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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.

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  • $\begingroup$ Did you already proved that $e$ is a minimal idempotent? $\endgroup$
    – J.-E. Pin
    Commented Jul 24, 2019 at 6:15
  • $\begingroup$ 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 $\endgroup$
    – Jakobian
    Commented Jul 24, 2019 at 14:15

1 Answer 1

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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.

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