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Given a finite semigroup $S$ containing a unique idempotent $e$, I can show that every element $s\in S$ has an ''inverse" $s^{-1}\in S$ in the sense that $s s^{-1} = s^{-1}s = e$:

Since $S$ is finite, every element $s$ has an idempotent power $s^{p_s}$ with $p_s\geq1$. It follows that $s^{2p_s}$ is also idempotent. Since $e$ is the only idempotent, we obtain $e=s^{2p_s}$. Since $2p_s\geq 2$, we can define $s^{-1} = s^{2p_s-1}$ in the sense stated above.

Now my question: Is $e$ actually a unit of $S$, i.e., does $es = se = s$ hold for every $s\in S$?

Jakobian
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Lemma 5
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    I'm not sure how you assert the existence of $p_s.$ There is certain a $p$ and $d$ such that $s^d=s^{p+d},$ but I can't see how you get $p_s$ from there. – Thomas Andrews Feb 26 '24 at 00:30
  • You can conclude $s^d(s^p)^2=s^ds^p,$ but I don't see how you get $(s^p)^2=s^p.$ – Thomas Andrews Feb 26 '24 at 00:34
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    @ThomasAndrews the existence of idempotent powers is known, I believe there are several proofs for that on SE. Here is my version:

    Let $a\in S$. By the pigeonhole principle, there are integers $m,p\geq1$ such that $a^m = a^{m+p}$, and therefore $a^m = a^{m+kp}$ for every $k\geq0$. Setting $b=a^m$ yields $b^{p+1} = a^{m+mp} = a^m = b$. It follows $$(a^{mp})^2 = b^{2p} = b^{p+1}b^{p-1} = bb^{p-1} = b^p = a^{mp},$$ hence $a^{mp}$ is idempotent.

     – Lemma 5 Feb 26 '24 at 00:40
  • The "one, two, many" additive semigroup has a unique idempotent, "many", but is not a group, as it is not cancellative: two+two = one+two = many. – Arturo Magidin Feb 26 '24 at 00:49
  • I think it does change your question, it seems to be true since left simple and right simple semigroups with unique idempotents are groups. While for simple semigroups there is bicyclic semigroup as a counter-example, its not finite – Jakobian Feb 26 '24 at 01:33
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    Also I'm sorry but you need to ask separate question if you want to ask about simple semigroups. Its a policy on math.stackexchange, would be unfair to people answering you otherwise. – Jakobian Feb 26 '24 at 01:35

1 Answers1

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What you're trying to prove is not true. Consider for example the two-element semigroup $\{e, a\}$ with the operation $\ast$ that always outputs $e$ ("the null semigroup"). It's clearly associative, and $e$ is the unique idempotent, but this is not a group.

Izaak van Dongen
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  • That is indeed true, thank you. I just noticed that in my context $S$ is a simple semigroup. As your example is not a simple semigroup, I wonder if that changes the outcome of my question. – Lemma 5 Feb 26 '24 at 01:01
  • @Lemma5, as Jakobian says, you should ask a new question about that! I don't know the answer off the top of my head. – Izaak van Dongen Feb 26 '24 at 03:56
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    Yeah I wasn't sure about that, but now I know :) In the meantime, I managed to solve it by myself. For anyone interested: If $S$ is simple, the statement is indeed true. To prove it, I used several results about Green's relations from J.E.Pin's "Mathematical Foundations of Automata Theory". You basically need to show that $S$ contains only one $\mathcal{H}$-class.. – Lemma 5 Feb 26 '24 at 14:10