Timeline for Maximal soluble subgroups in a parabolic subgroup of finite classical simple group

Current License: CC BY-SA 3.0

11 events

when toggle format	what		by	license	comment
Mar 23, 2013 at 20:31	history	edited	Geoff Robinson	CC BY-SA 3.0	typos
Nov 26, 2012 at 2:56	vote	accept	Binzhou Xia
Nov 25, 2012 at 5:29	vote	accept	Binzhou Xia
Nov 25, 2012 at 5:29
Nov 25, 2012 at 1:24	comment	added	Geoff Robinson		@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.
Nov 25, 2012 at 1:22	history	edited	Geoff Robinson	CC BY-SA 3.0	added 6 characters in body; added 3 characters in body
Nov 24, 2012 at 22:48	comment	added	Geoff Robinson		I am talking here about maximal solvable subgroup containing a Borel subgroup
Nov 24, 2012 at 20:45	history	edited	Geoff Robinson	CC BY-SA 3.0	amended comments on field size. Discussed the reduction to the rank 1 case
Nov 24, 2012 at 20:34	comment	added	Geoff Robinson		Yes, it is probably limited to $q = 2,3.$
Nov 24, 2012 at 19:27	comment	added	user29283		I meant "$q = 2$ and $G \ne {\rm{SL}}_2, {\rm{Sp}}_4$" at the end of the 2nd sentence in the preceding comment.
Nov 24, 2012 at 19:25	comment	added	user29283		@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 $G(k)$?
Nov 24, 2012 at 17:29	history	answered	Geoff Robinson	CC BY-SA 3.0