Subgroups of free abelian groups are free: a topological proof? - MathOverflow

Subgroups of free abelian groups are free: a topological proof?

2009-11-08T02:26:40Z

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:

1) 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.

2) 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.

3) The covering space of any graph is again a graph.

4) Any graph has the homotopy type of a wedge of circles, so the fundmamental 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 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?

Answer by Tyler Lawson for Subgroups of free abelian groups are free: a topological proof?

2009-11-08T13:56:08Z

The "free group" proof rests on proving that that the fundamental group of a graph is free. For the analogue we'd need to essentially prove that the fundamental group of a "torus" (something that looks like a quotient of a vector space by a discrete subgroup) is free abelian. A sketch:

Given a real vector space V, we can put the direct limit topology on it (so that subsets are closed if and only if their intersection with any finite dimensional subspace is closed). This is a contractible topological group.

If A is a free abelian group, then A is a discrete subgroup of the associated real vector space (ℝ ⊗ A) and the quotient space has fundamental group A. Any covering space is a quotient of (ℝ ⊗ A) by a discrete subgroup B of A.

So the question boils down to showing: Any discrete subgroup of a vector space (with the direct limit topology) is free abelian.

Let's say that a partial basis is a set S of elements of B such that

S is linearly independent, and
S generates B ∩ Span(S).

Then partial bases are a partial order under containment, and Zorn's lemma implies that there is a maximal element S. I claim that S is a basis of B as a free abelian group.

S is linearly independent by construction, so it generates a free abelian group, and hence it suffices to show that it generates all of B. If b in B is not in S, then it is not in Span(S). Let S' be (S ∪ {b}). Then Span(S')/Span(S) is a 1-dimensional vector space and the image of B ∩ Span(S') must be discrete, because otherwise Span(S') would contain an element (rb + v) for v in Span(S) that we could use to generate a non-discrete subset of B. (If v is a combination of w₁...w_n in S, then it suffices to check that any subgroup of the finite-dimensional space Span(w₁...w_n,b) requiring more than n generators is indiscrete.)

Thus any lift of a generator of B ∩ Span(S') would extend to a larger generating set, contradicting maximality.

(My apologies for the comment last night, which this morning looks snarkier than I intended. I'm a fan of using this topological reasoning for free groups myself, because it compartmentalizes the proof into much more understandable pieces. In particular, I don't think I'd really understand a purely algebraic proof that an index n subgroup of a free group on m generators is free on nm - n + 1 generators.)

Answer by HW for Subgroups of free abelian groups are free: a topological proof?

2009-11-08T16:02:52Z

An obvious point, but I hope worth making. The free group proof rests on the fact that graphs can be characterized locally. As tori can't be characterized locally by their topology, there's no hope of a truly analogous proof for free abelian groups using tori.

Answer by Andrew S for Subgroups of free abelian groups are free: a topological proof?

2009-11-17T03:39:00Z

For a purely topological proof of the statement "Every subgroup of a free abelian group is free abelian.", why not just continue to use a wedge of circles, but use the first simplicial homology and the Hurewicz theorem? If you take "free abelian" to mean "a direct limit of finte rank free abelian", it seems to follow from the statement about free groups.