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.