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It is known (given CFSG) that all non-abelian finite simple groups have small outer automorphism groups. However, it's quite tedious to list all the possibilities. Does anyone know a reference for a statement of the following form?

Let G be a non-abelian finite simple group. Then |Out(G)| < f(|G|) (where f is something straightforward that gives a good idea asymptotically of the worst case).

Also, a related question: How good a bound can be obtained in this case without using the classification? Can one do much better than the bounds one might obtain for the outer automorphism group of an arbitrary finite group?

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One person's "tedious" is another person's "fascinating". Much of the appeal in the classification of finite simple groups lies in the individuality of the groups which mysteriously appear in this list. –  Jim Humphreys Apr 9 '10 at 18:20
    
I agree that it's a very interesting feature of finite simple groups that their outer automorphism groups are so small, and the 'exceptional' automorphisms are also interesting. What I meant was that if one is proving some general and fairly crude statement about finite groups, it's often not very enlightening for the reader to see a classification trawl in the proof, especially if it is one that has been repeated several times in the existing literature. –  Colin Reid Apr 11 '10 at 9:24

3 Answers 3

up vote 9 down vote accepted

Check article "Probabilistic generation of wreath products of non-abelian finite simple groups" by Martyn Quick. In Section 3.1 he consider this question and get $|Out G|\leq |G|/30$ for every non-abelian finite simple group $G$, which was enough for his needs.

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You can derive bounds in a straightforward way from the complete list of finite simple groups in Wikipedia, which has references (particularly to ATLAS). If you look through all of the cases, you find that the worst case is $A_1(2^k)$ where the order of the group is $8^k-2^k$, and the order of the outer automorphism group is $k$. This gives you an upper bound of $\lfloor \frac{\log (16|G|/15)}{\log 8} \rfloor$, which is sharp for $A_5$.

Regarding your last question, this bound is far better than what you get for finite groups in general. For example, the elementary abelian group $(\mathbb{Z}/2\mathbb{Z})^n$ has order $2^n$, but its outer automorphism group is $GL_n(\mathbb{Z}/2\mathbb{Z})$, which has order $\prod_{k=0}^{n-1} (2^n-2^k)$. In other words, a group of order $m$ can have outer automorphism group with size about $m^{\log m}$ instead of $\log m$.

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I recommending listing all the possibilities. A number of sources already do this, so it is fairly easy to do again; the table in the ATLAS is quite reasonable. Basically, the outer automorphism group is ridiculously small "most" of the time, so you might care about the details.

I think you'll get |Out(G)| ≤ C*log(|G|) as worst case, but this is pretty pessimistic most of the time.

The outer automorphism group of an alternating group has order at most 4, and almost always has order 2. There are finitely many sporadic groups, and so will not matter asymptotically, but you can quickly check over the list to see they have outs of size at most 2.

The groups of Lie type have a 3-part outer-automorphism group; the diagonal, the field, and the diagram parts. The diagram part has order at most 6 (and only for D4). The field part is cyclic, but can be "large", as in, if "q" of your group is p^f, then it is cyclic of order f. The diagonal part is usually small (order at most 4, or even order at most 2), but can be larger for PSL(n,q) and PSU(n,q). Even there it is cyclic of order at most n.

So basically you handle the case of PSL/PSU a little more carefully, then the case of a general group of Lie type using bounds of 4 and 6 for diagonal and diagram so getting something like O(log(|G|)), then you handle the rest which are bounded by a constant.

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