There is a well-known topological proof of the fact that subgroups of free groups are free. Many people, myself included, think it is easier and more natural than the purely algebraic proofs which had been given earlier by (IIRC) Nielsen and Schreier. It goes as follows:

If $S$ is any set, then the CW-complex $X$ obtained as the wedge of $|S|$ circles is a graph whose fundamental group is isomorphic to $F(S)$, the free group on $S$.

If $H$ is a subgroup of $F(S)$, then by covering space theory $H$ is the fundamental group of a covering space $Y$ of $X$.

The covering space of any graph is again a graph.

Any graph has the homotopy type of a wedge of circles, so the fundamental group of $Y$ is again free.

My question is: to what extent is there an analogous proof of the result with "free" replaced everywhere by "free abelian"?

In the case of a finitely generated free abelian group -- say $G \cong \mathbb{Z}^n$ -- there is at least an evident topological *interpretation*. Namely, we can take $X$ to be the $n$-torus
(product of $n$ copies of $S^1$), and then observe that any covering space of a torus is homeomorphic to a torus of dimension $d$ cross a Euclidean space of dimension $n-d$, hence homotopy equivalent to a torus of rank $d <= n$.

Even in this case though I'd like some assurance that the proof of this topological fact does not use the algebraic fact we're trying to prove. (Is for instance some basic Lie theory relevant here?)

Then, what happens if the free group has arbitrary rank? Can we take $X$ to be a direct limit over a family of finite-dimensional tori? Does the proof go through?

`group-theory`

tag (the question should be of interest to people in group theory). Feel free to revert or discuss on Meta. $\endgroup$