# Subgroups of automorphism groups

Let $G$ be some (infinite) group, and let $Aut(G)$ be its automorphism group. Assume $H\leq Aut(G)$.

Under what conditions can I construct another group (or, say, graph), $\hat{G}$, such that $Aut(\hat{G})=H$? If so, is there an algorithm to do so?

I am pretty sure that this is not always possible - there are some groups which never occur as automorphism groups of other groups. I would therefore be interested to know if we can apply conditions on either $G$ or $H$ to get this to work.

I presume $H\lhd G$ is not sufficient, as then if $H$ is a centerless group which never occurs as an automorphism group of another group then we would have a counter-example (as $Inn(G) \cong G$ if the centre of $G$ is trivial). So...what about characteristic?

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Every group is the automorphism group of a graph (mathworld.wolfram.com/GraphAutomorphism.html). This question seems to be phrased at a strange level of generality to me. What do you actually want to do? –  Qiaochu Yuan Aug 4 '11 at 15:37
Well...basically, I would like the algorithm. But what I actually want to do is, well, pretty complicated...(and would take more than the 407 characters I have left in this comment to explain!) –  user1729 Aug 4 '11 at 15:41
What I mean is, why are you not interested in the more general problem "what groups are automorphism groups of groups"? Is $\hat{G}$ supposed to be related to $G$ in some way? –  Qiaochu Yuan Aug 4 '11 at 15:44
Well, "what groups are automorphism groups of groups" is ridiculously general! –  user1729 Aug 4 '11 at 15:47
Okay, but I'm not sure how being a subgroup of the automorphism group of another group is supposed to help. That's not a restrictive condition. Every group embeds in the automorphism group of another group (say a sufficiently large vector space over $\mathbb{F}_2$). It seems to me that conditions on $H$ would be much more useful than conditions on how $H$ sits in $\text{Aut}(G)$ (although maybe the latter is more restrictive than I think it is). –  Qiaochu Yuan Aug 4 '11 at 15:50