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\ihead{Errata to ``On ... finite $\mathcal{J}$-trivial monoids''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{On the representation theory of finite }$\mathcal{J}$\textbf{-trivial
monoids}

\textit{Tom Denton, Florent Hivert, Anne Schilling and Nicolas M. Thi\'{e}ry}

\href{https://arxiv.org/abs/1010.3455v3}{arXiv:1010.3455v3}

\texttt{denton hivert schilling thiery - rep theory of J-trivial monoids -
1010.3455v3.pdf}

version of 4 March 2011

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the article \textquotedblleft On the
representation theory of finite $\mathcal{J}$-trivial
monoids\textquotedblright\ by the numbers under which they appear in this
article (specifically, in its version of 4 March 2011, posted on arXiv under
the identifier arXiv:1010.3455v3).

\setcounter{section}{5}

\section{Errata}

\begin{itemize}
\item \textbf{Page 3:} Replace \textquotedblleft and illustrates
them\textquotedblright\ by \textquotedblleft and illustrate
them\textquotedblright.

\item \textbf{Page 4:} \textquotedblleft satistfies\textquotedblright\ should
be \textquotedblleft satisfies\textquotedblright.

\item \textbf{Page 4:} Here it is claimed that \textquotedblleft Groups are an
example of a variety of monoids, as are all of the classes of monoids
described in this paper\textquotedblright. The latter part of this sentence is
not completely true: The class of all ordered monoids with $1$ on top is not a
variety, since a quotient of such a monoid can fail to be in this class
(unless, I assume, you regard it as a variety of ordered monoids). An example
is provided by Examples 2.4 and 2.5 in this very paper (since the monoid $M$
there is $\mathcal{J}$-trivial and thus is a quotient of an ordered monoid
with $1$ on top, but itself is not ordered).

\item \textbf{Page 5:} Replace \textquotedblleft$1$ is the largest element of
these (pre)-orders\textquotedblright\ by \textquotedblleft in every of these
preorders, we have $x\leq1$ for every $x\in M$\textquotedblright. (Speaking of
\textquotedblleft the largest element\textquotedblright\ is mildly ambiguous,
because a preorder can have several elements each of which is
\textquotedblleft the largest\textquotedblright.)

\item \textbf{Page 5, Proposition 2.2:} Replace \textquotedblleft the partial
order $\leq$ is finer than $\leq_{\mathcal{K}}$\textquotedblright\ by
\textquotedblleft the partial order $\leq_{\mathcal{K}}$ is finer than $\leq
$\textquotedblright.

\item \textbf{Page 6, \S 2.1:} Here it is claimed that \textquotedblleft any
$\mathcal{R}$-trivial monoid can be represented as a monoid of regressive
functions on some finite poset $P$\textquotedblright. This is correct if one
uses the convention that functions act on values \textbf{from the right}
(i.e., the value of an element $x$ under a function $f$ is written $x.f$
rather than $f\left(  x\right)  $) and compose accordingly (i.e., the
composition $fg$ of two functions $f$ and $g$ is the function which sends
every $x$ to $\left(  x.f\right)  .g$). This is a nonstandard convention, and
ought to be explained early on in the paper. (It is explained in \S 2.5, but
it is used in \S 2.1 already.)

\item \textbf{Page 8, \S 2.5:} In the sentence starting with \textquotedblleft
When $P$ is a chain on $N$ elements\textquotedblright, as well as in the next
sentence, replace each \textquotedblleft$N$\textquotedblright\ by
\textquotedblleft$n$\textquotedblright.

\item \textbf{Page 11, proof of Lemma 3.6:} This proof can be simplified:

Assume that $e\leq_{\mathcal{J}}y$. Then, $e=ayb$ for some $a,b\in M$.
Applying Lemma 3.5 to $yb$ instead of $b$, we find $e=eyb$. Hence,
$e\leq_{\mathcal{J}}ey\leq_{\mathcal{J}}e$, which entails $e=ey$ since $M$ is
$\mathcal{J}$-trivial. Moreover, applying Lemma 3.5 to $ay$ instead of $a$, we
find $e=aye$. Hence, $e\leq_{\mathcal{J}}ye\leq_{\mathcal{J}}e$, thus $e=ye$.
The converse implications hold by the definition of $\leq_{\mathcal{J}}$.

\item \textbf{Page 12, proof of Corollary 3.8:} It also needs to be proved
that $\mathcal{C}\subseteq\operatorname*{rad}\mathbb{K}M$. This is a
consequence of Corollary 3.7: Indeed, Corollary 3.7 shows that the quotient
algebra $\mathbb{K}M/\operatorname*{rad}\mathbb{K}M$ is commutative (being
isomorphic to the semigroup algebra $\mathbb{K}E\left(  M\right)  $ of the
commutative monoid $\left(  E\left(  M\right)  ,\star\right)  $), and thus the
commutator ideal $\mathcal{C}$ of $\mathbb{K}M$ must be annihilated by the
canonical projection $\mathbb{K}M\rightarrow\mathbb{K}M/\operatorname*{rad}%
\mathbb{K}M$. But this means that this ideal $\mathcal{C}$ is contained in
$\operatorname*{rad}\mathbb{K}M$.

\item \textbf{Page 12, Example 3.9:} Replace \textquotedblleft$\mathbb{C}%
$\textquotedblright\ by \textquotedblleft$\mathbb{K}$\textquotedblright%
\ throughout this example.

\item \textbf{Page 12, Example 3.9:} Replace \textquotedblleft algebra
morphisms from $H_{0}\left(  W\right)  \rightarrow H_{0}\left(  W_{I\setminus
\left\{  i\right\}  }\right)  $\textquotedblright\ by \textquotedblleft
algebra morphisms from $\mathbb{K}H_{0}\left(  W\right)  \rightarrow
\mathbb{K}H_{0}\left(  W_{I\setminus\left\{  i\right\}  }\right)
$\textquotedblright.

\item \textbf{Page 13, (3.7):} It is worth saying that $g_{e}$ is understood
to lie in $\mathbb{K}M$ (not in $\mathbb{K}M/\operatorname*{rad}\mathbb{K}M$).

\item \textbf{Page 14:} \textquotedblleft For any number $a$ denote by
$\left\lceil a\right\rceil $ the smallest integer larger than $a$%
.\textquotedblright\ should be \textquotedblleft For any number $a$ denote by
$\left\lceil a\right\rceil $ the smallest integer larger than or equal to
$a$.\textquotedblright. (Otherwise, the statement \textquotedblleft there is
an $N$ such that $u_{N}=1$\textquotedblright\ in the proof of Proposition 3.12
does not hold.)

\item \textbf{Page 14, proof of Proposition 3.12:} After \textquotedblleft
Define $u_{n+1}=\left\lceil \dfrac{u_{n}}{2}\right\rceil $\textquotedblright,
add \textquotedblleft, so that Lemma 3.14 yields $\left(  y_{n}\left(
y_{n}-1\right)  \right)  ^{u_{n}}=0$ by induction on $n$\textquotedblright.

\item \textbf{Page 15, proof of Proposition 3.15:} In the first sentence
(\textquotedblleft First it is clear that the $f_{i}$ are pairwise orthogonal
idempotents\textquotedblright), remove the word \textquotedblleft
idempotents\textquotedblright. Indeed, the idempotency of the $f_{i}$ will
only be shown later.

Namely, the idempotency of the $f_{i}$ follows from the equality (3.17) proved
in the next paragraph. Indeed, this equality shows that the element
\newline$\phi\left(  \left(  1-\sum_{j<i}f_{i}\right)  g_{j}\left(
1-\sum_{j<i}f_{i}\right)  \right)  $ is idempotent (since $g_{j}$ is
idempotent). Therefore, $P\left(  \left(  1-\sum_{j<i}f_{i}\right)
g_{j}\left(  1-\sum_{j<i}f_{i}\right)  \right)  $ must be idempotent as well
(since $P\left(  x\right)  $ is idempotent whenever $\phi\left(  x\right)  $
is idempotent).

\item \textbf{Page 15, proof of Proposition 3.15:} In the last sentence, the
claim that \textquotedblleft the coefficient of $e_{i}$ in $f_{i}$ must be
$1$\textquotedblright\ doesn't look that obvious to me. I understand why this
coefficient equals the coefficient of $e_{i}$ in $P\left(  g_{i}\right)  $
(because we have $\left(  1-\sum_{j<i}f_{i}\right)  g_{j}\left(  1-\sum
_{j<i}f_{i}\right)  \equiv g_{i}$ modulo the ideal $\operatorname*{span}%
\left\{  x\ \mid\ x<_{\mathcal{J}}e_{i}\right\}  $) and thus also equals the
coefficient of $e_{i}$ in $P\left(  e_{i}\right)  $ (since $g_{i}\equiv e_{i}$
modulo the same ideal). But in order to see that the latter coefficient is
$1$, I need to use the fact that $P\left(  x\right)  =x$ whenever $x$ is
idempotent. This is itself quite easy, but should be stated as a lemma.

\item \textbf{Page 16, Proposition 3.17:} I think \textquotedblleft
decreasing\textquotedblright\ should be \textquotedblleft(weakly)
increasing\textquotedblright\ both times here.

\item \textbf{Page 17, proof of Theorem 3.23:} \textquotedblleft are two
ideals\textquotedblright\ $\rightarrow$ \textquotedblleft are two right
ideals\textquotedblright. (Or is there a reason why they are two-sided ideals?)

\item \textbf{Page 21:} \textquotedblleft An algebra is called \textit{split
basic}\textquotedblright\ $\rightarrow$ \textquotedblleft An algebra $A$ is
called \textit{split basic}\textquotedblright.
\end{itemize}

\section{Addenda and remarks}

\subsection{Page 6, \S 2.1.}

Let us prove some of the claims that are left unproven in this section.

First of all, the following very easy fact is used without proof:

\begin{lemma}
\label{lem.p6.xomega}Let $M$ be a monoid\footnotemark. Let $x\in M$. Let
$N\in\mathbb{N}$ be such that $x^{N}=x^{N+1}$. Then, $x^{N}=x^{N+1}%
=x^{N+2}=\cdots$.
\end{lemma}

\footnotetext{We follow the conventions of the paper. In particular,
\textquotedblleft monoid\textquotedblright\ means \textquotedblleft finite
monoid\textquotedblright\ for us.}

\begin{proof}
[Proof of Lemma \ref{lem.p6.xomega}.]We have
\begin{equation}
x^{N+m}=x^{N+m+1}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}.
\label{pf.lem.p6.xomega.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.p6.xomega.1}):} We shall prove
(\ref{pf.lem.p6.xomega.1}) by induction over $m$:
\par
\textit{Induction base:} We have $x^{N+0}=x^{N}=x^{N+1}=x^{N+0+1}$ (since
$N+1=N+0+1$). In other words, (\ref{pf.lem.p6.xomega.1}) holds for $m=0$. This
completes the induction base.
\par
\textit{Induction step:} Let $\mu$ be a positive integer. Assume that
(\ref{pf.lem.p6.xomega.1}) holds for $m=\mu-1$. We now need to prove that
(\ref{pf.lem.p6.xomega.1}) holds for $m=\mu$ as well.
\par
We know that (\ref{pf.lem.p6.xomega.1}) holds for $m=\mu-1$. In other words,
$x^{N+\left(  \mu-1\right)  }=x^{N+\left(  \mu-1\right)  +1}$. Now,
$x^{N+\left(  \mu-1\right)  }x=x^{N+\left(  \mu-1\right)  +1}=x^{N+\mu}$
(since $N+\left(  \mu-1\right)  +1=N+\mu$). Hence,%
\[
x^{N+\mu}=\underbrace{x^{N+\left(  \mu-1\right)  }}_{\substack{=x^{N+\left(
\mu-1\right)  +1}=x^{N+\mu}\\\text{(since }N+\left(  \mu-1\right)
+1=N+\mu\text{)}}}x=x^{N+\mu}x=x^{N+\mu+1}.
\]
In other words, (\ref{pf.lem.p6.xomega.1}) holds for $m=\mu$. This completes
the induction step. Thus, the induction proof of (\ref{pf.lem.p6.xomega.1}) is
complete.} Combining these equalities for all $m\in\mathbb{N}$, we obtain
$x^{N}=x^{N+1}=x^{N+2}=\cdots$. This proves Lemma \ref{lem.p6.xomega}.
\end{proof}

Next, we shall prove a fact which is almost obvious and is often used without
explicit mention:

\begin{lemma}
\label{lem.p6.implications}Let $M$ be a monoid.

\textbf{(a)} If $M$ is $\mathcal{J}$-trivial, then $M$ is $\mathcal{R}$-trivial.

\textbf{(b)} If $M$ is $\mathcal{J}$-trivial, then $M$ is $\mathcal{L}$-trivial.

\textbf{(c)} If $M$ is $\mathcal{R}$-trivial, then $M$ is $\mathcal{H}$-trivial.

\textbf{(d)} If $M$ is $\mathcal{L}$-trivial, then $M$ is $\mathcal{H}$-trivial.

\textbf{(e)} If $M$ is $\mathcal{J}$-trivial, then $M$ is $\mathcal{H}$-trivial.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.p6.implications}.]\textbf{(a)} Assume that $M$ is
$\mathcal{J}$-trivial. Then, all $\mathcal{J}$-classes are of cardinality one.
In other words, any two $\mathcal{J}$-equivalent elements of $M$ are
identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{J\ }y\nonumber\\
&  \text{ must satisfy }x=y. \label{pf.lem.p6.implications.a.1}%
\end{align}


Now, let $x$ and $y$ be two elements of $M$ satisfying $x\ \mathcal{R}\ y$.
Then, $xM=yM$ (since $x\ \mathcal{R}\ y$ holds if and only if $xM=yM$). Hence,
$M\underbrace{xM}_{=yM}=MyM$. In other words, $x\ \mathcal{J}\ y$ (since
$x\ \mathcal{J}\ y$ holds if and only if $MxM=MyM$). Therefore, $x=y$ (by
(\ref{pf.lem.p6.implications.a.1})).

Let us now forget that we fixed $x$ and $y$. We thus have shown that every two
elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{R\ }y$ must satisfy $x=y$.
In other words, any two $\mathcal{R}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{R}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{R}$-trivial. This proves Lemma
\ref{lem.p6.implications} \textbf{(a)}.

\textbf{(b)} Assume that $M$ is $\mathcal{J}$-trivial. Then, all $\mathcal{J}%
$-classes are of cardinality one. In other words, any two $\mathcal{J}%
$-equivalent elements of $M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{J\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.implications.b.1}%
\end{align}


Now, let $x$ and $y$ be two elements of $M$ satisfying $x\ \mathcal{L}\ y$.
Then, $Mx=My$ (since $x\ \mathcal{L}\ y$ holds if and only if $Mx=My$). Hence,
$\underbrace{Mx}_{=My}M=MyM$. In other words, $x\ \mathcal{J}\ y$ (since
$x\ \mathcal{J}\ y$ holds if and only if $MxM=MyM$). Therefore, $x=y$ (by
(\ref{pf.lem.p6.implications.b.1})).

Let us now forget that we fixed $x$ and $y$. We thus have shown that every two
elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{L\ }y$ must satisfy $x=y$.
In other words, any two $\mathcal{L}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{L}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{L}$-trivial. This proves Lemma
\ref{lem.p6.implications} \textbf{(b)}.

\textbf{(c)} Assume that $M$ is $\mathcal{R}$-trivial. Then, all $\mathcal{R}%
$-classes are of cardinality one. In other words, any two $\mathcal{R}%
$-equivalent elements of $M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{R\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.implications.c.1}%
\end{align}


Now, let $x$ and $y$ be two elements of $M$ satisfying $x\ \mathcal{H}\ y$.
Then, $x\ \mathcal{R}\ y$ and $x\ \mathcal{L}\ y$ (because $x\ \mathcal{H}\ y$
holds if and only if $\left(  x\ \mathcal{R}\ y\text{ and }x\ \mathcal{L}%
\ y\right)  $). Hence, $x=y$ (by (\ref{pf.lem.p6.implications.c.1})).

Let us now forget that we fixed $x$ and $y$. We thus have shown that every two
elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{H\ }y$ must satisfy $x=y$.
In other words, any two $\mathcal{H}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{H}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{H}$-trivial. This proves Lemma
\ref{lem.p6.implications} \textbf{(c)}.

\textbf{(d)} Assume that $M$ is $\mathcal{L}$-trivial. Then, all $\mathcal{L}%
$-classes are of cardinality one. In other words, any two $\mathcal{L}%
$-equivalent elements of $M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{L\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.implications.d.1}%
\end{align}


Now, let $x$ and $y$ be two elements of $M$ satisfying $x\ \mathcal{H}\ y$.
Then, $x\ \mathcal{R}\ y$ and $x\ \mathcal{L}\ y$ (because $x\ \mathcal{H}\ y$
holds if and only if $\left(  x\ \mathcal{R}\ y\text{ and }x\ \mathcal{L}%
\ y\right)  $). Hence, $x=y$ (by (\ref{pf.lem.p6.implications.d.1})).

Let us now forget that we fixed $x$ and $y$. We thus have shown that every two
elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{H\ }y$ must satisfy $x=y$.
In other words, any two $\mathcal{H}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{H}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{H}$-trivial. This proves Lemma
\ref{lem.p6.implications} \textbf{(d)}.

\textbf{(e)} Assume that $M$ is $\mathcal{J}$-trivial. Then, $M$ is
$\mathcal{L}$-trivial (by Lemma \ref{lem.p6.implications} \textbf{(b)}).
Hence, $M$ is $\mathcal{H}$-trivial (by Lemma \ref{lem.p6.implications}
\textbf{(d)}). This proves Lemma \ref{lem.p6.implications} \textbf{(e)}.

Now, Lemma \ref{lem.p6.implications} is proven.
\end{proof}

Next, we show an auxiliary lemma that is nearly trivial:

\begin{lemma}
\label{lem.p6.uxv}Let $M$ be a monoid. Let $x$, $u$ and $v$ be three elements
of $M$ such that $x=uxv$. Then,
\begin{equation}
x=u^{m}xv^{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}.
\label{eq.lem.p6.uxv.claim}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.p6.uxv}.]We will prove (\ref{eq.lem.p6.uxv.claim}) by
induction over $m$:

\textit{Induction base:} We have $\underbrace{u^{0}}_{=1}x\underbrace{v^{0}%
}_{=1}=x$, thus $x=u^{0}xv^{0}$. In other words, (\ref{eq.lem.p6.uxv.claim})
holds for $m=0$. This completes the induction base.

\textit{Induction step:} Let $k\in\mathbb{N}$. Assume that
(\ref{eq.lem.p6.uxv.claim}) holds for $m=k$. We now need to prove that
(\ref{eq.lem.p6.uxv.claim}) holds for $m=k+1$.

We know that (\ref{eq.lem.p6.uxv.claim}) holds for $m=k$. In other words, we
have $x=u^{k}xv^{k}$. Thus,%
\[
x=u^{k}\underbrace{x}_{=uxv}v^{k}=\underbrace{u^{k}u}_{=u^{k+1}}%
x\underbrace{vv^{k}}_{=v^{k+1}}=u^{k+1}xv^{k+1}.
\]
In other words, (\ref{eq.lem.p6.uxv.claim}) holds for $m=k+1$. This completes
the induction step. Thus, the induction proof of (\ref{eq.lem.p6.uxv.claim})
is complete. Hence, Lemma \ref{lem.p6.uxv} is proved.
\end{proof}

Next, let us show a slightly less trivial (but still easy) fact. In fact, we
are going to prove the claim made in \S 2.1 that \textquotedblleft The class
of $\mathcal{H}$-trivial monoids coincides with that of \textit{aperiodic}
monoids\textquotedblright. This fact is the equivalence of the assertions
$\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ in the following lemma:

\begin{lemma}
\label{lem.p6.acyclic}Let $M$ be a monoid. Then, the following four assertions
$\mathcal{A}_{1}$, $\mathcal{A}_{2}$, $\mathcal{A}_{3}$ and $\mathcal{A}_{4}$
are equivalent:

\begin{itemize}
\item \textit{Assertion }$\mathcal{A}_{1}$\textit{:} The monoid $M$ is
$\mathcal{H}$-trivial.

\item \textit{Assertion }$\mathcal{A}_{2}$\textit{:} The monoid $M$ is aperiodic.

\item \textit{Assertion }$\mathcal{A}_{3}$\textit{:} For every four elements
$x$, $y$, $b$ and $v$ of $M$ satisfying $x=yv$ and $y=bx$, we have $x=y$.

\item \textit{Assertion }$\mathcal{A}_{4}$\textit{:} For every four elements
$x$, $y$, $a$ and $u$ of $M$ satisfying $x=uy$ and $y=xa$, we have $x=y$.
\end{itemize}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.p6.acyclic}.]Let us recall the definition of the
relation $\mathcal{H}$: If $x$ and $y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{H}\ y\text{ holds if and only if }\left(  x\ \mathcal{R}\ y\text{
and }x\ \mathcal{L}\ y\right)  . \label{pf.lem.p6.acyclic.defH}%
\end{equation}


Let us also recall the definition of the relation $\mathcal{L}$: If $x$ and
$y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{L}\ y\text{ holds if and only if }Mx=My.
\label{pf.lem.p6.acyclic.defL}%
\end{equation}


Finally, let us recall the definition of the relation $\mathcal{R}$: If $x$
and $y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{R}\ y\text{ holds if and only if }xM=yM.
\label{pf.lem.p6.acyclic.defR}%
\end{equation}


Now, we shall prove the implications $\mathcal{A}_{1}\Longrightarrow
\mathcal{A}_{2}$, $\mathcal{A}_{2}\Longrightarrow\mathcal{A}_{3}$,
$\mathcal{A}_{3}\Longrightarrow\mathcal{A}_{4}$ and $\mathcal{A}%
_{4}\Longrightarrow\mathcal{A}_{1}$.

\textit{Proof of the implication }$\mathcal{A}_{1}\Longrightarrow
\mathcal{A}_{2}$\textit{:} Assume that Assertion $\mathcal{A}_{1}$ holds. We
shall show that Assertion $\mathcal{A}_{2}$ holds.

We know that Assertion $\mathcal{A}_{1}$ holds. In other words, the monoid $M$
is $\mathcal{H}$-trivial. In other words, all $\mathcal{H}$-classes are of
cardinality one. In other words, any two $\mathcal{H}$-equivalent elements of
$M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{H\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.acyclic.1to2.ass}%
\end{align}


Now, let $x\in M$. We know that $M$ is finite. Thus, $\left\vert M\right\vert
\in\mathbb{N}$. Let $\phi$ be the map%
\begin{align*}
\left\{  1,2,\ldots,\left\vert M\right\vert +1\right\}   &  \rightarrow M,\\
i  &  \mapsto x^{i}.
\end{align*}
There exist two distinct elements $p$ and $q$ of $\left\{  1,2,\ldots
,\left\vert M\right\vert +1\right\}  $ satisfying $\phi\left(  p\right)
=\phi\left(  q\right)  $\ \ \ \ \footnote{\textit{Proof.} Assume the contrary.
Thus, there exist no two distinct elements $p$ and $q$ of $\left\{
1,2,\ldots,\left\vert M\right\vert +1\right\}  $ satisfying $\phi\left(
p\right)  =\phi\left(  q\right)  $. In other words, any two distinct elements
$p$ and $q$ of $\left\{  1,2,\ldots,\left\vert M\right\vert +1\right\}  $
satisfy $\phi\left(  p\right)  \neq\phi\left(  q\right)  $. In other words,
the map $\phi$ is injective. Hence, there exists an injective map $\left\{
1,2,\ldots,\left\vert M\right\vert +1\right\}  \rightarrow M$ (namely, $\phi
$). Consequently, $\left\vert M\right\vert \geq\left\vert \left\{
1,2,\ldots,\left\vert M\right\vert +1\right\}  \right\vert =\left\vert
M\right\vert +1>\left\vert M\right\vert $. This is absurd. This contradiction
proves that our assumption was wrong, qed.}. Consider these $p$ and $q$. We
WLOG assume that $p\leq q$ (otherwise, we can simply switch $p$ with $q$).
Thus, $p<q$ (since $p$ and $q$ are distinct), and therefore $p\leq q-1$ (since
$p$ and $q$ are integers). Hence, $\left(  q-1\right)  -p\in\mathbb{N}$. Thus,
$q-\left(  p+1\right)  =\left(  q-1\right)  -p\in\mathbb{N}$. Hence,
$x^{q-\left(  p+1\right)  }$ is a well-defined element of $M$.

We have $\phi\left(  p\right)  =x^{p}$ (by the definition of $\phi$) and
$\phi\left(  q\right)  =x^{q}$ (by the definition of $\phi$). From
$\phi\left(  p\right)  =x^{p}$, we obtain $x^{p}=\phi\left(  p\right)
=\phi\left(  q\right)  =x^{q}$.

But $q=\left(  q-\left(  p+1\right)  \right)  +\left(  p+1\right)  $ and thus%
\begin{align*}
x^{q}  &  =x^{\left(  q-\left(  p+1\right)  \right)  +\left(  p+1\right)
}=\underbrace{x^{q-\left(  p+1\right)  }}_{\in M}x^{p+1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }q-\left(  p+1\right)  \in
\mathbb{N}\right) \\
&  \in Mx^{p+1}.
\end{align*}
Hence, $x^{p}=x^{q}\in Mx^{p+1}$, so that $M\underbrace{x^{p}}_{\in Mx^{p+1}%
}\subseteq\underbrace{MM}_{\subseteq M}x^{p+1}\subseteq Mx^{p+1}$. Combined
with $M\underbrace{x^{p+1}}_{=xx^{p}}=\underbrace{Mx}_{\subseteq M}%
x^{p}\subseteq Mx^{p}$, this yields $Mx^{p}=Mx^{p+1}$. But $x^{p}%
\ \mathcal{L}\ x^{p+1}$ holds if and only if $Mx^{p}=Mx^{p+1}$ (because of
(\ref{pf.lem.p6.acyclic.defL}), applied to $x^{p}$ and $x^{p+1}$ instead of
$x$ and $y$). Thus, we have $x^{p}\ \mathcal{L}\ x^{p+1}$ (since
$Mx^{p}=Mx^{p+1}$).

Also, $q=\left(  p+1\right)  +\left(  q-\left(  p+1\right)  \right)  $ and
thus%
\begin{align*}
x^{q}  &  =x^{\left(  p+1\right)  +\left(  q-\left(  p+1\right)  \right)
}=x^{p+1}\underbrace{x^{q-\left(  p+1\right)  }}_{\in M}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }q-\left(  p+1\right)  \in
\mathbb{N}\right) \\
&  \in x^{p+1}M.
\end{align*}
Hence, $x^{p}=x^{q}\in x^{p+1}M$, so that $\underbrace{x^{p}}_{\in x^{p+1}%
M}M\subseteq x^{p+1}\underbrace{MM}_{\subseteq M}\subseteq x^{p+1}M$. Combined
with $\underbrace{x^{p+1}}_{=x^{p}x}M=x^{p}\underbrace{xM}_{\subseteq
M}\subseteq x^{p}M$, this yields $x^{p}M=x^{p+1}M$. But $x^{p}\ \mathcal{R}%
\ x^{p+1}$ holds if and only if $x^{p}M=x^{p+1}M$ (because of
(\ref{pf.lem.p6.acyclic.defR}), applied to $x^{p}$ and $x^{p+1}$ instead of
$x$ and $y$). Thus, we have $x^{p}\ \mathcal{R}\ x^{p+1}$ (since
$x^{p}M=x^{p+1}M$).

Finally, $x^{p}\ \mathcal{H}\ x^{p+1}$ holds if and only if $\left(
x^{p}\ \mathcal{R}\ x^{p+1}\text{ and }x^{p}\ \mathcal{L}\ x^{p+1}\right)  $
(because of (\ref{pf.lem.p6.acyclic.defH}), applied to $x^{p}$ and $x^{p+1}$
instead of $x$ and $y$). Hence, we have $x^{p}\ \mathcal{H}\ x^{p+1}$ (since
$x^{p}\ \mathcal{R}\ x^{p+1}$ and $x^{p}\ \mathcal{L}\ x^{p+1}$). Thus,
$x^{p}=x^{p+1}$ (by (\ref{pf.lem.p6.acyclic.1to2.ass}), applied to $x^{p}$ and
$x^{p+1}$ instead of $x$ and $y$). Thus, there exists some positive integer
$N$ such that $x^{N}=x^{N+1}$ (namely, $N=p$).

Now, let us forget that we fixed $x$. We thus have shown that for every $x\in
M$, there exists some positive integer $N$ such that $x^{N}=x^{N+1}$. In other
words, the monoid $M$ is aperiodic. In other words, Assertion $\mathcal{A}%
_{2}$ holds. This proves the implication $\mathcal{A}_{1}\Longrightarrow
\mathcal{A}_{2}$.

\textit{Proof of the implication }$\mathcal{A}_{2}\Longrightarrow
\mathcal{A}_{3}$\textit{:} Assume that Assertion $\mathcal{A}_{2}$ holds. We
shall show that Assertion $\mathcal{A}_{3}$ holds.

Assertion $\mathcal{A}_{2}$ holds. In other words, the monoid $M$ is
aperiodic. In other words, for every $x\in M$,%
\begin{equation}
\text{there exists some positive integer }N\text{ such that }x^{N}=x^{N+1}.
\label{pf.lem.p6.acyclic.2to1.ass}%
\end{equation}


Now, let $x$, $y$, $b$ and $v$ be four elements of $M$ satisfying $x=yv$ and
$y=bx$. We are going to prove that $x=y$.

There exists some positive integer $N$ such that $b^{N}=b^{N+1}$ (according to
(\ref{pf.lem.p6.acyclic.2to1.ass}), applied to $b$ instead of $x$). Consider
this $N$.

We have $x=\underbrace{y}_{=bx}v=bxv$. Hence, Lemma \ref{lem.p6.uxv} (applied
to $u=b$) shows that%
\begin{equation}
x=b^{m}xv^{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}.
\label{pf.lem.p6.acyclic.2to1.bmxvm}%
\end{equation}
Applying (\ref{pf.lem.p6.acyclic.2to1.bmxvm}) to $m=N$, we obtain
$x=\underbrace{b^{N}}_{=b^{N+1}=bb^{N}}xv^{N}=bb^{N}xv^{N}$. Compared with
$y=b\underbrace{x}_{=b^{N}xv^{N}}=bb^{N}xv^{N}$, this yields $x=y$.

Let us now forget that we fixed $x$, $y$, $b$ and $v$. We thus have proven
that for every four elements $x$, $y$, $b$ and $v$ of $M$ satisfying $x=yv$
and $y=bx$, we have $x=y$. In other words, Assertion $\mathcal{A}_{3}$ holds.
This proves the implication $\mathcal{A}_{2}\Longrightarrow\mathcal{A}_{3}$.

\textit{Proof of the implication }$\mathcal{A}_{3}\Longrightarrow
\mathcal{A}_{4}$\textit{:} Assume that Assertion $\mathcal{A}_{3}$ holds. We
shall show that Assertion $\mathcal{A}_{4}$ holds.

Let $x$, $y$, $a$ and $u$ be four elements of $M$ satisfying $x=uy$ and
$y=xa$. Recall that Assertion $\mathcal{A}_{3}$ holds. Hence, Assertion
$\mathcal{A}_{3}$ (applied to $y$, $x$, $u$ and $a$ instead of $x$, $y$, $b$
and $v$) yields $y=x$. In other words, $x=y$.

Let us now forget that we fixed $x$, $y$, $a$ and $u$. We thus have shown that
for every four elements $x$, $y$, $a$ and $u$ of $M$ satisfying $x=uy$ and
$y=xa$, we have $x=y$. In other words, Assertion $\mathcal{A}_{4}$ holds. This
proves the implication $\mathcal{A}_{3}\Longrightarrow\mathcal{A}_{4}$.

\textit{Proof of the implication }$\mathcal{A}_{4}\Longrightarrow
\mathcal{A}_{1}$\textit{:} Assume that Assertion $\mathcal{A}_{4}$ holds. We
shall show that Assertion $\mathcal{A}_{1}$ holds.

Let $x$ and $y$ be two elements of $M$ such that $x\ \mathcal{H}\ y$. Recall
that $x\ \mathcal{H}\ y$ holds if and only if $\left(  x\ \mathcal{R}\ y\text{
and }x\ \mathcal{L}\ y\right)  $ (because of (\ref{pf.lem.p6.acyclic.defH})).
Hence, we must have $\left(  x\ \mathcal{R}\ y\text{ and }x\ \mathcal{L}%
\ y\right)  $ (since we have $x\ \mathcal{H}\ y$). Thus, $x\ \mathcal{R}\ y$
and $x\ \mathcal{L}\ y$.

We know that $x\ \mathcal{L}\ y$ holds if and only if $Mx=My$ (according to
(\ref{pf.lem.p6.acyclic.defL})). Thus, we must have $Mx=My$ (since
$x\ \mathcal{L}\ y$ holds). Hence, $x=\underbrace{1}_{\in M}x\in Mx=My$. In
other words, there exists an $u\in M$ such that $x=uy$. Consider this $u$.

We know that $x\ \mathcal{R}\ y$ holds if and only if $xM=yM$ (according to
(\ref{pf.lem.p6.acyclic.defR})). Thus, we must have $xM=yM$ (since
$x\ \mathcal{R}\ y$ holds). Hence, $y=y\underbrace{1}_{\in M}\in yM=xM$. In
other words, there exists an $a\in M$ such that $y=xa$. Consider this $a$.

Now, Assertion $\mathcal{A}_{4}$ yields $x=y$ (since $x=uy$ and $y=xa$).

Let us now forget that we fixed $x$ and $y$. We thus have proven that every
two elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{H\ }y$ must satisfy
$x=y$. In other words, any two $\mathcal{H}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{H}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{H}$-trivial. In other words, Assertion
$\mathcal{A}_{1}$ holds. This proves the implication $\mathcal{A}%
_{4}\Longrightarrow\mathcal{A}_{1}$.

We have thus proven the four implications $\mathcal{A}_{1}\Longrightarrow
\mathcal{A}_{2}$, $\mathcal{A}_{2}\Longrightarrow\mathcal{A}_{3}$,
$\mathcal{A}_{3}\Longrightarrow\mathcal{A}_{4}$ and $\mathcal{A}%
_{4}\Longrightarrow\mathcal{A}_{1}$. Combined, these implications yield the
equivalence $\mathcal{A}_{1}\Longleftrightarrow\mathcal{A}_{2}%
\Longleftrightarrow\mathcal{A}_{3}\Longleftrightarrow\mathcal{A}_{4}$. This
proves Lemma \ref{lem.p6.acyclic}.
\end{proof}

Next, let us prove another elementary result, which is used in \S 2.3 (in the
sentence \textquotedblleft Since $M$ is finite, this implies that $M$ is
$\mathcal{J}$-trivial (see [Pin10a, Chapter V, Theorem 1.9])\textquotedblright):

\begin{lemma}
\label{lem.p6.RLJ}Let $M$ be a monoid which is $\mathcal{R}$-trivial and
$\mathcal{L}$-trivial. Then, $M$ is $\mathcal{J}$-trivial.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.p6.RLJ}.]Let us recall the definition of the relation
$\mathcal{J}$: If $x$ and $y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{J}\ y\text{ holds if and only if }MxM=MyM.
\label{pf.lem.p6.RLJ.defJ}%
\end{equation}


Let us also recall the definition of the relation $\mathcal{L}$: If $x$ and
$y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{L}\ y\text{ holds if and only if }Mx=My.
\label{pf.lem.p6.RLJ.defL}%
\end{equation}


Finally, let us recall the definition of the relation $\mathcal{R}$: If $x$
and $y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{R}\ y\text{ holds if and only if }xM=yM.
\label{pf.lem.p6.RLJ.defR}%
\end{equation}


We know that $M$ is $\mathcal{R}$-trivial. Thus, all $\mathcal{R}$-classes are
of cardinality one. In other words, any two $\mathcal{R}$-equivalent elements
of $M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{R\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.RLJ.Rtriv}%
\end{align}


We know that $M$ is $\mathcal{L}$-trivial. Thus, all $\mathcal{L}$-classes are
of cardinality one. In other words, any two $\mathcal{L}$-equivalent elements
of $M$ are identical. In other words,%
\begin{align}
&  \text{every two elements }x\text{ and }y\text{ of }M\text{ satisfying
}x\ \mathcal{L\ }y\nonumber\\
&  \text{must satisfy }x=y. \label{pf.lem.p6.RLJ.Ltriv}%
\end{align}


But the monoid $M$ is $\mathcal{R}$-trivial, and thus $\mathcal{H}$-trivial
(according to Lemma \ref{lem.p6.implications} \textbf{(c)}). In other words,
the Assertion $\mathcal{A}_{1}$ of Lemma \ref{lem.p6.acyclic} holds. Hence,
the Assertion $\mathcal{A}_{2}$ of Lemma \ref{lem.p6.acyclic} holds as well
(since the Assertions $\mathcal{A}_{1}$, $\mathcal{A}_{2}$, $\mathcal{A}_{3}$
and $\mathcal{A}_{4}$ of Lemma \ref{lem.p6.acyclic} are equivalent (according
to Lemma \ref{lem.p6.acyclic})). In other words, the monoid $M$ is aperiodic.
In other words, for every $x\in M$,%
\begin{equation}
\text{there exists some positive integer }N\text{ such that }x^{N}=x^{N+1}.
\label{pf.lem.p6.RLJ.aper}%
\end{equation}


Let $x$ and $y$ be two elements of $M$ such that $x\ \mathcal{J}\ y$. Recall
that $x\ \mathcal{J}\ y$ holds if and only if $MxM=MyM$ (because of
(\ref{pf.lem.p6.RLJ.defJ})). Hence, we must have $MxM=MyM$ (since we have
$x\ \mathcal{J}\ y$).

Now, $x=\underbrace{1}_{\in M}x\underbrace{1}_{\in M}\in MxM=MyM$. In other
words, there exist $u\in M$ and $v\in M$ such that $x=uyv$. Consider these $u$
and $v$.

We also have $y=\underbrace{1}_{\in M}y\underbrace{1}_{\in M}\in MyM=MxM$
(since $MxM=MyM$). In other words, there exists $a\in M$ and $b\in M$ such
that $y=axb$. Consider these $a$ and $b$.

Now, $x=u\underbrace{y}_{=axb}v=uaxbv=\left(  ua\right)  x\left(  bv\right)
$. Thus, Lemma \ref{lem.p6.uxv} (applied to $ua$ and $bv$ instead of $u$ and
$v$) shows that%
\begin{equation}
x=\left(  ua\right)  ^{m}x\left(  bv\right)  ^{m}\ \ \ \ \ \ \ \ \ \ \text{for
every }m\in\mathbb{N}. \label{pf.lem.p6.RLJ.uamxbvm}%
\end{equation}


Now, there exists some positive integer $N$ such that $\left(  ua\right)
^{N}=\left(  ua\right)  ^{N+1}$ (according to (\ref{pf.lem.p6.RLJ.aper}),
applied to $ua$ instead of $x$). Let us denote this $N$ by $\alpha$. Thus,
$\alpha$ is a positive integer such that $\left(  ua\right)  ^{\alpha}=\left(
ua\right)  ^{\alpha+1}$. Now, (\ref{pf.lem.p6.RLJ.uamxbvm}) (applied to
$m=\alpha$) yields $x=\underbrace{\left(  ua\right)  ^{\alpha}}_{=\left(
ua\right)  ^{\alpha+1}=\left(  ua\right)  \left(  ua\right)  ^{\alpha}%
}x\left(  bv\right)  ^{\alpha}=\left(  ua\right)  \left(  ua\right)  ^{\alpha
}x\left(  bv\right)  ^{\alpha}$. Compared with $\left(  ua\right)
\underbrace{x}_{=\left(  ua\right)  ^{\alpha}x\left(  bv\right)  ^{\alpha}%
}=\left(  ua\right)  \left(  ua\right)  ^{\alpha}x\left(  bv\right)  ^{\alpha
}$, this yields $x=\left(  ua\right)  x$. Thus, $x=\left(  ua\right)
x=\underbrace{u}_{\in M}ax\in Max$, so that $M\underbrace{x}_{\in
Max}\subseteq\underbrace{MM}_{\subseteq M}ax\subseteq Max$. Combined with
$\underbrace{Ma}_{\subseteq M}x\subseteq Mx$, this yields $Max=Mx$. But
$ax\ \mathcal{L}\ x$ holds if and only if $Max=Mx$ (according to
(\ref{pf.lem.p6.RLJ.defL}), applied to $ax$ and $x$ instead of $x$ and $y$).
Hence, $ax\ \mathcal{L}\ x$ (because $Max=Mx$). Hence,
(\ref{pf.lem.p6.RLJ.Ltriv}) (applied to $ax$ and $x$ instead of $x$ and $y$)
yields $ax=x$.

Furthermore, there exists some positive integer $N$ such that $\left(
bv\right)  ^{N}=\left(  bv\right)  ^{N+1}$ (according to
(\ref{pf.lem.p6.RLJ.aper}), applied to $bv$ instead of $x$). Let us denote
this $N$ by $\beta$. Thus, $\beta$ is a positive integer such that $\left(
bv\right)  ^{\beta}=\left(  bv\right)  ^{\beta+1}$. Now,
(\ref{pf.lem.p6.RLJ.uamxbvm}) (applied to $m=\beta$) yields $x=\left(
ua\right)  ^{\beta}x\underbrace{\left(  bv\right)  ^{\beta}}_{=\left(
bv\right)  ^{\beta+1}=\left(  bv\right)  ^{\beta}\left(  bv\right)  }=\left(
ua\right)  ^{\beta}x\left(  bv\right)  ^{\beta}\left(  bv\right)  $. Compared
with $\underbrace{x}_{=\left(  ua\right)  ^{\beta}x\left(  bv\right)  ^{\beta
}}\left(  bv\right)  =\left(  ua\right)  ^{\beta}x\left(  bv\right)  ^{\beta
}\left(  bv\right)  $, this yields $x=x\left(  bv\right)  $. Thus, $x=x\left(
bv\right)  =xb\underbrace{v}_{\in M}\in xbM$, so that $\underbrace{x}_{\in
xbM}M\subseteq xb\underbrace{MM}_{\subseteq M}\subseteq xbM$. Combined with
$x\underbrace{bM}_{\subseteq M}\subseteq xM$, this yields $xbM=xM$. But
$xb\ \mathcal{R}\ x$ holds if and only if $xbM=xM$ (according to
(\ref{pf.lem.p6.RLJ.defR}), applied to $xb$ and $x$ instead of $x$ and $y$).
Hence, $xb\ \mathcal{R}\ x$ (because $xbM=xM$). Hence,
(\ref{pf.lem.p6.RLJ.Rtriv}) (applied to $xb$ and $x$ instead of $x$ and $y$)
yields $xb=x$.

Now, $ax=x$ and $xb=x$. Recall now that $y=\underbrace{ax}_{=x}b=xb=x$. Hence,
$x=y$.

Let us now forget that we fixed $x$ and $y$. We thus have shown that every two
elements $x$ and $y$ of $M$ satisfying $x\ \mathcal{J\ }y$ must satisfy $x=y$.
In other words, any two $\mathcal{J}$-equivalent elements of $M$ are
identical. In other words, all $\mathcal{J}$-classes are of cardinality one.
In other words, $M$ is $\mathcal{J}$-trivial. This proves Lemma
\ref{lem.p6.RLJ}.
\end{proof}

Finally, let us prove yet another elementary result about the equivalence
relations $\mathcal{R}$ and $\mathcal{L}$ (which is, to my knowledge, not used
in the paper, but still interesting):\footnote{Proposition \ref{prop.RL=LR} is
one part of Proposition 1.6 in Chapter V of [Pin10a] (in the case of monoids,
rather than arbitrary semigroups). (It is arguably the harder part.)}

\begin{proposition}
\label{prop.RL=LR}Let $M$ be a monoid. Let $a$ and $b$ be two elements of $M$.
Assume that there exists a $c\in M$ such that $a\ \mathcal{R}\ c$ and
$c\ \mathcal{L}\ b$. Then, there exists a $d\in M$ such that $a\ \mathcal{L}%
\ d$ and $d\ \mathcal{R}\ b$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.RL=LR}.]Let us recall the definition of the
relation $\mathcal{L}$: If $x$ and $y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{L}\ y\text{ holds if and only if }Mx=My.
\label{pf.prop.RL=LR.defL}%
\end{equation}


Also, let us recall the definition of the relation $\mathcal{R}$: If $x$ and
$y$ are two elements of $M$, then%
\begin{equation}
x\ \mathcal{R}\ y\text{ holds if and only if }xM=yM.
\label{pf.prop.RL=LR.defR}%
\end{equation}


Now, we have assumed that there exists a $c\in M$ such that $a\ \mathcal{R}%
\ c$ and $c\ \mathcal{L}\ b$. Consider this $c$.

We know that $a\ \mathcal{R}\ c$ holds if and only if $aM=cM$ (by
(\ref{pf.prop.RL=LR.defR}), applied to $x=a$ and $y=c$). Thus, we have $aM=cM$
(since $a\ \mathcal{R}\ c$ holds).

We know that $c\ \mathcal{L}\ b$ holds if and only if $Mc=Mb$ (by
(\ref{pf.prop.RL=LR.defL}), applied to $x=c$ and $y=b$). Thus, we have $Mc=Mb$
(since $c\ \mathcal{L}\ b$ holds).

We have $a=a\underbrace{1}_{\in M}\in aM=cM$. Thus, there exists an $x\in M$
such that $a=cx$. Consider this $x$.

We have $c=\underbrace{1}_{\in M}c\in Mc=Mb$. Thus, there exists a $y\in M$
such that $c=yb$. Consider this $y$.

We have $c=c\underbrace{1}_{\in M}\in cM=aM$ (since $aM=cM$). Thus, there
exists an $x^{\prime}\in M$ such that $c=ax^{\prime}$. Consider this
$x^{\prime}$.

We have $b=\underbrace{1}_{\in M}b\in Mb=Mc$ (since $Mc=Mb$). Thus, there
exists a $y^{\prime}\in M$ such that $b=y^{\prime}c$. Consider this
$y^{\prime}$.

Now, define an element $f$ of $M$ by $f=bx$. Then, $f=\underbrace{b}%
_{=y^{\prime}c}x=y^{\prime}\underbrace{cx}_{\substack{=a\\\text{(since
}a=cx\text{)}}}=y^{\prime}a$. Thus, $M\underbrace{f}_{=y^{\prime}%
a}=\underbrace{My^{\prime}}_{\subseteq M}a\subseteq Ma$.

On the other hand, $a=\underbrace{c}_{=yb}x=y\underbrace{bx}_{=f}=yf$. Hence,
$M\underbrace{a}_{=yf}=\underbrace{My}_{\subseteq M}f\subseteq Mf$. Combined
with $Mf\subseteq Ma$, this yields $Ma=Mf$.

We know that $a\ \mathcal{L}\ f$ holds if and only if $Ma=Mf$ (by
(\ref{pf.prop.RL=LR.defL}), applied to $x=a$ and $y=f$). Thus, $a\ \mathcal{L}%
\ f$ holds (since $Ma=Mf$).

Furhermore, $b=y^{\prime}\underbrace{c}_{=ax^{\prime}}=\underbrace{y^{\prime
}a}_{\substack{=f\\\text{(since }f=y^{\prime}a\text{)}}}x^{\prime}=fx^{\prime
}$. Thus, $\underbrace{b}_{=fx^{\prime}}M=f\underbrace{x^{\prime}M}_{\subseteq
M}\subseteq fM$. Combined with $\underbrace{f}_{=bx}M=b\underbrace{xM}%
_{\subseteq M}\subseteq bM$, this yields $fM=bM$.

We know that $f\ \mathcal{R}\ b$ holds if and only if $fM=bM$ (by
(\ref{pf.prop.RL=LR.defR}), applied to $x=f$ and $y=b$). Thus, $f\ \mathcal{R}%
\ b$ holds (since $fM=bM$).

Now, we know that $a\ \mathcal{L}\ f$ and $f\ \mathcal{R}\ b$. Thus, there
exists a $d\in M$ such that $a\ \mathcal{L}\ d$ and $d\ \mathcal{R}\ b$
(namely, $d=f$). This proves Proposition \ref{prop.RL=LR}.
\end{proof}

Note that Proposition \ref{prop.RL=LR} does not require the finiteness of $M$.
\end{document}