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\title{On Locally Finite Semigroups}
\author{T. C. Brown}
\date{}
\maketitle
\begin{center}{\small {\bf Citation data:} T.C. {Brown}, \emph{On locally finite semigroups} (In Russian), Ukraine Math. J.
  \textbf{20} (1968), 732--738.}\bigskip\end{center}

\begin{abstract} 
The classical theorem of Schmidt on locally finite group extensions may be stated as follows: If $\varphi: G \mapsto H$ is a homomorphism of the group $G$ onto the locally finite group $H$ with locally finite kernel, then $G$ is locally finite.

In this paper we prove the exact analogue of this theorem for semigroups. Then in the last section we give several consequences, including the well-known theorem of Shevrin which states that a band of locally finite semigroups is locally finite, and the theorem of Green and Rees on the equivalence of Burnside's problem for groups with ``Burnside's problem for semigroups".
\end{abstract} 

\section{Introduction} \label{sec: 1}

The whole of this paper is based on Lemma~\ref{lemma: 1} below, which is essentially combinatorial. The methods are completely elementary in nautre, and the paper is self-contained, except for a few facts used in the last two sections. The theorem we wish to prove is the following.

\begin{thm} \label{thm: 1}
If $\varphi: S \mapsto T$ is a homomorphism of the semigroup $S$ onto the locally finite semigroup $T$ such that $e\varphi^{-1}$ is a locally finite subsemigroup of $S$ for each idempotent element $e$ of $T$, then $S$ is locally finite.
\end{thm}

First we note that it is sufficient to consider the case where $T$ is finite. For suppose the theorem is true in this case, and let $\varphi: S \mapsto T'$ be a homomorphism with all the required properties onto an arbitrary (possibly infinite) locally finite semigroup $T'$. Let $A$ be a finite subset of $S$, and let $\id{A}$ denote the subsemigroup of $S$ generated by $A$. It is required to show that $\id{A}$ is finite. Now $\id{A \varphi} = T$ is a finite subsemigroup of $T'$, since $T'$ is locally finite, hence all we have to do is to restrict $\varphi$ to $T\varphi^{-1}$ to get a homomorphism $\varphi': T\varphi^{-1} \mapsto T$ onto a finite semigroup $T$; furthermore $\varphi'$ has all the required properties. Hence by our assumption, $T\varphi^{-1}$ is locally finite. But $A \subset T\varphi^{-1}$, hence $\id{A}$ is finite, as required.

In Sections~\ref{sec: 2} throug~\ref{sec: 4} below, we shall prove the theorem for the special cases where $T$ is a (finite) group, group with zero, null semigroup, simple semigroup, or 0-simple semigroup. (For definitions see below). In Section~\ref{sec: 5} we give some consequences to the theorem.

Assuming the truth of the theorem for the special cases listed above, the general case follows by induction on $|T|$, the order of $T$, as follows. For $|T| = 1$, the theorem is trivial. Let $T$ be a finite semigroup and suppose the theorem holds for all semigroups $T'$ with $|T'| < |T|$. If $T$ has no proper non-zero ideals, then $T$ is either null, simple, or 0-simple, and $S$ is then locally finite by the appropriate special case. If $T$ has a proper non-zero ideal $M$, then let $T' = T/M$, the Rees factor semigroup (for definition, see below), let $\psi$ be the natural mapping of $T$ upon $T'$, and let $\sigma = \varphi \psi$. Since $\sigma $is a homomorphism of $S$ upon $T'$, and $|T'| < |T|$, all we have to verify is that $e\sigma^{-1}$ is locally finite for each idempotent element $e$ of $T'$; for then by the inductive assumption $S$ is locally finite. Thus let $e$ be an idempotent element of $T'$. If $e\ne 0$, then $e\sigma^{-1} = (e\psi^{-1})\varphi^{-1} = e\varphi^{-1}$, and $e\varphi^{-1}$ is locally finite by hypothesis. If $e = 0$, then $0\sigma^{-1} = (0\psi^{-1})\varphi^{-1} = M\varphi^{-1}$, and $M\varphi^{-1}$ is locally finite by the inductive assumption. 

In the remainder of the proof (and above), the following definitions are used. If a semigroup $S$ contains an element 0 such that $0s = s0 = 0$ for all $s \in S$, $S$ is a \emph{semigroup with zero}. In case $S \setminus \{0\}$ is a group, $S$ is a \emph{group with zero}. A subsemigroup $M$ of $S$ is an \emph{ideal} is $SMS \cup MS \cup SM \subset M$, and is a \emph{proper ideal} if $M \ne S$. A semigroup without proper ideals is \emph{simple}. A semigroup $S$ with zero is \emph{0-simple} if $S^2 \ne\{0\}$ and $\{0\}$ is the only proper ideal of $S$. $S$ is \emph{null} if $S^2 = \{0\}$. Let $M$ be an ideal of $S$; the \emph{Rees factor semigroup of $S$ modulo $M$}, denoted by $S/M$, is defined as follows. As a set, $S/M = (S\setminus M) \cup \{0\}$. For $x,y \in S$, let $xy$ denote their product in $S$ and $x\ast y$ their product in $S/M$. Then, by definition, $x\ast y = 0$ if $xy \in M$, $x \ast y = xy $ if $xy \notin M$, and $0\ast z = z\ast 0 = 0$ for all $z \in S/M$. The mapping $x \mapsto 0$, $x \in M$, $x \mapsto x$, $x \notin M$, is the \emph{natural homomorphism} of $S$ upon $S/M$. 

We shall also require the following:

Let $A$ be a subset of the semigroup $S$. If $x \in \id{A}$, then we say that $x$ has \emph{length} $m$ (with respect to $A$) if $x$ can be written as the product of $m$ elements (not necessarily distinct) from $A$ and cannot be written as the product of any smaller number of elements from $A$. Thus, $x$ has length $m$ if and only if 

(1) $x = x_1 x_2 \cdots x_m$ ($x_i \in A, 1 \leq i \leq m$)

(2) $x = y_1 y_2 \cdots y_n$ ($y_i \in A, 1 \leq i \leq n$) implies $n \geq m$.

If $x$ has length $m$ with respect to some set $A$, we simply write $|x| = m$; the paricular set $A$ upon which $|x|$ depends will be clear from the context.

If $x,y \in \id{A}$ and $|x| + |y| = |xy|$, we say that $x$ is a \emph{left segment} of $xy$. (Here we do \emph{not} allow $|x| = 0$.) If $x,y,z \in \id{A}$ and $|x| + |y| + |z| = |xyz|$, we say that $y$ is a \emph{segment} of $xyz$. (Here we \emph{do} allow $|x| = 0$ or $|z| = 0$.)

\section{} \label{sec: 2} 

In this section we prove the theorem for the case where $T$ is a finite group. The proof is by a rather curious contradiction. We deny the theorem and then apply a kind of sieve to produce an element whose image under $\varphi$ is not in the range of $\varphi$.

\begin{lemma} \label{lemma: 1}
Let $T$ be a group with identity $e$ and let $\varphi: S \mapsto T$ be a homomorphism of the semigroup $S$ onto $T$ such that $e\varphi^{-1}$ is locally finite. Let $A$ be any finite subset of $S$ and let $m$ be any positive integer. Then there exists an integer $k = k(m,A)$ with the following properties: If $x \in \id{A}$, $|x| = km$ (length with repect to $A$), and $g \in T$, then $x = P(g)Q(g)R(g)$, where $P(g), Q(g),R(g) \in \id{A}$, $|Q(g)| = m$, and $(P(g)U)\varphi \ne g$ for every left segment $U$ of $Q(g)$.
\end{lemma}

\begin{proof} Let $x \in \id{A}$, $|x| = km$, $x = A_1A_2 \cdots A_k$, $A_i \in \id{A}$, $|A_i| = m$, $1 \leq i \leq k$, $g \in T$. We will show that if $k$ is taken large enough, then some $A_i$ may be chosen as $Q(g)$. For suppose the contrary. Then for arbitrarily large $k$ we can find $x \in \id{A}$ such that $|x| = km$, $x = A_1A_2 \cdots A_k$, $A_i \in \id{A}$, $|A_i| = m$, $A_i = B_iC_i$, $|B_i| + |C_i|= m$, $1 \leq i \leq k$, and $(B_1)\varphi = (B_1C_1B_2)\varphi = \cdots = (B_1C_1\cdots B_{k-1}C_{k-1}B_k) \varphi = g$. Then $C_1B_2, C_2B_3, \dots, C_{k-1}B_k \in e\varphi^{-1}$, and since $|B_i| + |C_i| = m$, we have $|C_iB_{i+1}| \leq 2m$.

Now let $H$ be the subsemigroup of $e\varphi^{-1}$ generated by the finite set 
\[ \{y \in \id{A} \cap e\varphi^{-1}: |y| \leq 2m\}. \]
Then $H$ is finite since $e\varphi^{-1}$ is locally finite. Furthermore, $H$ depends only on $A$ and $m$. Thus there exists a number $M = M(A,m)$ such that 
\[ z \in H \Rightarrow |z| \leq M.\]
(Note we are still writing lengths with respect to the set $A$.)

The point of constructing $H$ is that 
\[ W = C_1B_2C_2B_3 \cdots C_{k-1}B_k \in H, \]
hence $|W| \leq M$. But then
\[ km = |x| = |B_1WC_k| \leq |B_1| + |W| + |C_k| \leq M + 2m, \]
or $(k - 2)m \leq M$.

Since $M$ does not depend on $k$, this is a contradiction for sufficiently large $k$. This proves Lemma~\ref{lemma: 1}. 
\end{proof}

\begin{lemma} \label{lemma: 2}
Let $T$ be a group with identity $e$, and let $\varphi: S \mapsto T$ be a homomorphism of the semigroup $S$ onto $T$ such that $e \varphi^{-1}$ is locally finite. Let $V \subset T$, $g \in T$, $g \notin V$. Let $A$ be a finite subset of $S$, and suppose that $x \in \id{A}$, $x = P_1Q_1R_1$, where $Q_1 \in \id{A}$, $|Q_1| = k(m,A).m$ (see Lemma~\ref{lemma: 1}), and $(P_1U_1)\varphi \notin V$ for every left segment $U_1$ of $Q_1$. 

Then $x = P_1P_2Q_2R_2R_1$, where $Q_2 \in \id{A}$, $|Q_2| = m$, and $(P_1P_2U_2) \varphi \notin V \cup \{g\}$ for every left segment $U_2$ of $Q_2$.
\end{lemma}

\begin{proof} Let $h = (P\varphi)^{-1}g$. Since $|Q_1| = k(m,A).m$, by Lemma~\ref{lemma: 1} we have $Q_1 = P_2Q_2R_2$, where $Q_2 \in \id{A}$, $|Q_2| = m$, and $(P_2U_2)\varphi \ne h$ for every left segment $U_2$ of $Q_2$. Now if $U_2$ is a left segment of $Q_2$, then $P_2U_2$ is a left segment of $Q_1$, therefore by the hypotheses of the present lemma $(P_1P_2U_2) \varphi \notin V$. But also $(P_1P_2U_2) \varphi \ne g$, for otherwise we would have $(P_2U_2)\varphi = (P_1\varphi^{-1})g = h$, a contradiction. Therefore $(P_1P_2U_2) \varphi \notin V \cup \{g\}$, as required.
\end{proof}

\begin{lemma} \label{lemma: 3}
Let $T$ be a finite group with identity $e$, and let $\varphi: S \mapsto T$ be a homomorphism of the semigroup $S$ onto $T$ such that $e \varphi^{-1}$ is locally finite. Then $S$ is locally finite.
\end{lemma}

\begin{proof} Let $T$ have $n$ elements $\{g_1,\dots,g_n\}$. Let $A$ be a finite subset of $S$. In the notation of Lemma~\ref{lemma: 1}, let $k_0 = 1, k_1 = k(k_0,A)$, $k_2 = k(k_0k_1, A),\dots,$ $k_n = k(k0k_1 \cdots k_{n - 1}, A)$. We shall show that $x \in \id{A}$ implies $|x| < k_0k_1 \cdots k_n$. This of course means that $\id{A}$ is finite, and so $S$ is locally finite.

To prove our assertion, suppose $x \in \id{A}$, $|x| \geq k_0k_1 \cdots k_{n-1} k_n$. We may as well assume that $|x| = k_0 k_1 \cdots k_{n -1} k_n$. Then 
\[ |x| = k_n.(k_0k_1 \cdots k_{n-1}) = k(k_0k_1 \cdots k_{n-1}, A).(k_0k_1 \cdots k_{n-1}), \]
so by Lemma~\ref{lemma: 1} $x = P_1Q_1R_1$, where $Q_1 \in \id{A}$, $|Q_1| = k_0k_1 \cdots k_{n-1}$, and $(P_1Q_1) \varphi \ne g$ for every left segment $U_1$ of $Q_1$.

Now suppose we have $x = P_1 \cdots P_m Q_mR_m \cdots R_1$, where $Q_m \in \id{A}$, $|Q_m| = k_0k_1 \cdots k_{n-m}$ $= k_{n-m} .(k_0k_1 \cdots k_{n - m - 1}) = k(k_0 k_1 \cdots k_{n - m - 1}, A) . (k_0k_1 \cdots k_{n - m})$, and 
\[ (P_1 \cdots P_mU_m) \varphi \notin \{g_1, \dots,g_m\}\]
for every left segment $U_m$ of $Q_m$. 

We now use Lemma~\ref{lemma: 2} to sieve out the element $g_{m + 1}$, and obtain 
\[ x = P_1 \cdots P_{m + 1} Q_{m + 1} R_{m + 1} \cdots R_1, \]
where $Q_{m + 1} \in \id{A}$, $|Q_{m + 1}| = k_0 k_1 \cdots k_{n - m - 1}$, and $(P_1 \cdots P_{m + 1}) \varphi \notin \{g_1,\dots,g_{m + 1}\}$ for every left segment $U_{m + 1}$ of $Q_{m + 1}$.

Thus after $n$ steps we have $x = P_1 \cdots P_n Q_n R_n \cdots R_1$, where $Q_n \in \id{A}$, $|Q_n| = k_0 = 1$ (so that $Q_n$ has exactly one left segment, namely $Q_n$), and $(P_1 \cdots P_n Q_n) \varphi \notin \{g_1, \dots,g_n\} = T$. Since $\varphi$ is after all a mapping of $S$ into $T$, this is a contradiction, and completes the proof.
\end{proof}

\section{} \label{sec: 3}

We now consider the cases wherer $T$ is either a finite group with zero or a finite null semigroup.

Let $S, \varphi, T$ be as in the theorem, and suppose also that $T$ is a finite group with zero. Let $A = (T\setminus \{0\})\varphi^{-1}$; then $A$ is locally finite by Lemma~\ref{lemma: 3}. Also, by assumption, $B = 0\varphi^{-1}$ is locally finite. Thus we have $S = A \cup B$, where $A,B$ are locally finite and $B$ is an ideal in $S$. It is easy to see in this case that $S$ is locally finite. In the case that $T$ is a finite null semigroup, again letting $B = 0\varphi^{-1}$, we have that $B$ is a locally finite ideal in $S$ and $S^2 \subset B$. Here again it is easy to see that $S$ is locally finite. We summarize these cases as

\begin{lemma} \label{lemma: 4}
Let $T$ be either a finite group with zero or a finite null semigroup, and let $\varphi: S \mapsto T$ be a homomorphism of the semigroup $S$ onto $T$ such that $e \varphi^{-1}$ is locally finite for each idempotent element $e$ of $T$. Then $S$ is locally finite.
\end{lemma}

\section{} \label{sec: 4}

In this section we consider the remaining cases, where $T$ is either a (finite) simple semigroup or 0-simple semigroup. 

Let $T$ be a finite simple of finite 0-simple semigroup. Then $T$ is completely simple or completely 0-simple, and it follows in a standard way (\cite{clifford+preston1961}) that if $a,b \in T$ and $TabT \ne \{0\}$, then $bTa$ is a finite group or a finite group with zero. This fact will be used in what follows.

The next lemma is a variation on Lemma~\ref{lemma: 1}. Its proof is briefly sketched.

\begin{lemma} \label{lemma: 5}
Let $T$ be a finite simple or finite 0-simple semigroup, and let $\varphi: S \mapsto T$ be a homomorphism or the semigroup $S$ onto $T$ such that $e \varphi^{-1}$ is locally finite for each idempotent element $e$ of $T$. Let $A$ be any finite subset of $S$ and let $m$ be any positive integer. Let $a,b$ be element of $A$ such that $|ab| = 2$ and $T((ab)\varphi)T \ne \{0\}$. Then there exists an integer $k = k(ab,m,A)$ with the following properties: If $x \in \id{A}$, $|x| = km$, then $x = PQR$ where $Q \in \id{A}$, $|Q| = m$, and $ab$ is not a segment of $Q$.
\end{lemma}

\begin{proof} Assume the contrary. Then for arbitrarily large $k$ we can find $x \in \id{A}$ such that $|x| = km$, $x = A_1A_2 \cdots A_k$, $A_i \in \id{A}$, $|A_i| = m$, $A_i = B_1abC_i$, $|B_i| + 2 + |C_i| = m$, $1 \leq i \leq k$. Thus $x = B_1aybC_k$, where
\[ y = \prod_{i = 1}^{k-1} (bC_iB_{i + 1}a). \]
Let $G = (b\varphi)T(a\varphi)$. Then $G$ is a finite group or a finite group with zero, and so $G\varphi^{-1}$ is locally finite by Lemmas~\ref{lemma: 3} and~\ref{lemma: 4}. But $y$ belongs to a finitely generated subsemigroup of $G\varphi^{-1}$, hence $|y|$ is bounded above by a number which depends only on $a,b,m,$ and $A$, hence $|x|$ is similarly bounded. For sufficiently large $k$ this contradicts $|x| = km$.
\end{proof}

\begin{lemma} \label{lemma: 6}
Let $T$ be a finite simple or finite 0-simple semigroup. Let $\varphi: S \mapsto T$ be a homomorphism of the semigroup $S$ onto $T$ such that $e \varphi^{-1}$ is locally finite for each idempotent element $e$ of $T$. The $S$ is locally finite.
\end{lemma}

\begin{proof} Let $A$ be a finite subset of $S$. Let 
\begin{align*} 
B &= \{xy: x,y \in A, |xy| = 2, T((xy)\varphi)T \ne\{0\}\}, \\
C &= \{xy: x,y \in A, |xy| = 2, T((xy)\varphi)T = \{0\}\}, \\
D &= \{a_1b_1, \dots,a_pb_p\}.
\end{align*}
(Note that if $T$ is simple then $C$ is empty.)

We now assume that $\id{A}$ is infinite and proceed in two steps:

(i) We show that $\id{A}$ must contain elements of arbitrarily large lengths which contain no element of $B$ as a segment. 

(ii) Using (i), we obtain a contradiction.

(i) Let $m$ be an arbitrary positive integer. Using the notation of Lemma~\ref{lemma: 5}, let 
\begin{align*} 
k_0 &= m, \;\;\; k_1 = k(a_1b_1, k_0, A), \\
k_2 &= k(a_2b_2, k_0k_1, A), \dots, \\
k_p &= k(a_pb_p, k_0 \cdots k_{p - 1}, A). \end{align*}
By finite induction, as in Lemma~\ref{lemma: 3}, it follows that if $x \in \id{A}$, $|x| = k_0 \cdots k_p$, then $x$ contains a segment $R$, $R \in \id{A}$, $|R| = m$, such that no element of $B$ is a segment of $R$.

(ii) First suppose that $T$ is simple, so that $C$ is empty. By (i), setting $m =2$, there is $R \in \id{A}$, $|R| = 2$, such that $R$ contains no element of $B$ as a segment, that is $R \notin B$. But $R \in B \cup C$, and $C$ is empty. This case is finished.

Now suppose that $T$ is 0-simple. By (i), for arbitrarily large $m$ we have $R \in \id{A}$, $|R| = m$, and no element of $B$ is a segment of $R$. Then we can write $R = R'y$, where $y = \prod_{i=1}^t (x_iy_iz_iw_i)$, $R' \in \id{A}$, $|R'| < 4$, $x_i,y_i,z_i,w_i \in A$, $|y_iz_i| = 2$, $1 \leq i \leq t$.

Then $y_iz_i \in C$, or $T((y_iz_i)\varphi)T = \{0\}$, therefore in particular $(x_iy_iz_iw_i)\varphi = 0$. Hence $y$ is an element of a finitely generated (hence finite) subsemigroup of the locally finite semigroup $0\varphi^{-1}$, and so $|y|$ is bounded above. This contradicts the statement that $|R|$ is not bounded above. This finishes Lemma~\ref{lemma: 6}, and the proof of the main theorem is complete. 
\end{proof} 

\section{} \label{sec: 5}

In this section we are concerned with bands of locally finite semigroups. Suppose that a semigroup $S$ is the disjoint union of certain subsemigroups $S_\alpha$ $(\alpha \in I)$, $I$ an index set. Suppose further that for any pair $\alpha, \beta$, of elements of $I$ there is an element $\gamma$ in $I$ such that $S_\alpha S_\beta \subset S_\gamma$. Then $S$ is a \emph{bnad} of the semigroups $S_\alpha$. Evidently $I$ becomes an idempotent semigroup if we define $\alpha \beta = \gamma$ if and only if $S_\alpha S_\beta \subset S_\gamma$,
and $\varphi: S \mapsto I$ is a homomorphism, where $x\varphi = \alpha$ if $x \in S_\alpha$. $S$ is then called an \emph{$I$-band} of the semigroups $S_\alpha$. If each $S_\alpha$ is locally finite, then $S$ is called simply an $I$-band of locally finite semigroups.

Several people (\cite{green+rees1952,lazerson1961,mclean1954}) have shown idependently that an idempotent semigroup is locally finite. Thus from this and our main theorem follows the important result of Shevrin~\cite{shevrin1965}.

\begin{thm} Any band of locally finite semigroups is locally finite.
\end{thm}

The author received in a personal communication from B. M. Schein an extremely short and direct proof of Shevrin's theorem which is outlined as follows: Let $A$ be the two-element right zero semigroup, let $B$ be the multiplicative semigroup $\{0,1\}$, let $C$ be a right zero or left zero semigroup, let $D$ be a rectangular band, and let $E$ be a semilattice. It is shown that an $X$-band of locally finite semigroups is locally finite, where $X$ is successively $A,B,C,D,E,$ and Shevrin's theorem then follows since any band of locally finite semigroups is an $E$-band of $D$-bands of locally finite semigroups (\cite{clifford1954,clifford+preston1961,lyapin1960}).

It is also true that any semigroup which is the union of disjoint locally finite groups is an $E$-band of $D$-bands of locally finite groups, and thus we have the next theorem.

\begin{thm} A semigroup which is the union of locally finite groups is locally finite.
\end{thm}

From this follows the theorem of Green and Rees (\cite{brown1964,green+rees1952}) on the equivalence of Burnside's problem for groups with ``Burnside's problem for semigroups":

\begin{thm} The following two statements are equivalent:

(1) Every group of exponent $n$ is locally finite.

(2) Every semigroup satisfying the identity $x^{n + 1} = x$ is locally finite.
\end{thm}


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