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Let $G$ be a group and $T$ the set of elements of finite order in $G$.
If $T$ is a subgroup of $G$, then $G/T$ is a torsion-free group.

Suppose $G$ is a compact Hausdorff topological group.
Is it true that $G/cl(T)$ is torsion-free? (where $cl(T)$ is the topological closure of $T$ in $G$).

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    Is T a subgroup for the c.h.t. group too? G/cl(T) is certainly a little weirder. The unit circle has a dense torsion-subgroup, so the result is torsion-free, but also torsion. If T is not a subgroup, then I think that groups like G=SU(n,C) also have T is a dense subset, so cl(T)=G. Again not a counterexample, but pretty weird.2011-03-30

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