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Let $G$ be a finitely generated group. Then the automorphism group $\text{Aut}(G)$ of $G$ need not be finitely generated.

However, there are classes of f.g. groups for which the automorphism group will always be f.g., such as polycyclic groups: it is a result by Auslander from 1969 that the automorphism group of a polycyclic group is even finitely presented.

Let $\text{Aut}^0(G) := G$, and for $n \geq 1$ let $\text{Aut}^n(G)$ be defined inductively as $\text{Aut}(\text{Aut}^{n-1}(G))$.

My question is: what are some examples of groups $G$ such that $\text{Aut}^n(G)$ is f.g. for all $n \geq 0$?

Now, while $\text{Aut}(G)$ is f.g. if $G$ is polycyclic by the above, the automorphism group of a polycyclic group need not be polycyclic, as far as I am aware, which would indicate that it is at least conceivable that $\text{Aut}^n(G)$ need not be f.g. for all $n \geq 1$ when $G$ is polycyclic. But I do not know of any counterexamples in this class.

Note that any finite group satisfies the question. Also, if $G$ is taken as $\mathbb{Z}$, then its automorphism group is $C_2$. Are there any infinite examples where every automorphism group is infinite f.g.?

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  • $\begingroup$ $\text{Aut}(\Bbb Z)=C_2$ and from there the tower is made of finite groups. $\endgroup$ Nov 6, 2019 at 9:25
  • $\begingroup$ @ArnaudMortier I was just writing an edit to address this! Thanks. $\endgroup$ Nov 6, 2019 at 9:26
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    $\begingroup$ The automorphism group of a free group of finite rank is finitely generated. $\endgroup$
    – Derek Holt
    Nov 6, 2019 at 16:54
  • $\begingroup$ @DerekHolt Yes, of course you're right -- even finitely presented. I had in mind a Baumslag-Solitar group as a counterexample, and free groups of finite rank as an example. Somehow they got swapped around in my mind when writing this up... $\endgroup$ Nov 7, 2019 at 0:42

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If $G$ is a complete group, that is, with trivial center and no outer automorphisms, then $$G\simeq \text{Aut}(G)$$ as every automorphism is the conjugation by some element, and the map $g\mapsto g\cdot g^{-1}$ has a trivial kernel.

For such a group, the tower is constant, so all you need is to find a finitely generated, infinite, complete group. You will find one in Victor S. Guba's A finitely generated complete group, 1987, DOI:10.1070/im1987v029n02abeh000969.

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    $\begingroup$ Guba's examples have properties much harder to obtain (uniquely divisible). A low-tech example of a complete group, in addition finitely presented, is $\mathrm{GL}_d(\mathbf{Z})\ltimes\mathbf{Z}^d$ for $d\ge 3$. $\endgroup$
    – YCor
    Nov 7, 2019 at 15:16

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