I recently read this article [*Syntactic semigroups*](http://www.liafa.jussieu.fr/~jep/PDF/HandBook.pdf). In page $8$, he speaks about a J class having an idempotent is called *regular*:



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> 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.
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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:



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



Idempotents in Green J classes
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. $S=G\cup 0$.


