If $\lambda\subseteq\mathscr{L}, \rho\subseteq\mathscr{R}$, then $\lambda\circ\rho=\rho\circ\lambda$.

Question

This is (the first part of) Exercise 2.4 of Howie's Fundamentals of Semigroup Theory.

The Details.

Quoting Howie (on page 22 of my copy):

Definition: A relation $R\subseteq S\times S$ on a set $S$ is called left compatible (with the operation [implicit] on $S$) if $$(\forall s,t,a\in S)\quad (s, t)\in R\implies (as, at )\in S,$$ and right compatible if $$(\forall s,t,a\in S)\quad (s, t)\in R\implies (sa, ta)\in R.$$ It is called compatible if $$(\forall s,t,s', t'\in S)\quad [(s, t)\in R\text{ and } (s', t')\in R]\implies (ss', tt')\in R.$$

A left [right] compatible equivalence is called a left [right] congruence. A compatible equivalence relation is a congruence.

By $\mathscr{L}, \mathscr{R},$ and $\mathscr{D}$, of course, I mean the corresponding Green's relations.

The Question.

Let $S$ be a semigroup. If $\lambda\subseteq\mathscr{L}$ is a right congruence and $\rho\subseteq\mathscr{R}$ is a left congruence, then $\lambda\circ\rho=\rho\circ\lambda$.

My Attempt.

Suppose $s(\lambda\circ\rho)t$. Then $\exists u\in S$ such that $s\lambda u$ and $u\rho t$, so $s\mathscr{L}u$ and $u\mathscr{R}t$. (Clearly $s\mathscr{D}t$ so $s(\mathscr{R}\circ\mathscr{L})t$. I guess I could use this information somehow.)

We have $\exists\ell, k\in S^1$ such that $s=\ell u$ & $u=ks$ and $\exists\tau, \sigma\in S^1$ such that $u=t\tau$ & $t=u\sigma$. I suspect that there's some way of using the right-, left- compatibility of $\lambda$ and $\rho$, respectively, with some cunning words on $\{s, t, u, \ell, k, \tau, \sigma\}$; in particular, of course, I need some $v\in S$ with $s\rho v\lambda t$.

Much to my amusement I accidentally proved $\mathscr{L}\circ\mathscr{R}=\mathscr{R}\circ\mathscr{L}$ again by letting $\psi=\ell u\sigma$.

I doubt that $s\rho\psi\lambda t$ since some information might be lost by replacing $\lambda$ with $\mathscr{L}$, $\rho$ with $\mathscr{R}$ earlier . . .

Please help :)

Answer 1

This is maybe a bit late but here's a solution.

Suppose that $(x,y)\in \lambda\circ\rho$. Then there exists $z\in S$ such that $(x,z)\in \lambda\subseteq \mathscr{L}$ and $(z,y)\in \rho\subseteq \mathscr{R}$. Hence there exist $u,v\in S^1$ such that $uz=x$ and $zv=y$ and so $$(xv, y)= (xv, zv)\in \lambda\qquad \text{and}\qquad (x, uy)=(uz, uy)\in \rho.$$ But $xv=uzv=uy$ and so $(x, uy)=(x,xv)\in \rho$ and $(xv,y)\in\lambda$ and so $(x,y)\in \rho\circ\lambda$.

We have shown that $\lambda\circ \rho\subseteq \rho\circ\lambda$ and a dual argument proves that $\rho\circ\lambda\subseteq \lambda\circ\rho$.