I am working through the exercises in Howie's 'Fundamentals of Semigroup Theory'. The following is from page 138:
Let $S$ be a commutative semigroup, and let $a,b \in S$. Write $a|b$ if $\exists x \in S$ such that $ax = b$. Define $\eta$ as the set of pairs $(a,b) \in S \times S$ for which $a | b^{m}$ and $b|a^{n}$ for some $m,n \in \mathbb{N}$.
(a) Show that $\eta$ is a congruence.
(b) Show that $S/\eta$ is a semilattice.
(c) Show that if $\rho$ is a congruence on $S$ such that $S/\rho$ is a semilattice, then $\eta \subseteq \rho$.
I have done parts (a) and (b), but I am completely stuck on the final part. If $S/\rho$ is a semilattice then $a^{2}\rho = a\rho$ for all $a \in S$, and so at the very least $\rho$ contains all pairs $(a,a^{2})$ and $(a^{2},a)$ in addition to the usual constraints that result from being a congruence. Is there something I'm missing?
First observe that the relation $|$ is a partial order on a semilattice. Moreover, if $\rho:S \to T$ is a semigroup morphism, then $a \mathrel{|} b$ implies $a\rho \mathrel{|} b\rho$.
Suppose now that $S/\rho$ is a semilattice and that $(a, b) \in \eta$. Then $a \mathrel{|} b^m$ and $b \mathrel{|} a^n$ for some $m,n > 0$. It follows that $a \mathrel{|} b^m \mathrel{|} a^{nm}$, whence $a\rho \mathrel{|} b^m\rho \mathrel{|} a^{nm}\rho$. But since $S/\rho$ is a semilattice, $a^{nm}\rho = a\rho$ and $b^m\rho = b\rho$. Therefore $a\rho = b\rho$ and thus $\eta \subseteq \rho$.