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

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 quasisimple 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
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 quasisimple. Of course, G need not be perfect as G ≅ A_{5} × 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 quasisimple. The finite quasisimple 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 quasisimple 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) noncyclic [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 2^{n} 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 noncyclic 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.

1$\begingroup$ Torsion abelian groups split into pcomponents, so you're really asking about abelian pgroups (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 2groups. 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
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) 58755887
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 wordhyperbolic 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 torsionfree, wordhyperbolic and does not split. So one can conclude that a 'randomly chosen' finitely presented group $G$ is of finite index in its automorphism group.

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

$\begingroup$ 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<GZ(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