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
____________________________________________________