From: "peter.neumann"To: Simon Thomas cc: group-pub-forum@maths.bath.ac.uk Subject: Re: Automorphism groups of countable groups Date: Fri, 18 Apr 1997 18:58:19 +0100 () On Mon, 14 Apr 1997 14:32:31 -0400 Simon Thomas wrote: > Does there exist a countably infinite centreless group $G$ such that > $Aut(G)$ is countable and $Aut(Aut(G))$ is uncountable? > > Simon Thomas Two points about this. First that it is easy to see that if $G$ is finitely generated (and centreless) then $Aut(Aut(G))$ is countable. This, and the fact that I cannot easily make a counterexample makes me *suspect* that if $Aut(G)$ is countable then also $Aut(Aut(G))$ is countable. But I have no adequate evidence. Secondly that on mattters like this I would automatically turn for information to that very clever chap Simon Thomas in Rutgers. All best wishes, $\Pi$eter Queen's: 18.iv.97 ____________________________________________________ From: "peter.neumann" To: Simon Thomas cc: group-pub-forum@maths.bath.ac.uk Subject: Re: Automorphism groups of countable groups Date: Sat, 19 Apr 1997 15:07:17 +0100 () On Fri, 18 Apr 1997 14:05:37 -0400 Simon Thomas wrote: > Dear Peter, > Jim Wiegold guesses the opposite answer; and suggests > Q*Q or something like it as a possible answer. > I find both answers plausible. David Evan's theorem > on countable structures with countable automorphism groups > certainly suports your guess. There is a finite subset $F$ > of $G$ such that each automorphism of $G$ is determined by its > restriction to $F$. But maybe $F$ might be "less rigidly embedded" > in $Aut(G)$? > All the best, > Simon Dear Simon, Although Jim's guessing is usually far more accurate than mine, I'd need to know much more what his `something like that' meant before I was prepared to start laying bets. If, as I suppose, $Q$ means something isomorphic to the additive group of rational numbers, then the free product $Q * Q$ is certainly not an example. Let me call those visible free factors $Q_1$ and $Q_2$ respectively and define $A := \Aut(Q_1 * Q_2)$. The centraliser of a non-identity element $u$ of the free product is cyclic unless $u$ is conjugate to an element of $Q_1$ or $Q_2$. It follows easily that if $\alpha \in A$ then there exists an inner automorphism $\phi$ such that $\alpha \phi^{-1}$ fixes $Q_1 \cup Q_2$ setwise, and then that $A$ is the semidirect product of $Q_1 * Q_2$ with the wreath product $R \Wr Z_2$, where $R := \Aut(Q)$. Now $Q_1 * Q_2$ contains and is generated by all the subgroups of this semidirect product (that is, of $A$) isomorphic to $Q$, and so $Q_1 * Q_2$ is characteristic in $A$. Therefore $A = \Aut(A)$, so that $\Aut(\Aut(Q_1 * Q_2)) = \Aut(Q_1 * Q_2)$ in this case. All best wishes, $\Pi$eter Queen's: 19.iv.97 ____________________________________________________ Dr Peter M. Neumann, Queen's College, Oxford OX1 4AW tel. +44-1865-279 178 (messages: 279 120/21/22) fax: +44-1865-790 819 ____________________________________________________