%%%% THIS FILE IS AUTOMATICALLY GENERATED %%%% DO NOT EDIT THIS FILE DIRECTLY, %%%% ONLY EDIT THE SOURCE, tom-64/document.tex. %%%% %% Standard package list \documentclass[letterpaper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[english]{babel} \usepackage[top=3cm, bottom=3cm, left=3.5cm, right=3.5cm]{geometry} \usepackage[onehalfspacing]{setspace} \usepackage{amsmath,amssymb,amsthm,wasysym} \usepackage{nicefrac,booktabs} \usepackage{mathptmx} \usepackage{cite} \usepackage[colorlinks=true]{hyperref} %% Various helpers for Tom's papers \newcommand{\gs}{\textnormal{gs}} \newcommand{\ord}{\textnormal{ord}} \newcommand{\Exp}{\textnormal{Exp}} \newcommand{\Log}{\textnormal{Log}} \newcommand{\lcm}{\textnormal{lcm}} \newcommand{\range}{\textnormal{range}} \newcommand{\NR}{\textnormal{NR}} \newcommand{\Mod}[1]{\left(\textnormal{mod}~#1\right)} \newcommand{\ap}[2]{\left\langle #1;#2 \right\rangle} \newcommand{\summ}[1]{\sum_{k=1}^m{#1}} \newcommand{\bt}[1]{{{#1}\mathbb{N}}} \newcommand{\fp}[1]{{\left\lbrace{#1}\right\rbrace}} \newcommand{\intv}[1]{{\left[1,{#1}\right]}} %% Lifted from http://stackoverflow.com/questions/2767389/referencing-a-theorem-like-environment-by-its-name %% This lets me do things like "Theorem A" and have the references work properly. \makeatletter \let\@old@begintheorem=\@begintheorem \def\@begintheorem#1#2[#3]{% \gdef\@thm@name{#3}% \@old@begintheorem{#1}{#2}[#3]% } \def\namedthmlabel#1{\begingroup \edef\@currentlabel{\@thm@name}% \label{#1}\endgroup } \makeatother % end lift \newtheoremstyle{namedthrm} {}{}{}{}{}{}{ } % This last space needs to be there {\bf\thmname{#1} \thmnote{#3}.} %% End reference hack %% Document start \date{} \begin{document} %% Content start \newtheorem*{cor}{Corollary} \newtheorem{lemma}{Lemma} \newtheorem{thm}{Theorem} \newtheorem{q}{Problem} \theoremstyle{definition} \newtheorem*{defn}{Definition} \newtheorem*{notation}{Notation} \newtheorem{remark}{Remark} \title{On the Finiteness of Semigroups in Which $x^r = x$} \author{Thomas C. Brown \\ Communicated by P. Hall} \date{} \maketitle \begin{center}{\small {\bf Citation data:} T.C. {Brown}, \emph{On the finiteness of semigroups in which $x^r = r$}, Proc. Cambridge Philos. Soc. \textbf{60} (1964), 1028--1029.}\bigskip\end{center} In this note we present a new proof of the following theorem (see~\cite{green+rees1952}). \begin{thm} \label{theorem 1} For each $r$, $r \geq 2$, every finitely generated semigroup in which $x^r = x$ holds identically is finite if and only if every finitely generated group of exponent $r - 1$ is finite. \end{thm} \begin{notation} Throughout, $r$ is fixed and $S_k$ denotes a semigroup on $k$ generators in which $x^r = x$. The elements of $S_k$ are regarded as equivalence classes of words in $k$ symbols $X_1, \dots, X_k$. Upper case letters will denote words, and lower case letters the elements of the semigroup $S_k$; thus if the word $W$ represents the element $w$ of $S_k$, we write $W \in w \in S_k$, and also say that $W$ is a word of $S_k$. $W = W'$ signifies that $W$ and $W'$ are identical, while $W \sim W'$ signifies that $W$ and $W'$ are equivalent, i.e., represent the same element of $S_k$. If $W$ is a word of $S_k$, we write $|W|$ for the \emph{length} of $W$ in the symbols $X_i$; e.g., $|X_1X_2X_3| = 3$. If each $X_i$ appears in $W$, $1 \leq i \leq k$, we say that $W$ is a \emph{complete} word of $S_k$; any word $W$ is \emph{minimal} if $W \sim W'$ implies $|W| \leq |W'|$. \end{notation} \begin{remark} \label{remark 1} Clearly $S_k$ is finite if and only if the lengths of minimal words of $S_k$ are bounded from above. \end{remark} \begin{lemma} \label{lemma 1} $W \sim AXB$ implies $W \sim W(XW)^{r -1}.$ \end{lemma} \begin{proof} \[ W(XW)^{r - 1} \sim AXB(XAXB)^{r - 1} \sim (AX)^{r - 2} A[XAXB(XAXB)^{r - 1}] \sim AXB \sim W. \qedhere \] \end{proof} \begin{lemma} \label{lemma 2} Let $T$ be a complete word of $S_k$. Then for any word $X$ of $S_k$, $T \sim T(XT)^{r - 1}$. \end{lemma} \begin{proof} By Lemma~\ref{lemma 1} we need only show that $T \sim AXB$ for some $A,B$. We show by induction on $|X|$ that $T \sim TXB$ for a suitable $B$. If $|X| = 0$, i.e., $X$ is the empty word, we may take $B = X$. Now let $X = X_0X_i$, where $|X_0| = n$, and suppose $T \sim TX_0B_0$. Since $T$ is complete we have $T = T'X_iT''$, and hence \begin{align*} T \sim T'(X_iT''X_0)B_0 &\sim (T'X_iT''X_0X_i)T''X_0(X_iT''X_0)^{r - 2}B_0 \\ &=TX[T''X_0(X_iT''X_0)^{r - 2}B_0]. \qedhere \end{align*} \end{proof} \begin{lemma} \label{lemma 3} Let $T \in t \in S_k$, where $T$ is complete. Then the set $tS_kt$ is a group. \end{lemma} \begin{proof} $tS_kt$ is a semigroup with identity $t^{r - 1}$, and the inverse of $txt$ is $(txt)^{r - 2}$. (For by Lemma~\ref{lemma 2} we have $t(xt)^{r - 1} = t$; hence $(txt)^{r - 1} = t(xtt)^{r - 2}xt = t(xtt)^{r - 1}t^{r - 2} = t^{r - 1}$.) \end{proof} \begin{proof}[Proof of Theorem] One direction is trivial. For the other, we use induction on $k$; $S_1$ is obviously finite, and we suppose that $S_{k - 1}$ is finite. By Remark~\ref{remark 1}, there is a number $m$ such that if $|X| = m$, where $X$ is a minimal word of $S_k$, then $X$ is complete in the $k$ symbols $X_1,\dots,X_k$. Let $W$ be any minimal word of $S_k$; then $W$ can be written in the form \begin{equation} W = W_1W_2 \cdots W_NW', \label{eq 1} \end{equation} where \[ |W_j| = m \quad (1 \leq i \leq N), \: |W'| < m. \] Since $W$ is minimal, each $W_j$ is minimal and hence complete. Let $T \in t \in S_k$, where $T$ is complete. Then for each $j$, $W_j \sim W_j(T^2W_j)^{r - 1}$; and making this substitution into (1) for each $j$, we obtain \begin{equation} W \sim W_1[T(TW_1T)^{r - 2}(TW_1W_2T)(TW_2T)^{r - 2} \cdots (TW_NT)^{r - 2}T] W_N W'. \end{equation} Now let $G$ be the subgroup of $tS_kt$ generated by all those elements with representatives of the form $TXT$, where $|X| \leq 2m$. Since $G$ is finitely generated and of exponent $r - 1$, $G$ is finite by assumption; hence every element of $G$ has a representative word whose length does not exceed some number $M$. Thus by (2), $W \sim W_1YW_NW'$, where $|Y| \leq M$, and by the minimality of $W$, $|W| \leq M + 3m$. Finally, since $M$ did not depend on the particular choice of $W$, $S_k$ is finite, and the proof is complete. \end{proof} \bibliographystyle{amsplain} \bibliography{tom-all} \end{document}