# When is the automorphism group of a finite p-group abelian?

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.

• 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 $2^9$. – Derek Holt Sep 22 '18 at 13:17

## 1 Answer


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 $$2$$.

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