epsilon \equiv 0$ or $\not\equiv 0$, +(XXII) gives $j \equiv 0 \pmod{p}$. + +\smallskip +$A_{23}$: (VII) gives $\alpha \equiv 0$, (XIV) gives $\epsilon \equiv 0$, +(XVIII) gives $\epsilon \equiv 0$, (XXII) gives $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{26}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +\smallskip +$A_{27}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. Elimination between (XIV) and (XXII) gives +$\epsilon g - \delta j \equiv 0 \pmod{p}$. + +\smallskip +$A_{29}$: (VII) $\alpha \equiv 0$, (XIV) $\epsilon \equiv 0$ or +$\not\equiv 0$, (XVIII) $\epsilon \equiv 0$, (XXII) $j \equiv 0$ or +$\not\equiv 0 \pmod{p}$. + +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%% Start of original license %%%% +\iffalse + +\newpage +\small +\pagenumbering{gobble} +\begin{verbatim} + + +End of the Project Gutenberg EBook of Groups of Order p^m Which Contain +Cyclic Subgroups of Order p^(m-3), by Lewis Irving Neikirk + +*** END OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M *** + +This file should be named 9930-t.tex or 9930-t.zip + +Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, +John Hagerson, and the Online Distributed Proofreading Team. + +Project Gutenberg eBooks are often created from several printed +editions, all of which are confirmed as Public Domain in the US +unless a copyright notice is included. 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