If $ef = fe = e$ for two distinct idempotents in finite monoid, then $MeM \ne MfM$

Question

Let $e,f$ be two distinct idempotent elements of some finite monoid such that $$ ef = fe = e. $$ Then of course $MeM \subset MfM$. But is $MeM = MfM$ possible?

Do you know a proof that $MeM \ne MfM$, i.e. the two-sided ideals are different, or is $MeM = MfM$ possible?

I see that $f \notin eM \cup Me$, for assume $f = em$, then $e = ef = eem = em = f$, similar $f = me$ is not possible. But $f = men$ for $m,n \in M$ would not give any contradiction so far...

Answer 1

No, it is not possible (not even if you replace "monoid" by "semigroup"). See Theorem 3 below. Let me first build up some (standard, I assume) arsenal.

Lemma 1. Let $M$ be a finite monoid with neutral element $1$. Let $x$ and $y$ be two elements of $M$ such that $xy=1$. Then, $yx=1$.

Proof of Lemma 1. Let $\alpha$ be the map $M\rightarrow M,\ m\mapsto mx$. Let $\beta$ be the map $M\rightarrow M,\ m\mapsto my$. Then, every $m\in M$ satisfies

$\left( \beta\circ\alpha\right) \left( m\right) =\beta\left( \underbrace{\alpha\left( m\right) }_{\substack{=mx\\\text{(by the definition of }\alpha\text{)}}}\right) =\beta\left( mx\right) =m\underbrace{xy}_{=1}$ (by the definition of $\beta$)

$=m=\operatorname*{id}\left( m\right) $.

In other words, $\beta\circ\alpha=\operatorname*{id}$. Hence, the map $\beta$ is right-invertible, therefore surjective. But the set $M$ is finite. Hence, every surjective map $M\rightarrow M$ is bijective. Applying this to the surjective map $\beta:M\rightarrow M$, we conclude that the map $\beta$ is bijective. Hence, the map $\beta$ is invertible. The inverse $\beta^{-1}$ of this map $\beta$ must be $\alpha$ (since $\beta\circ\alpha=\operatorname*{id} $). Thus, $\alpha\circ\beta=\operatorname*{id}$. Now, the definition of $\alpha$ yields $\alpha\left( y\right) =yx$. The definition of $\beta$ yields $\beta\left( 1\right) =1y=y$. But

$\left( \alpha\circ\beta\right) \left( 1\right) =\alpha\left( \underbrace{\beta\left( 1\right) }_{=y}\right) =\alpha\left( y\right) =yx$,

so that $yx=\underbrace{\left( \alpha\circ\beta\right) } _{=\operatorname*{id}}\left( 1\right) =\operatorname*{id}\left( 1\right) =1$. This proves Lemma 1. $\blacksquare$

Lemma 2. Let $M$ be a finite monoid with neutral element $1$. Let $e$ be an idempotent element of $M$ such that $1\in MeM$. Then, $e=1$.

Proof of Lemma 2. We have $1\in MeM$. In other words, there exist $u\in M$ and $v\in M$ such that $1=uev$. Consider these $u$ and $v$.

We have $uev=1$. Thus, Lemma 1 (applied to $x=u$ and $y=ev$) yields $evu=1$. But $e$ is idempotent, so that $e^{2}=e$. Now, $\underbrace{e^{2}} _{=e}vu=evu=1$, so that $1=e^{2}vu=e\underbrace{evu}_{=1}=e$. This proves Lemma 2. $\blacksquare$

Theorem 3. Let $M$ be a finite semigroup. Let $e$ and $f$ be two idempotent elements of $M$ such that $MeM=MfM$ and $ef=fe=e$. Then, $e=f$.

Proof of Theorem 3. Since $f$ is idempotent, we have $f=f\underbrace{f} _{=ff}=fff$. Let $M^{\prime}$ be the subsemigroup $fMf$ of $M$. Then, $f = f\underbrace{f}_{\in M}f \in fMf = M^\prime$. Moreover, the element $f$ of $M^\prime$ is a neutral element with respect to multiplication (since each $m \in M$ satisfies $f \left(fmf\right) = \underbrace{ff}_{=f} mf = fmf$ and similarly $\left(fmf\right) f = fmf$). Hence, the subsemigroup $M^{\prime}$ of $M$ is a monoid with neutral element $f$.

We have $e=f\underbrace{e}_{=ef}=f\underbrace{e}_{\in M}f\in fMf=M^{\prime}$.

We have $f=\underbrace{f}_{\in M}f\underbrace{f}_{\in M}\in MfM=MeM$. But $f=f\underbrace{f}_{\in MeM}f\in fM\underbrace{e}_{=fef}Mf = \underbrace{fMf}_{=M^{\prime}}e\underbrace{fMf}_{=M^{\prime}}=M^{\prime }eM^{\prime}$. Hence, Lemma 2 (applied to $M^{\prime}$, $f$ and $e$ instead of $M$, $1$ and $e$) yields that $e=f$. This proves Theorem 3. $\blacksquare$