Cfrgtkky

Let $(M, cdot)$ be a finite monoid with identity $e$. It is easy to see that $gMg = { gxg : x in M }$ forms a monoid with identity $geg = g$ if $g$ is an idempotent. If $gMg$ contains no idempotent other than $g$, it must be a group, since it is finite.

Suppose that $g, h in M$ are distinct idempotents in $M$ such that both $gMg$ and $hMh$ are groups. Is it true that $gMg$ and $hMh$ are isomorphic?

Yes. You do not even need the finiteness of $M$.

Theorem 1. Let $left( M,cdotright) $ be a finite monoid with identity
$e$. For every idempotent $gin M$, the set $gMg=left{ gxgmid xin
Mright} $ is a monoid (with respect to the operation $cdot$) with neutral
element $geg=gg=g$. Let $g$ and $h$ be two idempotents of $M$ such that the
monoids $gMg$ and $hMh$ are groups. Then, these groups $gMg$ and $hMh$ are isomorphic.

Let us first show the following simple fact:

Lemma 2. Let $left( A,cdotright) $ and $left( C,cdotright) $ be
two groups. Let $varphi:Arightarrow C$ be a map. Assume that $varphileft(
aright) varphileft( bright) =varphileft( abright) $ for all $ain
A$ and $bin A$. Then, $varphi$ is a group homomorphism from $left(
A,cdotright) $ to $left( C,cdotright) $.

Proof of Lemma 2. Let $e$ denote the neutral element of any group. We have
assumed that
begin{equation}
varphileft( aright) varphileft( bright) =varphileft(
abright) qquad text{for all $ain A$ and $bin A$.}
label{darij.pf.l2.1}
tag{1}
end{equation}
Applying this to $a=e$ and $b=e$, we obtain $varphileft( eright)
varphileft( eright) =varphileft( underbrace{ee}_{=e}right)
=varphileft( eright) $. Since $C$ is a group, we can cancel
$varphileft( eright) $ from this equality, and obtain $varphileft(
eright) =e$. Combined with our assumption eqref{darij.pf.l2.1}, this shows that $varphi$
is a monoid homomorphism from $left( A,cdotright) $ to $left(
C,cdotright) $. Thus, $varphi$ is also a group homomorphism (since every
monoid homomorphism between groups is a group homomorphism). Lemma 2 is proven. $blacksquare$

Proof of Theorem 1. The elements $g$ and $h$ are idempotents; thus, we have
$gg=g$ and $hh=h$.

Let $p$ be the inverse of the element $ghg$ in the group $gMg$. Then,
$pleft( ghgright) =left( ghgright) p=g$ (since $g$ is the neutral
element of the group $gMg$).

Let $q$ be the inverse of the element $hgh$ in the group $hMh$. Then,
$qleft( hghright) =left( hghright) q=h$ (since $h$ is the neutral
element of the group $hMh$).

The element $p$ belongs to the group $gMg$, but $g$ is the neutral element of
this group. Hence, $pg=gp=p$. Similarly, $qh=hq=q$.

We have
begin{align}
gqhghgp=gunderbrace{qleft( hghright) }_{=h}gp=ghgp=left( ghgright)
p=g ,
end{align}
so that
begin{align}
g
&= gqhghgp
= gqh underbrace{left( ghgright) p}_{=g}
= gunderbrace{qh}_{=q}g
label{darij.pf.thm1.2}
tag{2}
\
&= gqg .
label{darij.pf.thm1.3}
tag{3}
end{align}
Similarly,
begin{equation}
h = hph .
label{darij.pf.thm1.4}
tag{4}
end{equation}
We define a map $alpha:gMgrightarrow hMh$ by setting
begin{align}
alphaleft( xright) =hxq qquad text{ for every $xin gMg$.}
end{align}
This is well-defined, because every $xin gMg$ satisfies $hxunderbrace{q}
_{in hMh}in hunderbrace{xhM}_{subseteq M}hsubseteq hMh$.

We define a map $beta:hMhrightarrow gMg$ by setting
begin{align}
betaleft( yright) =pyg qquad text{ for every $yin hMh$.}
end{align}
This is well-defined, because every $yin hMh$ satisfies $underbrace{p}_{in
gMg}ygin gunderbrace{Mgy}_{subseteq M}gsubseteq gMg$.

Let us now prove that
begin{align}
alphaleft( aright) alphaleft( bright) =alphaleft(
abright) qquad text{ for all $ain gMg$ and $bin gMg$.}
label{darij.pf.thm1.5}
tag{5}
end{align}

[Proof of eqref{darij.pf.thm1.5}: Let $ain gMg$ and $bin gMg$.

We have $ain gMg$; thus, there exists some $xin M$ such that $a=gxg$.
Consider this $x$. Thus, $underbrace{a}_{=gxg}g=gxunderbrace{gg}_{=g}=gxg=a$.

We have $bin gMg$; thus, there exists some $yin M$ such that $b=gyg$.
Consider this $y$. Thus, $gunderbrace{b}_{=gyg}=underbrace{gg}_{=g}yg=gyg=b$.

The definition of $alpha$ yields $alphaleft( aright) =hunderbrace{a}
_{=ag}q=hagq$ and $alphaleft( bright) =hunderbrace{b}_{=gb}q=hgbq$.
Multiplying these two equalities, we obtain
begin{align}
alphaleft( aright) alphaleft( bright) =haunderbrace{gqhg}
_{substack{=g\text{(by eqref{darij.pf.thm1.2})}}}bq=hunderbrace{ag}_{=a}bq=habq .
label{darij.pf.thm1.6}
tag{6}
end{align}
On the other hand, the definition of $alpha$ yields $alphaleft( abright)
=habq$. Compared with eqref{darij.pf.thm1.6}, this yields $alphaleft( aright)
alphaleft( bright) =alphaleft( abright) $. Thus, eqref{darij.pf.thm1.5} is proven.]

Now, Lemma 2 (applied to $A=gMg$, $C=hMh$ and $varphi=alpha$) yields that
$alpha$ is a group homomorphism from $left( gMg,cdotright) $ to $left(
hMh,cdotright) $. We shall now focus on proving that $alpha$ is invertible.

Indeed, let us first show that $betacircalpha=operatorname*{id}$. Indeed,
let $xin gMg$. Then, $xg=gx=x$ (since $x$ belongs to the group $gMg$, but $g$
is the neutral element of this group). We have
begin{align}
left( betacircalpharight) left( xright)
&=betaleft(
underbrace{alphaleft( xright) }_{=hxq}right) =betaleft( hxqright) \
&=underbrace{p}_{=pg}hunderbrace{x}_{=xg}qg
qquad text{(by the definition of $beta$)} \
&=pghunderbrace{x}_{=gx}underbrace{gqg}_{substack{=g\text{(by
eqref{darij.pf.thm1.3})}}}=underbrace{pghg}_{=pleft( ghgright) =g}underbrace{xg}
_{=x}=gx=x=operatorname*{id}left( xright) .
end{align}
Thus, we have shown that $left( betacircalpharight) left( xright)
=operatorname*{id}left( xright) $ for every $xin gMg$. This proves
$betacircalpha=operatorname*{id}$.

Now, let us show that $alphacircbeta=operatorname*{id}$. Indeed, let $yin
hMh$. Then, $hy=yh=y$ (since $y$ belongs to the group $hMh$, but $h$ is the
neutral element of this group). We have
begin{align}
left( alphacircbetaright) left( yright)
&=alphaleft(
underbrace{betaleft( yright) }_{=pyg}right) =alphaleft( pygright) \
&= hpunderbrace{y}_{=hy}gunderbrace{q}_{=hq}
qquad text{(by the definition of $alpha$)} \
&=underbrace{hph}_{substack{=h\text{(by eqref{darij.pf.thm1.4})}}}underbrace{y}
_{=yh}ghq=underbrace{hy}_{=y}underbrace{hghq}_{=left( hghright)
q=h}=yh=y=operatorname*{id}left( yright) .
end{align}
Thus, we have shown that $left( alphacircbetaright) left( yright)
=operatorname*{id}left( yright) $ for every $yin hMh$. This proves
$alphacircbeta=operatorname*{id}$.

The maps $alpha$ and $beta$ are mutually inverse (since $alphacirc
beta=operatorname*{id}$ and $betacircalpha=operatorname*{id}$). Thus,
the map $alpha$ is invertible. Since $alpha$ is a group homomorphism, this
shows that $alpha$ is a group isomorphism. Thus, there exists a group
isomorphism $gMgrightarrow hMh$ (namely, $alpha$). This proves Theorem 1. $blacksquare$

This proof was obtained by lots of experimentation and iterative
simplification. Do not ask me for the intuition behind it, for I have none. I
suspect it could still be shortened twice, but I am happy enough that I was
able to dispose of the finiteness condition and at least some of the ugliest
computations. Feel free to improve!

This is true for any semigroup (even if it is not a monoid) and is easy to prove if you know about Green's relations.

If $S$ be a semigroup and $e$ is an idempotent of $S$, then $eSe$ is a monoid
(with identity $e = eee$). Suppose that $e$ and $f$ are idempotents such that $eSe$ and $fSf$ are groups. Then in particular, $e mathrel{mathcal H} efe$ and $f mathrel{mathcal H} fef$, which implies that $e mathrel{mathcal R} ef$ and
$ef mathrel{mathcal L} f$, whence $e mathrel{mathcal D} f$. It follows by Green's Lemma, than the two groups $eSe$ and $fSf$, which are maximal groups of a regular $mathcal D$-class, are isomorphic.