9
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ There are some examples at en.wikipedia.org/wiki/Outer_automorphism_group . $\endgroup$ – darij grinberg Aug 20 '10 at 20:25
  • 4
    $\begingroup$ One more remark: "if and only if" is wrong. $G=S_n$ for $n\neq 6$ is not simple, yet Inn = Aut. $\endgroup$ – darij grinberg Aug 20 '10 at 20:39
  • 9
    $\begingroup$ 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. $\endgroup$ – Jack Schmidt Aug 20 '10 at 20:42
  • $\begingroup$ @Jack: how do you know G/Z(G) is complete? $\endgroup$ – Steve D Aug 20 '10 at 21:01
  • 1
    $\begingroup$ @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. $\endgroup$ – Jack Schmidt Aug 20 '10 at 23:15
19
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Steve D Aug 21 '10 at 12:42
  • $\begingroup$ Thanks! I had only been considering multiplication on one factor (a "diagonal" automorphism) and so missed the key property of central inversion. $\endgroup$ – Jack Schmidt Aug 21 '10 at 15:56
  • $\begingroup$ $M_{24}$ 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. $\endgroup$ – DavidLHarden Jun 17 '13 at 16:44
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Just to clarify: in these examples, $G$ (and hence $\mathrm{Aut}(G)$), is never simple. $\endgroup$ – HJRW Aug 20 '10 at 20:57
-3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Karl Schwede Jul 11 '13 at 16:10
  • $\begingroup$ 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 ) $\endgroup$ – mesel Jul 11 '13 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.