I am trying to prove this statement:
Every finitely generated group $G$ that is infinite has a normal subgroup $K$ such that $G/K$ is infinite and has only finite proper quotients.
This is a problem from a group theory work sheet. The problem text also suggests that the following lemma (due to Poincaré) should be used:
All subgroups of finite index of a finitely generated group are also finitely generated.
I feel like the lemma suggests that some sort of induction should be used here, together with the third isomorphism theorem. I am having trouble seeing how we even need the condition that $G$ be finitely generated here, though. Here is my proof attempt, which does not use this property at all:
If $G$ has only finite proper quotients, we are done. Therefore, let $G$ be a group with infinite proper quotients. We then have a nonempty set $L$ of normal subgroups $K$ of $G$ such that $G/K$ is infinite. Every chain in $L$ has an upper bound in $L$ (i.e. the union of all the elements in the chain), thus by Zorn's Lemma we have maximal elements of $L$. Let $K \in L$ be maximal.
We know $G/K$ is infinite. If $G/K$ has only trivial normal subgroups ($1$, $G/K$ itself), we are done. Therefore let $M/K \lhd G/K$ be a nontrivial normal subgroup (we know that each normal subgroup of $G/K$ has this form). We then have $M \lhd G$ and $K < M$, therefore $M \notin L$, thus $G/M$ is finite. By the third isomorphism theorem, $$ G/M \cong (G/K)/(M/K),$$ thus every proper quotient of $G/K$ is finite.
I must be missing something. Perhaps the union of chain elements might not actually be contained in $L$ is $G$ is not finitely generated, but I don't see why that would be the case.