Such groups (where every automorphism is inner, i.e., all automorphisms are of the form g -> xgx^1 for some x) are called "complete." A marvelous fact is that the group of permutations on n elements is complete --- unless n=6. In other words, there is an automorphism of S_6 that, for instance, sends transpositions to triplets of transpositions. I have a wonderful little toy that illustrates this, which I will take a picture of if I remember... Another marvelous fact is that if you take any finite group G, and go to Aut(G), then Aut(Aut(G)), and so on, you reach a complete group after a finite number of steps. I have no idea how to prove this, but I believe it involves group cohomology. As you say, if the elements of an irreducible representation themselves have a field automorphism, then this is an outer automorphism (at least some of the time). The simplest case of this is the cyclic group, where the representations (a.k.a. characters) are roots of unity. All nontrivial representations of the cyclic group of prime order are related to each other by an outer automorphism, which is induced by the affine group. - Cris On May 20, 2014, at 6:52 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:

Are there theorems that characterize groups all of whose automorphisms are inner? For example, for a finite group, the characters of a matrix representation must be algebraic integers, and the presence of a non-rational integer in the character table betrays the existence of an outer automorphism.

-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun