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Is there a simple proof of the following fact:

Fact. Let $S$ be a completely simple semigroup with cancellations, i.e. each of the equalities $xa=xb$, $ax=bx$ implies $a=b$. Prove that $S$ is a group.

Using Sushkevich-Rees Theorem, I can prove it, but my proof is not elegant.

Can you prove this fact using only simple arguments? Probably, you know a paper, where this result was proved at the first time?

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A completely simple semigroup has by definition a (primitive) idempotent $e$ and $SxS=S$ for all $x\in S$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. Then $xyuxyu=xyu$ and so $xyu=1$ as $1$ is the unique idempotent. Similarly, $yuxyux=yux$ and hence $yux=1$ because each idempotent is the identity. Thus $x$ is invertible with inverse $yu$. So $S$ is a group.

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  • $\begingroup$ I simplified the argument getting rid of the bicyclic monoid. Now it is elementary and works for simple cancellative semigroups with an idempotent. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 7, 2017 at 14:34
  • $\begingroup$ A simple semigroup with cancellations is not necessarily a group? $\endgroup$
    – user6976
    Commented Nov 30, 2019 at 2:09
  • $\begingroup$ @MarkSapir, no. See Trotter, P. G. Cancellative simple semigroups and groups. Semigroup Forum 14 (1977), no. 3, 189–198. They do not even have to embed in groups $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2019 at 2:29
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    $\begingroup$ OK. Thank you ! $\endgroup$
    – user6976
    Commented Nov 30, 2019 at 2:50

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