# Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?

Let $$G$$ be a compact Lie group. An Abelian Lie subgroup $$A \leq G$$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $$A'$$ such that $$A \leq A' \leq G$$, then $$A' = A$$.

Of course any maximal torus of $$G$$ (there is only one, up to conjugacy classes) is a maximal Abelian Lie subgroup, but there are other ones too, for example the Klein 4-group in $$\mathrm{SO}(3)$$.

What I'm wondering is if the number of conjugacy classes of maximal Abelian Lie subgroups of any compact Lie group $$G$$ is always finite?

Neighboring subgroups theorem  Any compact subgroup $$H$$ of an arbitrary Lie group $$G$$ admits a neighborhood $$O$$ in $$G$$ such that any subgroup contained in $$O$$ is $$G$$-conjugate to a subgroup of $$H$$.
Now let by contradiction $$(A_n)$$ be a sequence of pairwise non-conjugate closed maximal abelian subgroups in a compact Lie group. Let $$A$$ be a limit point in the Hausdorff topology; this is a compact abelian subgroup. By the above result, for $$n$$ large enough $$A_n$$ is conjugate to a subgroup of $$A$$. By maximality, this means that for $$n$$ large enough, $$A_n$$ is conjugate to $$A$$. This contradicts the non-conjugation.
• Another approach: any abelian (or even nilpotent) subgroup of $G$ is contained in the normalizer of a maximal torus. – abx May 23 at 13:21
• @abx I didn't know this result, which seems to provide an alternating approach reducing the problem to the case when $G^0$ is abelian. – YCor May 23 at 13:41
• @abx thanks. Anyway some further argument is needed to deal with $G^0$ abelian. – YCor May 23 at 14:09
• Just to emphasize a possible difficulty: while every abelian subgroup is contained in a maximal abelian one (obvious and true in every group), the same is false for nilpotent/maximal nilpotent subgroups in the compact Lie group $(\mathbf{R}/\mathbf{Z})\rtimes_\pm(\mathbf{Z}/2\mathbf{Z})$. – YCor May 23 at 23:11