7
$\begingroup$

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 $.

$\endgroup$
6
  • 1
    $\begingroup$ What's the Lie group version of the $A_1^7.PSL_2(7)$ in $E_7$? ($PSL_2(7)$ is a complex reflection group, but this is not in the reflection representation.) Or the $A_1^8.AGL_3(2)$ in $E_8$? ($AGL_3(2)$ is not a CRG, I think.) I assume such a thing exists, as it does in the algebraic group. $\endgroup$ Commented Feb 20, 2023 at 20:17
  • $\begingroup$ @DavidA.Craven I asked Jeff Adams and those are the exact same two examples he gave and they should exist in the compact group as well. Do you want to just post this as an answer? They are both very interesting examples. Also how is $ PSL_2(7) $ a complex reflection group? I thought that complex reflection groups were never (non-abelian) simple? And I agree that $ AGL_3(2) $ does not seem like a CRG, although I have no idea how to show that. $\endgroup$ Commented Feb 22, 2023 at 15:39
  • $\begingroup$ @DavidA.Craven Isn't ST-24 the direct product of cyclic 2 with $ PSL_2(7) $? The size is 336 which is twice the size of $ PSL_2(7) $? $\endgroup$ Commented Feb 22, 2023 at 16:08
  • $\begingroup$ @DavidA.Craven Oh that's very interesting! Do you have a particular example of a perfect CRG in mind ? $\endgroup$ Commented Feb 22, 2023 at 16:15
  • $\begingroup$ OK, I keep writing nonsense. There are no perfect examples either. There is going to be some geometry reason that I can't remember right now. I should dig out my notes of CRGs and check. You do get things like $2\times SL_2(9)$, which is again a direct product, but the $SL_2(9)$ has a centre so there is a non-simple component. (ST 32 has no subgroup of index $2$.) $\endgroup$ Commented Feb 22, 2023 at 16:21

1 Answer 1

1
+50
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .