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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.

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

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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.

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  • $\begingroup$ That's great, thanks. I'd love to see a CFSG free proof just for simple groups, but I think that won't happen. $\endgroup$ Apr 21, 2021 at 7:02
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    $\begingroup$ 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! $\endgroup$
    – Derek Holt
    Apr 21, 2021 at 7:50
  • $\begingroup$ @DerekHolt: Thanks and I agree, edited now. $\endgroup$ Apr 21, 2021 at 8:08

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