TITLE: A torsion-free quotient?
QUESTION [7 upvotes]: 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 $G$).

REPLY [2 votes]: 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.