If S is finite and $x/\rho$ is an idempotent of $S/\rho$ then $x/ \rho$ contains an idempotent

Question

Currently working on the following problem, need a little help with the solution to the last part, any hints?

Q: Let S be a semigroup, and let ρ be a congruence on S. Prove that if e ∈ S is an idempotent, then its equivalence class e/ρ is a subsemigroup of S, and is an idempotent in the quotient S/ρ. Also, prove that if S is finite and x/ρ is an idempotent of S/ρ then x/ρ contains an idempotent.

A: e\ $\rho$ = $ x \in S : (e,x) \in \rho $ ie the set of all things related to e. This is a subset of S and thus still associative so to show it is a subsemigroup we just need to show closure which is easy using the fact $e^{2} = e$ and that the relation is a congruence.

Showing $e / \rho$ is an idempotent in the quotient $S / \rho$ is also fine by considering e/$\rho$ * e/ $\rho$ = $e^{2} / \rho$ = $e / \rho$ as e is an idempotent so thats dandy.

However the last line is the bit thats confusing me "prove that if S is finite and $x/\rho$ is an idempotent of $S/\rho$ then $x/ \rho$ contains an idempotent." Surely if $x/\rho$ is an idempotent, clearly $x/\rho$ $\in$ $x/\rho$? I dont really get what this is asking or why the finiteness condition is needed... Any help would be appreciated.

Answer 1

It is simpler to state your question in terms of homomorphisms. Let $\rho: S \to T$ be a semigroup homomorphism. If $e$ is idempotent in $S$, then $\rho(e)$ is idempotent in $T$. Furthermore, the set $$ \{s \in S \mid \rho(s) = \rho(e) \} $$ is equal to $\rho^{-1}(\rho(e))$ and hence is a subsemigroup of $S$. Finally if $\rho(x)$ is equal to an idempotent $f$ of $T$, then $$ R = \{s \in S \mid \rho(s) = \rho(x) \} = \{s \in S \mid \rho(s) = f\} = \rho^{-1}(f) $$ is a nonempty subsemigroup of $S$. If $S$ is finite, then so is $R$ and thus $R$ contains an idempotent.