Criteria for Aut(G) to be simple

Question

It is well known that the automorphisms of a group $G$ form a group under composition, and that the group of inner automorphisms $\phi (x)=gxg^{-1}$ forms a normal subgroup of $\mbox{Aut}(G)$. Thus, $\mbox{Aut}(G)$ is simple if and only if either $\mbox{Inn}(G)=\mbox{Aut}(G)$ or $\mbox{Inn}(G)$ is trivial. In the second case, since $G/Z(G)=\mbox{Inn}(G)$, $G$ must be abelian. My question is, when does $\mbox{Inn}(G)=\mbox{Aut}(G)$? Or, as it is unlikely that the general case is not fully understood, are there nice classes of groups for which there are a nice set of criteria for $\mbox{Inn}(G)=\mbox{Aut}(G)$.

Answer 1

Here is an approximation of an answer to "For what finite groups is Aut(G) simple?"

As Daniel Miller mentioned, Inn(G) is a normal subgroup of Aut(G), so for Aut(G) to be simple either Inn(G) = 1, in which case G is abelian, or Inn(G) = Aut(G) is simple. The former case should be somewhat easy to handle assuming G is finite. In the latter case, we have that G/Z(G) is simple. If G is also perfect, then G is called quasi-simple. Of course, G need not be perfect as G ≅ A₅ × 2 shows. However, I believe this is the only obstruction, so ignoring a possible cyclic direct factor of order 2, G/Z(G) is simple, and G is quasi-simple. The finite quasi-simple groups and their automorphism groups are classified, but the classification is a bit long. For a fixed simple group, X = G/Z(G), there are only finitely many isomorphism classes of quasi-simple groups D such that D/Z(D) = X. In fact there is a unique largest one called the Schur cover, that I'll call D. If Z(D) is cyclic, then in fact Aut(G) = Aut(X) = Aut(D) does not pay any attention to the center. So all we need to do is find all X with Aut(X) = X [and each one works], and all X with Z(D) non-cyclic [and check which ones work].

Having done most, but not all, of that, I thought it might help to record the basic result:

If G = H×T where T=1 if H is abelian and T is cyclic of order dividing 2 otherwise, and where H is on the following list, then Aut(G) is simple:

• cyclic of order 3, 4, or 6
• elementary abelian of order 2ⁿ for n ≥ 3
• M11, 2.Sz(8), J1, 2.Sp(6,2), M23, M24, Ru, 2.Ru, Co3, Co2, Ly, Th, Fi23, Co1, 2.Co1, J4, B, 2.B, E7(2), M
• Ω(2n+1,2) for n ≥ 3
• Sp(2n,2) for n ≥ 3
• E8(p) for any prime p
• F4(p) for any prime p
• G2(p) for any prime p ≥ 5

Additionally if Aut(G) is simple, then G = H×T as above, except possibly H/Z(H) is on the following list:

• L3(4), U4(3), U6(2), 2E6(2)
• Ω⁺(4n,q) for certain q

These are groups with non-cyclic multiplier other than Sz(8) [definitely an example] and Ω⁺(8,2) [not an example]. The Ω⁺(4n,q) case should be mostly easy, as there are too many automorphisms to kill. The others would be easy in an ideal world, but as far as I know our computational knowledge of these groups is limited and/or flawed. Of course, I also need to check the abelian case carefully, but I think 3,4,6 and 2^n are the only abelian examples.

It would make another good answer: For what torsion abelian groups G is Aut(G) simple? This would handle the abelian groups here, as well as some of the original poster's interest, without delving into the nastier aspects of abelian groups.

Answer 2

Obraztsov has shown that if $p$ is a sufficiently large prime, then there exists a finitely generated infinite simple complete group $G$, all of whose proper subgroups are cyclic of order $p$. In particular, $G$ is an example of a group such that $Aut(G)$ is an infinite simple group. The relevant reference is:

V. N. OBRAZTSOV, `On infinite complete groups', Comm. Algebra 22 (1994) 5875--5887

Answer 3

This is not an answer to your exact question (which I interpreted to be 'When does $\mathrm{Inn}(G)=\mathrm{Aut}(G)$?'---as pointed out in the comments, this is not the same as asking for $\mathrm{Aut}G$ to be simple), and is only really interesting if you care about examples where $G$ is infinite.

If you do care about $G$ infinite, then a natural slight weakening is to ask for criteria for $\mathrm{Out}(G)$ to be finite. One such criterion is provided by Paulin's Theorem.

Theorem. If $G$ is word-hyperbolic and $\mathrm{Out}(G)$ is infinite then $G$ splits (as an amalgamated free product or HNN extension) over a virtually cyclic subgroup.

It is known that, using some suitable definition of 'randomly chosen', a randomly chosen finitely presented group is torsion-free, word-hyperbolic and does not split. So one can conclude that a 'randomly chosen' finitely presented group $G$ is of finite index in its automorphism group.

Answer 4

let $G$ be non ableian group and $A$ be set of all groups including $Z(G)$. for All H in A,send H to Z(H)(Notice that this is a map from A to A). notice if Z(H)=Z(G) all H in A, it cause a contradiction(easy to show) if Inn(G)=Aut(G) then there is a uniqe proper group with Z(H)=Z(G) in A.

M.Y.K