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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 $\mbox{Inn}(G)=\mbox{Aut}(G)$.

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There are some examples at en.wikipedia.org/wiki/Outer_automorphism_group . –  darij grinberg Aug 20 '10 at 20:25
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One more remark: "if and only if" is wrong. $G=S_n$ for $n\neq 6$ is not simple, yet Inn = Aut. –  darij grinberg Aug 20 '10 at 20:39
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Inn(G) = Aut(G) does not imply Aut(G) is simple. For instance G nonabelian of order 6 is not simple, but Inn(G) = Aut(G). If G is centerless, then Inn(G) = Aut(G) is called being a complete group. If Aut(G) is simple, then Inn(G) = Aut(G) is simple, so G/Z(G) is simple. Roughly speaking G is quasi-simple and G/Z(G) is simple and complete. Modulo some A5 x 2 silliness, this is more or less a classification of G with Aut(G) simple. –  Jack Schmidt Aug 20 '10 at 20:42
    
@Jack: how do you know G/Z(G) is complete? –  Steve D Aug 20 '10 at 21:01
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@Steve D: In point of fact 2.Sz(8) has Sz(8) as its automorphism group, and Sz(8) has Sz(8):3 as its, so no G/Z(G) need not be complete. –  Jack Schmidt Aug 20 '10 at 23:15

5 Answers 5

up vote 19 down vote accepted

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.

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Torsion abelian groups split into p-components, so you're really asking about abelian p-groups (ignoring a C_2 factor). But if the group has exponent higher than 2, inversion is a central automorphism. So we really only care about elementary abelian 2-groups. In other words, you got all of them. –  Steve D Aug 21 '10 at 12:42
    
Thanks! I had only been considering multiplication on one factor (a "diagonal" automorphism) and so missed the key property of central inversion. –  Jack Schmidt Aug 21 '10 at 15:56
    
$M_{24}$ is also a sporadic group with trivial outer automorphism group, so it needs to be added (without any covers, since it has trivial Schur multiplier) to your third family of groups. –  DavidLHarden Jun 17 '13 at 16:44

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

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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.

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Just to clarify: in these examples, $G$ (and hence $\mathrm{Aut}(G)$), is never simple. –  HJRW Aug 20 '10 at 20:57

Does anyone know of any infinite simple groups which are the Automorphism groups of other groups? E.g. some $G$ with $Aut(G)$ infinite and simple?

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You should post this as a question, not as an answer here. –  Arturo Magidin Aug 26 '10 at 20:39
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I don't know about Aut(G), but I can do Out(G). Now Bumagina and Wise proved that every countable group arises as the outer automorphism group of some finitely generated group. –  HJRW Aug 26 '10 at 21:56

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

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I'm having a hard time reading this and figuring out exactly what you are saying. When you say "set of all groups", do you mean subgroups of $G$? Also, you may want to use dollar signs (like in LaTeX) around your math phrases. I've done it for your first sentence. –  Karl Schwede Jul 11 '13 at 16:10
    
A={H<G|Z(G)<H} i.e A is the set of all subgroups of G including in Z(G). and first notice that Z(H)£A for all H in A. so, let f:A->A be function which send H to Z(H). first show that if f(H)=Z(G) for all H, it cause a contradiction since center of G is properly contained in a abelian subgroup of G if G is nonabelian. claim: if Inn(G)=Aut(G) then f(H)=Z(G) for only a uniqe element (G,Z(G) is trivially satisfy this,I mean except them ) –  mesel Jul 11 '13 at 22:44

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