Let $ H $ be a connected subgroup of $ G=\mathrm{SU}(n) $ such that $ N_G(H)/H $ is finite. Is $ N_G(H)/H $ always a subgroup of the symmetric group $ \mathrm{S}_n $?
I just noticed this from the examples in https://math.stackexchange.com/a/4398724/758507
https://math.stackexchange.com/a/4535587/758507
and
Maximal Closed Subgroups of $ SU_4 $
and
Is the irreducible $ SU(3) $ subgroup of $ SU(6) $ maximal?
In general I'm interested in the case where $ G $ is simple and $ H $ is a connected subgroup of $ G $ such that $ N_G(H)/H $ is finite. What can we say about the finite group $ N_G(H)/H $?
For $ G=\mathrm{SU}(2) $ the only nontrivial finite group $ N_G(H)/H $ is cyclic $ 2 $.
For $ G=\mathrm{SU}(3) $ the nontrivial finite groups $ N_G(H)/H $ are cyclic $ 3 $ and the symmetric group $ \mathrm{S}_3 $
For $ G=\mathrm{SO}(5) $ the only nontrivial finite $ N_G(H)/H $ are cyclic $ 2 $ and $ \mathrm{D}_4 $.
For $ G=\mathrm{G}_2 $ the only nontrivial finite groups $ N_G(H)/H $ are cyclic $ 2 $ and $ \mathrm{D}_6 $.
For $ G=\mathrm{SU}(4) $ the only nontrivial finite $ N_G(H)/H $ are cyclic $ 2 $, cyclic $ 4 $ and the symmetric group $ \mathrm{S}_4 $
For $ G=\mathrm{SU}(6) $ the finite $ N_G(H)/H $ include cyclic $ 2 $, cyclic $ 3 $ and the symmetric groups $ \mathrm{S}_3 $ and $ \mathrm{S}_6 $
In general if $ G $ is a simple Lie group and $ H $ is a subgroup such that $ N_G(H)/H $ is finite then $ N_G(H)/H $ always seems to be a subgroup of the Weyl group of $ G $ (same thing if $ G $ is a simple algebraic group). Is that true?
