Groups acting on two-ended graphs

a dominated end which yields a contradiction.

Theorem 6.1.7. LetΓbe a two-ended quasi-transitive graph without dominated ends. Then each end ofΓ is thin.

Proof. By Lemma 6.1.2 we can find a type 1 separation (A, A^∗) of Γ. Sup-pose that the diameter of Γ[A∩A^∗] is equal to d. LetC be a big component of Γ\A∩A^∗. By Lemma6.1.4we can pick a vertexriof the rayRwith distance greater thandfromS. As Γ is quasi-transitive andA∩A^∗contains an element from of each orbit we can find an automorphismgsuch thatri∈g(A∩A^∗). By the choice ofri we now have that

(A∩A^∗)∩g(A∩A^∗) =∅.

Repeating this process yields a defining sequence of vertices for the end living in Ceach of the same finite size. This implies that the degree of the end living in Cis finite.

For a two-ended quasi-transitive graph Γ without dominated ends let s(Γ) be the maximal number of disjoint double rays in Γ. By Theorem6.1.7this is always defined. With a slight modification to the proof of Theorem 6.1.7 we obtain the following corollary:

Corollary 6.1.8. Let Γ be a two-ended quasi-transitive graphs without domi-nated ends. Then the degree of each end ofΓ is at mosts(Γ).

Proof. Instead of starting the proof of Theorem6.1.7with an arbitrary separa-tion of finite order we now start with a separasepara-tion (B, B^∗) of order s(Γ) sep-arating the ends of Γ which we then extend to a connected separation (A, A^∗) containing an element of each orbit. The proof then follows identically with only one additional argument. After finding the defining sequence as images of (A, A^∗), which is too large compared tos(Γ), we can reduce this back down to the separations given by the images of (B, B^∗) because (B∩B^∗)⊆(A∩A^∗) and because (B, B^∗) already separated the ends of Γ.

It is worth mentioning that Jung [53] proved that if a connected locally finite quasi-transitive graph has more than one end then it has a thin end.

Woess for vertex cuts [95, Proposition 4.2] with a proof which is quite closely related to their proof.

Lemma 6.1.9. LetΓbe a two-ended graph without dominated ends then for any vertexv∈V(Γ) there are only finitely manyk-tight separations containingv. Proof. We apply induction onk. The casek= 1 is trivial. So letk≥2 and letv be a vertex contained in the separator of a k-tight separation (A, A^∗). LetC1

and C2 be the two big components of Γ\(A∩A^∗). As (A, A^∗) is a k-tight separation we know thatvis adjacent to bothC1andC2. We now consider the graph Γ⁻:= Γ−v. Asvis not dominating any ends we can find a finite vertex set S1 ( C1 and S2 (C2 such that Si separates v from the end living in Ci

fori∈ {1,2}.² For each pairx, yof vertices withx∈S1andy∈S2we now pick ax−ypathPxyin Γ⁻. This is possible ask≥2 and because (A, A^∗) isk-tight.

LetP be the set of all those paths and letVP be the set of vertices contained in the path contained inP. Note thatVP is finite because each path Pxy is finite and bothS1 andS2are finite. By the hypothesis of the induction we know that for each vertex inVPthere are only finitely (k−1)-tight separations meeting that vertex. So we infer that there are only finitely many (k−1)-tight separations of Γ⁻ meeting VP. Suppose that there is a k-tight separation (B, B^∗) such that v∈B∩B^∗ andB∩B^∗ does not meetVP. As (B, B^∗) isk-tight we know thatvis adjacent to both big components of Γ\B∩B^∗. But this contradicts our choice ofSi. Hence there are only finitely manyk-tight separations containingv, as desired.

In the following we extend the notation of diameter from connected graphs to not necessarily connected graphs. Let Γ be a graph. We denote the set of all subgraphs of Γ by P(Γ). We define the function ρ:P(Γ) → Z∪ {∞} by settingρ(X) =sup{diam(C)|C is a component ofX}.³

Lemma 6.1.10. LetΓbe a quasi-transitive two-ended graph without dominated ends with |Γv| < ∞ for every vertex v of Γ and let (A, A^∗) be a tight separa-tion of Γ. Then for infinitely many g∈Aut(Γ) either the number ρ(A∆gA) orρ(A∆gA)^c is finite.

Proof. It follows from Lemma6.1.9that (A, A^∗) andg(A, A^∗) are nested for all but finitely many g∈Aut(Γ). Letg∈Aut(Γ) such that

(A∩A^∗)∩g(A∩A^∗) =∅.

2A finite vertex setSseparates a vertexv /∈Sfrom an endω1ifvis not contained in the componentG\Swhichω1lives.

3If the componentCdoes not have finite diameter, we say its diameter is infinite.

By definition we know that either A∆gAor (A∆gA)^c contains a ray. Without loss of generality we may assume the second case. The other case is analogous.

We now show that the number ρ(A∆gA) is finite. Suppose that C1 is the big component of Γ\(A∩A^∗) which does not meetg(A∩A^∗) andC2is the big compo-nent of Γ\g(A∩A^∗) which does not meet (A∩A^∗). By Lemma6.1.4we are able to find type 1 separations (B, B^∗) and (C, C^∗) in such a way thatB∩B^∗(C1

andC∩C^∗(C2and such that theB∩B^∗andC∩C^∗each have empty intersec-tion withA∩A^∗ andg(A∩A^∗). Now it is straightforward to verify thatA∆gA is contained in a rayless component X of Γ\((B∩B^∗)S

(C∩C^∗)). Using Lemma 6.1.5we can conclude thatX has finite diameter and hence ρ(A∆gA) is finite.

Assume that an infinite groupGacts on a two-ended graph Γ without dom-inated ends with finitely many orbits and let (A, A^∗) be a tight separation of Γ.

By Lemma6.1.10we may assumeρ(A∆gA) is finite. We set H :={g∈G|ρ(A∆gA)<∞}.

We callH theseparation subgroup induced by (A, A^∗).⁴ In the sequel we study separations subgroups. We note that we infer from Lemma 6.1.10 that H is infinite.

Lemma 6.1.11. Let G be an infinite group acting on a two-ended graph Γ without dominated ends with finitely many orbits such that with |Γv|<∞ for every vertex v of Γ. Let H be the separation subgroup induced by a tight sepa-ration(A, A^∗)of Γ. Then H is a subgroup ofGof index at most 2.

Proof. We first show that H is indeed a subgroup of G. As automorphisms preserve distances it is that forh∈H, g∈Gwe have

ρ(g(A∆hA)) =ρ(A∆hA)<∞.

As this is in particular true forg=h⁻¹ we only need to show thatH is closed under multiplication and this is straightforward to check as one may see that

A∆h1h2A= (A∆h1A)∆(h1A∆h1h2A)

= (A∆h1A)∆h1(A∆h2A).

Sinceρ(A∆hiA) is finite fori= 1,2, we conclude that h1h2belongs toH. Now we only need to establish thatH has index at most two inG. Assume that H is a proper subgroup of G and that the index of H is bigger than

4See the proof of Lemma6.1.11for a proof thatHis indeed a subgroup.

two. Let H and Hgi be three distinct cosets for i= 1,2. Furthermore by Lemma 6.1.10we may assumeρ((A∆giA)^c) is finite fori= 1,2 . Note that

A∆g1g⁻₂¹A= (A∆g1A)∆g1(A∆g⁻₂¹A).

On the other hand we already know that

A∆g1g₂⁻¹A= (A∆g1A)^c∆(g1(A∆g⁻₂¹A))^c.

We notice that the diameter of A∆giA is infinite for i = 1,2. Since g2 ∈/ H we know that g₂⁻¹ ∈/ H and so ρ(g1(A∆g⁻¹₂ A)) is infinite. By Lemma 6.1.10 we infer thatρ(g1(A∆g₂⁻¹A)^c) is finite. Now as the two numbersρ((A∆g1A)^c) and ρ(g1(A∆g₂⁻¹A)^c) are finite we conclude that ρA∆g1g⁻₂¹A <∞. Thus we conclude thatg1g₂⁻¹belongs toH. It follows thatH =Hg1g₂⁻¹and multiplying byg2 yieldsHg1=Hg2 which contradictsHg16=Hg2.

Theorem 6.1.12. Let Gbe a group acting with only finitely many orbits on a two-ended graphΓwithout dominated ends such that|Γv|<∞for every vertexv of Γ. Then Gcontains an infinite cyclic subgroup of finite index.

Proof. Let (A, A^∗) be a tight separation and let ( ¯A,A¯^∗) be the type 2 separation given by Corollary6.1.3. Additionally letH be the separation subgroup induced by (A, A^∗). We now use Lemma 6.1.6 on ( ¯A,A¯^∗) to find an elementh∈G of infinite order. It is straightforward to check that h∈H. Now it only remains to show thatL:=hhihas finite index inH.

Suppose for a contradiction thatLhas infinite index inH and for simplicity set Z :=A∩A^∗. This implies thatH =F

i∈NLhi. We have the two following cases:

Case I:There are infinitely many i∈Nandji∈Nsuch thathiZ=h^jiZ and so Z=h^−jihiZ. It follows from Lemma 6.1.9 that there are only finitely many f-tight separations meeting Z where |Z| = f. We infer that there are infinitely many k ∈ N such that h<sup>−j`</sup>h`Z = h^−jkhkZ for a specific ` ∈ N. Since the size of Z is finite, we deduce that there is v ∈ Z such that for a specificm∈Nwe haveh^−jmhmv=h^−jnhnv for infinitely manyn∈N. So we are able to conclude that the stabilizer ofv is infinite which is a contradiction.

Hence forni ∈Nwhere i= 1,2 we have to have

(h^−jmh⁻_m¹)h^−jn1hn₁= (h^−jmhm)⁻¹h^−jn2hn₂.

The above equality implies thatLhn1 =Lhn2 which yields a contradiction.

Case II: We suppose that are only finitely many i∈Nand ji∈N such that hiZ=h^jiZ. We define the graphX:= Γ[A∆hA]. We conclude that Γ =∪ⁱ∈ZhⁱX.

We can assume that hiZ ⊆h^jiX for infinitely many i∈N and ji ∈N and so we have h^−jihiZ ⊆ X. Let p be a shortest path between Z and hZ. For every vertex v of p, by Lemma 6.1.9 we know that there are finitely many tight separation gZ forg ∈G meeting v. So we infer that there are infinitely manyk∈Nsuch that h<sup>−j`</sup>h`Z =h^−jkhkZ for a specific`∈N. Then with an analogue method we used for the preceding case, we are able to show that the stabilizer of at least one vertex of Z is infinite and again we conclude that (h^−jmh⁻_m¹)h^−jn1hn1 = (h^−jmhm)⁻¹h^−jn2hn2 for n1, n2 ∈ N. Again it yields a contradiction. Hence each case gives us a contradiction and it proves our theorem as desired.