5.2 Applications
5.2.1 Two-ended groups
The above theorem with conjunction of Theorem 5.1.12 implies the fol-lowing corollary immediately:
Corollary 5.2.2. Let G be an infinite group acting with only finitely many orbits on a two-ended graph Γ without dominated ends. Then G is two-ended.
Before we can prove Theorem 5.2.1 we have to collect some tools and concepts used in the proof of Theorem 5.2.1. For the sake of simplicity, we introduce the following shorthand. We call
HomZ(ZG,Z2) and HomZ(ZG,Z2)/Z2G
byZ2G andZg2G, respectively. We notice that those groups can be regarded as Z2-vector spaces. We start with the following lemma which is known as Shapiro’s Lemma.
Lemma 5.2.3. [6, Proposition 6.2] Let H be a subgroup of a group G and let A be an RH-module. Then Hi(H, A) = Hi(G,HomRH(RG, A)).
Lemma 5.2.4. Let G be a finitely generated group. Then dimH0(G,Zg2G) = 1 +dimH1(G,Z2G).
Proof. First of all, we note that the short exact sequence 0→Z2G ,→Z2GZg2G→0 gives rise to the following long sequence:
0→H0(G,Z2G)→H0(G,Z2G)→H0(G,Zg2G)→H1(G,Z2G)→0 We notice that G acts on Z2G by g.f(x) := gf(g−1x) and it follows from Lemma 5.2.3 that Hi(G,Z2G) = 0 for every i ≥ 1. But H0(G, A) is an invariant subset of A under the group action ofG. Thus we deduce that
H0(G,Z2G) = 0 and H0(G,Z2G) =Z2.
Hence we have
dimH0(G,Zg2G) = 1 +dimH1(G,Z2G).
Lemma 5.2.5. Let G=hSi be a finitely generated group and Γ := Γ(G, S).
Then the spaces PΓ and FΓ can be identified by Z2G and Z2G, respectively.
Proof. Suppose that f ∈ Z2G. We define Af := {g ∈ G | f(g) = 1}. Now it is straightforward to check that there is a one to one correspondence between Z2G and PΓ. The second case is obvious.
Lemma 5.2.5 directly yields the following corollary.
Corollary 5.2.6. Let G=hSi be a finitely generated group and let Γ be the Cayley graph of G with respect to S. Then dimension of QΓ/FΓ is equal to dimH0(Γ,Zg2G).
Before we can start the proof of Theorem 5.2.1 we cite some well known facts we use proof of Theorem 5.2.1.
Lemma 5.2.7. [62, Theorem 15.1.13] Let G be a finitely generated group such that [G:Z(G)] is finite. Then G0 is finite.
Lemma 5.2.8. [39, Proposition 4.8] Let G be a finitely generated group and let H and K be subgroups of G such that HK is also a subgroup of G.
Then [HK :H] = [K :H∩K].
Lemma 5.2.9 (N/C Theorem). [62, Theorem 3.2.3] Let G be a group and let H ≤G then NG(H)/CG(H) is isomorphic to a subgroup of Aut(H).
Lemma 5.2.10. [70, Proposition 4.1] Let Γ be an infinite graph, let e be an edge of Γ, and let k be a natural number. Then G has only finitely many k-tight cuts containing e.
Lemma 5.2.11. [22, Theorem 1.1]LetΓbe a connected graph with more than one end. Then there exists a k-tight cut(A, A∗)such that for anyg ∈Aut(Γ) either (A, A∗)≤g(A, A∗) or vice versa.
Let us now have a precise look at an HNN-extension.
Remark 5.2.12. Let C =hS |Ri be a finite group. Every automorphism φ of C gives us an HNN-extension G=∗φC. We can build an HNN-extension from an automorphism φ: C →C. Therefore C is a normal subgroup of G with the quotient Z, as the presentation of HNN-extension G=∗φC is
hS, t|R, t−1ct=φ(c)∀c∈Ci.
Hence G can be expressed by a semidirect product Co Z which is induced by φ.
We now are in the position to prove the main theorem of this section.
Theorem 5.2.1. We illustrate the strategy to proof Theorem 5.2.1 in the following diagram, see Figure 5.1.
(i)
(ii) (iii) (iv) (v) (vi) (vii)
⇒ ⇒ ⇒
⇒
⇒
⇒
⇒
⇐=
Figure 5.1: Structure of the proof of Theorem 5.2.1
Proof of Theorem 5.2.1. (i) ⇒ (ii) Let Γ be a Cayley graph of G and thus G acts on Γ transitively. Now it follows from Theorem 5.1.12 that G has an infinite cyclic subgroup of finite index.
(ii) ⇒ (iii) Suppose that H = hgi and we may assume that H is normal, otherwise we replace H by Core(H). Let K = CG(H) and since [G : H] is finite, we deduce that [K :Z(K)] is finite , because H is contained in Z(K) and the index of H in G is finite. In addition, we can assume that K is a finitely generated group, as [G:K]<∞ we are able to apply Lemma 2.3.7.
We now invoke Lemma 5.2.7 and conclude that K0 is a finite subgroup. On the other hand K/K0 must be a finitely generated abelian group. Since K is infinite, one may see that K/K0 ∼= Zn⊕K0, where K0 is a finite abelian group and n ≥1.
We now claim that n = 1. Since [G : H] < ∞ and H ⊆ K, we infer that [K :H]<∞. But Lemma 2.3.7 implies thate(K) =e(H ∼=Z). ThusK is two-ended and if n≥2, thenZn⊕Ris one-ended which is a contradiction.
Hence the claim is proved. Next we define a homomorphism ψ: K →Zwith the finite kernel K0. Since K0 is finite subgroup of K such that K/K0 ∼=Z, we deduce that K0charK. It follows from Lemma 5.2.9, that G/CG(H) is isomorphic to a subgroup of Aut(Z)∼=Z2 and so we may assume that K is a normal subgroup ofG. IfK =G, then we are done. We suppose thatK < G.
We notice that K0charKCG and so K0 is a finite normal subgroup of G.
We claim that G/K0 is not an abelian group. Since K is a proper sub-group of G, we are able to find g ∈ G\K such that g does not commutate with h∈H ⊆K and we have h−1ghg−1 ∈H. So gK0 and hK0 do not com-mutate and the claim is proved. LetaK0 generateK/K0 ∼=Zand we pick up an element bK0 in (G/K0)\(K/K0). We can see that G/K0 = haK0, bK0i. We note thatK/K0EG/K0 and sobab−1K0 =aiK0for somei∈Z. SinceK0 is a finite group andG/K0is not abelian, we conclude thatbab−1K0 =a−1K0. We already know that [G : K] = 2 and so b2K0 ∈K/K0. We assume that b2K0 =ajK0 for some j ∈ Z. With bab−1K0 =a−1K0 and we deduce that j = 0. Thus b2K0 =K0 and we conclude that G/K0 =K/K0hbK0i. In other words one can see that G/K0 =ZZ2, whereZ is a normal subgroup.
(iii) ⇒ (iv) Let G=KN such that N is a finite normal subgroup of G and K ∼=Z or K ∼=D∞ and moreover K∩N = 1. If K ∼= Z, then by Re-mark 5.2.12 we get an HNN-extension of∗ψN whereψ ∈Aut(N). So we may assume that φ: G/N → hai ∗ hbi, where hai ∼=hbi ∼=Z2. Let Aand B be the pull-backs of hai and hbi by h, respectively. We note that the index ofK in both of A and B is two. Let us define a homomorphism Φ : A∗C B → G, by setting Φ(X) =X, where X ∈ {A, B}. It is not hard to see that Φ is an isomorphism.
(iv) ⇒ (v) Assume that G is isomorphic to either A∗CB where C is finite and [A:C] = [B :C] = 2 or ∗φC with C is finite and φ ∈ Aut(C). If we consider a canonical generating set S for G, then one may see that Γ(G, S) is a two-ended graph. So by Theorem 5.1.1 we are done.
(v) ⇒ (vi) Since the Cayley graph is quasi-isometric to the double ray, we conclude thatGis a two-ended group. We choose a generating setSforGand consider Γ := Γ(G, S). We now construct a “structure tree”6 R of Γ, which will be the double ray, in such a way that G acts onR and all stabilizers are finite with exactly one edge orbit. It follows from Lemma 5.2.11 that there is a finite cut C = (A, A∗) of Γ such that the set S :={g(A, A∗)|g ∈G} is a nested set. As S is nested, we can consider S as a totally ordered set. Let g ∈ G be such that g(A, A∗) is the predecessor of (A, A∗) in this order. We may assume that A(gA. This implies that Γ\(A∪gA∗) is finite.
Letg0 ∈Gsuch thatg0(A, A∗) is the predecessor ofg(A, A0). We can conclude thatg−1g0(A, A∗) is the predecessor of (A, A∗) and as predecessors are unique we can conclude that g0 =g2. Hence we can decompose Γ byg into infinitely many finite subgraphs such that between any two of these subgraphs there are finitely many edges. We now contract each finite subgraph to a vertex and for every finite cut between two consecutive subgraphs we consider an edge. Thus we obtain the double ray R in such way that G acts onR. It is straightforward to check that there is only one edge orbit. So we only need to establish that the stabilizers are finite. Let e be an edge of R. Then e corresponds to a k-tight cut C. It follows from Lemma 5.2.10 that there are finitely many k-tight cuts meeting C. So it means that the edge stabilizer of R is finite. With an analogous argument one can show that the vertex stabilizer of R is finite as well.
(vi) ⇒ (iv) Since G acts on the double ray, we are able to apply the Bass-Serre theory. So it follows from Lemma 2.3.3 that G is either a free product with amalgamation over a finite subgroup or an HNN-extension of finite subgroup. More precisely, the group G is isomorphic to G1∗G2 G3 or ∗φG1, where Gi is finite subgroup for i = 1,2,3 and φ ∈ Aut(G2). On the other hand, Theorem 5.1.12 implies that G must be two-ended. Now we show that [G1 : G2] = [G1 : G3] = 2. We assume to contrary [Gi : G2] ≥ 3 for some i ∈ {1,3}. Then G1 ∗G2 G3 has infinitely many ends which yields a contradiction. One may use a similar argument to show that G1 = G2 for
6For more details about the structure tree see [52].
the HNN-extension.
(vi) ⇒ (vii) Since Γ = Γ(G, S) ∼QI R, where R is the double ray, we conclude that G is a two-ended group. It follows from Lemma 5.2.4 that we only need to computedimH0(G,Zg2G) in order to calculatedimH1(G,Z2G).
By Corollary 5.2.6, it is enough to show that the dimension ofQΓ/FΓ is two.
Let {e1, . . . , en} be an independent vector of QΓ. Since the co-boundary of each ei is finite, we are able to find finitely many edges of G containing all co-boundaries, say K. We note that Γ is a locally finite two-ended graph and so we have only two components C1 and C2 of Γ\K. Every ei corresponds to a set of vertices of Γ. We notice that each ei takes the same value on each Ci. In other words, ei contains both ends of an edge e ∈ Ci or none of them. We first assume that 2 ≤ n. Then there are at least two vectors of {e1, . . . , en} which take the same value on a component C1 and it yields a contradiction with independence of these vectors. Hence we have shown that n ≥ 2. Let K be a finite set of vertices of Γ such that C1 and C2 are the infinite components of Γ\K. Since the co-boundary of eachCi is finite, each Ci can be regarded as an element of QΓ/FΓ and it is not hard to see that they are independent.
(vii) ⇒ (i) As we have seen in the last part the dimension of QΓ/FΓ is exactly the number of ends. Hence Lemma 5.2.4 and Corollary 5.2.6 complete the proof.
Remark 5.2.13. It is worth remarking that by Part (iii) of Theorem 5.2.1 every two-ended group can be expressed by a semi-direct product of a finite group with Z or D∞.