If two elements of an abelian Sylow P are conjugate in G, then they are conjugate in N

_{G}(P), but in GL(3,q) it can require conjugation in two separate normalizers. Is there a group that requires three?

So far GL(4,2) and GL(5,2) only require two, and in fact any two maximal normalizers (called maximal parabolic subgroups) work.

I am also confused, because the normalizers of the essential subgroups of the fusion system are NOT maximal parabolic. I would have though maximal normalizers and essential subgroups would be very, very closely related.

Ok, so in GL(4,2), the normalizers of the essential subgroups are parabolic subgroups, just not maximal. And using those normalizers, conjugacy of elements requires all 3! However, using maximal parabolics, it only requires 2 (any 2).

Ok, so in GL(5,2) the maximal subgroups come in two varieties: the “coordinate” ones that just omit a single simple root generator, and the “others”. They can be detected group-theoretically as the ones that have larger normalizers: if Q ≤ P is a maximal subgroup, then it is an “other” iff N_{G}(Q) = N_{G}(P) is a Borel subgroup. Otherwise N_{G}(Q) is a “minimal parabolic”, that is, it has exactly one 2×2 block on the diagonal. To get the other parabolics, one forms Ferrers looking diagrams as Q. Maximal parabolics correspond to rectangles.

Even stranger: in GL(n,2) for n in {2,3,4,5,6}, the normalizer N_{G}(Q) of a normal subgroup Q of P is always a parabolic subgroup! This is not true in GL(3,3) or GL(2,4), but maybe it is close to true. The extra normalizers are contained in, but do not contain parabolics. The index in the minimal parabolic covers for:

- GL(2,3) — none
- GL(3,3) — 2
- GL(4,3) — 2, 4
- GL(5,3) — 2, 4, 8
- GL(2,4) — 3
- GL(3,4) — 3, 9
- GL(4,4) — 3, 9, 27

Probably then the normalizers are just missing some diagonal elements (maximally split toral elements)? That seems strange (especially only getting the maximally split torus), but I was surprised the answer was nice, so I guess I shouldn’t be too surprised when the answer is also not nice.

Ok, so maybe this makes sense: if a proper subgroup of GL(n,q) contains a Sylow p-subgroup (q=p^{k}), then it must sit inside some parabolic subgroup. If it is a full normalizer of a normal subgroup Q of a Sylow p-subgroup P, then it has a normal p-subgroup Q and contains P, and so it should be contained in some parabolic. The idea is to squeeze it between the parabolic and something sane, but I’m not sure what the lower bound should be. Root subgroups are normal and normalized by the full torus. MAYBE the issue is that not all normal subgroups of P are normalized by T. Aha! That fixes it for GL(3,4)!

Ok, so we have Q ≤ P, Q normal in B = N_{G}(P), and we want to show that N_{G}(Q) is a parabolic subgroup. It contains B, and perhaps there is a theorem that the subgroups containing B are exactly the parabolics?

Proof: Suppose B ≤ M ≤ GL(n,q) where B is the Borel subgroup consisting of all upper-triangular matrices. We want to show that M consists of all block upper-triangular matrices with the same fill pattern as M. I guess that means that if M contains a matrix A with a nonzero i,j entry with j ≤ i, then it contains all matrices that are upper triangular except for k,l with j ≤ l < k ≤ i. B contains enough to bring A into lower triangular form, I believe (at least into a permuted such state), and then I think the commutator with E_{k,i} and E_{j,l} moves the entry. Be nice to have a more convincing proof, but that is basically how to pollute the other entries.

Ah, so A is in some coset BwB=BnB, and so the subgroup generated by A and B contains B and the permutation mat n, so also block upper with the same coset rep. I guess basically we need to write down what BwB looks like for each permutation mat, and recognize it as the “full” ish matrices in the parabolic subgroup. So Bw is easy to describe, it is just B with the columns permuted by w. BwB appears to be the filled-in version of this. So if w=(1,3), then BwB fills in the entire 1–3 block. if w = (1,3)(7,10), then we get blocks of size 3,1,1,1,4,1,1,….

OK, a very important observation: while any two maximal parabolics (probably) determine the fusion of elements, they do NOT do so by acting on their p-cores. In other words we have to cheat and allow them to act on their full Sylow p-subgroups (which they can have many of). Now of course these maximal parabolics are smaller than the original group, but in a lot of ways they are just as complicated, so it is sort of circular to use them. On the other hand, the minimal parabolics are the minimal collection of parabolics that control fusion of elements using only their action on their p-cores! Here we really do have pretty trivial action: something like GL(2,q) acting on a direct power of its natural module. So it is just GL(2,q), not GL(n-1,q), but also it is acting on the natural module, not on its “Bouc complex” or “Frobenius category” or what have you, that is, not on its set of Sylow subgroups.