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A power automorphism maps every subgroup of a group to within itself, with equality if the group is finite. More specifically, for a subgroup $H$ of $G$, a power automorphism $f$ has $f(H) \subseteq H$. If $G$ is finite then $f(H) = H$.

There are lots of these of the form $f:G \to G$, $f(x) = x^n$, especially for finite abelian groups. This is a universal power automorphism.

I want to find an example of a non-trivial power automorphism of a finite group that isn't universal, particularly with some insight in how it is constructed. Can someone give me an example?

abnry
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    So a power automorphism of a finite group is just any old automorphism? You haven't really given a definition for the non-universal case. – Steve D Mar 06 '18 at 22:41
  • @SteveD: A power automorphism is not just any old automorphism, it's one that maps any subgroup to itself. The OP's explanation isn't very clear but the wikipedia discussion of a power automorphism is. The OP doesn't need to give the definitions because they are well known (but not, apparently, by you). – Rob Arthan Mar 06 '18 at 22:53
  • It is called a power automorphism because any element $g \in G$ of order $n$ must be mapped to a power of $g$, as ${1, g, g^2, \dots, g^{n-1}}$ is a cyclic group, and is preserved by the power automorphism. The thing is, the powers may vary for each $g$. I think it is kind of strange definition. I think a power automorphism should be what is called a universal power automorphism. – abnry Mar 06 '18 at 23:08
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    I added some clarification of what precisely a power automorphism is. – abnry Mar 06 '18 at 23:10
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    Small typo: you mean it maps every subgroup to itself, not every subset. – Lazward Mar 06 '18 at 23:26

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Here's an example: consider the automorphism of the quaternion group $Q_8$ whose action on its generators $i$, $j$, $k$ is as follows: $$i \mapsto i \hspace{10pt} j \mapsto -j \hspace{10pt} k \mapsto -k$$

(Of course, this also works for $i \mapsto -i , j \mapsto j , k \mapsto -k$ and $i \mapsto -i \hspace{10pt} j \mapsto -j \hspace{10pt} k \mapsto k$.)

Alexander Gruber
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