Condition implying $N(H)/H$ a Coxeter group

coxeter-groupsfinite-groupsgroup-theorylie-groupsrepresentation-theory

I'm interested in which finite groups can arise as
$$
N(H)/H
$$

for $ H $ a connected subgroup of a compact connected simple Lie group $ G $.

One obvious family of examples is take $ H $ to be the maximal torus then $ N(H)/H $ is the Weyl group of $ G $. For some other examples I looked at $ N(H)/H $ is just cyclic 2. Also all the $ N(H)/H $ given in the second column of tables 5,6,7,8 of [https://arxiv.org/abs/math/0605784] seem to be Coxeter groups or at least complex reflection groups.

Is it possible that $ N(H)/H $ is always a Coxeter group? (again I'm assuming $ H $ a connected subgroup of a connected Lie group $ G $)

Seems like a bit of a crazy conjecture, but mostly I'm just interested in understanding the structure of the finite group $ N(H)/H $.

Best Answer

Inside the algebraic group $E_8(\mathbb C)$ there is a subgroup $A_1^8\cdot \mathrm{AGL}_3(2)$. This will have a corresponding subgroup in at least one of the real forms of $E_8$, but I am not sure which, hopefully the compact form.

The group $H=\mathrm{AGL}_3(2)$ is not a complex reflection group. To see this easily, note first it does not appear on the Shephard--Todd list, so can only be of the form $G(m,p,n)$, which has order $m^nn!/p$. It has rank $n$, and so since $H$ has no faithful representations of degree less than $7$, $n\geq 7$. But $5\nmid |H|$, so $5\mid p$. Since $p\mid m$, this means $5\mid m$, so again $5$ divides $m^nn!/p$, a contradiction.

The facts about complex reflection groups I used can all be found on the Wikipedia page.

Edit: One should not leave this subject without mentioning a related result by Raphael Rouquier, which talks about automizers of Sylow $p$-subgroups of finite groups. It is related because the automizer of a Sylow $p$-subgroup $P$ of a group of Lie type is closely connected to the component group of the centralizer of that group, if $P$ is abelian.

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