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This addresses the second question "What is known about finite subgroups of $SU(n)$". A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$ such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4 of Thurston's book). Thus, there is a normal abelian subgroup of index at most $m!$. So one may make a statement: there are finitely many finite groups so that any finite subgroup of $SU(n)$ is an abelian extension (of rank at most $n-1$) of one of these finitely many groups. It would be quite interesting to obtain an estimate of the function $m(n)$, which should be possible by giving an effective proof of Margulis' theorem. I did a literature search once to see if anyone had attempted this, but I didn't find anything, and I would be curious if someone knows something.

Addendum: Working backwards from Weisfeiler's paper referenced in Keivan's comment, I found a result of Collins implies that a finite linear subgroup of $GL(n,C)$ has an abelian normal subgroup of index at most $(n+1)!$ when $n\geq 71$ (and gives the bound for all $n$). Since finite subgroups of $GL(n,C)$ are conjugate into $U(n)$, this bound works for $SU(n)$. See also Collins paper on primitive representations, which has some historical discussion of this problem.

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This addresses the second question "What is known about finite subgroups of $SU(n)$". A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$ such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4 of Thurston's book). Thus, there is a normal abelian subgroup of index at most $m!$. So one may make a statement: there are finitely many finite groups so that any finite subgroup of $SU(n)$ is an abelian extension (of rank at most $n-1$) of one of these finitely many groups. It would be quite interesting to obtain an estimate of the function $m(n)$, which should be possible by giving an effective proof of Margulis' theorem. I did a literature search once to see if anyone had attempted this, but I didn't find anything, and I would be curious if someone knows something.

Addendum: Working backwards from Weisfeiler's paper referenced in Keivan's comment, I found a result of Collins implies that a finite linear subgroup of $GL(n,C)$ has an abelian normal subgroup of index at most $(n+1)!$ when $n\geq 71$ (and gives the bound for all $n$). Since finite subgroups of $GL(n,C)$ are conjugate into $U(n)$, this bound works for $SU(n)$.

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This addresses the second question "What is known about finite subgroups of $SU(n)$". A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$ such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4 of Thurston's book). Thus, there is a normal abelian subgroup of index at most $m!$. So one may make a statement: there are finitely many finite groups so that any finite subgroup of $SU(n)$ is an abelian extension (of bounded rank ) at most $n-1$) of one of these finitely many groups. It would be quite interesting to obtain an estimate of the function $m(n)$, which should be possible by giving an effective proof of Margulis' theorem. I did a literature search once to see if anyone had attempted this, but I didn't find anything, and I would curious if someone knows something.

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