Torsion-free quotient group of an abelian group Let $G$ be an abelian group, and let $H\leq G$. Prove that if $G/H$ is torsion-free, then $H$ contains the torsion group of $G$.
Proof:
Let $x\neq1$ be an element in the torsion group. Thus there exist $k\in \mathbb{N} $ with $x^k=1$.
Now we look at $(xH)^k = x^kH = H $
$G/H$ is torsion-free, so $xH$ must be equal to $H$ and therefore $x\in H $, as we wanted.
My question is that: In this proof I didn't really use the fact that $G$ is abelian. I could assume only that $H\lhd G$ and the proof still stands. Is it true? 
 A: As David Wheeler says in the comments, we cannot generally speak of the torsion subgroup of groups that are not abelian, because the torsion points need not be closed under multiplication; an obvious example being the infinite dihedral group. On the other hand the proof you've written does demonstrate that all torsion points are contained in $H$.
It is possible for non-abelian groups to have torsion subgroups, of course. If $G$ is finite then the torsion points are just all of $G$, for instance, and Wikipedia states that the torsion subset of any nilpotent group forms a normal subgroup (so this would include infinite groups). The orders of elements do not behave predictably in nonabelian settings, however: whereas in abelian groups we have that the order of $ab$ is the $\rm lcm$ of the orders of $a$ and $b$, we can actually pick any integers $r,s,t$ and there will be a group $G$ with elements $a,b,ab$ of those orders respectively (this is Thm 1.64 in Milne's freely available group theory course notes), specifically the $\rm PSL_2$s of some finite fields.
