8
$\begingroup$

I met the following problem:

Let $G$ be a finite simple group (non-commutative, otherwise trivial). Let $p$ be a prime number not dividing $|G|$. Prove that any Sylow $p$ subgroup of $\mathrm{Aut}(G)$ is cyclic.

I know nearly nothing about advanced examples/theories of finite groups, so forgive me if this question is too naive for experts. Is this related to any big theorem in finite group theory? I tried to search it in some textbooks, but I didn't find a satisfactory answer. (I'm also interested in the case $p$ dividing $|G|$.) Can any veteran give me some clue? Thanks a lot in advance!

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ I believe you should change to a more neutral username. $\endgroup$
    – YCor
    May 31, 2023 at 8:08
  • 4
    $\begingroup$ Try math.stackexchange.com $\endgroup$
    – Alan
    May 31, 2023 at 8:37
  • $\begingroup$ You'll find a lot of information about the automorphism groups of the finite simple groups (including their orders) in the ATLAS of finite groups. You may check the statement case by case for each infinite series of simple groups and also for each sporadic group. I guess that only a few difficult cases remain after checking, if any. $\endgroup$
    – Martin Seysen
    May 31, 2023 at 18:40
  • 5
    $\begingroup$ I find it a bit weird that this was directed to stackexchange, if the known proofs of this fact rely on the classification of finite simple groups, surely this question is fine here (and probably would not have received much attention on stackexchange)? $\endgroup$
    – Achim Krause
    Jun 1, 2023 at 7:49

1 Answer 1

Reset to default
11
$\begingroup$

This property goes by the name of "outer $p$-cyclic" and can be found in Definition 2.5 of Gorenstein, Lyons and Solomon, "The classification of the finite simple groups, II". It is a consequence of the classification of finite simple groups, and the proof that the known finite simple groups have this property can be found in Theorem 7.1.9 in part III of the same series.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I believe the terminology originates in Patrick McBride's paper, "Nonsolvable signalizer functors on finite groups" - J Algebra 78 (1982), 215-238. But I haven't searched for earlier references. $\endgroup$
    – Dave Benson
    Jun 1, 2023 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.