Existence of an idempotent element in a finite monoid.

Question

I'm trying to show that if $E$ is a finite monoid, then there exists $s$ such that $s²=s$, i.e, there exists an idempotent element in $E$.

What I did is this: Take an element $a$ of $E$, and consider the application $F_a :E\to E$ such that for $x$ in $E$, $F_a(x)=ax$. Now, if $E$ was a group, it's easy to see that $F_a$ is a permutation of $E$ (consider $F_{sym}(a)$) since $E$ is finite. Then there exists at least one element $x$ such that $F_a(x)=x$. Take $x=a$ we'll have $a²=a$. But what about the initial case, where $E$ is just a finite monoid?

Thanks.

Monoids have identity elements, so you can take $s$ to be the identity element of $E$. — Derek Holt, Oct 18 '16 at 16:17
I just reread the definition of a monoid, never paid attention to the presence of an identity element ! Thank you so much for this simple answer. — Zakaria Oussaad, Oct 18 '16 at 16:29
You could replace monoid by semigroup in the question. Then Hagen von Eitzen's answer applies. — Derek Holt, Oct 18 '16 at 17:21

score 5 · Answer 1 · answered Oct 18 '16 at 16:18

5

The sequence $a,a^2,a^3,a^4,\ldots$ is eventually periodic, i.e., there exists $N\in\Bbb N$ and $p>0$ suvch that $a^{n+p}=a^n$ for all $n>N$. This is weaker than what we have for groups (immediate periodicity), but it suffices to find idempotent $s$. For example $s=a^m$ for $m$ a multiple of $p$ that is $>N$.

answered Oct 18 '16 at 16:18

Hagen von Eitzen

382,479

Thank You so much for this concise answer ! – Zakaria Oussaad Oct 18 '16 at 16:23

Existence of an idempotent element in a finite monoid.

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