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Let $G$ be the group $SO(n)$, $SU(n)$ or $Sp(n)$.

Let $F_m$ be the collection of finite subgroups of $G$ such that its order is bounded by $m$. Two elements in $F_m$ are identified if they are conjugate in $G$.

Can we prove that $\#F_m \le C(n,m)$ for some finite number $C(n,m)$?

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  • $\begingroup$ See this question: mathoverflow.net/questions/17072/the-finite-subgroups-of-sun $\endgroup$
    – user158834
    Commented Jun 9, 2020 at 0:55
  • $\begingroup$ I am sure this was asked earlier; indeed, there are only finitely many finite subgroups of the given order in each connected Lie group up to conjugation. $\endgroup$
    – Moishe Kohan
    Commented Jun 9, 2020 at 4:56
  • $\begingroup$ It amounts to showing that for every finite group $F$ there are finitely many conjugacy classes of homomorphisms $F\to G$. This is a standard fact, at least in the OP's setting. For instance this is explicit in [D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694-699], but already follows from Weil's rigidity (1964), and was possibly known earlier. (For the general case: use conjugacy of maximal compact subgroups —due to Iwasawa— to reduce to the compact case.) $\endgroup$
    – YCor
    Commented Jun 9, 2020 at 7:24

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