4
$\begingroup$

Let $G$ be a finite nonabelian simple group. Write $\mathrm{Out}(G)$ the outer automorphism group of $G$. For a finite group $H$, let $\pi(H)$ be the prime divisors of the order of $H$. By check the Atlas of Finite Groups, $\pi(G)-\pi(\mathrm{Out}(G)) \neq \varnothing$ when $G$ is among the Tits group, sporadic groups or alternating groups.

My question is: if $G$ is simple group of Lie type, is it always true that $\pi(G)-\pi(\mathrm{Out}(G)) \neq \varnothing$?

Any explanation, references suggestion and examples are appreciated.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ For a simple group $G = G(p^d)$ of Lie type, the order of $Out(G)$ is often something like $d$ or $\gcd(p-1,2) d$. While $|G|$ has factors of the form $p^{ed}-1$. Then you can probably use Zsigmondy's theorem in most cases to conclude that what you ask is true. $\endgroup$
    – spin
    Commented Sep 3, 2022 at 4:44
  • $\begingroup$ Thanks very much! I get your point. $\endgroup$
    – user44312
    Commented Sep 5, 2022 at 0:04

1 Answer 1

Reset to default
3
$\begingroup$

Since outer automorphism groups of finite simple groups of Lie type are rather small solvable groups, and outer automorphisms are products of graph, field, and diagonal automorphisms in general, it should be possible to check this case by case, as @spin suggested in comments.

Using heavier machinery, it is possible to give a positive answer for "most", but not all, finite simple groups of Lie type.

If $G$ is any non-Abelian simple group whose order is divisible by the prime $p$ to the first power only, then it follows by a 2020 Theorem of Glauberman, Guralnick, Lynd and Navarro ("Centers of Sylow subgroups and automorphisms", Israel Journal of Mathematics 240, 253-266, (2020)) that the outer automorphism group of $G$ has order prime to $p$.

However, there are finite simple groups which have no Sylow subgroup of prime order, though examples are rather hard to find. It is at least known that every finite simple group has at least one cyclic Sylow subgroup, but I don't see how to apply the above result of Glauberman et al when the Sylow is cyclic of non-prime order.

Later edit: Well, I think that Glauberman et al's result shows that if a non-Abelian finite simple group $G$ has a cyclic Sylow $p$-subgroup of order $p^{n}$, then ${\rm Out}(G)$ has a cyclic Sylow $p$-subgroup of order at most $p^{n-1}$, but that is weaker than you wish for.

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ Some examples of simple groups with square order: link $\endgroup$
    – spin
    Commented Sep 4, 2022 at 1:48
  • $\begingroup$ Thanks very much for your detailed explanation @Geoff Robinson. I agree with spin, I had checked the Atlas of the finite groups carefully. $B_2(7)$ is one of the example with square order. I am interested with the fact that every finite simple group has at least one cyclic Sylow subgroup, could you please inform me the reference? Many thanks! $\endgroup$
    – user44312
    Commented Sep 5, 2022 at 0:38
  • 1
    $\begingroup$ For example: Theorem 4.9 in "Kimmerle, Wolfgang; Lyons, Richard; Sandling, Robert; Teague, David N.; Composition factors from the group ring and Artin's theorem on orders of simple groups. Proc. London Math. Soc. (3) 60 (1990), no. 1, 89-122." $\endgroup$
    – Mikko Korhonen
    Commented Sep 5, 2022 at 2:15
  • 1
    $\begingroup$ Or earlier, Proposition 3 in "Hiss, Gerhard; A converse to the Fong-Swan-Isaacs theorem. J. Algebra 111 (1987), no. 1, 279-290." $\endgroup$
    – Mikko Korhonen
    Commented Sep 5, 2022 at 2:30
  • $\begingroup$ @MikkoKorhonen Thanks very much for the references! They are very helpful. $\endgroup$
    – user44312
    Commented Sep 5, 2022 at 6:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .