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For every finite group $G$, if $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$.

I know that if $S$ is a semigroup, by adjoining an identity element to $S$, clearly $S$ is a subsemigroup of $S^1$. So in general, the order of a subsemigroup does not necessarily divide the order of the semigroup.

My question is a more special case:

For an automorphism $\phi$ of a finite semigroup $S$, the set of fixed points of $\phi$ is a subsemigroup of $S$. Is there any relation between the order of $S$ and the cardinality of the set of fixed points of $\phi$? Furthermore, is there any upper bound on $Aut(S)$, the cardinality of the automorphism group of $S$?

khers
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1 Answers1

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Consider the finite semigroup on the set $\{z_1,\dots,z_n\}$ with multiplication $z x=z:$ all the elements are left zeroes. Any permutation is an automorphism, so there is no better upper bound than $|\mathrm{Aut}(S)|\leq |S|!$ in terms of $|S|.$ And for any number $k\in\{0,1,\dots,n-2,n\}$ there is an automorphism with exactly $k$ fixed points: for $k=n$ use the identity, and for $k\leq n-2$ take a cycle of length $n-k.$

Dap
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