I'm ultimately concerned with a proof showing that every subgroup of a finitely generated abelian group is also finitely generated by covering spaces.
These thoughts are premature, but my curiosity could not stop me from asking:
Can we show that every subgroup of $\pi_1(T) \cong \mathbb Z^2$ is a finitely generated abelian group?
We know that subgroups of $\pi_1(T)$ are in correspondence with normal covering spaces (since they are all normal.)
Taking the universal cover $p:X \to T$, we can see that anu subgroup $H$ acts on the quotient by the group action $X/H$.
The problem here, is that I don't know at all what kind of restriction being abelian puts on the structure of a covering space so there is no way for me to proceed here.
On the other hand, we could study instead the subgroup $\langle(1,0) \rangle \subset \mathbb Z^2$, and try to look at a quotient by it (which will be finitely generated abelian) and use the correspondence/iso theorems to lift back up. If there is a nice geometric interpretation of this procedure, I am interested, but otherwise I know this proof already.