Consider the compact group $ G=\operatorname{SO}_3(\mathbb{R}) $. The closed subgroups of $ G $ (other than the trivial group 1 and the whole group $ G $) are exactly $ O_2$, $\operatorname{SO}_2 $ and the finite groups $ C_n$, $D_{2n}$, $T \cong A_4$, $O \cong S_4$, $I \cong A_5 $ (cyclic groups with $ n $ elements, dihedral groups with $ 2n $ elements and the three symmetry groups of the platonic solids). The normalizers of these groups are as follows:

\begin{align*} G&=N_G(G)=N_G(1) \\ O_2&=N_G(O_2)=N_G(\operatorname{SO}_2)=N_G(C_n) \\ I&= N_G(I) \\ O&=N_G(O)=N_G(T)=N_G(D_4) \\ D_{4n} &= N_G(D_{2n}) \end{align*} where in the last equation $ n \geq 3 $. We say a (closed) subgroup is maximal if it is maximal among all proper closed subgroups of $ G $.

Observe that in the example above the maximal subgroups exactly coincide with the self-normalizing subgroups. Namely, $$ O_2, I,O. $$ That the maximal subgroups are all self-normalizing is not too surprising. The normalizer of a closed subgroup is always closed. Thus a maximal subgroup is always either normal or self-normalizing. Since $ G $ is simple, adjoint (i.e. center-free), and connected that means the maximal subgroups must be self-normalizing. However I am a bit surprised that the reverse holds. That is, that every self-normalizing subgroup of $ G $ is maximal. That inspires my question:

**For a closed subgroup of a compact Lie group does self-normalizing imply maximal?**

This is true for the compact Lie group $ \operatorname{SO}_3(\mathbb{R}) $ and thus also true for $ \operatorname{SU}_2 $. What about the generic case? I am especially interested in $ \operatorname{SU}_3 $.

alwaysclosed, whether of closed or non-closed subgroups, or even of arbitrary subsets.) $\endgroup$