Periodic Automorphism Towers - MathOverflow

Periodic Automorphism Towers

2010-06-11T19:51:36Z

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.

Answer by Guntram for Periodic Automorphism Towers

2010-06-12T09:02:05Z

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.