# Right ideal $eS$ in a simple semigroup $S$ without zero, with idempotent $e$ and minimal left ideal $L$ is a minimal right ideal of $S$.

Suppose that $$S$$ is like in the title, a simple semigroup without zero, with a minimal left ideal $$L$$ and idempotent $$e$$. We can always assume that $$L = Se$$.

I've already proved few facts about $$S$$. I know that $$S$$ is a sum of it's minimal left ideals, that $$Se$$ is a left group, and that $$H_e = eSe$$ is a group. This was also a part of the exercise.

How do I prove that $$eS$$ is a minimal right ideal? If possible, I'd like a hint before a full solution. It's an exercise from a subsection '$$0$$-minimal ideals and $$0$$-simple semigroups' if that helps.

Suppose that $$J$$ is a right ideal of $$S$$ such that $$JS\subseteq J\subseteq eS$$. If $$e\in J$$, then $$J = eS$$ is obvious. I want to draw a contradiction in the case when $$e\notin J$$, so far no luck.

Note that $$Sa = L$$ [$$aS = L$$] for any $$a\in L$$ if $$L$$ is left [right] minimal ideal.

• Did you already proved that $e$ is a minimal idempotent? Commented Jul 24, 2019 at 6:15
• If trusting this article, minimal idempotent would seem to be an idempotent such that $Se$, $eS$ are minimal left and right ideals and $H_e = eSe$ is a maximal subgroup of $S$. This is much stronger than what I want to prove Commented Jul 24, 2019 at 14:15

We will use that $$eSe$$ is a group.
Note that $$eS$$ being minimal right ideal of $$S$$ is equivalent to $$eS$$ being right simple. This means that every principal right ideal of $$eS$$ is equal to $$eS$$, i. e. $$\forall_{s\in S} eseS = eS$$.
Because $$eSe$$ is a group, for any $$s\in S$$ there exists $$r\in S$$ such that $$(ese)(ere) = e$$. Hence $$esereS = eS \subseteq eseS \implies eS = eseS.$$ Hence $$eS$$ is right simple, and so a minimal right ideal.