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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?

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Yes it's true. It essentially follows from a lemma quoted in the linked answer by user Qayum Khan, which I quote verbatim:

Neighboring subgroups theorem [1942] 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.

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    $\begingroup$ Another approach: any abelian (or even nilpotent) subgroup of $G$ is contained in the normalizer of a maximal torus. $\endgroup$ – abx May 23 at 13:21
  • $\begingroup$ @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. $\endgroup$ – YCor May 23 at 13:41
  • $\begingroup$ Bourbaki's Lie IX, §5, Cor. 4 of Théorème 1. $\endgroup$ – abx May 23 at 13:51
  • $\begingroup$ @abx thanks. Anyway some further argument is needed to deal with $G^0$ abelian. $\endgroup$ – YCor May 23 at 14:09
  • $\begingroup$ 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 $(\mathbf{R}/\mathbf{Z})\rtimes_\pm(\mathbf{Z}/2\mathbf{Z})$. $\endgroup$ – YCor May 23 at 23:11

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