$eSe$ in a finite semigroup

Question

Where is the following argument going wrong?

Let $S$ be a finite semigroup. There exists $e\in S$ such that $ee=e$. The subsemigroup $eSe = \{ese \mid s\in S\}\subseteq S$ is a monoid with the identity $e$. The map $ese\mapsto s$ is an injection from $eSe \to S$. Therefore $eSe = S$. Thus, every finite semigroup is a monoid ?! What?!

That map is not well defined. Plus not all semigroups have such an $e$. — , Jun 17 '20 at 17:33
@TokenToucan: Every finite semigroup does have an idempotent: see, e.g., here — Arturo Magidin, Jun 17 '20 at 17:44

score 2 · Accepted Answer · answered Jun 17 '20 at 17:40

2

The reason is, formula $ese\mapsto s$ does not always define a function, as there might be two different $s$ and $s'$ such that $ese = es'e$.

answered Jun 17 '20 at 17:40

JSch

56

1

As an explicit example, take the semigroup that maps every product to the first factor, $ab=a$ for all $a,b$. Then $ese=e$ for all $s\in S$. – Arturo Magidin Jun 17 '20 at 17:46
That's correct. We ought to show $ese = es'e$ implies $s=s'$, but this is false, in general, as seen from the previous comment. – AlvinL Jun 17 '20 at 17:58

$eSe$ in a finite semigroup

1 Answers1