The adjunction problem over torsion-free groups

A long-standing problem in group theory is the adjunction problem of deciding when a given group injects into the group obtained by adjoining new generators and relators (1). This note solves the adjunction problem over torsion-free groups in the special case in which new generators and relators are added in pairs and such that the exponent matrix is invertible. We prove that, in this case, the group does inject. The case of one such pair was proved by Klyachko (2). The extension uses our previous article (3) and an additional result that follows from a theorem of Bogley and Pride (4), which in turn is based on an old theorem of Serre (5). We shall give a direct elementary proof of the result that we need.

Let (L, K) be a relative 2-complex (a CW-pair such that L – K is at most two-dimensional). We say that (L, K) is relatively aspherical if the map is surjective. As shown in results 3.1–3.3 of ref. 3, this occurs if and only if conclusions a and b of Theorem 1 hold. This is the natural topological notion of asphericity, but it should be noted that it differs from the combinatorial notion introduced in ref. 4. The difference concerns the definition of irreducibility of diagrams representing elements of π ₂(L, K) (see ref. 3).

The fundamental group of L is obtained from G = π ₁(K) by adding generators {t_i} and relators {r_j} corresponding to the 1-cells and 2-cells, respectively, of L – K. The relators r_j ∈ G * 〈t ₁,..., t_n〉 can then be viewed as a system of equations in the variables {t_i} with coefficients in G. It is well known [see Howie (6), for example] that the map π ₁(K) → π ₁(L) is injective if and only if the system has a solution in an overgroup of G.

The exponent matrix of the system [or of the pair (L, K)] has entries m_ij equal to the exponent sum of t_i in the relator r_j. In topological terms, it is the two-dimensional boundary map in the relative cellular chain complex of (L, K).

A long-standing conjecture (6) states that for any relative 2-complex (L, K), if the exponent matrix is nonsingular, then π ₁(K) → π ₁(L) is injective. If we assume further that π ₁(K) is torsion-free and the exponent matrix is invertible over , then we conjecture that (L, K) is also relatively aspherical. (As shown in ref. 3, this conclusion can fail if either of the additional hypotheses is omitted.) Our main result proves this in a special case:

Theorem 1. Let (L, K) be a layered relative 2-complex with π ₁(K) torsion-free. If the exponent matrix is invertible over , then (a) π ₁(K) → π ₁(L) is injective, and (b) the inclusion-induced map is an isomorphism.

Here, (L, K) is layered if L – K has equal numbers of 1- and 2-cells and L is formed from K by alternately adding 1- and 2-cells. In terms of the associated relative presentation, it means that the generators and relators can be added alternately.

A special case of Theorem 1 was proved in ref. 3: The theorem was proved when L – K consists of one 1-cell and one 2-cell. In this note, we observe that the special case can be applied inductively. It is worth stressing that part b of Theorem 1, for the case of one new generator and one new relator, is a nontrivial extension of Klyachko's theorem, and it is the key to allowing the inductive argument of this note to proceed.

Proof: The layered hypothesis implies that there is a nested sequence of subcomplexes K = K ₀ ⊂ K ₁ ⊂ ··· ⊂ K_n = L where K_{i +1} – K_i has one 1-cell and one 2-cell for each i. Note that the exponent matrix for (L, K) is triangular with diagonal entries equal to ±1, and that these diagonal entries represent the 1 × 1 exponent matrices for the pairs (K_{i +1}, K_i). In particular, each pair (K_{i +1}, K_i) is amenable in the sense of Fenn and Rourke (7). Then, by the main theorem of ref. 3, the pair (K_{i +1}, K_i) is relatively aspherical provided π ₁(K_i) is torsion-free.

We are given that π ₁(K ₀) is torsion-free, and so (K ₁, K ₀) is relatively aspherical. By Proposition 1 below, π ₁(K ₁) is then torsion-free. Proceeding inductively, using ref. 3 and Proposition 1, we find that every pair (K_{i +1}, K_i) is relatively aspherical.

It remains to be verified that relative asphericity is transitive. Given K ⊂ L ⊂ M with M – K at most two-dimensional, relative asphericity of (M, L) and (L, K) implies so condition b holds for (M, K). Condition a for (M, K) is clear.

Remark: The proof shows that the exponent matrix hypothesis can be relaxed to allow layered relative 2-complexes for which each pair (K_{i +1}, K_i) is amenable, i.e., that the relator given by the new 2-cell has an “amenable t-shape” in terms of the new generator (see ref. 3 or 7). The result also solves the adjunction problem for systems of generators and relators that can be transformed, by a change of variables, into a layered amenable system.

Proposition 1. If (L, K) is relatively aspherical and π ₁(K) is torsion-free, then π ₁(L) is also torsion-free.

Proof: By adding cells of dimension ≥3, we can arrange that all of the homotopy groups of K vanish in dimensions 2 and above. This does not change the fact that (L, K) is relatively aspherical. The easiest way to see this is to use the diagram interpretation used in ref. 3: Relative asphericity means that there are no irreducible diagrams over π ₁(K) using the cells of L – K. This only depends on π ₁(K) and the form of the added relators and, hence, is unchanged by a change in the higher homotopy groups of K. After adding the new cells, π ₂(L) is trivial.

Let L̃ be the universal cover of L and K̃ the preimage of K in L̃. Let L̂ be the 2-complex obtained from L̃ by collapsing each connected component of K̃ to a vertex. Since each of these components is contractible, the map L̃ → L̂ is a homotopy equivalence, and so π ₁(L̂) and π ₂(L̂) are trivial. Then, since L̂ is two-dimensional, it is contractible.

Note that the induced action of π ₁(L) on L̂ is free away from the 0-skeleton, and that the vertices have stabilizers equal to the conjugates of π ₁(K) in π ₁(L). We claim that any element g ∈ π ₁(L) of prime order fixes a vertex of L. It follows that it lies in a conjugate of π ₁(K), which is torsion-free and, hence, must be trivial. Thus, π ₁(L) has no elements of prime order and, hence, is torsion-free.

Suppose that g has prime order and fixes no vertex. Then, we have acting freely on Q = L̂, which is a contractible 2-complex. This is well known to be impossible, as would then be a finite-dimensional space.

Remark: Proposition 1 can be extended to show that if (L, K) is relatively aspherical, then every finite subgroup of π ₁(L) is contained in a unique conjugate of π ₁(K). This result (in a more general form, based on the weaker notion of relative asphericity) was first proved by Bogley and Pride (4), using a theorem of Serre (5), whose proof depends on Tate cohomology. In ref. 8, we give elementary geometric proofs (along the lines of the proof above) of the Bogley–Pride result and of Serre's theorem.