# Is there a name for a semigroup whose idempotents form a subsemigroup?

For a semigroup $S,$ I will denote by $E(S)$ the set of all idempotents of $S$. For $X\subseteq S,$ let $X^2$ mean $\{xy\,|\,x,y\in X\}.$

Is there a name for the class of semigroups $S$ such that $$\left(E(S)\right)^2\subseteq E(S)?$$

To have an example, in every inverse semigroup, the idempotents form a subsemigroup. More generally, as rschwieb points out in a comment, any semigroup such that the idempotents commute with each other satisfies this condition.

I need a name to be able to search for information about such semigroups. So any contribution besides the name will be welcomed.

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We can start with abelian semigroups meaning "commutative". In ring theory sometimes "abelian" is used to mean "central idempotents", and that would work here too. –  rschwieb Jun 28 '12 at 20:48
@rschwieb Do you mean that idempotents form a subsemigroup iff they commute with everything? –  user23211 Jun 28 '12 at 20:55
No, sorry, I did not even attempt to classify, only to give more examples. I mean that if idempotents commute with each other, then they are obviously closed under multiplication: $efef=eeff=ef$ if $e,f$ are idempotents. –  rschwieb Jun 28 '12 at 20:56
@rschwieb Oh, OK. Yes, I know about it -- I should have put it in the question. This is actually the case in inverse semigroups: an inverse semigroup is a regular semigroup whose idempotents commute. –  user23211 Jun 28 '12 at 21:00
I've just found out that the answer to my question in the comment is no: in the bicyclic semigroup, the idempotents do not commute with everything. –  user23211 Jun 28 '12 at 21:01

According to wikipedia, we have

• A regular semigroup whose idempotents forms a subsemigroup is called an orthodox semigroup.
• A completely regular semigroup whose idempotents forms a subsemigroup is called an orthogroup.

Every semigroup can be embedded into a regular semigroup. Perhaps subsemigroup of an orthodox semigroup comes closest to the condition you are looking for. I don't know whether there exists a semigroup which satisfies your condition without being a subsemigroup of an orthodox semigroup.

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Thank you, Thomas. I haven't seen these terms before, so this is very helpful. I will accept this answer if no full answer comes up. –  user23211 Jun 29 '12 at 22:56

In a very important paper, C.J. Ash, 'Finite semigroups with commuting idempotents', J. Austra. Math. Soc. 43 (1987), 81-90, Chris Ash proved that a finite semigroup has commuting idempotents if and if it divides (that is, is a homomorphic image of a subsemigroup) of an inverse semigroup.

There are examples of finite semigroups that have commuting idempotents, but do not embed in any inverse semigroup.

Birget, Margolis, Rhodes, `Semigroups whose idempotents form a subsemigroup' Bull. Austral. Math. Soc. 41 (1990) [161-184] prove that a finite semigroup has idempotents that form a semigroup if and only if it divides (but not necessarily embeds into) an orthodox finite semigroup, that is a regular semigroup with idempotents a subsemigroup. You can preserve the variety of bands in which the idempotents of the original semigroup reside.

These were all preliminaries for the so called Rhodes Type II conjecture which was also proved by Ash.

For details in book form, see Rhodes-Steinberg, "The q-theory of finite semigroups", Springer, 2009. u

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A semigroup whose idempotents form a subsemigroup is called an E-semigroup in a bunch of papers, e.g. J. Almeida, J.-E. Pin and P. Weil which was originally published in 1992, Gomes and Howie (1998), Weipoltshammer (2002), Gigoń (2012), or Fountain and Hayes (2014). Note that all of these were published after Birget, Margolis, and Rhodes' paper (1990), mentioned by Margolis himself above, which did not use this E-semigroup terminology. If the E-semigroup is also regular, then it is called an orthodox semigroup. (This latter terminology is much older dating to 1969.) I have updated the Wikipedia page on orthodox semigroup to mention E-semigroup, but E-semigroup is still a "red link" over there. I don't think the E-semigroup terminology has made it in any semigroup textbooks though and it seems to actually clash with the same name chosen for something else in Arveson's textbook. (Do note that Gomes & Howie's paper on the topic came out 3 years after Howie's textbook was published though.)

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