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Suppose $G$ is a finite $p$-group with odd $p$. Is it true, that $Aut(G)$ is abelian iff $G$ is cyclic?

When $G$ is cyclic, $Aut(G)$ is indeed abelian.

However, I do not know how to prove the statement that if $Aut(G)$ is abelian, then $G$ is cyclic. Nor do I possess any counterexamples.

The answer to the similar question about nilpotent groups is negative: When is the automorphism group of a finite $p$-group nilpotent?

Also, as any group with abelian automorphism group is metabelian, any $p$-group with abelian automorphism group is nilpotent of degree 2.

Any help will be appreciated.

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    $\begingroup$ There are nonabelian groups with abelian automorphism groups. See here A specific example is ${\mathtt{ SmallGroup}}(64,68)$, which has automorphism group elementary abelian of order $2^9$. $\endgroup$
    – Derek Holt
    Commented Sep 22, 2018 at 13:17

1 Answer 1

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$\newcommand{\Hom}[0]{\mathrm{Hom}}$$\newcommand{\Aut}[0]{\mathrm{Aut}}$$\newcommand{\Inn}[0]{\mathrm{Inn}}$As per your other question, the groups of this paper provide concrete examples of finite $p$-groups $G$ of nilpotence class $2$ such that $\Aut(G)$ is isomorphic to the additive group $\Hom(G/G', Z(G))$, and thus $\Aut(G)$ is abelian.

Note that if $G$ is a group such that $\Aut(G)$ is abelian, then so is $G/Z(G)$, as it is isomorphic to the subgroup $\Inn(G) \le \Aut(G)$ of the inner automorphisms of $G$. Thus $G$ is nilpotent, of nilpotence class at most $2$.

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  • $\begingroup$ Could you please expand this answer so that it is more self-contained. What is the relevance of this "other" question, and what is in that paper. $\endgroup$
    – quid
    Commented Sep 22, 2018 at 16:48
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    $\begingroup$ @quid The question from yesterday. $\endgroup$ Commented Sep 22, 2018 at 18:11
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    $\begingroup$ @quid will edit as requested. $\endgroup$ Commented Sep 23, 2018 at 17:26

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