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

2 Answers 2

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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 $n$.

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    What about the inverse? I see that we have the inverse for any cyclic group.2012-11-12
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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 $\gamma‎_{n+1} (G)‎=e$