Timeline for Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$
Current License: CC BY-SA 3.0
15 events
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| Mar 19, 2014 at 6:58 | answer | added | Marla | timeline score: 3 | |
| Dec 15, 2012 at 8:16 | vote | accept | Binzhou Xia | ||
| Dec 15, 2012 at 8:08 | comment | added | Binzhou Xia | Of course least power of $p$ greater than or equal to $n$... | |
| Dec 15, 2012 at 8:06 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 14, 2012 at 10:18 | answer | added | Nick Gill | timeline score: 4 | |
| Dec 13, 2012 at 17:14 | answer | added | Jim Humphreys | timeline score: 7 | |
| Dec 13, 2012 at 10:07 | comment | added | Someone | "largest power of $p$ greater than or equal to $n$"? | |
| Dec 13, 2012 at 3:25 | comment | added | user29720 | @Binzhou: Write the positive roots as sums of simple positive roots. Look at the positive integers which show up as coefficients, and let $e$ be the largest of these. Then $p^{e-1}$-power kills the unipotent radical of the Borel subgroup (as an algebraic group, let alone on rational points over a finite field); for GL$_n$ we have $e=n$. Indeed, you can make a composition series with successive vector group quotients by taking the terms to be the subgroups directly spanned by the root groups for positive roots whose maximal coefficient against a simple root is $\ge i$ for $i = 1,2,\dots,e$. | |
| Dec 13, 2012 at 2:00 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 12, 2012 at 16:05 | comment | added | Nick Gill | I recommend Gorenstein, Lyons, Solomon vol 3 which gives a full description of the Sylow p-subgroups of the classical groups. (I don't have a copy in front of me so can't be more precise.) Off the top of my head, I think they all contain an element with one Jordan block, hence will have the same exponent as a Sylow-$p$ of $GL_n(q)$... But I could be wrong... | |
| Dec 12, 2012 at 14:34 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 12, 2012 at 13:38 | comment | added | Binzhou Xia | What is the exact formulation of the upper bound? | |
| Dec 12, 2012 at 13:26 | comment | added | user29720 | Although root system considerations give an upper bound on $m$ for which $p^m$ kills the group, with the bound on $m$ depending only on the Killing-Cartan "type" of the classical group (not on the size of the finite field or its characteristic), the exact exponent may be sensitive to peculiar features of small finite fields or small characteristic (e.g., surprising commutation of some root groups in symplectic and odd special orthogonal groups in characteristic 2). | |
| Dec 12, 2012 at 5:28 | comment | added | user29720 | According to the theorem of Borel and Tits discussed in mathoverflow.net/questions/104201/…, if $G$ is a connected reductive group over a perfect field $k$ of characteristic $p > 0$, any finite $p$-subgroup of $G(k)$ is contained in $U(k)$ where $U$ is the ($k$-split) unipotent radical of a minimal parabolic $k$-subgroup $P$ of $G$. All such $P$ are $G(k)$-conjugate, so such $U(k)$ are the Sylow $p$-subgroups of $G(k)$ when $k$ is finite. The general context for your question is to compute the exponent of $U(k)$. Try more to settle GL$_n$ by yourself. | |
| Dec 12, 2012 at 3:21 | history | asked | Binzhou Xia | CC BY-SA 3.0 |