# Is it true that if $\mathrm{Aut}(G)$ is nilpotent then $G$ is also nilpotent?

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

-

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 $n$.

-
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