Let $G := SO_n(R)$ be equipped with the Euclidean metric on vectors of length $n^2$. Is it true that for any $\epsilon >0$, there is a finite subgroup of $G$ which intersects every metric ball of radius $< \epsilon$ in $G$?
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Generalizing Robin's answer to arbitrary $n$: Jordan's theorem implies that for any $n$ there is an integer $J(n)$ such that the index of a normal abelian subgroup of a finite subgroup of $GL(n,\mathbf{C})$ and hence $SO(n)$ is $\leq J(n)$. A theorem by Boris Weisfeiler (based on the classification of finite simple groups) implies that there are real $a,b$ such that $J(n)<(n+1)!n^{b+a\log n}$. See http://www.pnas.org/content/81/16/5278.short?related-urls=yes&legid=pnas;81/16/5278 So any finite subgroup of $SO(n)$ is included in $\leq J(n)$ copies of the maximal torus and so if one takes $n\geq 3$ and small enough $\varepsilon>0$, then for any finite subgroup $G$ of $SO(n)$ there will there elements $\varepsilon$ or further away from $G$ (with respect to any say left-invariant metric) |
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The only finite subgroups of $G=SO_3(\mathbf{R})$ are cyclic, dihedral or of order 12, 24 or 60. The latter three can't work for small enough $\epsilon$ but neither can the cyclic or dihedral groups as each one of the these lie inside a $1$-dimensional Lie subgroup of $G$ which is not dense in $G$. |
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