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\title{A Semigroup Union of Disjoint Locally Finite Subsemigroups Which is Not Locally Finite}
\author{T. C. Brown}
\date{}
\maketitle
\begin{center}{\small {\bf Citation data:} T.C. {Brown}, \emph{A semigroup union of disjoint locally finite subsemigroups which
is not locally finite}, Pacific J. Math. \textbf{22} (1967), 11--14.}\bigskip\end{center}
\begin{abstract}
The semigroup $S$ of the title is the free semigroup $F$ on four generators factored by the congruence generated by the set of relations $\{w^2 = w^3: w \in F\}$. The following lemma is proved by examining the elements of a given congruence class of $F$:
\textsc{Lemma.} If $x,y \in S$ and $x^2 = y^2$, then either $xy = x^2$ or $yx = x^2$.
From the Lemma it then easily follows that the (disjoint) subsemigroups $\{y \in S: y^2 = x^2\}$ of $S$ are locally finite.
\end{abstract}
This note answers in the negative a question raised by Shevrin in~\cite{shevrin1965}.
\begin{thm} \label{thm: 1}
There exists a semigroup $S$ with disjoint locally finite subsemigroups $S_e$ such that $S = \bigcup S_e$ and $S$ is not locally finite.
\end{thm}
Let $F$ be the free semigroup with identity on four generators. Let $\sim$ denote the smallest congruence on $F$ containing the set $\{(x^2,x^3): x \in F\}$. That is, for $w,w' \in F$, $w \sim w'$ if and only if a finite sequence of ``transitions", of either of the types $ab^2c \to ab^3c$ or $ab^3c \to ab^2c$, transforms $w$ into $w'$.
The equivalence classes of $F$ with respect to $\sim$ are taken as the elements of $S$, and multiplication in $S$ is defined in the natural way.
There is given in~\cite{dean1965} a sequence on four symbols in which no block of length $k$ is immediately repeated, for any $k$. Thus the left initial segments of this sequence give elements of $F$ containing no squares. Since no transition of the form $ab^2c \to ab^3c$ or $ab^3c \to ab^2c$ can be applied to an element of $F$ contining no squares, the equivalence classes containing these elements consist of precisely one element each; thus the semigroup $S$ is infinite, and hence not locally finite.
In what follows, the symbols $\alpha, \alpha_1,\alpha_2, \dots$ refer to transformations (on elements of $F$) of the form
\[ ab \to ayb, \quad \textup{where $a \sim ay$,} \quad \textup{and} \quad a,b,y \in F. \]
The symbols $\beta,\beta_1,\beta_2,\dots$ refer to transformations of the type
\[ axb \to ab, \quad \textup{where $a \sim ax$,} \quad \textup{and} \quad a,b,x \in F. \]
Note that $ab^2c \to ab^3c$ is an $\alpha$, and $ab^3c \to ab^2c$ is a $\beta$.
\begin{lemma} \label{lemma: 1}
If $w,w' \in F$, and $w\beta\alpha = w'$, then there are $\alpha_1,\beta_1$ such that $w\alpha_1\beta_1 = w'.$
\end{lemma}
\begin{proof}
Let
\[ w = axb, \quad w\beta = ab, \quad \textup{where $a \sim ax$.} \]
Let
\[ w\beta = AB, \quad w\beta\alpha = AyB, \quad \textup{where $A \sim Ay$.} \]
There are two cases:
(i) $A$ is contained in $a$. That is,
\[ a = Aa' \quad \textup{and} \quad w' = wb\alpha = a\beta \alpha = Aa'b\alpha = Aya'b. \]
Here let
\[ w\alpha_1 = axb\alpha_1 = Aa'xb\alpha_1 = Aya'xb. \]
Now since
\[ Aya' \sim Aa' = a \sim ax = Aa'x \sim Aya'x, \]
we may let
\[ w\alpha_1\beta_1 = Aya'xb\beta_1 = Aya'b = w'. \]
(ii) $A$ is not contained in $a$. That is,
\[ b = b_1b_2, \quad A = ab_1, \quad A \sim Ay, \]
and
\[ w' = w\beta \alpha = ab\alpha = ab_1b_2\alpha = Ab_2\alpha = Ayb_2 = ab_1yb_2. \]
Since
\[ axb_1 \sim ab_1 = A \sim Ay = ab_1y \sim axb_1y, \]
we may let
\[ w\alpha_1 = axb\alpha_1 = axb_1b_2\alpha_1 = axb_1yb_2, \]
and
\[ w\alpha_1\beta_1 = axb_1yb_2\beta_1 = ab_1yb_2 = w'. \qedhere \]
\end{proof}
\begin{lemma} \label{lemma: 2}
If $w,w' \in F$, $w\gamma_1\gamma_2\cdots \gamma_m = w'$, where each $\gamma_i$ is either an $\alpha$ or a $\beta$, then there are $\alpha_1,\dots,\alpha_n, \beta_1,\dots,\beta_k$ such that $w\alpha_1 \cdots \alpha_n \beta_1 \cdots \beta_k = w'$.
\end{lemma}
\begin{proof} This follows immediately from Lemma~\ref{lemma: 1} by induction.
\end{proof}
\begin{lemma} \label{lemma: 3}
The word $ab\alpha$ contains a left initial segment which is equivalent to $a$.
\end{lemma}
\begin{proof} Let $ab = AB$, $ab\alpha = AyB$, where $A \sim Ay$. Again there are two cases:
(i) $A$ is contained in $a$. That is, $a = Aa'$, $ab\alpha = Aa'b\alpha = Aya'b$. Since $A \sim Ay$, the left initial segment $Aya'$ of $ab\alpha$ is equivalent to $a$.
(ii) $A$ is not contained in $a$. That is, $b = b_1b_2$, $A = ab_1$, $ab\alpha = ab_1b_2\alpha = ab_1yb_2$. Here, $a$ itself is a left initial segment of $ab\alpha$, and is certainly equivalent to $a$.
\end{proof}
\begin{lemma} \label{lemma: 4}
If $x,y \in F$ and $x^2 \sim y^2$, then either $y \sim xa$ for some $a \in F$, or $x \sim yb$ for some $b \in F$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma: 2}, thee are $\alpha_i$ and $\beta_j$ such that $xx\alpha_1 \cdots \alpha_m \beta_1 \cdots \beta_n = yy$. Let $w = xx\alpha_1\cdots \alpha_m= yy\beta_n^{-1} \cdots \beta_1^{-1}$. By Lemma~\ref{lemma: 3}, $w$ contains a left initial segment $A$ equivalent to $x$. Similarly, since each $\beta_i^{-1}$ is an $\alpha$, $w$ also contains a left initial segment $B$ equivalent to $y$. Depending on which segment contains the other, either $B = Aa$ for some $a$, or $A = Bb$ for some $b$. In the first case, $y \sim B = Aa \sim xa$; in the second, $x \sim A = Bb \sim yb$.
\end{proof}
\begin{lemma} \label{lemma: 5}
In this lemma, ``=" will denote equality in $S$. Let $e$ be an idempotent element of $S$: $e = e^2$. Let $S_e = \{x \in S: x^2 = e\}$. Then $S_e$ is a locally finite subsemigroup of $S$.
\end{lemma}
\begin{proof} First, we note that $z \in S_e$ implies $ez = ze = e$. For $ez = z^2z = z^2 = e$, and similarly $ze = e$. Now let $x,y \in S_e$, that is, $x^2 = y^2 = e$. By Lemma~\ref{lemma: 4}, either $y = xa$ or $x = yb$. In the first case, we obtain
\[ xy = x^2a = x^3a = x^2y = ey = e. \]
In the second case, we obtain similarly that $yx = e$. Thus $x, y \in S_e$ implies $xy = e$ or $yx = e$. In either case, $(xy)^2 = e$, that is, $xy \in S_e$. Thus $S_e$ is a semigroup.
Now let $x_1,\dots,x_n \in S_e$, and let $\langle x_1,\dots,x_n \rangle$ denote the subsemigroup of $S_e$ generated by $x_1,\dots,x_n$. If $n = 1$, then $\langle x_1 \rangle$ is clearly finite; so suppose $n > 1$. Then every element of $\langle x_1,\dots,x_n \rangle$ may be expressed as a product of not more than $n$ of the $x_i$'s. For any product $z$ of more than $n$ $x_i$'s must contain some $x_i$ twice: $z = ax_ibx_ic$, where $a,b,c \in S_e$. Since either $x_ib = e$ or $bx_i = e$, it follows that $x_ibx_i = e$ and $z = aec = ec = e = x_1x_1$. This shows that $\langle x_1,\dots,x_n \rangle$ is finite, and hence that $S_e$ is locally finite.
\end{proof}
The theorem follows immediately from Lemma~\ref{lemma: 5}, since clearly $e \ne e'$ implies that $S_e$ and $S_{e'}$ are disjoint, and
\[ S = \cup S_e. \]
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