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I recently read this article Syntactic semigroups. In page $8$, he speaks about a J class having an idempotent is called regular:

A $\mathcal J$-class containing an idempotent is called regular. One can show that in a regular $\mathcal J$-class, every $\mathcal R$-class and every $\mathcal L$-class contains an idempotent.

As all Green J classes make a partition for given semigroup $S$, so if we accept $S$ has idempotents so they are included in some J classes. My question is:

Is there any reference that, I can find if J classes have just one idempotent inside or not? In fact, how can one find out, in a given semigroup (say, finite semigroup), whether a J class cannot have two idempotents?

Thank you for your time and the hints or any references.

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  • $\begingroup$ Perhaps, the most interesting things around this are finitely generated simple and bisimple semigroups -- you may find quite a lot of beautiful papers by Byleen about this $\endgroup$
    – Victor
    Aug 7, 2013 at 8:01
  • $\begingroup$ @Victor: Thanks for your suggestion. I follow the Byleen's papers now. $\endgroup$
    – Mikasa
    Aug 7, 2013 at 8:44
  • $\begingroup$ This question should probably be migrated to math.stackexchange.com since it is not a research question. $\endgroup$
    – J.-E. Pin
    Aug 29, 2013 at 0:06
  • $\begingroup$ @J.-E.Pin: There are few ones who are interested doing semigroups as I have seen up to now and moreover I am working on quasicommutative semigroups. And this question came to my mind and ask it here. Anyway, if they decide to make this migrated, That'll be Ok. :-) Thanks for your consideration. $\endgroup$
    – Mikasa
    Aug 29, 2013 at 1:01

1 Answer 1

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Any regular $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

Moreover, if a 0-simple semigroup $S$ with 1 has no other idempotents, then it is a group with 0.

Proof: Let $G$ be its subgroup of invertible elements. For every $a\in S\setminus 0$ there are such $x,y\in S$ that $xay=1$. Then $(ayx)^2=ayx$ whence $ayx=1$. Since $xay=ayx=1$, hence $x\in G$. But then $a=x^{-1}y^{-1}\in G$, i.e. $S=G\cup 0$.

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  • $\begingroup$ I guess, finiteness of $S$ is not that important -- we anyway will have $J^0$ a completely $0$-simple semigroup -- we do have a minimal idempotent $\endgroup$
    – Victor
    Aug 7, 2013 at 7:58
  • $\begingroup$ @Victor "we anyway will have $J^0$ a completely 0-simple semigroup" - what about 0-simple semigroups wich are not completely 0-simple? $\endgroup$ Aug 7, 2013 at 8:01
  • $\begingroup$ Well, in our setting, when $J^0$ has only two idempotents, we do have a minimal idempotent $\endgroup$
    – Victor
    Aug 7, 2013 at 8:02
  • $\begingroup$ Ah, ok, I see what you mean, I didn't read carefully what you wrote in the first paragraph $\endgroup$
    – Victor
    Aug 7, 2013 at 8:04
  • $\begingroup$ @Victor OK, nothing. $\endgroup$ Aug 7, 2013 at 8:05

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