# unitary subset of a semigroup $S$

• Let $S$ be a semigroup and $A$ be a subset of a semigroup $S$. Then $A$ is said to be right unitary $$(\forall a \in A) (\forall s \in S) \; sa \in A \; \Rightarrow s \in A$$
• Let $S$ be a semigroup and $A$ be a subset of a semigroup $S$. Then $A$ is said to be left unitary $$(\forall a \in A) (\forall s \in S) \; as \in A \; \Rightarrow s \in A$$ and unitary if it is both left and right uniitary.

• An element $a \in S$ is said to be regular if $\exists x \in S$ such that $axa = a$. A semigroup $S$ is said to be regular if every element of $S$ is regular.

• Let $a \in S$. Then $x \in S$ is said to be an inverse of $a$ if $axa= a$ and $xax = x$ and $V(a)$ is the collection of all inverse element of $a$.

Question:

Let $E$ be the set of all idempotents of a regular semigroup $S$, and suppose that $E$ is right unitary subset of $S$. Then $E$ is left unitary and satisfy the closure property.

My attempt is : Suppose $es \in E$, where $e \in E, \; s \in S$, then $$(sess')(sess') = (se)(ss's)(ess') = sesess'= s(es)(es)s' = sess'$$ for all $s' \in V(s).$

Thus $sess' \in E$ . if we show that $ess' \in E$, then $s \in E$, since $E$ right unitary. How to show that $ess' \in E$.

I think it's easier to prove the second part of the statement first; namely $e,f\in E$ implies $ef\in E$.
Suppose $e,f\in E$ and pick $z\in V(ef)$. Then $zef\in E$, so $ze\in E$, so $z\in E$. Also $efz\in E$, so $ef\in E$ as required.
Now we can complete your argument; $e,ss'\in E$ implies $ess'\in E$.