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Why is it so, that a finite semigroup, say $(S, \circ)$ has $a^m=a^n$ for positive integers $m$ and $n$ with $m>n$ for $a\in S$? Does it imply some sort of periodicity in the binary composition?

Martin Sleziak
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Subhasis Biswas
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    It follows from the pigeonhole principle: $\left\{a,a^2,\dots\right\}$ is a subset of $S$, hence it must be finite; this proves your claim. – Guy Sep 02 '17 at 16:15
  • Could you please post it as an answer??? I am having trouble reading it. – Subhasis Biswas Sep 02 '17 at 16:19
  • @Guy oh, I see. The first condition of being a binary operation is to be closed. Hence, all the repeated operations must lie within the set and their values may coincide after certain repetitions??? – Subhasis Biswas Sep 02 '17 at 16:24
  • Yes, you are right. – Guy Sep 02 '17 at 16:37

2 Answers2

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If no two of $a^1, a^2, a^3, \ldots$ were the same, then the semigroup would not be finite.

Michael Hardy
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You have that $a^{m-n}=e$ the identity, where $m-n>0$ is an integer. So that, there exist a minimal integer $k>0$ such that $a^k=e$. Then we have that the elements of the group are $\{a,a^2,\ldots,a^{k-1},a^k=e\}$.

Michael Hardy
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José Alejandro Aburto Araneda
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