Prerequisites:
Let $S$ be a semigroup.
An element $a \in S$ is regular if there is an element $x \in S$ such that $axa = a$, and $S$ is regular if every element of $S$ is regular.
We let $Reg(S)$ denote the set of regular elements of $S$.
Question:
Suppose $S$ is a regular semigroup, and $e$ is an idempotent of $S$. Show that $Reg(eS)$ is a subsemigroup of $S$.
Thoughts:
So we want to show that if $ea, eb \in Reg(eS)$, then $eaeb \in Reg(eS)$ also. To do this, we need to find an element $ez \in eS$ such that $eaeb(ez)eaeb = eaeb$. My instinct would be to say that since $ea, eb \in Reg(eS)$, there are elements $ex$ and $ey$ such that $ea(ex)ea = ea$ and $eb(ey)eb = eb$.
I can't see an obvious way to proceed from this point, though. Clearly we expect to use the fact that $S$ is regular at some stage...any suggestions or hints would be very welcome!
I think I am doing something wrong because I am not using the assumption that $e$ is an idempotent.
Since $S$ is regular, there is a $w\in S$ such that $eaeb=eaebweaeb$. Since $eb\in Reg(eS)$, there is a $x\in S$ such that $eb=ebexeb$. Thus, $eaeb=eaeb(exebw)eaeb$. Since $xebw\in S$, the result follows.