# The minimal ideal of a finite semigroup whose idempotents commute

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.

• This is in Howie's book on semigroups. Feb 14 at 14:33
• @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? Feb 15 at 8:31

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.

• 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. Feb 19 at 21:11