## Outer automorphisms

I have just put on the arXiv a paper I wrote with Sam Tarzi ten years ago. I want to say here something about the context, the contents of the paper, and the reason for posting it now.

Outer automorphisms of a group are automorphisms which are not induced by conjugation by elements of the group itself.

If a group G naturally occurs as a permutation group, then its automorphism group contains as a subgroup those automorphisms which are induced by conjugations in the symmetric group. If this subgroup is the entire automorphism group, then we can understand the outer automorphisms by looking just within the symmetric group. What I am concerned with here is whether or not this condition holds.

For the symmetric group, we have the remarkable Schreier–Ulam theorem: if n is any cardinal number, finite or infinite, except 6, then the symmetric group on n letters has no outer automorphisms. The exception, which I have described here, arguably is the key to understanding many sporadic phenomena in group theory and beyond. For example, the existence of an outer automorphism of S6 means that this group acts in two different ways on sets of size 6; taking the union of two such sets, we can enlarge the group to the Mathieu group M12. This group also has an outer automorphism not induced by permutations, and so has two actions on sets of size 12, which similarly give rise to the large Mathieu group M24.

As I described here, there is no similar phenomenon for transformation semigroups. The automorphism group of the full transformation semigroup on any set is the symmetric group on that set (a theorem of Sullivan, and maybe others); and this extends to various “large” subsemigroups, in particular, those whose units form a synchronizing permutation group. It is conjectured that it extends to subsemigroups whose units form a primitive permutation group; we took our first steps towards this in the paper just cited.

Now to the paper with Sam Tarzi. To warm up, consider the random graph R (discussed here). Let G be the automorphism group of R, which we regard as a permutation group on R. Now G has an outer automorphism induced by a permutation. For R is isomorphic to its complement, and a permutation inducing such an isomorphism must be an automorphism of G (since the group preserving R is identical with the group preserving the complement of R). Now it is possible to show, using results of Hrushovski and of Hodges and others, the following facts:

• The outer automorphism group of G has order 2; that is, any automorphism of G is induced by a permutation of the vertex set of R, which is either an automorphism or an anti-automorphism of R.
• The automorphism group of G does not split over G; in other words, there is no subgroup H such that GH = Aut(G) and GH = 1.

The second fact is very easy to show. Suppose that such a subgroup H exists. Then the non-identity element h of H would be an isomorphism from R to its complement whose square is the identity. Take a 2-cycle of h. This is fixed, as a set, by h; so if it is an edge, then its image is also an edge, and similarly for non-edge. But this contradicts the fact that h interchanges edges with non-edges.

Now the paper that Sam Tarzi and I wrote concerns a slightly more general situation. We replace R by Rm, the randomly m-coloured complete graph. (Similar to R, if we take a palette of m colours, and for each edge of the complete graph on a countable set we apply a random colour to it, then we obtain an object isomorphic to Rm with probability 1. If m = 2 and the colours are “solid” and “transparent”, then the edges coloured with the solid colour form the random graph.

Let Gm be the automorphism group of Rm. (We assume that m > 1.) Our theorem has three parts.

• The automorphism group of Gm is induced by permutations of the vertices; so the outer automorphism group is isomorphic to the symmetric group Sm.
• The automorphism group splits over Gm if and only if m is odd.
• If m is even and not divisible by 8, then the smallest supplement of Gm in its automorphism group (a subgroup H such that GmH = Aut(Gm)) has order 2·m! (that is, just twice that of a hypothetical complement).

(For m = 2, this means that the random graph has an anti-automorphism which is a permutation of order 4.)

A corollary is that the groups Gm are pairwise non-isomorphic, since they have different outer automorphism groups. (John Truss showed that all these groups are simple.)

There is an interesting unsolved problem here: what happens if m is divisible by 8? Is there a finite supplement, and if so, what is the smallest such?

Anyway, we wrote up this result, which I think is nice, and submitted the paper to a journal (not by any means a top journal). They very quickly rejected it, and for one reason or another we never got round to re-submitting it.

But there was a preprint on my web page at Queen Mary. This was found by Greg Cherlin, who talked at my birthday conference in Lisbon on a very wide generalisation inspired by this. I described it briefly here. So I decided to put the paper on the arXiv, where it now resides for the foreseeable future. If Greg or anyone else needs to refer to it, the arXiv is less likely to disappear than my Queen Mary webpage. Also, someone might be inspired to look at the paper and tackle the unsolved problem above.  I count all the things that need to be counted.
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### 2 Responses to Outer automorphisms

1. David Roberts says:

>the key to understanding many sporadic phenomena in group theory

How much of the happy family is made possible by this iterated process? Does it impinge on any constructions in the second generation?

• Peter Cameron says:

Good question. We have S_6 -> M_12 -> M_24 -> Leech, Conway -> Monster, which looks like a chain. But one can regard the branch M_24 -> Leech, Conway as going off to the side, and then the problem is to “complete the square”. Some of this can be done. But I don’t know how to push these things on to later parts of the story.

Unfortunately, since these things are sporadic, there is no reason why a beautiful trick that works once should ever work again!

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