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 $\mbox{Inn}(G)=\mbox{Aut}(G)$.
[Math] Criteria for Aut(G) to be simple
gr.group-theory
Related Solutions
Amongst the finitely generated abelian groups, those with commutative endomorphism ring are exactly the cyclic groups.
Torsion abelian groups with commutative endomorphism rings are exactly the locally cyclic groups, that is, the subgroups of Q/Z. They were classified in:
- Szele, T.; Szendrei, J. "On abelian groups with commutative endomorphism ring." Acta Math. Acad. Sci. Hungar. 2, (1951). 309–324 MR51835 DOI:10.1007/BF02020735
This papers also gives more complicated examples of mixed groups with commutative endomorphism ring.
The mixed case was completed in:
- Schultz, Ph. "On a paper of Szele and Szendrei on groups with commutative endomorphism rings." Acta Math. Acad. Sci. Hungar. 24 (1973), 59–63. MR316598 DOI:10.1007/BF01894610
This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").
In the literature the following terminology is common. If $H$ is a group such that $H \cong G/Z(G)$ for some $G$, we say that $H$ is capable. As you mention, non-trivial cyclic groups are not capable. Another example is the quaternion group of order $8$, which is not capable.
The earliest paper on this seems to be the one by Baer: "Groups with preassigned central and central quotient group. Trans. Amer. Math. Soc. 44 (1938), no. 3, 387–412." He gives a complete classification of finite abelian $H$ that are capable.
In general the description of groups that are isomorphic to some $G/Z(G)$ is not known and seems to be a difficult problem. But you will find many papers on the subject for example by looking up "capable groups" on google, or by looking up on MathSciNet which papers cite the 1938 paper of Baer.
Best Answer
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:
Additionally if Aut(G) is simple, then G = H×T as above, except possibly H/Z(H) is on the following list:
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.