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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

1 Answer 1

up vote 5 down vote accepted

No, a characteristic subgroup of an automorphism group of a group need not be isomorphic to the automorphism group of any group.

Here are two very well known infinite families of examples:

The cyclic group of order p (p an odd prime) is not the automorphism group of any group (such a group is abelian since its inner automorphism group is cyclic, but an abelian group either has inversion as an automorphism of order 2, or is an elementary abelian 2-group, and an elementary abelian 2-group either has trivial automorphism group or has a coordinate swap as an order 2 automorphism). However, the cyclic group of order p is a characteristic Sylow p-subgroup of AGL(1,p), which is the automorphism group of the dihedral group of order 2p and of itself. AGL(1,p) is the normalizer of a Sylow p-subgroup of the symmetric group on p-points.

The alternating group of degree n (n ≥ 9) is a characteristic, index 2 subgroup of its automorphism group, but is not itself the automorphism group of any group by Robinson (1982, MR678545).


At least as far as I understand it, automorphism groups of groups tend to be big and "full", and so it should not be surprising that many of their subgroups are not themselves automorphism groups of groups since they are "missing" something. For instance, a simple group cannot be an automorphism group unless it is complete; M12 is incomplete. Odd order cyclic groups do not work, since one is missing inversion (and indeed, the rest of AGL1).

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