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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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I believe the answer is no. Choose an element $\theta$ of infinite order in $\mathbb R/\mathbb Z=S^1$. Consider the group $G_n=(\mathbb Z\times S^1)/\langle(n,\theta)\rangle$. This is a finite extension of the compact topological group $S^1$, so it is compact. Also $\langle (n,\theta) \rangle$ is a closed subgroup, so the quotient is Hausdorff. The torsion subgroup consists of all torsion elements in the second factor: any element of the form $(k,\mu)$ where $k\neq 0$ cannot be torsion. Now if we mod out by the closure of torsion elements, as Jack Schmidt notes, this kills the entire $S^1$, leaving $\mathbb Z/\mathbb n\mathbb Z$ coming from the first factor.

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    Actually the group $G_n$ is homeomorphic to $n$ disjoint copies of $S^1$ so is clearly compact Hausdorff.2011-03-31