# Proof of Lallement’s Lemma

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?

Your proof is correct and I just have two remarks. First, when dealing with a single congruence, it is often simpler to think of it in terms of morphism. Secondly, you just need to know that $a^2$ has an inverse, so the following more general statement holds:
Let $\rho: S \to T$ be a semigroup morphism and let $a$ be an element of $S$ such that $\rho(a)$ is an idempotent. If $a^2$ has an inverse $x$, then $axa$ is idempotent and $\rho(axa) = \rho(a)$.
Proof. Since $x$ is an inverse of $aa$ we get $aaxaa = aa$ and $xaax = x$. Therefore $(axa)(axa) = axaaxa = a(xaax)a = axa\$ and thus $axa$ is idempotent. Since $\rho(a)$ is idempotent, we get $\rho(aa) = \rho(a)\rho(a) = \rho(a)\$. Thus $$\rho(axa) = \rho(a)\rho(x)\rho(a) = \rho(aa)\rho(x)\rho(aa) = \rho(aaxaa) = \rho(aa) = \rho(a)\ .$$ As you can see, this is just a rewriting of your proof.