TITLE: Finitely-generated group such that all (non-trivial) normal subgroups have finite index implies all (non-trivial) subgroups have finite index?
QUESTION [5 upvotes]: Let $G$ be a finitely generated group such that every non-trivial normal subgroup has finite index. Does it follow that every non-trivial subgroup of $G$ has finite index?

This question arose as a side novelty from another problem I was working on. I think it's interesting in its own right, but I cannot for the life of me prove it or find a counter-example. This would be cut-and-dry if I knew that every non-trivial subgroup contained a non-trivial normal subgroup, but I don't see how to get that. As far as I can tell, I don't have a guarantee the normal core isn't trivial.
Any help is appreciated.

REPLY [6 votes]: As mentioned, groups for which every nontrivial normal subgroup has finite index are very common (if infinite, they're known as "just-infinite" and include simple groups and many more).
On the other hand, groups with all non-trivial subgroups of finite index are very rare: either $G$ is finite, or is isomorphic to $\mathbf{Z}$.
Indeed, suppose that such a group $G$ is infinite. Let $C$ be a nontrivial cyclic subgroup. The assumption implies that $C$ has finite index, hence is infinite. Thus $G$ has a finite index subgroup isomorphic to $\mathbf{Z}$; but the argument also shows that $G$ is torsion-free. Now it is known that a torsion-free group with an infinite cyclic subgroup of finite index is always infinite cyclic (general fact about 2-ended groups).