Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it possible for a finitely generated, recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

It has been known for some time (1989, I believe) that every group $Q$ can be realised as the outer automorphism group of some group $G_Q$. Indeed, $G_Q$ can be taken to have certain properties, such as simple or torsion-free metabelian with trivial centre. If $Q$ is assumed to be countable then $G_Q$ can be taken to be a locally finite $p$-group, or finitely generated. If $Q$ is assumed to be finitely presentable then $G_Q$ can be taken to be finitely generated and residually finite. If $Q$ is assumed to be finite then $G_Q$ can be taken to be the fundamental group of a closed hyperbolic $3$-manifold.

Hand-wavingly: As $Q$ acquires more finiteness properties then so can $G_Q$.

I cannot find anything which talks about recursive presentability though. I have a proof that if $Q$ is recursively presented then $G_Q$ can be taken to be finitely generated and recursively presented (and some other nice properties which are what I am really interested in but they are irrelevant to this question). I am (essentially) wondering if it is reasonable to try and take this further: will I ever be able to prove that $G_Q$ can be taken to be recursively presented even though $Q$ is not recursively presented?

My initial instinct was that this is silly - obviously a recursively presented group cannot have anything to do with a non-recursively presented group. I mean, they cannot be their subgroups, so, you know, wave your hands a bit and you have a proof...On the other hand, outer automorphism groups are very different from subgroups. I cannot think of any valid reason why finitely generated, recursively presented $G$ must have $\operatorname{Out}(G)$ recursively presented.

So, again, (just to reiterate) does the following hold,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

share|improve this question
2  
You should specify that your group is finitely generated (more precisely has a presentation with finitely many generators and a recursive family of relators), otherwise Out could be uncountable. –  Yves Cornulier Apr 3 '13 at 16:28
2  
... while I guess you allow Out(G) on the other hand to be infinitely generated. So in any case you should clarify. –  Yves Cornulier Apr 3 '13 at 16:32
2  
For example, the free abelian group on countably many generators is computably presentable, but the outer automorphism group has size continuum, as it includes all those arising from any permutation of the generators, and is therefore not computably presentable. –  Joel David Hamkins Apr 3 '13 at 16:59
    
Sorry, you are right, I was assuming the groups $G$ were finitely generated. I have edited the question accordingly. –  user6503 Apr 4 '13 at 9:43
    
Fearful of being too glib a second time round, it seems to me that an easy argument shows that, if G is recursively presented, then Out(G) admits an infinitely generated recursive presentation. Indeed, you can enumerate elements of Aut(G), and also relations between them, since both of these just involve checking identities in G. But then you can add the inner automorphisms to your list of relations easily enough. –  HJRW Apr 4 '13 at 13:37
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.