Grushko theorem


In the mathematical subject of group theory, the Grushko theorem or the Grushko–Neumann theorem is a theorem stating that the rank of a free product of two groups is equal to the sum of the ranks of the two free factors. The theorem was first obtained in a 1940 article of Grushko and then, independently, in a 1943 article of Neumann.

Statement of the theorem

Let A and B be finitely generated groups and let A∗B be the free product of A and B. Then
It is obvious that rank ≤ rank + rank since if X is a finite generating set of A and Y is a finite generating set of B then X∪Y is a generating set for A∗B and that |X∪Y|≤|X| + |Y|. The opposite inequality, rank ≥ rank + rank, requires proof.
Grushko, but not Neumann, proved a more precise version of Grushko's theorem in terms of Nielsen equivalence. It states that if M = is an n-tuple of elements of G = A∗B such that M generates G, <g1, g2,..., gn> = G, then M is Nielsen equivalent in G to an n-tuple of the form

History and generalizations

After the original proofs of Grushko and Neumann, there were many subsequent alternative proofs, simplifications and generalizations of Grushko's theorem. A close version of Grushko's original proof is given in the 1955 book of Kurosh.
Like the original proofs, Lyndon's proof relied on length-functions considerations but with substantial simplifications. A 1965 paper of Stallings
gave a greatly simplified topological proof of Grushko's theorem.
A 1970 paper of Zieschang gave a Nielsen equivalence version of Grushko's theorem and provided some generalizations of Grushko's theorem for amalgamated free products. Scott gave another topological proof of Grushko's theorem, inspired by the methods of 3-manifold topology Imrich gave a version of Grushko's theorem for free products with infinitely many factors.
A 1976 paper of Chiswell gave a relatively straightforward proof of Grushko's theorem, modelled on Stallings' 1965 proof, that used the techniques of Bass–Serre theory. The argument directly inspired the machinery of foldings for group actions on trees and for graphs of groups and Dicks' even more straightforward proof of Grushko's theorem.
Grushko's theorem is, in a sense, a starting point in Dunwoody's theory of accessibility for finitely generated and finitely presented groups. Since the ranks of the free factors are smaller than the rank of a free product, Grushko's theorem implies that the process of iterated splitting of a finitely generated group G as a free product must terminate in a finite number of steps. There is a natural similar question for iterating splittings of finitely generated groups over finite subgroups. Dunwoody proved that such a process must always terminate if a group G is finitely presented but may go on forever if G is finitely generated but not finitely presented.
An algebraic proof of a substantial generalization of Grushko's theorem using the machinery of groupoids was given by Higgins. Higgins' theorem starts with groups G and B with free decompositions G = ∗i Gi, B = ∗i Bi and f : G → B a morphism such that f = Bi for all i. Let H be a subgroup of G such that f = B. Then H has a decomposition H = ∗i Hi such that f = Bi for all i. Full details of the proof and applications may also be found in

Grushko decomposition theorem

A useful consequence of the original Grushko theorem is the so-called Grushko decomposition theorem. It asserts that any nontrivial finitely generated group G can be decomposed as a free product
where each of the groups Ai is nontrivial, freely indecomposable and not infinite cyclic, and where Fs is a free group of rank s;
moreover, for a given G, the groups A1,..., Ar are unique up to a permutation of their conjugacy classes in G and the numbers s and r are unique as well.
More precisely, if G = B1∗...∗Bk∗Ft is another such decomposition then k = r, s = t, and there exists a permutation σ∈Sr such that for each i=1,...,r the subgroups Ai and Bσ are conjugate in G.
The existence of the above decomposition, called the Grushko decomposition of G, is an immediate corollary of the original Grushko theorem, while the uniqueness statement requires additional arguments.
Algorithmically computing the Grushko decomposition for specific classes of groups is a difficult problem which primarily requires being able to determine if a given group is freely decomposable. Positive results are available for some classes of groups such as torsion-free word-hyperbolic groups, certain classes of relatively hyperbolic groups, fundamental groups of finite graphs of finitely generated free groups and others.
Grushko decomposition theorem is a group-theoretic analog of the Kneser prime decomposition theorem for 3-manifolds which says that a closed 3-manifold can be uniquely decomposed as a connected sum of irreducible 3-manifolds.

Sketch of the proof using Bass–Serre theory

The following is a sketch of the proof of Grushko's theorem based on the use of foldings techniques for groups acting on trees.
Let S= be a finite generating set for G=A∗B of size |S|=n=rank. Realize G as the fundamental group of a graph of groups Y which is a single non-loop edge with vertex groups A and B and with the trivial edge group. Let be the Bass–Serre covering tree for Y. Let F=F be the free group with free basis x1,....,xn and let φ0:F → G be the homomorphism such that φ0=gi for i=1,...,n. Realize F as the fundamental group of a graph Z0 which is the wedge of n circles that correspond to the elements x1,....,xn. We also think of Z0 as a graph of groups with the underlying graph Z0 and the trivial vertex and edge groups. Then the universal cover of Z0 and the Bass–Serre covering tree for Z0 coincide. Consider a φ0-equivariant map so that it sends vertices to vertices and edges to edge-paths. This map is non-injective and, since both the source and the target of the map are trees, this map "folds" some edge-pairs in the source. The graph of groups Z0 serves as an initial approximation for Y.
We now start performing a sequence of "folding moves" on Z0 to construct a sequence of graphs of groups Z0, Z1, Z2,...., that form better and better approximations for Y. Each of the graphs of groups Zj has trivial edge groups and comes with the following additional structure: for each nontrivial vertex group of it there assigned a finite generating set of that vertex group. The complexity c of Zj is the sum of the sizes of the generating sets of its vertex groups and the rank of the free group π1. For the initial approximation graph we have c=n.
The folding moves that take Zj to Zj+1 can be of one of two types:
One sees that the folding moves do not increase complexity but they do decrease the number of edges in Zj. Therefore, the folding process must terminate in a finite number of steps with a graph of groups Zk that cannot be folded any more. It follows from the basic Bass–Serre theory considerations that Zk must in fact be equal to the edge of groups Y and that Zk comes equipped with finite generating sets for the vertex groups A and B. The sum of the sizes of these generating sets is the complexity of Zk which is therefore less than or equal to c=n. This implies that the sum of the ranks of the vertex groups A and B is at most n, that is
rank+rank≤rank, as required.

Sketch of Stalling's proof

' proof of Grushko Theorem follows from the following lemma.

Lemma

Let F be finitely generated free group, with n generators. Let G1 and G2 be two finitely presented groups. Suppose there exists a surjective homomorphism, then there exists two subgroups F1 and F2 of F with and such that.
Proof:
We give the proof assuming that F has no generator which is mapped to the identity of, for if there are such generators, they may be added to any of or.
The following general results are used in the proof.
1. There is a one or two dimensional CW complex, Z with fundamental group F. By Van Kampen theorem, the wedge of n circles is one such space.
2. There exists a two complex where is a point on a one cell of X such that X1 and X2 are two complexes with fundamental groups G1 and G2 respectively. Note that by the Van Kampen theorem, this implies that the fundamental group of X is.
3. There exists a map such that the induced map on the fundamental groups is same as
For the sake of convenience, let us denote and .
Since no generator of F maps to identity, the set has no loops, for if it does, these will correspond to circles of Z which map to, which in turn correspond to generators of F which go to the identity. So, the components of are contractible.
In the case where has only one component, by Van Kampen's theorem, we are done, as in that case, :.
The general proof follows by reducing Z to a space homotopically equivalent to it, but with fewer components in, and thus by induction on the components of.
Such a reduction of Z is done by attaching discs along binding ties.
We call a map a binding tie if it satisfies the following properties
1. It is monochromatic i.e. or
2. It is a tie i.e. and lie in different components of.
3. It is null i.e. is null homotopic in X.
Let us assume that such a binding tie exists. Let be the binding tie.
Consider the map given by. This map is a homeomorphism onto its image. Define the space as
Note that the space Z' deformation retracts to Z
We first extend f to a function as
Since the is null homotopic, further extends to the interior of the disc, and therefore, to .
Let i = 1,2''.
As and lay in different components of, has one less component than.

Construction of binding tie

The binding tie is constructed in two steps.
Step 1: Constructing a null tie:
Consider a map with and in different components of. Since is surjective, there exits a loop based at γ' such that and are homotopically equivalent in X.
If we define a curve as for all, then is a null tie.
Step 2: Making the null tie monochromatic:
The tie may be written as where each is a curve in or such that if is in, then is in and vice versa. This also implies that is a loop based at p in X. So,
Hence, for some j.
If this is a tie, then we have a monochromatic, null tie.
If is not a tie, then the end points of are in the same component of. In this case, we replace by a path in, say. This path may be appended to and we get a new null tie
, where.
Thus, by induction on m, we prove the existence of a binding tie.

Proof of Grushko theorem

Suppose that is generated by. Let be the free group with -generators, viz.. Consider the homomorphism given by, where.
By the lemma, there exists free groups and with such that and. Therefore, and.
Therefore,