Is the group of outer class-preserving automorphisms of a finite group soluble?

This came from an old idea of mine from when I was an undergraduate for how to approach the Schreier conjecture. I (obviously) never gave it much serious thought, but wondered if much was known about it.

Let $$G$$ be a finite group. The group $$\mathrm{Out}(G)$$ has a normal subgroup $$\mathrm{Out}_c(G)$$ of class-preserving outer automorphisms, i.e., the group $$\mathrm{Aut}_c(G)/\mathrm{Inn}(G)$$, where $$\mathrm{Aut}_c(G)$$ is the subset of $$\mathrm{Aut}(G)$$ that leave invariant each conjugacy class of $$G$$. Note that all primes dividing $$|\mathrm{Out}_c(G)|$$ divide $$|G|$$, and if $$G$$ is simple then $$\mathrm{Out}_c(G)=1$$ (Feit-Seitz).

One approach to Schreier without CFSG is to slice $$\mathrm{Out}(G)$$ into two, and prove that $$\mathrm{Out}_c(G)$$ and $$\mathrm{Out}(G)/\mathrm{Out}_c(G)$$ are both soluble. Each of these looks hard, but the obvious question, even with CFSG, is:

Is $$\mathrm{Out}_c(G)$$ soluble for any finite group $$G$$?

The obvious candidates for groups with non-inner class-preserving automorphisms are $$p$$-groups, where this statement clearly holds. So perhaps it has a chance of being true generally. I couldn't find much progress at all on class-preserving automorphisms, so perhaps this question is too far out of reach at the moment.

• One may ask the same question for infinite groups, but I'd be amazed if it were true. Apr 20, 2021 at 23:19

See

Sah, Chih-han Automorphisms of finite groups. J. Algebra 10 (1968), 47–68.

Erratum: Sah, Chih Han "Automorphisms of finite groups'' (J. Algebra 10 (1968), 47–68): addendum. J. Algebra 44 (1977), no. 2, 573–575.

Theorem 2.10 in this paper states the following (erratum has a correction to the original proof):

Theorem: Suppose that $$G$$ is a group which has a composition series. If $$\operatorname{Out}(F)$$ is solvable for every composition factor $$F$$ of $$G$$, then $$\operatorname{Out}_c(G)$$ is solvable.

So it follows from the Schreier conjecture that $$\operatorname{Out}_c(G)$$ is solvable for any finite group $$G$$.

For infinite groups, in the paper above Sah points out the following example. Let $$S$$ be the symmetric group on an infinite countable set, and let $$G$$ be the normal subgroup formed by the permutations with finite support. In this case $$\operatorname{Aut}_c(G) = S$$, you can prove this the same way you prove that a transposition-preserving automorphism of the finite symmetric group is inner. Therefore $$\operatorname{Out}_c(G) \cong S/G$$, which is not solvable.

• That's great, thanks. I'd love to see a CFSG free proof just for simple groups, but I think that won't happen. Apr 21, 2021 at 7:02
• You have misquoted the theorem. It says "for each composition factor $F$ of $G$", not "for any composition factor $F$ of $G$", which is completely ambiguous. (I was genuinely unsure what it meant when I read it for the first time.) I really wish people would stop using the word "any" in formal mathematics! Apr 21, 2021 at 7:50
• @DerekHolt: Thanks and I agree, edited now. Apr 21, 2021 at 8:08