In J. E. Pin: On Reversible Automata, LATIN 92, Springer LNCS 584, 1992 the author states that "it is a well-known fact of semigroup theory that the minimal ideal of a semigroup in which the idempotents commute is actually a group whose identity is an idempotent $f$". Where can I find a proof of this fact in the literature?

This is a slightly stronger statement than the one in The intersection of all ideals of a monoid because it is referring to a semigroup instead of a commutative monoid.

I have found a proof for the case of (non-commutative) monoids in the literature: Proposition 4.1. in S. W. Margolis, J. E. Pin: Inverse Semigroups and Varieties of Finite Semigroups, Journal of Algebra 110, 306-323, 1987 states that the minimal ideal of a finite monoid whose idempotents commute is a group. However, I cannot follow that proof. It lets $G$ be the minimal ideal of $M$ and defines $S := E(M) \cap G$ where $E(M)$ is the set of idempotents of $M$. It then claims that $S$ is a simple semigroup. I do see that it is a semigroup. But why is it simple? Once we know that $S$ is a simple semigroup I also see how to finish the proof.

  • $\begingroup$ This is in Howie's book on semigroups. $\endgroup$
    – Randall
    Feb 14 at 14:33
  • $\begingroup$ @Randall: I have looked through Howie's book "Fundamentals of Semigroup Theory". The statement most directly related to my question is Proposition 3.1.4. which states that the kernel of a semigroup is a simple semigroup. In the notation of my above question this shows that G is a simple semigroup. But, as far as I can tell, it does not answer my original question how to prove that S is simple. Or am I missing something? $\endgroup$ Feb 15 at 8:31

1 Answer 1


Let $S$ be a finite semigroup and let $I$ be its minimal ideal. You already seem to know that $I$ is a simple semigroup. If $e$ and $f$ are two idempotents of $I$, then $e \mathrel{\mathcal J} f$ (since $I$ is simple) and there exists an idempotent $g$ such that $e \mathrel{\mathcal R} g$ and $g \mathrel{\mathcal L} f$. Thus $eg = g$ and $ge = e$, but since idempotents commute, $e = g$. Similarly, $g = f$ and finally $e = f$. Consequently, $I$ contains only one idempotent and hence is a group.

  • $\begingroup$ Thank you for your reply! For future readers: it took me a while to figure out why g is idempotent. It follows from the Location theorem. $\endgroup$ Feb 19 at 21:11

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