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In Scott's classic textbook on Group Theory, he asks:

Suppose that $G$ is a finite group. Is the sequence of isomorphism types of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?

Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence $|Aut^{(n)}(G)|$ always bounded above?

It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:

Does there exists a finite group such that $Aut(G) \not \cong G$ but $Aut^{(n)}(G) \cong G$ for some $n \geq 2$?

Edit: Joel has pointed out that my question is perhaps even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.

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You might want to look at the answers to – David Speyer Jun 11 '10 at 20:10
@David: I hadn't noticed that this question had already been asked in But it's a good question so it's worth asking again! – Simon Thomas Jun 11 '10 at 20:30
How are the answers for 5635 not adequate? – Kevin O'Bryant Jun 11 '10 at 21:21
For everyone's information, some of the best answers to the other question amount to providing links to Simon Thomas' articles and book on the subject. – Joel David Hamkins Jun 11 '10 at 21:26
Mmmm ... maybe the infinite case is worth thinking about! – Simon Thomas Jun 12 '10 at 0:12

1 Answer 1

I think this anwers the question for infinite groups:

MR0470091 (57 #9858) Collins, Donald J. The automorphism towers of some one-relator groups. Proc. London Math. Soc. (3) 36 (1978), no. 3, 480--493. 20F55

Theorem (ii) states that if $G=\langle a,b \mid a^{-1}b^ra=b^s \rangle$ is a Baumslag-Solitar group with $r-s$ even, then $Aut(Aut(G))$ is isomorphic to $G$ and $G$ has an outer automorphism.

Moreover, when $r=1$, $G$ is the semidirect product $\mathbf Z \ltimes \mathbf Z[\frac 1 s]$, where $\mathbf Z$ acts via multiplication by $\frac 1 s$. Then $G$ is torsionfree, but $Aut(G)$ has an element of order 2 (see his lemma 3). If $G$ is represented as a matrix group, $(a,b) \mapsto \begin{pmatrix} s^a & b \\ 0 & 1 \end{pmatrix}$, then this outer automorphism is explicitely given by conjugation by $diag(i,-i)$, where $i$ is a square root of -1.

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This looks like an example but there is one part of your argument that doesn't seem correct. By Lemma 1 of Collins, when $r=1$, the group $G$ is centreless. This means that $Aut(G)$ is also centreless and hence embeds in $Aut(Aut(G)) \cong G$. Thus $Aut(G)$ cannot have an element of finite order. Of course, given the statement of Collins' Theorem, it seems almost certain that one of his groups does provide an answer to the infinite case of my question. – Simon Thomas Jun 12 '10 at 12:26
Indeed, after reading the paper more closely I think Theorem (ii) should read that Aut(Aut(G)) is isomorphic to Aut(G), so this paper doesn't seem to provide an example. Sorry about that. – Guntram Jun 12 '10 at 13:15
Looking more carefully at Collins' paper, I believe there is a typo in the statement of Theorem 1. Combining Proposition A and Proposition B, we see that $Aut(Aut(G)) \cong Aut(G)$ instead of $Aut(Aut(G)) \cong G$. And my previous comment shows that $Aut(Aut(G)) \not\cong G$. So I no longer believe that Collins' paper answers the infinite case of my question. – Simon Thomas Jun 12 '10 at 13:23
I see that we spotted the typo simultaneously! – Simon Thomas Jun 12 '10 at 13:24

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