A finite subgroup $ \Gamma $ of a Lie group $ G $ is called Lie primitive if it is not contained in any proper positive dimensional closed subgroup.
Must a connected Lie group have only finitely many Lie primitive subgroups?
(the exception is $ S^1 $ where every finite subgroup is Lie primitive since there are no proper positive dimensional closed subgroups, but I think it is the only exception)
For example this is true for $ SO_3(\mathbb{R}) $ which has only 3 Lie primitive subgroups: the exceptional rotation groups of the platonic solids $ A_4,S_4,A_5 $.
Also
https://math.stackexchange.com/a/4469654/758507
shows that $ SU_n $ only has finitely many Lie primitive subgroups, since every Lie primitive subgroup of $ SU_n $ is also a primitive subgroup in the matrix group sense.
Interesting to note that some of these exceptional subgroups are simple like $ A_5 $ in $ SO_3 $ or at least quasisimple like $ SL(2,5) $ in $ SU_2 $. And by Jordan Schur theorem there are only finitely many isomorphism types of quasisimple subgroups of any closed subgroup of any Lie group see for example
"First, we indicate roughly what to expect from a classification of subgroups of an algebraic group. An old theorem of C. Jordan ((14.12), [I]) says that there exists a function f(n) so that if F is a finite subgroup of GL(n, C), there exists a normal abelian subgroup A of F such that |F : A| ≤ f(n). A corollary is that GL(n, C) contains only finitely many isomorphism types of finite quasisimple groups. The same is therefore true of any algebraic subgroup of GL(n, C)."
from
https://www.ams.org/journals/bull/1999-36-01/S0273-0979-99-00771-5/S0273-0979-99-00771-5.pdf
