Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it true that if automorphism group of $G$ is nilpotent then $G$ is also nilpotent?

share|cite|improve this question

Yes, if the automorphism group of $G$ is nilpotent, then so is the subgroup of inner automorphisms, which is isomorphic to $G/Z(G)$. But if $G/Z(G)$ is nilpotent, then so is $G$ (standard exercise).

To answer the comment: The converse does not hold. For example, if we take the abelian (and thus nilpotent) group $\mathbb{Z}/2\times \mathbb{Z}/2$ then the automorphism group is isomorphic to $S_3$ which is not nilpotent. If we take some larger still but similar examples, ie $(\mathbb{Z}/p)^n$ for a prime $p$ and any natural number $n$, we get that the automorphism group is isomorphic to $GL_n(\mathbb{F}_p)$ which is never nilpotent (unless $n = 1$) and in fact only solvable for small values of $p$ and $n$.

share|cite|improve this answer
What about the inverse? I see that we have the inverse for any cyclic group. – Babak S. Nov 12 '12 at 14:08
Thanks for the answer +1. – Babak S. Nov 12 '12 at 14:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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