Let $G$ be a hyperbolic group. I know that it is an open problem whether $G$ has a torsion-free subgroup of finite index. But if we let $N$ be the subgroup of $G$ generated by its non-torsion elements, then is $G/N$ necessarily finite?


If not, $G/N$ would be a finitely generated infinite torsion group with finitely many conjugacy classes.



Subgroup of hyperbolic group generated by non-torsion elements
Yes. The following answer is inspired by Andy Putman's comment. Let $N\_\infty(G)$ be the subgroup generated by elements of infinite order, in a group $G$.


Every non-elementary hyperbolic group $G$ with trivial finite radical has a useful property, which bears the ridiculous name "Pnaive", namely that for every finite subset $F$ there exists a $x$ such that the natural homomorphism from $F\ast\mathbf{Z}$ to $G$ mapping $F$ identically to itself and the generator of $\mathbf{Z}$ to $x$, is injective. This is a result of Arzhantseva and Minasyan: *G. Arzhantseva, A. Minasyan, Relatively hyperbolic groups are C\*-simple. 
J. Funct. Anal. 243 (2007), no. 1, 345–351.*


This applies to $F$ equal to an arbitrary singleton $\{y\}$, $y\in G$: then both $x$ and $yx$ have infinite order, hence $y\in N\_\infty(G)$. So Pnaive for a group $G$ implies $G=N\_\infty(G)$, that is, $G/N\_\infty(G)=\{1\}$.


Moreover, the property $N\_\infty(G)=G$ for nontrivial (!) groups, is also stable under taking extensions with torsion kernel. Hence it is satisfied by all non-elementary hyperbolic groups, since these have a maximal finite normal subgroup and we inherit Pnaive from the nontrivial quotient.


For completeness, the elementary cases: if $G$ is finite then $N\_\infty(G)=\{1\}$ and the quotient $G/N\_\infty(G)$ is $G$ itself. If $G$ is 2-ended, then, denoting by $W(G)$ the finite radical, $G/W(G)$ is infinite cyclic or infinite dihedral according to whether $G$ acts trivially or not on its 2-element boundary, and then the desired quotient $G/N\_\infty(G)$ is either trivial, or cyclic of order 2. 


Thus for an arbitrary hyperbolic group, $G/N\_\infty(G)$ is finite.


(Note that we use much much less than the full result of Arzhantseva-Minasyan! we only need it for a singleton $\{y\}$, and we need much less than injectivity: only injectivity on the cyclic subgroup $\langle xy\rangle$.)


