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Let $G$ be a classical simple group over a finite field $GF(q)$ and $P$ a parabolic subgroup of $G$ stabilizing an isotropic subspace. Is the Borel subgroup of $G$ maximal soluble in $P$ and is there any maximal soluble subgroup in $P$ besides the Borel subgroup? Here by maximal soluble I mean a maximal one among soluble subgroups of $P$.

Any feedback is appreciated!

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Solvable subgroups not containing all points of a Borel are out of control (so best to focus of $B$ being maximal solvable in $G$ for "large" $p$, without mentioning $P$). To see this, for any $p$ and any finite solvable $\Gamma$ you can put $\Gamma$ into ${\rm{GL}}_n(\mathbf{F}_p)$ for large $n$ (e.g. permutations) and view ${\rm{GL}}_n$ as a proper parabolic in ${\rm{SL}}_{n+1}$ in an evident way. If $\Gamma$ doesn't have normal $p$-Sylows or does yet quotients by them are not abelian then $\Gamma$ has "nothing" to do with Borel subgroups (e.g., not contained in one) when $G$ is split. –  user29283 Nov 24 '12 at 19:04
@Binzhou: Geoff has answered your basic question by pointing out that the rational points of a larger parabolic than $B$ can itself be solvable. This occurs only for some $P$, of course. (The surrounding discussions add other interesting points, but of course the determination of all maximal solvable subgroups in finite groups of Lie type is a much more open-ended question.) –  Jim Humphreys Nov 25 '12 at 14:37
@Jim: Many thanks to you and Geoff! I'm new to algebraic groups, so maybe I should read some text book or papers for fundation about this subject. By the way, if I mean a parabolic subgroup by the full stabilizer of an isotropic subspace, should I use the term maximal parabolic subgroup? –  Binzhou Xia Nov 26 '12 at 3:11
@Binzhou: I'm not sure about the current terminology used in finite classical groups, but it needs to be consistent with the general language of algebraic groups to avoid confusion. (In either case, by the way, "parabolic" is a term which originates far away from the application to subgroups here.) –  Jim Humphreys Nov 26 '12 at 22:22
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2 Answers

up vote 3 down vote accepted

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One can occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behavior perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

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@Geoff: Is this phenomenon limited to $q = 2, 3$ (and not more general $q$ with $p \in \{2, 3\}$, let alone larger $p$)? More specifically, consider a split connected semisimple $G$ over $k = \mathbf{F}_q$ so that $G$ is simply connected and $k$-simple, with either: $q > 3$, $q = 3$ and $G \ne {\rm{SL}}_2$, or $q = 3$ and $G \ne {\rm{SL}}_2, {\rm{Sp}}_4$. By BN-pair stuff, $G(k)$ has center $\mu(k)$ where $\mu$ is the center of $G$, and $G(k)/\mu(k)$ is simple as an abstract group. For a Borel $k$-subgroup $B$ of $G$, is $B(k)$ maximal as a solvable subgroup of $G(k)$? –  user29283 Nov 24 '12 at 19:25
I meant "$q = 2$ and $G \ne {\rm{SL}}_2, {\rm{Sp}}_4$" at the end of the 2nd sentence in the preceding comment. –  user29283 Nov 24 '12 at 19:27
Yes, it is probably limited to $q = 2,3.$ –  Geoff Robinson Nov 24 '12 at 20:34
I am talking here about maximal solvable subgroup containing a Borel subgroup –  Geoff Robinson Nov 24 '12 at 22:48
@xuhan: you can also have parabolics $P$ such that $P/U$ is isomorphic to a product of several copies of ${\rm GL}(2,3)$ when $p = 3$, for example. –  Geoff Robinson Nov 25 '12 at 1:24
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EDIT: This responds to Geoff's comment (and my carelessness), but also adds a couple of other remarks.

Geoff's answer and the comments might be clarified a little as follows. You are looking at rational points of a parabolic subgroup over a finite field (where it's not important whether the algebraic group is of classical type or not). In the split (Chevalley group) case, it's easy to check orders of the various finite groups involved. Since a Levi factor of a proper parabolic group is just the product of a central algebraic torus and a lower rank semisimple group, the rational points sometimes yield a solvable group even if the original parabolic is not itself a solvable algebraic group (a Borel subgroup). This also involves the (nilpotent) group of rational points of the unipotent radical, so it gets complicated. To assemble a complete list of possibilities, look at the simple algebraic groups having solvable groups of rational points over a field of small characteristic. The torus part is already commutative, so the product and therefore the Levi subgroup of a parabolic can be solvable.

Naturally it gets more complicated to treat all types of twisted groups as well as Suzuki or Ree groups, but in principle it looks reasonable to assemble a complete list of solvable finite parabolics. (If there is enough motivation to do so.)

Concerning the determination of all maximal solvable subgroups of a finite simple group of Lie type, that's asking for a lot. The BN-structure is a valuable organizing tool, but the groups have considerably richer subgroup structure than can be identified from Lie theory alone. Even in the study of linear algebraic groups, older work of V.P. Platonov shows that it is tricky to pin down all their not necessarily connected maximal closed solvable subgroups.

[Terminology: for me language like "the Borel subgroup" is a bit jarring, since there are many conjugate choices starting with the algebraic group.]

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@Jim: Do you really mean " a proper parabolic is just the product of a central algebraic torus and a lower rank semisimple group", or are you talking about the Levi factor after the unipotent radical is factored out? –  Geoff Robinson Nov 24 '12 at 22:36
@Geoff: I was multi-tasking at the time. See my revised version. By the way, thanks for adapting to the weird American spelling "solvable". For us, most powders are soluble while some problems are solvable. –  Jim Humphreys Nov 24 '12 at 23:09
@Jim: Yes, I have always preferred "solvable" myself, partly a rebellious reaction of a postgraduate student against the Oxford tendency to use "soluble"- though Burnside and P. Hall also used "soluble" and they were in Cambridge long ago. I remember once David Sibley gave some lectures in Oxford about the Character Theory of the Odd Order Paper, which he started with " A finite group is soluble if and only if it can be dissolved in water"! I envy the ability to multitask, and I note your (correct) comment about THE Borel subgroup. –  Geoff Robinson Nov 25 '12 at 1:12
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