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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 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?

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    $\begingroup$ For the record, a topological proof that the fundamental group of a graph is free usually involves something like finding a maximal spanning tree and applying Seifert-van Kampen. This may look easier than an algebraic proof, but it hides the same details - algebraic proofs generally choose a set of coset representatives (spanning tree), and then reduce any word to a canonical product of elements (those corresponding to edges not in the spanning tree). All these details are there in the (gory) proof of the Seifert-van Kampen theorem itself. $\endgroup$
    – Tyler Lawson
    Commented Nov 8, 2009 at 3:06
  • $\begingroup$ @Pete, Ben I retagged the question editing out some tags that don't seem convey any information (per meta.stackexchange.com/questions/18878/…) and adding the group-theory tag (the question should be of interest to people in group theory). Feel free to revert or discuss on Meta. $\endgroup$
    – Ilya Nikokoshev
    Commented Nov 8, 2009 at 18:18

3 Answers 3

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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...wn in S, then it suffices to check that any subgroup of the finite-dimensional space Span(w1...wn,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.)

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  • $\begingroup$ This is a great answer -- just what I was looking for. Thanks. $\endgroup$
    – Pete L. Clark
    Commented Nov 8, 2009 at 20:41
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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.

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    $\begingroup$ I think it is the geometry that plays a role, not the topology. Locally graphs are spaces of curvature minus infinity, while tori are spaces of curvature zero. Both curvature conditions are therefore inherited by the covering spaces. $\endgroup$
    – Igor Belegradek
    Commented Nov 8, 2009 at 19:32
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    $\begingroup$ True. However, spaces of curvature 0 have fundamental groups that are virtually free abelian, which isn't quite good enough (spaces of curvature -infty, ie graphs, have fundamental groups that are free on the nose). Moreover, the proof that spaces of zero curvature have virtually free abelian fundamental groups (which is part of Bieberbach's theorem) is quite a bit deeper than the fact that graphs have free fundamental groups. I'd have to think about it a bit, but it wouldn't surprise me if the fact that subgroups of free abelian groups are free abelian is already used in the proof. $\endgroup$
    – Andy Putman
    Commented Nov 8, 2009 at 22:58
  • $\begingroup$ It is hard to be sure when talking about assumptions like this that are always unspoken, but I think you could prove Bieberbach without this. Thinking about the proof, the following statement seems attainable: if G is a lattice in Isom(R^n), then the translational part of G is finite index. This subgroup, of course, is a discrete subgroup of R^n, so by Tyler's argument above it's free abelian. This wouldn't yield another proof of the result, it would just say that Bieberbach's theorem is independent of it. $\endgroup$
    – Tom Church
    Commented Nov 9, 2009 at 1:07
  • $\begingroup$ I agree completely, Igor - it's clearly the geometry, rather than the topology, that's important. $\endgroup$
    – HJRW
    Commented Nov 9, 2009 at 1:21
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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.

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    $\begingroup$ Because the subgroups of homology don't correspond to subgroups of the free group in the way you want. If V is a subgroup of Z^n, the corresponding subgroup G of F_n has infinite-rank first homology whenever V has infinite index. $\endgroup$
    – Tom Church
    Commented Nov 17, 2009 at 4:31

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