$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$A closed subgroup $ \Gamma $ of a Lie group $ G $ is called Lie primitive if it is not contained in any proper positive dimensional closed subgroup.


Let $ G $ be a compact group. Then a Lie primitive subgroup $ \Gamma $ of $ G $ must be finite.


Must a compact simple Lie group $ G $ have only finitely many isomorphism types of Lie primitive subgroups?


Context:


This is true for $ \SO\_3(\mathbb{R}) $ which has only $ 3 $ Lie primitive subgroups up to conjugacy: the exceptional rotation groups of the platonic solids $ A\_4$, $S\_4$, $A\_5 $.


This is also true for $ \SU\_3 $ and $ \SU\_4 $ due to work by Blichfeldt. For an explicit list for $ \SU\_4 $ for example see [Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4)](https://arxiv.org/abs/hep-th/9905212).


Since it is true for $ \SU\_4 $ then it must be true for $ \SO\_6 $ by the double cover $ \SU\_4 \to \SO\_6 $.


Also this seems relevant: [Primitive subgroup of $ SU\_n $ contained in maximal finite subgroup?](https://math.stackexchange.com/questions/4468936/primitive-subgroup-of-su-n-contained-in-maximal-finite-subgroup)


Some comments about the case when $ G $ is not compact:


A closed subgroup not contained in any positive dimensional subgroup must be dimension $0$ i.e. it must be discrete. Peter McNamara [points out](https://mathoverflow.net/questions/436711/compact-lie-group-has-finitely-many-lie-primitive-subgroups#comment1132858_436711) that the only proper positive dimensional closed subgroups of $ \SL\_2(R) $ are solvable and thus can't contain a free group. Since a free group on any number of letters is a subgroup of $ \SL\_2(R) $ that means there are infinitely many non isomorphic Lie primitive subgroups of $ \SL\_2(R) $.



Compact Lie group has finitely many Lie primitive subgroups
Yes, it is true. Let us show that, in a compact Lie group $G$ with $G^0$ nonabelian, finite subgroups that are not contained in any proper subgroup of positive dimension have bounded cardinal.


By contradiction, let finite subgroups $G\_n$ have cardinal tending to infinity, and not contained in any proper subgroup of positive dimension.


Let $B$ be a neighborhood of 0 in the Lie algebra of $G$ such that $\exp|\_B$ is a homeomorphism onto its image, with inverse denoted by $\log$ (we will let $B$ vary, but can assume it is contained in a single such neighborhood $B\_0$ once and for all, hence the function $\log$ will not depend on $B$).


Define $T\_n^B=\exp\big(\mathrm{span}\big(\log(G\_n\cap\exp(B))\big)\big)$.


The first point is that for each $B\subset B\_0$ and for $n$ large enough, $G\_n\cap\exp(B)$ consists of commuting elements. This is close to Jordan's theorem and is a particular case of Zassenhaus' theorem (see for instance [Theorem 1.3 here](https://arxiv.org/abs/1402.0962), with in mind that connected nilpotent subgroups of compact groups are abelian). We can extract once and for all to suppose that this holds for all $n$.


The next point is that $T\_n^B$ is a torus. For an arbitrary commutative subset, it might have been dense in a torus. But the point is that elements of finite subgroups are well-behaved. More precisely: for this point, we can suppose that $G$ is embedded in the unitary group $\mathrm{U}(d)$. Every element of $G$ can be diagonalized with eigenvalues $e^{i\pi k\_1/m},\dots,e^{i\pi k\_d/m}$. We can arrange $B\_0$ so that it avoids elements with eigenvalues $-1$ and define the log accordingly, so assume all $k\_i$ to be in $]-m,m[$. Hence the log of such an element will be, in the same diagonal basis, the matrix with eigenvalues $i\pi k\_1/m,\dots i\pi k\_d/m$, and the exp of its span will be the 1-dimensional torus of diagonal matrices with eigenvalues $\theta^{k\_1},\dots,\theta^{k\_d}$ for $\theta$ in the unit circle. Now when we consider all $G\_n\cap \exp(B)$, we just consider the torus generated by several such (commuting) tori.


Now consider $\liminf\_n \dim(T\_n^B)$. This is positive (because $G\_n$ is infinite, so that $G\_n\cap\exp(B)$ is never reduced to $\{1\}$. This liming decreases when $B$ decreases, and hence we can find $B\_1$ for which its value, say $c>0$, is minimal. After extracting once again, we an suppose that $\lim\_n\dim(T\_n^{B\_1})=c$. Define $T\_n=T\_n^{B\_1}$. Hence, for every neighborhood $B\subset B\_1$, there exists $n\_0$ (depending on $B$) such that for all $n\ge n\_0$ we have $T\_n^B=T\_n$ (because $T\_n^B\subset T\_n$ and these are tori of the same dimension).


This observation allows to deduce that $T\_n$ is, for large $n$, normalized by $G\_n$. Indeed, for $g\_n\in G\_n$, we have $$g\_nT\_n^Bg\_n^{-1}=\exp(\mathrm{span}\log(G\_n\cap \exp(g\_nBg\_n^{-1})));$$
since $1$ has a basis of conjugation-invariant neighborhoods, we can choose $B$ such that $\bigcup\_g gBg^{-1}$ is contained in $B\_1$. We deduce that $g\_nT\_ng\_n^{-1}\subseteq T\_n$ for large $n$, hence this is an equality.


Then $G\_nT\_n$ is a positive-dimensional closed subgroup containing $G\_n$. By assumption, it equals $G$, and this is a contradiction since $G^0$ is non-abelian.


