Timeline for Maximal soluble subgroups in a parabolic subgroup of finite classical simple group
Current License: CC BY-SA 3.0
10 events
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| Nov 26, 2012 at 22:22 | comment | added | Jim Humphreys | @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.) | |
| Nov 26, 2012 at 3:11 | comment | added | Binzhou Xia | @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? | |
| Nov 26, 2012 at 2:56 | vote | accept | Binzhou Xia | ||
| Nov 25, 2012 at 14:37 | comment | added | Jim Humphreys |
@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.)
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| Nov 25, 2012 at 5:29 | vote | accept | Binzhou Xia | ||
| Nov 25, 2012 at 5:30 | |||||
| Nov 25, 2012 at 5:29 | vote | accept | Binzhou Xia | ||
| Nov 25, 2012 at 5:29 | |||||
| Nov 24, 2012 at 21:54 | answer | added | Jim Humphreys | timeline score: 1 | |
| Nov 24, 2012 at 19:04 | comment | added | user29283 | 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. | |
| Nov 24, 2012 at 17:29 | answer | added | Geoff Robinson | timeline score: 3 | |
| Nov 24, 2012 at 13:11 | history | asked | Binzhou Xia | CC BY-SA 3.0 |