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Are all finite monoids groups?

If I have a monoid $M$ such that $|M|=c$ for an integer $c$, then for all $x\in M$, we should have $x^k=e$ for some minimal $k$. Then, we have $x^{-1}=x^{k-1}$.

I don't see anything wrong with my proof, but I haven't found an answer online, and it doesn't feel like a `trivial' result, so I'm wondering if this is true.

Samuel Han
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    It's definitely not true: consider the monoid with underlying set ${0,1}$ and operation $\max$ (so that $0$ is the identity and $1$ is an annihilator). Using this example, do you see where your argument breaks down? – Noah Schweber Aug 23 '23 at 02:44
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    I see. I can have $x^n=x$ for some $n$, but I can't extend this argument to $x^m=e$ for $m=n-1$. – Samuel Han Aug 23 '23 at 03:04

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You argument is definitely wrong, but it can be modified to get a property related to groups. Take an element $x$ of a finite monoid $M$. First show that $x$ has an idempotent power, that is, for some $k > 0$, $x^k = x^{2k}$ (or see here for a proof). Next show that there is a maximal subsemigroup of $M$ which is a group with $x^k$ as identity.

J.-E. Pin
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