TITLE: Is it true that if $\mathrm{Aut}(G)$ is nilpotent then $G$ is also nilpotent?
QUESTION [6 upvotes]: Is it true that if automorphism group of $G$ is nilpotent then $G$ is also nilpotent?

REPLY [2 votes]: Yep. Also if aut(G) be a nilpotent of dgree n the G be a nilpotent of degree n+1. Since
Let G be a group. Let  Aut(G) be nilpotent then G is nilpotent also. Since
$\frac{G}{Z(G)}‎\cong‎ Inn(G)‎\unlhd‎  Aut(G)$
Since Aut(G) is nilpotent then $\frac{G}{Z(G)}$ is nilpotent. We have
$‎\gamma‎_n(\frac{G}{Z(G)})=Z(G)$
We can write
$‎\gamma‎_n(\frac{G}{Z(G)})=\frac{\gamma‎_n (G)Z(G)}{Z(G)}=Z(G)$
So, $\gamma‎_n (G)Z(G)‎\subseteq‎ Z(G)$ and $\gamma‎_n (G)‎\subseteq Z(G)$. Then G is nilpotent and $\gamma‎_{n+1} (G)‎=e$