2
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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 Sep 22 '18 at 13:17
3
$\begingroup$

$\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 $2$.

$\endgroup$
  • $\begingroup$ 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 Sep 22 '18 at 16:48
  • 1
    $\begingroup$ @quid The question from yesterday. $\endgroup$ – Dietrich Burde Sep 22 '18 at 18:11
  • 1
    $\begingroup$ @quid will edit as requested. $\endgroup$ – Andreas Caranti Sep 23 '18 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.