This is (most of) Exercise 2.6.6 of Howie's "Fundamentals of Semigroup Theory". The first part is here.
The Details:
Let $S$ be a semigroup.
Definition 1: We call $S$ right simple if $\mathcal R=S\times S$, where $\mathcal R$ is Green's $R$-relation.
Definition 2: We call $S$ left cancellative if $(\forall a, b, c\in S) ca=cb\implies a=b$.
Definition 3: If $S$ is right simple and left cancellative, then $S$ is by definition a right group.
Definition 4: If for all $a, z\in S$, $az=z$, then $S$ is a right zero semigroup.
The Question(s):
This question is in 6 parts. The aim of the exercise is to show that, with the question linked to above, (f) a semigroup is a right group if and only if it is isomorphic to the direct product of a group and a right zero semigroup. I'll try each part separately.
Suppose $S$ is a right group.
(a) Show that the set $E$ of idempotents in $S$ is non-empty.
Since $S$ is right simple, $aS=S$ for all $a\in S$. Thus for a fixed $s\in S$, there exists $e\in S$ such that $se=s$, so that $se^2=se$ gives $e^2=e$. Hence $E\neq\emptyset$.$\square$
(b) Show that $E$ is a right zero subsemigroup of $S$.
Let $e,f\in E$. Then $ef=eef$ gives $f=ef\in E\subseteq S$ by left cancellation.$\square$
(c) Show that, for $e\in E$, $eb=b$ for every $b\in S$.
Let $b\in S, e\in E$. Then $eb=eeb$ gives $b=eb$ by left cancellation.$\square$
(d) Show that $Se$ is a subgroup of $S$ for every $e$ in $E$.
I suppose it would suffice to show that $aSe=Se=Sea$ for all $a\in Se$ but I don't know how. Please help.
(e) Let $f$ be a fixed element of $E$, and denote the group $Sf$ by $G$. Show that the map $\phi: G\times E\to S$ defined by $$(a, e)\phi=ae$$ is an isomorphism.
I'm stuck. Please help.
(f) Deduce that a semigroup is a right group if and only if it is isomorphic to the direct product of a group and a right zero semigroup.
This is a combination of part (d) and part (e), right? It is shown here that the direct product of a group with a right zero semigroup is a right group.
(d) The identity of $Se$ will be $e$: by (c) it is left identity in whole $S$, and by idempotency it is right identity in $Se$. And, every element $a\in S$ has a right inverse with respect to $e$, by applying $aS=S\ni e$.
(e) Can we find an inverse for $\phi$?
Given an $s\in S$, by (a) there is an $e\in E$ with $s=se$. This $e$ must be unique by left cancellation.
Then, prove that $s\mapsto (sf, e)$ is the inverse of $\phi$.
(f) Well, all of (a)-(e) are needed for this. Considering the set of idempotents and proving it's nonempty (a) is crucial.