Normalizer of normalizer of maximal torus in a Lie group

Question

I'm stuck at this problem,

Let $G$ be a compact connected Lie group and $T$ a maximal torus. Let $H$ be a closed subgroup of $G$. Let $N(T)$ and $N(H)$ denote the normalizers of $T$ and $H$ respectively. Show that if $N(T) \subset H $then $N(H) = H$.

I was able to show that $N(H)/H$ should be finite. But showing this only used the fact that $T \subset H$. Because this question can no longer be reduced to a question about Lie algebras, I am not able to see how to proceed.

Any hints would be appreciated.

score 1 · Accepted Answer · answered Feb 15 '16 at 15:51

1

Does this work? Suppose $n \in G$ normalizes $H$. Then $nTn^{-1}$ is a maximal torus of $H$. All maximal tori of $H$ are conjugate, so this means the new torus is $h^{-1}Th$ for some $h \in H$. Multiplying through to remove inverses, we see $hn \in G$ normalizes $T$. By assumption, then, $hn$ is in $H$, so $n$ must be too.

answered Feb 15 '16 at 15:51

jdc

4,757

"All maximal tori of $H$ are conjugate, so this means the new torus is $h^{-1}Th$ for some $h \in H$." How? Note that $T$ isn't a maximal torus in $H$. – Neeraj Kumar Apr 14 '17 at 04:33
If we use $N(T)\subset H $, then that follows. – Neeraj Kumar Apr 14 '17 at 04:35

Normalizer of normalizer of maximal torus in a Lie group

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