Is there any result about maximal abelian p-subgroups of the exceptional group E_6(q), where q=p^a is prime power?
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$\begingroup$ The formulation is unclear, since this group doesn't have abelian Sylow $p$-subgroups. $\endgroup$– Jim HumphreysCommented Nov 26, 2014 at 1:57
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$\begingroup$ I'm sorry, I mean maximal abelian p-subgroups. $\endgroup$– daryaCommented Nov 26, 2014 at 3:18
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3$\begingroup$ You can use Chevalley's commutator formula to figure out which of the root groups commute - it's conceivable I suppose that all maximal abelian unipotents are conjugates of these.... Also, the abelian unipotent subgroups of maximal order were classified by Mal'cev I believe. See the references in this paper: math.nsc.ru/~vdovin/evdavg.ps $\endgroup$– Nick GillCommented Nov 26, 2014 at 16:23
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Aspects of this question have been thoroughly treated in The Classification of Finite Simple Groups, Number 3, by Gorenstein-Lyons-Solomon, AMS, 1994: see especially their Table 3.3.1 for the Chevalley groups. In your notation (which differs somewhat from theirs), the $p$-rank is $16a$. This gives only the rank of a maximal elementary abelian $p$-subgroup, however.
P.S. Concerning maximal abelian $p$-subgroups of $E_6$, the relevant table in Vdovin's thesis (linked by Nick Gill) seems to give the same answer $p^{16a}$. Probably the point here is that the 16 "commuting" positive roots yield the only possible maximal abelian $p$-subgroups in a Chevalley group, automatically elementary abelian because of the structure of root groups. The emphasis on $p$-rank comes mainly from the connection with cohomological support varieties and such. Of course, Sylow $p$-subgroups are all conjugate, so their subgroup structure is what one needs to know.
answered Nov 26, 2014 at 21:34
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$\begingroup$ Well once p is sufficiently large, all unipotent elements of E_6(F_q) have order p so the notion of maximal abelian subgroup and maximal elementary abelian subgroup coincide. $\endgroup$ Commented Nov 27, 2014 at 1:48
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$\begingroup$ Jim, one query. I don't have GLS3 in front of me but, from memory, their results describe the *maximal possible rank of an elementary abelian $p$-subgroup, as opposed to the possible ranks of a maximal elementary abelian $p$-subgroup. (In theory there could be different maximal elementary abelian $p$-subgroups with different ranks, couldn't there?) $\endgroup$ Commented Nov 27, 2014 at 17:29
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$\begingroup$ @Nick: In this case the structure of the root groups (and the simple multiplication/commutation rules independent of $q$) come into play, along with the fact that 16 is the maximum number of "commuting" positive roots (earlier work of Mal'cev). I don't know a more explicit reference beyond Vdovin and GLS, but there may be one. $\endgroup$ Commented Nov 28, 2014 at 14:40