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It is a well-known and deep${}^\ast$ theorem that if a group $G$ has cohomological dimension one then it must be free. This was proved in the late 60's by Stallings (for finitely generated groups) and Swan.

The proofs in the original articles are well-written and informative, bringing together a lot of ideas from topology, algebra and set theory. However there seems to be some scope for the proofs to be shortened, for instance by a clever choice of projective resolution, or the use of non-abelian cohomology.

My questions then is

Have any alternative (shorter) proofs of the Stallings-Swan theorem appeared since 1969?

${}^\ast$I am prepared to accept that the answer may be 'no', but in that case I wonder if someone could offer an explanation of why this theorem is so "deep", ie why there cannot exist some "quick trick" proof.

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2 Answers 2

up vote 23 down vote accepted

The heart of the matter is the Stallings' "ends of groups" theorem: A finitely-generated group with infinitely many ends splits as graph of groups with finite edge groups. In addition, one also has to show that the decomposition process terminates for your group (this property is called accessibility). Neither one has a quick an dirty proof and for a good reason.

a. Dunwoody has shown that accessibility fails for some finitely-generated groups with torsion, so something interesting (Grushko's theorem) is going on even in the easy part of the proof.

b. The ends of groups deals with the key problem of geometric group theory: Relating geometric properties of a group and its algebraic structure. Each time one manages to recover algebraic structure of a group from geometric information about the group, some minor (or major) miracle has to happen and, to the best of my knowledge (with few trivial exceptions) there are no easy proofs of the results of this type.

The "shortest" proof of the "Ends of groups theorem" is due to Gromov, see pages 228-230 of his essay on hyperbolic groups. The trouble with Gromov's proof is that it relies on a compactness property ("obvious" to Gromov) for a certain family of harmonic functions (in addition, a construction of the tree was missing in his proof, and this requires some trickery if one uses harmonic functions for finitely-generated groups). This compactness result (as far as I know) has no easy proof. I wrote a (somewhat long) proof in http://arxiv.org/pdf/0707.4231.pdf , Bruce Kleiner managed to shorten it to about 7 pages (this is not published), but his proof is still not quick.

Dunwoody's proof (see John Klein's excellent comments) improved on the Stallings' proof, but his proof is still quite complicated.

Niblo's proof in http://eprints.soton.ac.uk/29820/1/Stallingstheorem.pdf provides another geometric argument using Sageev's complex, but Niblo's paper is still 20 pages long.

Just to indicate how nontrivial Stallings' theorem is, consider the question: Is it true that every finitely generated group $G$ of homological dimension 1 is free? This is false for infinitely generated groups (like $G={\mathbb Q}$) and is true for finitely-presented groups (simply since in this case cohomological dimension is also 1). Otherwise, this problem is open since Stallings' theorem (and not for the lack of trying!).

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Thank you Misha for this informative answer. –  Mark Grant May 8 '12 at 10:54

This is really a comment rather than an answer, but perhaps the answer to your question follows from doing a mathscinet search? I did one and I found the following reference:

Dunwoody, M. J. Accessibility and groups of cohomological dimension one. Proc. London Math. Soc. (3) 38 (1979), no. 2, 193–215.

In this important and exciting paper the author obtains the structure of groups of cohomological dimension one over an arbitrary (commutative) ring with unity. Theorem 1.1: cdRG≤1 if and only if G is isomorphic to the fundamental group of a graph of groups in which every vertex group is finite with no R-torsion.(A finite group has no R-torsion if its order is invertible in R.) This extends the results of J. R. Stallings [Ann. of Math. (2) 88 (1968), 312--334; MR0228573 (37 #4153)] and R. G. Swan [J. Algebra 12 (1969), 585--610; MR0240177 (39 #1531)] that if cdRG≤1 and G is torsion-free then G is free, and also the results of various authors on free-by-finite groups.

The methods are an ingenious combination of Bass-Serre theory with a relative version of the theory of accessible groups due to C. Bamford and the author [J. Pure Appl. Algebra 7 (1976), no. 3, 333--346; MR0399271 (53 #3122)], and the reviewer's approach to the Stallings-Swan theory by means of almost invariant subsets [Groups of cohomological dimension one, Lecture Notes in Math., Vol. 245, Springer, Berlin, 1972; MR0344359 (49 #9098)].

In the course of the paper the author gives what is likely to be the best possible proof of Stallings' structure theorem for groups with more than one end [Stallings, op. cit.]. It has been known for some time that this theorem can be expressed by saying that the group acts on a tree with suitable properties. The reviewer spent much time trying to find the relevant tree; the set of edges of the tree was fairly obvious but he had no success in finding the vertex set. The author's solution is brilliantly simple. There is no need to determine the vertex set! His tree theorem (Theorem 2.1) gives necessary and sufficient conditions for a set to be the set of edges of a tree. These conditions, for a group with more than one end, are little more than previously known results about almost invariant sets.


The following might also have something to do with your question:

Cohen, Daniel E. Groups of cohomological dimension one. Lecture Notes in Mathematics, Vol. 245. Springer-Verlag, Berlin-New York, 1972. v+99 pp

The object of these well-written notes is to give a completely self-contained account of the following theorems. Theorem A: A torsion-free group of cohomological dimension one over some ring with unit is free. Theorem B: A torsion-free group containing a free subgroup of finite index is free. Theorem C: Let H be a subgroup of finite index in a torsion-free group G; then G and H have the same cohomological dimension. Theorem D: Let H be a subgroup of a free group G; then H is a free factor of G if and only if IHG is a direct summand in IG, the augmentation ideal of G. Theorems A and B were proved by J. R. Stallings [Ann. of Math. (2) 88 (1968), 312--334; MR0228573 (37 #4153)] for finitely generated groups and by R. G. Swan [J. Algebra 12 (1969), 585--610; MR0240177 (39 #1531)] in the general case. Theorem C is attributed to Serre; Theorem D is a stronger version of a result of Swan [op. cit.]. The presentation of the material differs in many significant details from the papers of Stallings and Swan; among them are the following: (i) the theory of ends, which plays an important rôle in the proof of Theorem A, is given in the algebraic form due to the author [Math. Z. 114 (1970), 9--18; MR0260877 (41 #5497)]; (ii) Stallings's structure theorem for groups with infinitely many ends [Applications of categorical algebra (Proc. Sympos. Pure Math., Vol. XVII, New York, 1968), pp. 124--128, Amer. Math. Soc., Providence, R.I., 1970; MR0255689 (41 #349)] is given a proof combining methods due to M. J. Dunwoody [J. Algebra 12 (1969), 339--344; MR0238931 (39 #291)] and P. C. Oxley [Math. Z. 127 (1972), 265--272; MR0332997 (48 #11322)]; (iii) some of Swan's homological arguments are replaced by more explicit discussions of the augmentation ideal of the group G. In an appendix proofs of the theorems of Kuroš and Gruško are included.

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Dunwoody's proof appears in his book with Dicks, Groups Acting on Graphs, and is explained in more detail there. –  Steve D May 4 '12 at 14:38
    
Thanks John (and Steve) for the references. –  Mark Grant May 8 '12 at 10:54

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