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Question: Let $G$ be a finite group. If the group automorphism $\sigma$ satisfies that $\sigma(H)=H$ for some nontrivial subgroup $H$, we call $\sigma$ has the preserving property.

Now, let $G$ be a non-abelian group. How to prove that any automorphism has the preserving property?

Trying: I have tried the abelian case, once we decomposite the abelian group into the product of cyclic groups with distinct prime powers, we can construct such automorphism by shifting each cyclic group. But I have no idea with the non-abelian case.

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    $\begingroup$ I expect you meant to exclude a lot of trivial cases. $H=e$, say. Or $\sigma = id_G$. $\endgroup$ – lulu Nov 14 '18 at 21:04
  • $\begingroup$ Also, if $G$ has a non-trivial center, and $\sigma$ is inner... $\endgroup$ – lulu Nov 14 '18 at 21:05
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    $\begingroup$ On the other hand, if $\sigma$ is inner and $G$ is simple, $H$ can only be $\{e\}$ or $G$. $\endgroup$ – Robert Israel Nov 14 '18 at 21:09
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    $\begingroup$ ... and there are simple groups whose automorphisms are all inner. $\endgroup$ – Robert Israel Nov 14 '18 at 21:11
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    $\begingroup$ Any automorphism of a group $G$ preserves the subgroups $\{e\}$ (where $e$ is the identity element) and $G$. But it need not preserve any other subgroups, for the trivial reason that there might not be any other subgroups (if $G$ is cyclic of prime order). But even if there are other subgroups, an automorphism need not preserve any of them. For example, the Klein 4-group (the non-cyclic group of order 4) has three subgroups other than $\{e\}$ and the whole group, but none of these is preserved by an automorphism that cyclically permutes the 3 non-$e$ elements. $\endgroup$ – Andreas Blass Nov 15 '18 at 1:45

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