centralizer of p-subgroups in almost simple groups of characteristic p

Question

Suppose $G$ is an almost simple group with normal simple group $S$. Suppose also that $S$ is of Lie type of characteristic $p$. If $P$ is a sylow $p$-subgroup $S$, we know that $C_S(P)\subseteq P$. My question is: Under which condition $C_G(P)\subseteq P$, where $P$ is a sylow $p$-subgroup $S$?

Answer 1

I believe this is true in general. Here's how I'd go about proving it. I'll consider only the case where $S$ is a Chevalley group - the twisted case is basically the same. All the background is in Carter's Simple groups of Lie type.

We write elements of $P$ as products of elements of root groups. So a typical element has form $$x_{r_1}(t_1)x_{r_2}(t_2) \cdots x_{r_k}(t_k),$$ where $r_1,\dots, r_k$ are roots and $t_1,\dots, t_k$ are field elements.

Let $G_0$ be the subgroup of $G$ generated by automorphisms of $S$ that are not graph automorphisms. Then $N_{G_0}(P)$ is generated by

• $P$
• $H$, a split torus (including inner-diagonal elements). An element $h(\chi)\in H$ acts on a root element via $$x_r(t)^{h(\chi)} = x_r(\chi(r)t)$$ where $\chi: \mathbb{Z}\Phi\to k^*$ is a character;
• $\delta$, a field automorphism where the action on a root element is via $$x_r(t)^\delta = x_r(t^\delta).$$ Now for a fixed $\chi$ and $r$ it is clear that $t^\delta \neq \chi(r) t$ for all $t\in K$. Since no non-central element of $P$ normalizes all root groups simultaneously you are done.

That gives the result whenever $S$ has no graph automorphisms. Notice, though, that all elements of $N_{G_0}(P)$ normalize $X_r(t)P^1$, i.e. the fundamental root groups, modulo the subgroup of $P$ generated by non-fundamental roots. Since elements in $N_G(P)\backslash N_{G_0}(p)$ do not do this, you are done.

(An alternative approach to deal with the graph aut, at least for classical groups, is to just write some matrices down and prove it directly.)

Remark To clarify Geoff's response to Jim's comment: An almost simple group is a group $G$ that has a unique normal subgroup $S$ such that $S\leq G \leq Aut(S)$. This is equivalent to Geoff's definition.

Answer 2

As Nick points out, you don't have to impose any extra conditions to get a positive answer. This suggests a formulation based primarioly on the Tits axioms for a $BN$-pair, refined somewhat for your situation. There is in fact a uniform development in the 2004 Cambridge book Representation Theory of Finite Reductive Groups by Marc Cabanes and Michel Enguehgard. They are primarily concerned with representations over fields of characteristic dividing the group order, emphasizing primes other than the natural one. But along the way they make good use of $BN$-pairs. The main price paid for doing this is the exclusion of rank 1 cases, which for groups of Lie type can be filled in directly.

Here is how their Theorem 2.31 is stated, assuming $G$ has a "strongly split" $BN$-pair:

Assume that $W$ is of irreduible type with $|W| \neq 2$. Then $C_G(U) =Z(G) Z(U)$.

Here $U$ is a Sylow $p$-subgroup involved in the splitting of $B$, and $W$ is the Weyl group. Their assumption permits arbitrary finite groups of Lie type having irreducible root systems, but of rank at least 2. In particular, this uniform approach doesn't omit the finite special linear groups or other classical types. The definition of "almost simple" implicitly used in the question (as clarified by Geoff and Nick) is convenient for the study of finite simple groups but is somewhat restrictive in requiring the given group to contain a normal simple subgroup and be included in its automorphism group.

As far as I can tell, Cabanes-Enguehard avoid detailed knowledge of graph automorphisms and the like for Lie-type groups, though their proof is built on some previous material in their Chapter 2 which needs to be checked carefully.

ADDED: Thinking further about the question, I'm somewhat puzzled about what is being asked. For a finite simple group $S$ of Lie type (Chevalley or twisted variant), Steinberg determined all possible automorphisms beyond inner ones to be composites of diagonal, field, and graph automorphisms: see sections 10-11 of his Yale lectures here.

Extending $S$ by a diagonal automorphism can be done in a larger group $G$ with strongly split $BN$-pair, though I'm less sure whether that can be done for other types of automorphisms. But it doesn't seem to matter, since obviously none of these types of non-inner automorphisms can fix pointwise a Sylow $p$-subgroup ($U$ in my notation above). Why is there a problem about the centralizer of $U$ in $G$ when $S$ is assumed simple?