Periodic Automorphism Towers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:12:07Z http://mathoverflow.net/feeds/question/27861 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27861/periodic-automorphism-towers Periodic Automorphism Towers Simon Thomas 2010-06-11T19:51:36Z 2010-06-12T17:16:16Z <p>In Scott's classic textbook on Group Theory, he asks:</p> <p>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?</p> <p>Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence $|Aut^{(n)}(G)|$ always bounded above?</p> <p>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:</p> <p>Does there exists a finite group such that $Aut(G) \not \cong G$ but $Aut^{(n)}(G) \cong G$ for some $n \geq 2$?</p> <p>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.</p> http://mathoverflow.net/questions/27861/periodic-automorphism-towers/27917#27917 Answer by Guntram for Periodic Automorphism Towers Guntram 2010-06-12T09:02:05Z 2010-06-12T09:29:16Z <p>I think this anwers the question for infinite groups:</p> <p>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 </p> <p>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. </p> <p>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, <code>$(a,b) \mapsto \begin{pmatrix} s^a &amp; b \\ 0 &amp; 1 \end{pmatrix}$</code>, then this outer automorphism is explicitely given by conjugation by $diag(i,-i)$, where $i$ is a square root of -1.</p>