Let $e$ be an idempotent of the monoid $M$, $x$, $y$ be two elements of $eMe$. Then, $(eMe)\,x\,(eMe) = (eMe)\,y\,(eMe)$ if and only if $MxM = MyM$

Question

I am studying semigroups. I saw a Lemma in the text that states:

Let $e$ be an idempotent of the monoid $M$, $x$, $y$ be two elements of $eMe$. Then, $(eMe)\,x\,(eMe) = (eMe)\,y\,(eMe)$ if and only if $MxM = MyM$.

If we have $MxM=MyM$, then by $x = exe$ and $y = eye$, it is not hard. I can't figure the other direction, but the text states the other direction is trivial.

@Shaun Representation Theory of Finite Monoids of Benjamin Steinberg. The Lemma 1.7 . — CJJ, Jul 11 '22 at 23:16

Levent Michael Dasar · Accepted Answer · 2022-07-12T10:44:41.757

1

Rightward proof:

Since $(eMe)x(eMe)=(eMe)y(eMe)$ then $(e1e)x(e1e)=(ee)x(ee)=exe=x$ can be expressed as an element of $(eMe)y(eMe)$. Since $eMe\subseteq M$, we have $x\in(eMe)y(eMe)\subseteq MyM$. By symmetry we can show $y\in MxM$.

Since $x\in MyM$ then $x=ayb$ and then for any $cxd\in MxM$ we have $cxd=(ca)y(bd)\in MyM$ so that $MxM\subseteq MyM$. By symmetry we can show $MyM\subseteq MxM$ to give us $MxM=MyM$.

$\square$

I think the text calls this trivial because we know for a semigroup $S$ that $x\mathcal{J}y\iff (x=s_1ys_2)\wedge(y=s_3xs_4)$ for $s_i\in S^1$. In this question we might be able to see quite quickly how to express our $x$ and $y$ in this way.

edited Jul 12 '22 at 10:44

answered Jul 12 '22 at 10:07

Levent Michael Dasar

58

How do you get $exe = x$ in the first line of your answer? – J.-E. Pin Jul 13 '22 at 20:48
Thanks, yes, $eMe\ni x=eme$ for some $m\in M$, then $exe=eemee=eme=x$. – Levent Michael Dasar Jul 13 '22 at 21:02

score 0 · Answer 2 · answered Jul 14 '22 at 16:17

Let $S$ be a submonoid of $M$ and $x,y\in S$. If $SxS=SyS$, it means you can find $a,b \in S$, such that $x=ayb$ because $S$ is monoid and let e be its identity, we get $exe=x$.

So, $MxM= M(ayb)M$. Since $a,b\in S\subset M$, we get $MxM= (Ma)y(bM)\subset MyM$. And similarly, we have the other direction.

score 0 · Answer 3 · answered Jul 14 '22 at 16:37

First of all, since $x\in eMe$, you have $exe=x$ (why?). Similarly $y=eye$.

Thus $(eMe)x(eMe)=eMxMe$ and $(eMe)y(eMe)=eMyMe$.

Clearly, $MxM=MyM$ implies $eMxMe=eMyMe$.

Conversely, suppose $eMxMe=eMyMe$. Then $x=exe=e1x1e\in eMxMe=eMyMe$, so $x=euyve$ for some $u,v\in M$. As a consequence $MxM=MeuyveM\subseteq MyM$. By symmetry, the reverse inclusion also holds.

Let $e$ be an idempotent of the monoid $M$, $x$, $y$ be two elements of $eMe$. Then, $(eMe)\,x\,(eMe) = (eMe)\,y\,(eMe)$ if and only if $MxM = MyM$

3 Answers3