Lie Theory: Included in normalizer of a maximal Torus

Question

Hello People of Mathoverflow, I am searching for a Theorem about Lie Theory:

Let $X_{l}(q)$ be a Group of Lie type, with Lie rank $l$ over the finite field with $q=r^{a}$ elements and $r$ is a prime. Let $K$ be a subgroup of $X_{l}(q)$, with the order of $K$ prime to $r$ (a $r$´ group), so there is a maximal torus $T < X_{l}(q)$ and $K$ is subgroup of the normalizer of $T$ in $X$ ( $K \leq N_{X_{l}(q)}(T)$ ).

It should be right for $K$ which only include semisimple elements, but where can I find these theorem? Is there another property for the elements of K which the Theorem is true?

Thank you much for your Answers.

Answer 1

There are some relevant older results in section 5 of part II (on semisimple elements), of the 1968-69 IAS seminar notes by Springer and Steinberg included in Lecture Notes in Mathematics 131 (Springer, 1970). In particular, their Corollary 5.17 shows that a supersolvable subgroup of a semisimple algebraic group consisting of semisimple elements must lie in the normalizer of some maximal torus (defined over the finite field in your set-up). [Here "supersolvable" generalizes the usual notion for a finite group by allowing subquotients to be either finite cyclic or algebraic tori.] I'm not sure whether there are any stronger results in the literature, but the discussion by Springer and Steinberg shows clearly that such questions are not straightforward to deal with in the algebraic group context.

As Marguax observes in a comment, being an $r'$-subgroup is the same as consisting entirely of semisimple elements.

Possibly more can be said in the context of finite groups of Lie type, using only finite group techniques, but it seems that the given subgroup $K$ consisting of semisimple (that is, $r'$-) elements must be of a rather special sort to embed in such a normalizer. ADDED: Keep in mind also that the normalizer of a maximal torus in the ambient algebraic group has a somewhat delicate structure in its own right, sometimes but not always a semidirect product of the torus and a copy of the Weyl group. (This has implications for the structure of the finite groups.) One standard reference is the paper by J. Tits, Normalisateurs de tores. I. Groupes de Coxeter etendus. J. Algebra 4 (1966), 96–116.

SUMMARY: In one direction, Springer-Steinberg give a uniform sufficient condition for a subgroup consisting of semisimple elements to embed in the normalizer: it's enough for it to be supersolvable. (This strengthens a classical result of Blichfeldt for finite nilpotent matrix groups.) In the other direction, there seems to be no useful necessary condition on the structure of such a subgroup of semisimple elements. Probably case-by-case study of the simple Lie types is needed to go further.

All of this is illustrated by comparing general linear and special linear groups. Here the Weyl group $S_n$ lives naturally in $\mathrm{GL}_n$ as the subgroup of permutation matrices; but only its rotation group $A_n$ consists of matrices of determinant 1. For large enough $r$ and $n$ this subgroup of the normalizer inside $\mathrm{SL}_n$ consists of $r'$-elements and is actually simple. Manguax's observation shows however that a finite nonabelian simple subgroup of $\mathrm{SL}_n$ (say over a finite field) may consist of semisimple elements but not lie in the normalizer. So there's unlikely to be any better general result than the one of Springer-Steinberg.

Answer 2

first thank you all for your kind help.

About the question of clarification which Marguax figures out at his first answer: I need the Theorem with the assumption at the last sentence of my 2nd paragraph. But I have heard only that the Theorem holds for the assumption mentioned at the first sentence of my 3rd paragraph (but I couldn´t find where to quote it – thanks for the answer of the Lecture Notes by Springer and Steinberg).

If it would be the same (K being a r´Group and K including only semisimple elements) it would be wonderful. But like Marguax mentioned it in his second answer, the assumption of being a r´Group is not sufficient to show the Theorem. Another example is the cyclic subgroup of the $SL_{3}(q)$ of order $q^{2}+q+1$ (the Singer Subroup). It is an r´ subgroup (when $q=r^{a}$, r prime) but is cannot be a subgroup of any normalizer of a maximal Torus, which order is $6(q-1)^{2}$ (for q not beeing 2).

I am confused. If it would be the same (K being a r´Group and K including only semisimple elements), then it should be contained in the normalizer of a maximal Torus.