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.
Best Answer
Counterexample: Let $G$ be the infinite dihedral group, i.e. $$ G = \langle a,x : x^2=e, xax=a^{-1} \rangle. $$ Note $\langle a\rangle \cong \mathbb{Z}$ is a normal subgroup of index $2$. Let $N$ be a nontrivial normal subgroup of $G$. If $a^k\in N$ for any $k\neq 0$, then $N$ contains a subgroup of $\langle a \rangle$ of finite index, hence $N$ must have finite index in $G$.
Now suppose $xa^k \in N$ for some $k$ (note $k$ may now be $0$). Then as $N$ is normal, $N$ must contain $$ a(xa^k)a^{-1} = xa^{k-2}. $$ Then $N$ must contain $$ (xa^k)(xa^{k-2}) = a^{-2}, $$ and hence $N$ has finite index by the argument above. But using the relation $xax=a^{-1}$ (via $xa=a^{-1}x$), every nontrivial element of $G$ can be written as $a^k$ for $k\neq 0$ or $xa^k$ for any $k$.
So every nontrivial normal subgroup of $G$ has finite index. But the subgroup $\langle x \rangle \cong \mathbb{Z}/2$ does not have finite index.