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Nov 29, 2019 at 5:55 comment added Shripad A very naive suggestion would be to allow $q = 1$. As is well-known, the reductive groups over the field of order 1 are the Weyl groups.
Aug 2, 2019 at 9:53 comment added Jesper Grodal @TheoJohnson-Freyd, sorry, the definition above of of $\mathcal L_3$ should be $G/H$ of $p'$ index (it is then automatically abelian it turns out). This gives the right definition also few exceptions classified by Tits, where $H_1(G;{\mathbb F}_p) \neq 0$, and e.g. rules out the example you mention. One can actually describe $\mathcal L_3$ slightly cleaner than what I wrote above, and I'll try to update my post to describe this, as the issue keeps popping up (the post is now almost 6 years old...).
Jul 20, 2019 at 16:55 comment added Theo Johnson-Freyd Wikipedia warns that the Tits group $^2F_4(2)'$ does not have a BN structure, even though, if I understand correctly, it is in your $\mathcal L_3$.
Jun 21, 2016 at 22:17 comment added Jim Humphreys P.S. What you've written is helpful and probably is the best way to answer this admittedly imprecise question.
Jun 21, 2016 at 22:15 vote accept Jim Humphreys
Nov 26, 2013 at 14:31 comment added Jim Humphreys Thanks for the detailed discussion (which I only noticed today). This may convince me that it's hopeless to reach a consensus on what the definition should be. While I'd want to include all the relevant simple groups and "closely" related ones arising naturally from Lie theory, I still can't quite draw a line. I guess I'd stay closer to $\mathcal{L}_1$ than to $\mathcal{L}_2$ (and probably give up on the Weyl group of $E_8$). But any choice seems to include too much or too little. In any case, I'd want there to be an underlying irreducible root system.
Sep 25, 2013 at 15:35 history answered Jesper Grodal CC BY-SA 3.0