Vertex Separators, Chordality and Virtually Free Groups

VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS

SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´ (1)

Abstract. In this paper we consider some results obtained for graphs using minimal vertex separators and generalized chordality and translate them to the context of Geometric Group Theory. Using these new tools, we are able to give two new characterizations for a group to be virtually free. Furthermore, we prove that the Baumslag-Solitar group BS(1, n) is k-chordal for some k if and only if |n| < 3 and we give an application of generalized chordality to the study of the word problem.

Keywords: vertex separator, chordality, hyperbolicity, quasi-isometry, tree, virtually free. 2020 AMS Subject Classiﬁcation Numbers. Primary: 20F65. Secondary: 20F10, 05C25

1. Introduction In [24] the second author studied some relations between vertex separator sets, certain chordality properties that generalize being chordal and conditions for a graph to be quasi-isometric to a tree. Some of these ideas can be easily translated to the language of Geometric Group Theory. Moreover, some of the definitions can be re-written in new terms given a group presentation. Thus, we introduce herein new tools in the context of Geometric Group Theory which can be used from either a more geometric or more algebraic approach. Let Γ be a graph. A subset P ⊆ V (Γ) is an ab-separator in Γ if the vertices a and b are in different connected components of Γ r P . A graph Γ satisfies the bottleneck property (BP) if there exists some constant ∆ > 0 so that given any two 1 distinct points x, y ∈ V (Γ) and a midpoint c ∈ Γ such that d(x, c) = d(y, c) = 2 d(x, y), then every xy-path intersects N∆(c). This is an equivalent definition to the original which was defined in [22], see Proposition 2 in [24]. Then we have the following result for graphs. Theorem 1.1 (Theorem 7, [24]). Let Γ be a uniform graph. If every minimal vertex separator in Γ has diameter at most m, then Γ satisfies (BP) (i.e., Γ is quasi-isometric to a tree). This is translated in Section 3 to Geometric Group Theory. As a result, we obtain in Theorem 3.6 a new characterization for a group to be virtually free which is somehow a re-writting of (BP). By a cycle in a graph we mean a simple closed curve γ, i.e., a path with different vertices, except for the arXiv:2106.07331v1 [math.GR] 14 Jun 2021 last one, wich is equal to the first vertex. Without loss of generality we will suppose that every cycle has length at least 3. A shortcut in γ is a path σ joining two vertices p, q in γ such that L(σ) < dγ (p, q), where L(σ) denotes the length of the path σ and dγ denotes the length metric on γ. A shortcut σ in γ between p, q is strict if σ ∩ γ = {p, q}. In this case, we say that p, q are shortcut vertices in γ associated with σ. A graph Γ is (k, m)-chordal if for every cycle γ in X with length L(γ) > k, there exists a shortcut σ such that L(σ) 6 m. The graph Γ is k-chordal if we make m = ∞. A graph Γ is ε-densely (k, m)-chordal if for every cycle γ in Γ with length L(γ) > k, there exist strict shortcuts σ1, . . . , σr with L(σi) 6 m ∀ i, and such that their associated shortcut vertices define an ε-dense subset in (γ, dγ ). The graph Γ is ε-densely k-chordal if we make m = ∞.

Date: June 15, 2021. (1) Supported in part by a grant from Ministerio de Ciencia, Innovaci´ony Universidades (PGC2018-098321-B-I00), Spain. 1 2 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

Remark 1.2. If Γ is ε-densely (k, m)-chordal, then Γ is dεe-densely (dke , dme)-chordal. Therefore, we can suppose always that ε, k, m ∈ N. We proved that this property characterizes being quasi-isometric to a tree for graphs: Theorem 1.3 (Corollary 3, [24]). A graph Γ is quasi-isometric to a tree if and only if it is ε-densely (k, m)-chordal. A group G is virtually free if it has a free subgroup of finite index. There are many characterizations of being virtually free in the literature. Let us recall here some of them. Theorem 1.4. Let Γ be the Cayley graph of a finitely generated group G, then the following are equivalent: (1) G is virtualle free. (2) The language of all words on S±1 representing the identity is context-free. (Muller-Schupp’s Theorem [10, 25]) (3) G is context-free. [10, 25] (4) There exist k > 0 such that every closed path in Γ is k-triangulable. [25] (5) The ends of Γ have finite diameter. [31] (6) Γ admits a uniform spanning tree. [31] (7) Γ is quasi-isometric to a Cayley graph of a free group. [10, 14] (8) Γ has finite strong tree-width. [20] (9) G is finitely presentable and adimΓ = 1. [12, 13, 17] (10) There exists a generating set of G and k > 0 such that every k-locally geodesic in G is a geodesic. [15] (11) There exists a generating set of G such that G admits a (regular) broomlike combing with respect to the generating set. [4] (12) G acts on a tree with finite stabilizers.[1, 7, 18] (13) G is the fundamental group of a finite graph of finite groups. [1, 7, 18] (14) Γ is quasi-isometric to a tree. [10, 29] (15) Γ satisfies (BP). [22] (16) G admits a finite presentation by some geodesic rewriting system. [15] (17) G is the universal group of a finite pregroup. [30] (18) The monadic second-order theory of Γ is decidable. [20, 26] (19) G admits a Stallings section. [28] (20) Γ is polygon hyperbolic. [2] (21) For any finite generating set, the corresponding Cayley graph is minor excluded. [19] For some alternative proofs of some of these see also [1]. In Section 4 we translate the definition of being ε-densely (k, m)-chordal to the language of Group Theory, see Theorem 4.8. In those terms, we give a new characterization of being virtually free in Corollary 4.12 which states the following:

Theorem 1.5. A ﬁnitely generated group G is virtually free if and only if G is (i0, k, m)-chordal.

In Section 5 we analyze the example of Baumslag-Solitar groups and prove that BS(1, n) is k-chordal for some k if and only if |n| < 3. The word problem in the group presentation G = hS|Ri consists in ﬁnding an algorithm that takes as its input a word w over S and answers if w represents the trivial element in G or not. For many groups this problem is undecidable. A group has solvable word problem if such an algorithm exists. For details about the word problem, see [9, 21]. In Section 6 we give an application of generalized chordality to the study of the word problem for groups. In particular we obtain Theorem 6.2 which states the following: VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 3

Theorem 1.6. Let G = hS|Ri be a finite group presentation. If G is k-chordal, then hS|Ri admits a recursive isoperimetric function, i.e., hS|Ri has a solvable word problem. Moreover, we obtain that if G is k-chordal, then there is an isoperimetric function f such that f(n) = c2n−k for some constant c. Some groups are known to have solvable word problem: free groups, see [9]; finitely presented residually finite groups, see [9]; hyperbolic groups, [16], among others. In particular, it is well known that the free abelian group ha, b | [a, b]i has solvable word problem, see [21]. We prove again this fact to illustrate an application of Theorem 6.2 in search of groups with solvable word problem.

2. Background

Let (X, d1), (Y, d2) be a metric spaces, f : X −→ Y and a, b > 0. The function f is an (a, b)-quasi- isometric embedding if 1 d (x, y) − b d (f(x), f(y)) ad (x, y) + b, ∀ x, y ∈ X. a 1 6 2 6 1

Two functions f, g : X −→ Y are said to be at finite distance if there exists ε > 0 such that d2(f(x), g(x)) < ε for all x ∈ X. The function f is a quasi-isometry if f is a quasi-isometric embedding (there exist a, b > 0 such that f is an (a, b)-quasi-isometric embedding) and there exists a quasi-isometric embedding g : Y −→ X such that f ◦ g and IdY are at finite distance and g ◦ f and IdX are also at finite distance. Let ε > 0 be a constant. A subset D ⊆ X is said to be ε-dense if given a point x ∈ X there exists p ∈ P such that d1(x, p) < ε. It is well known that f is a quasi-isometry if and only if f is a quasi-isometric embedding and f(X) is ε-dense for some ε > 0, see [21] for details. It is said that two metric spaces are quasi-isometric if there is a quasi-isometry between both of them. Given a metric space (X, d), a geodesic from x to y is an isometry γ : [0,L] −→ X such that γ(0) = x and γ(L) = y. Abusing the notation, we will identify the geodesic γ with its image and refer to this image as a geodesic. A metric space (X, d) is geodesic when there is a geodesic from x to y for every pair of points x, y ∈ X. It is important to note that geodesics do not have to be unique, for convenience, [xy] will denote any such geodesic. There are several definitions of Gromov δ-hyperbolic space which are equivalent although the constant δ may appear multiplied by some constant, see [6] for details. Since we will use geodesic spaces, it is natural to use the Rips condition on the triangles. Given three points x, y, z in a geodesic space (X, d), a geodesic γ1 from x to y, a geodesic γ2 from y to z and a geodesic γ3 from z to x, the union of these geodesics is a geodesic triangle and will be denoted by T = {x, y, z} or T = {γ1, γ2, γ3} when we want to emphasize geodesics. T is δ-thin if any side (any geodesic γi) of T is containded in the δ-neighborhood of the union of the two other sides. A geodesic metric space (X, d) is δ-hyperbolic if every geodesic triangle in (X, d) is δ-thin. A geodesic metric space (X, d) is hyperbolic if (X, d) is δ-hyperbolic for some δ > 0. A graph Γ can be interpreted as a metric space with a length metric dΓ by considering every edge e ∈ E(Γ) as isometric to the interval [0, 1]. Therefore, the distance between any pair of points x, y ∈ Γ will be the length of the shortest path in Γ joining x and y. In this paper we will assume that every graph is locally finite and connected. Thus, every graph is a geodesic metric space. For any vertex v ∈ V (Γ) and any constant ε > 0, let us denote:

Sε(v) := {w ∈ V (Γ) | d(v, w) = ε},

Bε(v) := {w ∈ V (Γ) | d(v, w) < ε},

Nε(v) := {w ∈ V (Γ) | d(v, w) 6 ε}. 4 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

Let G be a group generated by a subset S ⊆ G and let S±1 = S ∪ S−1 r {e}. S is a generating set if every element of G can be written as a finite product of elements in S±1. G is finitely generated if there is a finite genarator set. The Cayley graph of G with respect to S is a graph Γ = Cay(G, S) whose set of vertices is G and whose set of edges is {{g, gs} | g ∈ G, s ∈ S±1}. Given two finite generating sets S and S0 of G, then 0 Cay(G, S) and Cay(G, S ) are quasi-isometric (via idG), see [21] for details. A finitely generated group G is quasi-isometric to a metric space (X, d) if there is a quasi-isometry between Γ and (X, d). This is well defined because it does not depend on the chosen generating sets. Hyperbolicity is a well-known invariant under quasi-isometries. Therefore, it can be said that a finitely generated group G is δ-hyperbolic if some Cayley graph Γ of G is δ-hyperbolic.

3. Vertex separators and virtually free groups Deﬁnition 3.1. Let G be a group generated by S. We say P ⊆ G is an ab-separator with respect to S if for ±1 all s1, . . . , sn ∈ S such that b = as1 ··· sn, then there exists 0 < i < n such that as1 ··· si ∈ P . A vertex separator with respect to S is an ab-separator with respect to S for some a, b ∈ G. A vertex separator with respect to S is a minimal vertex separator with respect to S if no proper subset is a vertex separator with respect to S. If Γ is the Cayley graph of G with respect to S and P ⊂ G = V (Γ), it is immediate that P is an ab-separator in Γ if and only if P is an ab-separator with respect to S. −1 Remark 3.2. Let G be a group generated by S. The equation b = as1 ··· sn can be read as a b = s1 ··· sn. Therefore, P is an ab-separator with respect to S if and only if a−1P is an e(a−1b)-separator with respect to S. Let G be a group generated by S and A ⊆ G. We denote the diameter of A (with respect to S) as diam(A) = sup{dΓ(x, y) | x, y ∈ A}. The set of vertex separators between e and the rest of elements of G will be denoted as follows. Sep(G) = {P | P is a minimal ex-separator with respect to S for some x ∈ G \{e}}. We denote D(G) = sup{diam(P ) | P ∈ Sep(G)}. Theorem 3.3. Let G be a group generated by S. If D(G) < ∞, then G is virtually free. Proof. Suppose D(G) < ∞. Let P be a minimal vertex separator between a and b. By Remark 3.2, a−1P ∈ Sep(G) and therefore diam(P ) = diam(a−1P ) < D(G) < ∞. By Theorem 1.1, Cay(G, S) satisﬁes (BP), i.e., G is virtually free. The following example shows that the converse is not true.

Example 3.4. Let G = Z × Z4 be a group generated by S = {(1, 0), (0, 1)}. Obviously, G is virtually free. However, P = Z × {1} ∪ Z × {3} is a minimal separator with inﬁnite diameter, see Figure 1. However, Theorem 3.3 can be improved to obtain a characterization for being virtually free in terms of minimal vertex separators. The result, Theorem 3.6 below, is quite a straightforward re-writing of Manning’s Bottleneck Property.

Deﬁnition 3.5. Let G be a group generated by S. F = {Px : x ∈ G r N1(e)} is a k-separating family with respect to S if:

(1) Px is a minimal vertex ex-separator with respect to S. (2) There is v ∈ Px such that d(v, c) 6 k, where c is a midpoint of a geodesic [ex]. 1 From Lemma 1 in [24] it follows that for every ﬁnitely generated group there exists a 2 -separating family 1 and, therefore, a k-separating family for every k ≥ 2 . Given a k-separating family F, we denote D(F) = sup{diam(P ) | P ∈ F} and Dˆ(G) = inf{D(F) | F is a k-separating family}. VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 5

(−1, 3) (0, 3) (1, 3)

(−1, 2) (0, 2) (1, 2)

(−1, 1) (0, 1) (1, 1)

(−1, 0) (0, 0) (0, 1)

Figure 1. The Cayley graph of G with respect to S (the cross in vertical lines are glued). In bold the minimal vertex separator P .

Theorem 3.6. Let G be a group generated by S. Then G is virtually free if and only if Dˆ(G) < ∞.

Proof. Suppose that Dˆ(G) < ∞. Then there is a k-separating family, F, such that D = sup{diam(Px) | Px ∈ F} < ∞. We claim that Γ = Cay(G, S) satisfies (BP ). Consider an element x ∈ G r N1(e) and a midpoint m ∈ Γ = Cay(G, S) such that dΓ(x, m) = dΓ(e, m) = 1 2 dΓ(e, x). If dΓ(e, x) = 2, then (BP) is trivial. Suppose dΓ(e, x) > 2. Then, there is some vertex v ∈ Px such that d(v, m) 6 k. Thus, since every ex-path contains a vertex in Px and diam(Px) 6 D, then every ex-path intersects ND(v) ⊂ ND+k(m). Hence, (BP) is satisfied and therefore G is virtually free. Suppose that G is virtually free. Then, (BP) is satisfied with some constant ∆ > 0. Thus, for every 1 x ∈ G\N1(e) and every midpoint m ∈ Γ such that dΓ(x, m) = dΓ(e, m) = 2 dΓ(e, x), every ex-path intersects N∆(m). Pick a vertex vx ∈ N 1 (m) and let Px = V (Γ) ∩ N 1 (vx). The family F = {Px : x ∈ G \ N1(e)} 2 ∆+ 2 1 ˆ is a 2 -separating family with respect to S such that D(F) 6 2∆ + 1. Thus, D(G) ≤ D(F) < ∞. 4. Chordality on groups ±1 Definition 4.1. Let G be a group generated by S, n > 2 and s1, . . . , sn ∈ S . We say that s1 ··· sn = e is a simple relation in S±1 when

spsp+1 ··· sq−1sq = e if and only if (p, q) = (1, n)

Remark 4.2. Let γ be a cycle in Γ defined by g0, . . . , gn = g0, then u1 ··· un = e is a simple relation in ±1 −1 ±1 S where uk = gk−1gk, k ∈ {1, . . . , n}. On the other hand, if s1 ··· sn = e is a simple relation in S , then g0 = e, gk = s1 ··· sk, k ∈ {1, . . . , n}, defines a cycle in Γ. Definition 4.3. Let G be a group generated by S and k, m ∈ N. We say that G is (k, m)-chordal with ±1 0 0 ±1 respect to S if for every simple relation s1 ··· sn = e in S with n > k, then there exist s1, . . . , sr ∈ S 0 0 such that si ··· sj = s1 ··· sr with r 6 min{m, j − i, n − j + i − 2} for some 1 6 i < j 6 n. We say that G is k-chordal if we make m = ∞. By Remark 4.2, it is easy to check that a group is (k, m)-chordal with respect to S if and only if the Cayley graph Γ is (k, m)-chordal. ±1 Remark 4.4. A group G is (k, 1)-chordal if for every simple relation s1 ··· sn = e in S with n > k, then ±1 there exists s ∈ S such that si ··· sj = s for some 1 6 i < j 6 n with j − i < n − 2. In [24] we proved the following: 6 SAMUEL G. CORREGIDOR AND ALVARO´ MARTÍNEZ-PEREZ´

Theorem 4.5. If Γ is a (k, 1)-chordal graph, then Γ is quasi-isometric to a tree. By Theorem 4.5 and 1.4 it is immediate the following: Corollary 4.6. Let G be a ﬁnitely generated group. If G is (k, 1)-chordal, then G is virtually free.

Definition 4.7. Let G be a group generated by S and i0, k, m ∈ N. We say that G is (i0, k, m)-chordal with ±1 0 0 ±1 respect to S if for every simple relation s1 ··· sn = e in S with n > k, then there exist s1, . . . , sr ∈ S 0 0 such that si ··· sj = s1 ··· sr with r 6 min{m, j − i, n − j + i − 2} for some 1 6 i 6 i0 and i < j 6 n. We say that G is [i0, k]-chordal if we make m = ∞. Theorem 4.8. Let G be a group generated by S, Γ = Cay(G, S) the Cayley graph and k, m > 0. Then, Γ is ε-densely (k, m)-chordal for some ε > 0 if and only if G is (i0, k, m)-chordal with respect to S for some i0 > 0. Proof. Let G be a group generated by S and suppose that the Cayley graph Γ is ε-densely (k, m)-chordal ±1 for some ε, k, m ∈ N. Let s1 ··· sn = e be a simple relation in S with n > k and consider i0 = 2ε. Then, define the cycle γ by g0 = e, gl = s1 ··· sl ∀ l ∈ {1, . . . , n}. Since L(γ) = n > k, then there exist strict shortcuts σ1, . . . , σt with L(σi) 6 m ∀ i, and such that their associated shortcut vertices define an ε-dense subset in (γ, dγ ). Therefore, there exists a shortcut σλ with r = L(σλ) 6 m and associated shortcut vertices gi−1, gj, i < j, such that 1 6 i 6 2ε = i0. Since σλ is a path from gi−1 to gj, then there exist 0 0 ±1 0 0 s1, . . . , sr ∈ S such that σλ is the path defined by h0 = gi−1, hl = gi−1s1 ··· sl, ∀ l ∈ {1, . . . , r} where r 6 min{j − i, n − j + i − 2}. Note that 0 0 gi−1si ··· sj = gj = hr = gi−1s1 ··· sr 0 0 Thus, si ··· sj = s1 ··· sr with r 6 min{m, j − i, n − j + i − 2} with i 6 i0 and G is (i0, k, m)-chordal.

Reciprocally, suppose that G is (i0, k, m)-chordal with respect to S for some i0, k, m ∈ N. Let γ be a cycle in Γ with length L(γ) = n > k and let ε = i0. Fix a point q ∈ γ and suppose p, p0 two adjacent vertices such that q ∈ [pp0] and q 6= p0. Let us label the 0 −1 ordered vertices in γ as g0, . . . , gn = g0 so that p = g0 and p = g1. If we deﬁne sl = gl−1gl, l ∈ {1, . . . , n}, 0 0 ±1 then s1 ··· sn = e is a simple relation. By hypothesis, there exist i, j, r ∈ N, s1, . . . , sr ∈ S such that 0 0 si ··· sj = s1 ··· sr with r 6 min{m, j − i, n − j + i − 2} for some 1 6 i 6 i0 and i < j 6 n. Let σ be the path 0 0 0 0 deﬁned by the vertices h0 = gi−1, hl = gi−1s1 ··· sl, l ∈ {1, . . . , r}. Since hr = gi−1s1 ··· sr = gi−1si ··· sj = gj and L(σ) = r 6 min{j − i, n − j + i − 2} = dγ (gi−1, gj), then σ contains an strict shortcut σ0 with an associated shortcut vertex gi−1. Since q ∈ [g0g1], dγ (q, gi−1) < i0 = ε and L(σ0) 6 L(σ) = r 6 m. Remark 4.9. In fact, from the proof we obtain that if Γ is ε-densely (k, m)-chordal, then G is (2ε, k, m)- chordal. On the other hand, if G is (i0, k, m)-chordal, then Γ is i0-densely (k, m)-chordal. Remark 4.10. Theorem 4.8 it is true for m0 = m = ∞. Hence, by Theorem 1.3, we get the following corollary.

Corollary 4.11. A ﬁnitely generated group G is quasi-isometric to a tree if and only if G is (i0, k, m)- chordal. Thus, by Theorem 1.4, we obtain the following result.

Corollary 4.12. A ﬁnitely generated group G is virtually free if and only if G is (i0, k, m)-chordal. Let us recall also the following results. Theorem 4.13 (Theorem 4, [23]). If a graph Γ is ε-densely (k, m)-chordal, then Γ is δ-hyperbolic with k δ 6 max{ 4 , ε + m}. Theorem 4.14 (Theorem 8, [23]). If a graph Γ is δ-hyperbolic, then Γ is ε-densely k-chordal, with ε = 2δ+1 and k = 4δ + 3. VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 7

By Theorem 4.8, Theorem 4.13 and Theorem 4.14 we obtain also the following results.

Corollary 4.15. If a ﬁnitely generated group G is (i0, k, m)-chordal, then G is δ-hyperbolic, with δ 6 k max{ 4 , m + i0}. Corollary 4.16. If a ﬁnitely generated group G is δ-hyperbolic, then G is [4δ + 2, 4δ + 3]-chordal. Many groups may satisfy being k-chordal. Let us see some examples.

Proposition 4.17. Z2 is 5-chordal.

Proof. Let γ be a cycle with length L(γ) > 5 and consider a simple relation s1 ··· sn = e associated with γ in the sense of Remark 4.2. Suppose without loss of generality that s1 = a, then there is 1 < i < n such that ε ε s1 = ... = si−1 = a and si = b , where ε ∈ {−1, 1}. Now, consider i < j < n such that si = ... = sj−1 = b −1 −1 and sj = a or sj = a . Since γ is a cycle, we can assume relabeling if necessary that sj = a . If we denote 0 0 ε q = ε(j − i), r = |q| and s1 = ... = sr = b , then we get the following equality, see Figure 2. q −1 q 0 0 si−1si ··· sj−1sj = ab a = b = s1 ··· sr

bq ap a−p

Figure 2. Shortcut abqa−1 = bq.

0 0 Obviously r < j − (i − 1). If s1 ··· sr deﬁnes a shortcut, we are done. Otherwise, if r > n − j + (i − 1) − 2, ε −1 −ε then γ is deﬁned by s1 = a, s2 = ... = sr+1 = b , sr+2 = a , sr+3 = ... = sn = b , therefore s1 ··· sn = q −1 −q 0 −1 ab a b = e. If we denote s1 = a , then we get the following equality, see Figure 3. ε −1 −ε −1 0 sr+1sr+2sr+3 = b a b = a = s1

0 Since L(γ) > 5, then |q| > 1 and therefore s1 is a shortcut. In the general case, since n-hypercubes are Hamiltonian graphs, we conjecture that Zn is (2n +1)-chordal, where 2n is the number of vertices of an n-hypercube.

5. Baumslag-Solitar groups BS(1, n) Deﬁnition 5.1. For integers m, n, let the Baumslag-Solitar group deﬁned by following presentation: BS(m, n) = a, b | bamb−1 = an

We will focus on Gn := BS(1, n). Let Γn denote the corresponding Cayley 2-complex built from rectangles homeomorphic to [0, 1] × [0, 1], with both vertical sides labeled by b upward, the top horizontal side labeled by an a to the right, and the bottom horizontal side split into n edges each labeled by a to the right. These rectangles can be stacked horizontally into leafs. For each leaf, n other leafs can be attached at the top, and 8 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

bq a a−1

Figure 3. Shortcut bεa−1b−ε = a−1. a

b b n a a

b b b b a n a a n a

Figure 4. One sheet of Gn one on the bottom. For any set of successive choices upward, then, these leafs can be stacked vertically to ﬁll the plane. This is called a sheet. See Figure 4. The Cayley complex is then homeomorphic to the Cartesian product R × Tn where Tn is a Cantor tree where every vertex has n descendants on the next level, this is, the regular tree with degree n + 1 at each vertex. Let us denote the horizontal projection as πn :Γn −→ Tn, see Figure 5. More details can be found in [11].

Proposition 5.2. Given s, t ∈ N and ε, k, m > 0 the following properties are equivalent. (1) BS(s, t) is ε-densely (k, m)-chordal. (2) BS(−s, t) is ε-densely (k, m)-chordal. (3) BS(s, −t) is ε-densely (k, m)-chordal. (4) BS(−s, −t) is ε-densely (k, m)-chordal. (5) BS(t, s) is ε-densely (k, m)-chordal. (6) BS(−t, s) is ε-densely (k, m)-chordal. (7) BS(t, −s) is ε-densely (k, m)-chordal. (8) BS(−t, −s) is ε-densely (k, m)-chordal. Proof. The corresponding Cayley 2-complex of the groups BS(s, t), BS(−s, t), BS(s, −t) and BS(−s, −t) are the same except for the orientation of some labels on the edges. Therefore, these groups have Cayley graphs quasi-isometric and the proof of (1) ⇔ (2) ⇔ (3) ⇔ (4) is ﬁnished. VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 9

πn

Figure 5. The Cayley graph of Gn projected onto the Cantor tree Tn

Since the corresponding Cayley 2-complex of BS(t, s) is a 180◦ rotation of the corresponding Cayley 2-complex of BS(s, t), these groups have Cayley graphs quasi-isometric and the proof of (1) ⇔ (5). The rest of equivalences are deduced by exchanging the role of the variables s, t.

Let us denote in Gn as a-edges the edges labelled by a and b-edges labelled by b.

Deﬁnition 5.3. Given ω ∈ Gn we say that hn(ω) ∈ Z is the height of ω if there exists k ∈ Z such that

k hn(ω) ω =Ab(Gn) a b , where Ab(Gn) denotes the abelianization of Gn. Given p any vertex in (Tn) we denote ehn(p) = hn(ω) for some ω ∈ Gn such that πn(ω) = p. Also, given a b-edge e = {v, w} ∈ Gn, let hn(e) := max{hn(v), hn(w)}.

Remark 5.4. If ω1, ω2 ∈ Gn are two points such that πn(ω1) = πn(ω2), then hn(ω1) = hn(ω2), i.e., ehn(p) is well deﬁned. Let us deﬁne

Lq := {x ∈ Γn | dTn (πn(x), q) = hn(x) − ehn(q)}. −1 −1 In addition, we will consider pq : Lq −→ πn (q) the vertical proyection onto πn (q) and we will consider an −1 λ order in πn (q) where x < y if y = xa for some λ > 0. See Figure 6.

Deﬁnition 5.5. Given x, y ∈ Γn we denote the geodesic path in Tn from πn(x) to πn(y) as σen(x, y). We say that the point q ∈ σen(x, y) such that ehn(q) = min ehn(σen(x, y)) is the branching point between x and y.

Remark 5.6. Let x, y ∈ Γn and γ be a path between x and y. If q ∈ Tn is the branching point between x −1 and y, then γ ∩ πn (q) 6= ∅. 10 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

−1 π2 pq

a q

−1 Figure 6. A sheet of G2 projected onto π2 (q)

Definition 5.7. Let γ be a path in Γn between two vertices x, y ∈ V (Γn). Suppose that γ is defined by the vertices x0 = x, xi = xs1 ··· si for i = 1, . . . , m. We define the horizontal length of γ as follows: ε H(γ) = |{i ∈ {1, . . . , m} : ∃ε ∈ {−1, 1}, si = a }| .

Notice that given two vertices in Gn, the horizontal length is not an invariant on the geodesics joining them as the following example shows. 3 −1 Example 5.8. In G3, a geodesic deﬁned by a and a geodesic deﬁned by bab conect the same pair of vertices but they do not have the same horizontal length. See Figure 7.

b b−1

a a a

Figure 7. The geodesic given by a3 has horizontal length 3 and the geodesic given by bab−1 has horizontal length 1.

Lemma 5.9. If σ is a geodesic in G2, then σ does not contain three b-edges e1, e2, e3 with hn(ei) = hn(ej) for every 1 ≤ i, j ≤ 3. −1 Proof. Consider that the geodesic σ is deﬁned by the vertices v0, v1, . . . , vn and let si = vi−1vi for every 1 ≤ i ≤ n. Suppose that there exist three b-edges e1 = {vi, vi+1}, e2 = {vj, vj+1}, e3 = {vk, vk+1} with i < j < k and hn(ei) = hn(ej) for every 1 ≤ i, j ≤ 3. Let us assume that these edges are consecutive −1 −1 with this property. Therefore, if vi+1 = vib, then vj+1 = vjb and vk+1 = vkb, and if vi+1 = vib , then −1 vj+1 = vjb and vk+1 = vkb . Case 1. Suppose vi+1 = vib. Then, it is immediate to check that

visj+2 ··· skbsi+2 ··· sj = vk+1 VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 11

(notice that it has the same b-edges and a-edges with the same orientations at the same heigths) and this deﬁnes a shortcut (two edges shorter) in σ between vi and vk+1 leading to contradiction since σ is geodesic. See Figure 8.

vi+1 vj vk+1

vi vj+1 vk

Figure 8. A shortcut in dashed line for Case 1.

−1 Case 2. Suppose vi+1 = vib . Then, it is immediate to check that −1 visj+2 ··· skb si+2 ··· sj = vk+1 and this deﬁnes a shortcut (two edges shorter) in σ between vi and vk+1 leading to contradiction. See Figure 9.

vi vj+1 vk

vi+1 vj vk+1

Figure 9. A shortcut in dashed line for Case 2.

Theorem 5.10. The group G2 is 6-chordal.

Proof. Let γ be a cycle in Γ2 with length L(γ) > 6. Case 1. Suppose that there are four consecutive vertices xi with 1 ≤ i ≤ 4 in γ such that x2 = x1b, −1 2 x3 = x2a and x4 = x3b . Then, v1 = x1a, v2 = x1a = x4 deﬁnes a shortcut between x1 and x4 and we are done. Case 2. Suppose, otherwise, that there are not such four consecutive vertices. Consider M = maxv∈V (γ){h2(v)}, m = minv∈V (γ){h2(v)}, qm ∈ π2(γ) such that eh2(qm) = m and the vertical proyection p := pqm : Lqm −→ −1 −1 π (qm). Let us denote Vq = V (γ) ∩ π2 (q) for some q ∈ T2. Then, since there exists at least one a-edge e = {v, w} with h2(v) = h2(w) = M and we are not in case 1, there exist three consecutive vertices in the −1 −1 cycle x0, x, x1 ∈ h2 (M) with p(x0) < p(x) < p(x1), i.e., x0 = xa , x1 = xa. Since γ is a cycle, then

P = {z ∈ γ | p(x) = p(z) and π2(x) 6= π2(z)}= 6 ∅, where z need not be a vertex. Let y ∈ P be such that dγ (x, y) = minv∈P dγ (x, v) and we will consider that γy is the shortest path in the cycle γ that joins x and y. Fix q0 ∈ T2 the branching point between x and y and m0 = eh2(q0). Case 2a. Suppose that y ∈ V (γ). Then x, y are in the same sheet and h2(y) = m0. Thus, the vertical path from x to xb−(M−m0) = y deﬁnes a shortcut between x and y. Case 2b. Suppose that y∈ / V (γ). We can suppose that y is in an horizontal edge with vertices y0, y1, see Figure 10. Suppose without loss of generality that x1, y1 ∈ γy. 12 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

x0 x x1

y0 y y1

−1 πn (q0)

Figure 10. Case 2b.

If γy is not a geodesic, since it is the shortest path in the cycle, then there is a shortcut between x and y and we are done. Suppose, otherwise that γy is a geodesic. Then, by Lemma 5.9, there are not three b-edges in γy with the same height. −1 Suppose γy is defined by the vertices x1 = v0, v1, . . . , vn = y1 and let si = vi−1vi for every 1 ≤ i ≤ n. Case 2b(i). Suppose h2(y) = h2(x). Then, since x1s2 ··· sn = y1 it is immediate to check that xs2 ··· sn = y0 (it has the same a-edges and b-edges with the same orientations at the same heigths) and dΓ(x, y) ≤ n − 1 + dΓ(y0, y) < n < L(γy) leading to contradiction since γy is a geodesic. Case 2b(ii). Suppose h2(y) < h2(x). Since there are not three b-edges in γy with the same height, it is −1 −1 k −1 k−2 immediate to see that all vertices in γy ∩ π (q0) must be consecutive. Also, since b a b = ab a b and −1 −k −1 −1 −k+2 −1 b a b = a b a b for every 2 < k ∈ Z it is readily seen that γy ∩ π (q0) is a single edge, {vi, vi+1} for some 0 < i < n (otherwise, these relations allow a shortcut in γy leading to contradiction). Since h2(x1) = M = maxv∈V (γ){h2(v)} and there are not three b-edges in γy with the same height it −1 follows that all b-edges {vj, vj+1} with j < i satisfy that vj+1 = vjb . Also, let us see that all b-edges {vj, vj+1} with j > i satisfy that vj+1 = vjb. Suppose, otherwise that −1 vj+1 = vjb for some j > i. If h2(vj) ≤ h2(y1) then necessarily there must be three b-edges with the same heigth leading to contradiction with Lemma 5.9. If h2(y1) < h2(vj) ≤ h2(x1), then there are at least three b-edges with heigth h2(y1) + 1, one between x1 = v0 and vi, one between vi and vj and one between vj and vn = y1, leading to contradiction. Therefore, L(γy) = H(γy) + h2(x1) + h2(y1) − 2m0. ε Let us build a new geodesic σ from x to y as follows. For each a-edge {vj, vj+1} with j > i and vj+1 = vja with ε ∈ {−1, 1}, consider any vertex vk with k < i such that h2(vk) = h2(vj). Then it is readily seen that ε the path xs1 ··· ska sk+1 ··· sjsj+2 ··· sn is also a path from x to y1 with the same length as the restriction of γy joining x to y1. Therefore, repeating this change for every a-edge betweeen the lowest a-edge and −1 y1, we build a path σ from x to y such that L(σ) = L(γy) and such that if V (σ) ∩ π2 (qm) = {wr, wr+1}, ws+1 = wsb for every s > r, this is, σ is a path that after its lowesy a-edge is just a vertical geodesic. Since γy is a geodesic, σ is also a geodesic from x to y containing x1 and y1. −l Let l = h2(x) − h2(y). Then, let z = xb a. By construction, notice that p(y1) < p(z) < p(x1). If there is a vertex w 6= x in σ such that p(w) = p(x) then the vertical geodesic from x to w is a shortcut in σ leading to contradiction. Otherwise, there is a vertex w in σ such that p(w) = p(z), see Figure 11. However, the path defined by x, xb−1, . . . , xb−l, xb−la = z followed by the vertical geodesic from z to w has length h2(x) − h2(w) + 1 and is strictly shorter than the restriction of σ. Thus, this defines a shortcut in σ leading to contradiction. By Proposition 5.2 and Theorem 5.10, we get the following result. Corollary 5.11. The groups BS(1, 2), BS(−1, 2), BS(1, −2), BS(−1, −2), BS(2, 1), BS(−2, 1), BS(−2, −1) and BS(−2, −1) are 6-chordal. VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 13

x 1 x1

1 z

p(y1)

Figure 11. If there is no vertex w 6= x in σ with p(w) = p(x), then there is a vertex w in σ with p(w) = p(z).

k Lemma 5.12. Let n > 2 and x, y ∈ V (Γn). If there exists x˜ ∈ V (Γn) such that x˜ = xb for some k ∈ N 0 and y˜ = ybk for some k0 ∈ N where y˜ =xa ˜ , then there is a geodesic σ between x and y such that x,˜ y˜ ∈ σ. 0 Proof. Let us denote σ the path between x and y deﬁned by bkab−k . L(σ) = k + k0 + 1. Let σ0 be other 0 0 0 0 path between x and y. Denote h = max{hn(x): x ∈ σ} and h = max{hn(x ): x ∈ σ }. 0 0 0 0 0 0 If h > h, then L(σ ) > h − hn(x) + h − hn(y) + H(σ ) > k + k + 1 = L(σ). 0 h−h0 0 0 x Otherwise, let us assume h < h, then n + 2h − hn(x) − hn(y) 6 L(σ ). Since n > 2, then 2x + 1 6 n 0 h−h0 for all x > 1. Therefore 2(h − h ) + 1 6 n . h−h0 0 2h + 1 6 n + 2h h−h0 0 2h − hn(x) − hn(y) + 1 6 n + 2h − hn(x) − hn(y) h−h0 0 0 L(σ) 6 n + 2h − hn(x) − hn(y) 6 L(σ ) 0 L(σ) 6 L(σ )

Theorem 5.13. Gn is not k-chordal for every n > 2, k > 0.

Proof. Fix N > 1. Let us deﬁne the cycle γN by his vertices as follow. x0 = e, x1 = a , x2 = ab, ..., N N N −1 N −N N −N −1 xN+1 = ab , xN+2 = ab a, xN+3 = ab ab , ..., x2N+2 = ab ab , x2N+3 = ab ab a , x2N+4 = N −N −1 N −N −1 N N −N −1 N −1 N −N −1 N −1 −1 ab ab a b, ..., x3N+3 = ab ab a b , x3N+4 = ab ab a b a , x3N+5 = ab ab a b a b , N −N −1 N −1 −N ..., x4N+4 = ab ab a b a b = x0. See Figure 12.

Figure 12. Example of γ2 in G3

Claim. γN has not shortcuts. Let xi, xj ∈ V (γ). Denote L1 = {1, ..., N + 1}, L2 = {N + 2, ..., 2N + 2}, L3 = {2N + 3, ..., 3N + 3}, L4 = {3N + 4, ..., 4N + 4}, O = πn(e). Case 1. Suppose i, j ∈ Lm for m ∈ {1, 2, 3, 4}. Since pO(xi) = pO(xj), then γ deﬁnes a vertical geodesic between xi and xj. Therefore, there is not a shortcut between xi and xj. 14 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

Case 2. Suppose i ∈ L1, j ∈ L2 (or i ∈ L3, j ∈ L4). By Lemma 5.12, γN deﬁnes a geodesic between xi and xj and therefore there is not a shortcut between xi and xj. Case 3. Suppose i ∈ L1, j ∈ L3 (or i ∈ L2, j ∈ L4). Let σ be a path between xi and xj. Suppose without loss of generality that hn(xi) > hn(xj). Denote h = max{hn(x): x ∈ σ}. If h > N, then L(σ) > hn(xj) + h + (h − hn(xi)) + H(σ) > 2h + 2 + hn(xj) − hn(xi) > 2N + 2 + hn(xj) − hn(xi) = dγN (xi, xj). N−h Otherwise, let us assume h < N. On one side n + 1 6 H(σ), on the other hand the number of edges N−h in σ labelled by b is at least H = hn(xi) + hn(xj) + 2(h − hn(xi)), therefore n + 1 + H 6 L(σ). Since x N−h n > 2, then 2x + 1 6 n for all x > 1. Therefore 2(N − h) + 1 6 n .

N−h 2(N − h) + 2 + H 6 n + 1 + H N−h dγN (xi, xj) = 2(N − h) + 2 + H 6 n + 1 + H 6 L(σ) dγN (xi, xj) 6 L(σ) Thus, there is not a shortcut between xi and xj. Case 4. Suppose i ∈ L1, j ∈ L4 (or i ∈ L2, j ∈ L3). In this case, dH (xi, xj) > 0. If σ is a path between xi and xj, then L(σ) > hn(xi) + hn(xj) + H(σ) > hn(xi) + hn(xj) + 1 = dγN (xi, xj) Therefore, σ is not a shortcut, and the proof of the claim is finished. Suppose that Gn is k-chordal for some k > 0. Then γk has a shortcut, which contradicts the claim. 6. Word problem Let S be a finite set, F (S) the free group generated by S and R ⊆ F (S) a finite set. We consider a finite group presentation G = hS|Ri. We will denote a =H b the equality a = b in the group H, for example, two different relations r1, r2 ∈ R are equal in G, i.e., r1 =G r2, but they are not equal in F (S), i.e., a 6=F (S) b. Let ω =F (S) s1 ··· sn ∈ F (S) such that s1 ··· sn =G e. It is defined the area of ω as follows:

ε1 −1 εm −1 A(ω) = min{m | ω =F (S) u1r1 u1 ··· umrm um , ui ∈ F (S), ri ∈ R, ε ∈ {−1, 1}} The Dehn function of the presentation hS|Ri is defined as follow: Dehn(n) = max{A(ω) | ω ∈ hhRii , |ω| 6 n}, QM εi −1 where hhRii = { i=1 uiri ui | ∀M ∈ N, ∀ui ∈ F (S), ∀ri ∈ R, ∀ε ∈ {−1, 1}}. See [27]. A function f : N −→ R is an isoperimetric function for the presentation hS|Ri if it is a non-decreasing function such that Dehn 6 f. Theorem 6.1 (Theorem 1.1, [3]). A finite presentation has a solvable word problem if and only if it admits a recursive isoperimetric function. Theorem 6.2. Let G = hS|Ri be a finite group presentation. If G is k-chordal, then hS|Ri admits a recursive isoperimetric function, i.e., hS|Ri has a solvable word problem. Proof. Suppose that G is k-chordal with respect to S. Since hS|Ri is a finite presentarion, then the set of words with length less than or equal to k is finite, and we can compute c = Dehn(k). We define the following recursive function:

f(n) = c if n = 1, . . . , k f(n) = 2f(n − 1) = c2n−k if n > k

We claim that Dehn(n) 6 2Dehn(n − 1) if n > k. In fact, let ω ∈ hhRii such that Dehn(n) = A(ω) and suppose that |ω| = n > k. ±1 Suppose ω = s1 ··· sn = e is a simple relation for some s1, . . . , sn ∈ S . Since G is k-chordal, then there 0 0 ±1 0 0 exist s1, . . . , sr ∈ S such that si ··· sj = s1 ··· sr with r 6 min{j − i, n − j + i − 2} for some 1 6 i < j 6 n. VERTEX SEPARATORS, CHORDALITY AND VIRTUALLY FREE GROUPS 15

0−1 0−1 0 0 We can splitω ¯ = ω1ω2 where ω1 = si+1 ··· sjsr ··· s1 , ω2 = s1 ··· srsj+1 ··· sns1 ··· si andω ¯ is a rotation of ω. Suppose that D1 and D2 deﬁne a van Kampen diagram for ω1 and ω2 over G (with base point g = s1 ··· si) respectively, see Figure 13.

sj+1 0 0−1 sj sr sr

D2 D1

sn 0 s0−1 s1 1

s1 si si+1

Figure 13. Van Kampen diagram for ω1 at right and Van Kampen diagram for ω2 at left.

0 0 If we join D1 and D2 through the common vertices s1, . . . , sr (see Figure 14), we obtain a van Kampen diagram D forω ¯ over G (with base point g).

sj+1 sj

s1 si si+1

Figure 14. Van Kampen diagram for ω.

If ω = e is not a simple relation, then we can splitω ¯ = ω1ω2 where ω1 = e, ω2 = e andω ¯ is a rotation of ω. Thus Dehn(n) = A(ω) = A(¯ω) 6 A(ω1) + A(ω2) 6 Dehn(n − 1) + Dehn(n − 1) = 2Dehn(n − 1). By induction, Dehn(n) 6 f(n) for all n ∈ N. We have shown that hS|Ri admits a recursive isoperimetric function, and by Theorem 6.1, the word problem is solvable. It is well known that ﬁnitely generated free abelian groups have solvable word problem. For the particular case of the group Z2, since it is 5-chordal (Proposition 4.17), it follows also from Theorem 6.2: Corollary 6.3. Z2 has solvable word problem.

We have seen in Section 5 that G2 is 6-chordal (Theorem 5.10). Therefore, by Theorem 6.2:

Corollary 6.4. The Baumslag-Solitar group G2 has solvable word problem. 16 SAMUEL G. CORREGIDOR AND ALVARO´ MART´INEZ-PEREZ´

Given two functions f, g : N −→ R, we will denote f 4 g if there exists C > 0 such that f(n) 6 Cg(Cn + C) + Cn + C for all n ∈ N. This gives an equivalence relation f ≈ g ⇐⇒ f 4 g and g 4 f. We n denote the exponential function in base m as expm(n) = m . This simpliﬁes the notation in some cases, for mn k m... example, expm(n) = m . Remark 6.5. Let G = hS|Ri be a ﬁnite group presentation. If G is k-chordal and c = Dehn(k), then the Dehn function is bounded by following function: f(n) = c if n = 1, . . . , k f(n) = 2f(n − 1) = c2n−k if n > k

Thus, if G is k-chordal, then Dehn 4 exp2. This idea provides us with a method to detect groups with solvable word problem that are not k-chordal for any k ∈ N. Example 6.6. We will consider the group −1 Γk = a1, . . . , ak, t, p | t ait = aiai−1, [t, a1] = 1, [p, ajt] = 1, i > 1, j > 0 n In [8], the authors prove that Dehn ≈ Ak, where A3(n) = exp2 (1) and Am+1 > Am. By Theorem 6.1, Γk has a solvable word problem. However, if Γk were l-chordal, by Remark 6.5, then Dehn 4 exp2, which is a contradiction for k > 3. Thus, we have shown that Γk is not chordal for k > 3. 7. Acknowledgments The authors are very grateful to Yago Antol´ınfor his helpful comments and suggestions.

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Facultad CC. Matematicas,´ Universidad Complutense de Madrid, Plaza de Ciencias, 3. 28040 Madrid, Spain Email address: [email protected]

Facultad CC. Sociales, Universidad de Castilla-La Mancha, Avda. Real Fabrica´ de Seda, s/n. 45600 Talavera de la Reina, Toledo, Spain Email address: [email protected]