i am a bit weak with my congruence manipulation when it comes to semigroup theory. Could you possibly check my proof and give me constructive criticism on it?
Lemma: Let $S$ be a regular semigroup and let $\rho$ be a congruence on $S$. Assume that $a \in S$ is such that $a/ρ$ is an idempotent in $S/ρ$. If $x$ is any inverse of $a^{2}$ then prove that $e = axa$ is an idempotent, and that it belongs to $a/ρ$.
Proof Firstly we recall some definitions;
A semigroup S is said to be regular if $\forall x \in S$ $\exists y \in S$ such that $xyx=x$
So take a congruence $\rho$ on S, and let $a \in S$ be an idempotent in $S / \rho$ that is to say $(a,a^{2}) \in \rho$ or $a^{2}/\rho = a/\rho$. also we let $x$ be an inverse of $a^{2}$ that is to say $xa^{2}x=x$ and $a^{2}xa^{2}=a^{2}$ (from the definition of a mutual inverse element of a semigroup)
first things first we prove that $e$ is an idempotent.
$e^{2}=axaaxa=a(xa^{2}x)a=axa=e$ thus $e$ is an idempotent.
Now to show that it belongs to $a/ \rho$ we need to show that $(a,axa) \in a/ \rho$ or that they have the same congruence class ie $axa / \rho = a / \rho$
for this we proceed as follows
$$axa/ \rho = (a / \rho)(x/ \rho)(a/ \rho)$$ using the fact that a is an idempotent in $s/ \rho$ we know that $a/ \rho = a^{2}/ \rho$ and so our equality above becomes;
$$(a^{2} / \rho)(x/ \rho)(a^{2}/ \rho)=(a^{2}xa^{2})/ \rho$$ then finally we use the fact that $x$ is an inverse of $a^{2}$ and so this is the same as $a^{2}/ \rho = a/ \rho$ due to the idempotency (potentially not a word) of $a$ in the congruence, and thus we are done.
Is this correct? Any comments on how to make it better or any errors?