#### Abstract

We give a short proof of the following theorem of Sang-hyun Kim: if is a right-angled Artin group with defining graph , then contains a hyperbolic surface subgroup if contains an induced subgraph for some , where denotes the complement graph of an -cycle. Furthermore, we give a new proof of Kim's cocontraction theorem.

#### 1. Introduction and Definitions

Suppose that is a simple finite graph with vertex set and edge set . We say that is the defining graph of the right-angled Artin group defined by the presentation Right-angled Artin groups are also called graph groups or partially commutative groups in the literature. All graphs in this paper are assumed to be simple and finite.

Right-angled Artin groups have been studied using both combinatorial and geometric methods. In particular, it is well known that these groups have simple solutions to the word and conjugacy problems. Moreover, each right-angled Artin group can be geometrically represented as the fundamental group of a nonpositively curved cubical complex called the Salvetti complex. For these and other fundamental results, we refer the reader to the survey article by Charney [1].

Let be a graph, and suppose that . The induced subgraph is the maximal subgraph of on the vertex set . A subgraph is called an induced subgraph if . In this case, the subgroup of generated by is canonically isomorphic to . This follows from the fact that given by if and if defines a retraction. Therefore, we identify with its image in .

In this paper, we study the following problem: find conditions on a graph which imply or deny the existence of hyperbolic surface subgroup in . Herein, we say that a group is a hyperbolic surface group if it is the fundamental group of a closed orientable surface with negative Euler characteristic.

Servatius et al. proved that if contains an induced -cycle, that is, the underlying graph of a regular -gon, for some , then has a hyperbolic surface subgroup [2]. In fact, they construct an isometrically embedded closed surface of genus in the cover of the Salvetti complex corresponding to the commutator subgroup of .

Kim and, independently, Crisp et al. gave the first examples of graphs without induced -cycles, , which define a right-angled Artin groups which, nonetheless, contain hyperbolic surface subgroups [3, 4]. We give one such example here to illustrate the main lemma of this paper.

Consider the graphs in Figure 1. The map sending to , to , and to for defines an injective homomorphism onto an index two subgroup of (see the discussion below). Since contains an induced circuit of length five (the induced subgraph on the vertices , and ), contains a hyperbolic surface subgroup; however, does not contain an -cycle for any .

**(a)**

**(b)**

That the map is injective can be seen from several perspectives. The approach of Kim and Crisp et al. is to use dissection diagrams; these are collections of simple closed curves which are dual to van Kampen diagrams on a surface over the presentation . The method was introduced in this context by Crisp and Wiest [5] and used with much success by Kim [3] and Crisp et al. [4].

The purpose of this paper is to demonstrate that classical methods from combinatorial group theory offer another perspective and simplify some of the arguments in the aforementioned articles. We will use the Reidemeister-Schreier rewriting process to give a direct proof that the map , above, is injective, and we also indicate how this can be proven using normal forms for splittings of groups. This in turn will lead to a short proof of Kim's theorem on cocontractions (in [3, Theorem 4.2]) alluded to in the abstract; see Theorem 3.5 in this paper.

Lemma 1.1. *Suppose that is a right-angled Artin group, and let be a positive integer. Choose a vertex , and define by if and . Then is a right-angled Artin group with defining graph obtained by gluing copies of to along , where and are the star and link, respectively. Moreover, the vertices of naturally correspond to the following generating set:
*

The proof is a straightforward computation using the Reidemeister- Schreier method. The details are given in Section 2. Applying Lemma 1.1 to the graphs in Figure 1 proves that injects into : if maps to 1 mod 2, then .

Another way to prove Lemma 1.1 is to take advantage of “visual” splittings of the groups and as an HNN extension or amalgamated free product. This second approach is stated in the article by Crisp et al. (see [4, Remark 4.1]).

We illustrate the utility of Lemma 1.1 by giving a short proof of the following theorem of Kim.

Theorem 1.2 (see Kim [3, Corollary 4.3(2)]). *Let denote the complement graph of an -cycle. For each , the group contains a hyperbolic surface subgroup.*

In fact, we give a new proof of Kim's more general theorem (in [3, Theorem 4.2]) on cocontractions of right-angled Artin groups in Section 3. Kim's proof used the method of dissection diagrams. Kim has also discovered a short proof using visual splittings (personal communication).

In preparing this paper, we found that the Reidemeister-Schreier method has been used previously to study certain Bestvina-Brady subgroups of right-angled Artin groups (see [6, 7]).

This work was inspired by a desire to better understand Crisp et al.'s very interesting classification of the graphs on fewer than nine vertices which define right-angled Artin groups with hyperbolic surface subgroups. We hope that this paper will help to clarify some aspects of the general problem.

#### 2. The Reidemeister-Schreier Method and Proof of Lemma 1.1

The Reidemeister-Schreier method solves the following problem: suppose that is a group given by the presentation , and suppose that is a subgroup; find a presentation for . The treatment below is brisk; see [8] for details and complete proofs.

Let be free with basis , and let extend the identity map on . Consider the preimage . Let be a right Schreier transversal for in , that is, is a complete set of right coset representatives that is closed under the operation of taking initial subwords (of freely reduced words over ). Given , let be the unique element of such that . For each and , let . Define . Then is a basis for the free group . Define a rewriting process on freely reduced words over by where . Then for every reduced word , and This rewriting process together with the resulting presentation for the given subgroup of is called the Reidemeister-Schreier Method.

*Proof of Lemma 1.1. *Let be a graph, the corresponding right-angled Artin group, and a distinguished vertex of . Let be given by if and .

Let be free on , and let be the set of defining relators corresponding to . Let , and let be the inverse image of in under the natural map . The set is a right Schreier transversal for . One verifies (directly) that the following equations hold:

Thus, we have a set of generators for ; however, many of these generators are redundant. Again, one verifies (using ) that the following equations hold:

Therefore, if is a relation in , then
hold in . It follows that is generated by , the vertices adjacent to in , and copies () of each vertex and not adjacent to in . Moreover, the relations are such that is presented as a right-angled Artin group where the defining graph is obtained from by taking the star of and copies of the complement of the star of and gluing these copies along the link of . This completes the proof of Lemma 1.1.

#### 3. A Short Proof of Two Theorems of Kim

Suppose that is a graph. The complement graph is the graph having the same vertices as but which has edges complementary to the edges of . Recall that an -cycle is the underlying graph of a regular -gon.

Theorem 1.2 follows from Kim's cocontraction theorem (see Theorem 3.5 below); however, we present a short independent proof here.

*Proof of Theorem 1.2. *Suppose that is a graph which contains an induced . Then contains a hyperbolic surface subgroup by [2]. Since , Theorem 1.2 follows from Lemma 3.1 below.

Lemma 3.1 (see Kim [3, Corollary 4.3(1)]). *For each , .*

* Proof. *Let . Define by and for . By Lemma 1.1, the defining graph of has vertex set , where and . Let . Consider the induced subgraph . The vertices and are not adjacent if and only if . The vertices and are not adjacent if and only if . Therefore, .

Kim proved a more general theorem about subgroups of a right-angled Artin group defined by “cocontractions.” Let , and let . If is connected, then the contraction of relative to is defined by taking the induced subgraph together with a vertex and declaring to be adjacent to if is adjacent in to some vertex in . The cocontraction is defined as follows:

Kim insists that be connected whenever he considers the contraction . This assumption is not necessary. Moreover, the following lemma shows that the structure of is immaterial; the proof follows directly from the definitions.

Lemma 3.2. *Suppose that is a graph and . Let be the graph obtained from by removing any edges joining two elements of . Then .*

Corollary 3.3. *Suppose that is a graph and . Let be a graph obtained from by adding or deleting any collection of edges with both of their vertices belonging to . Then and .*

Lemma 3.4. *Suppose that is a graph and . Suppose that . Let and . Then .*

* Proof. *It suffices to compare the collection of vertices which are adjacent to in and ; in the latter case, we are identifying the vertex with .

A vertex in not belonging to is adjacent to if and only if is adjacent to every in . A vertex in not equal to nor is adjacent to if and only if is adjacent to and in ; but this, in turn, means that is adjacent to every in . (Note that the case of is trivial since .)

A collection of vertices is said to be anticonnected if is connected. (Note that .)

Theorem 3.5 (see Kim [3, Theorem 4.2]). * Suppose that is a graph and is an anticonnected subset. Then embeds in .*

* Proof. *First consider the case when consists of two nonadjacent vertices . Define by , and if . Let . Let
We claim that via if and .

If and are distinct from and , then and are adjacent in if and only if they are adjacent in .

On the other hand, a vertex is adjacent to in if and only if is adjacent to and , whereas a vertex is adjacent to in if and only if belongs to the link of and to the link of , that is, is adjacent to and . Therefore, and, hence, embeds in .

Now we prove the general statement by induction on . Suppose that is anticonnected, and suppose that we have chosen the ordering so that is also anticonnected. (This is always possible: choose so that it is not a cut point of .) Let . Suppose that embeds in . By the case of the two vertices above, embeds in . (Note that and are not adjacent in for, otherwise, would be adjacent to every , , which would contradict the hypothesis that is anticonnected.) This proves the inductive step. The proof of the theorem is completed by applying Lemma 3.4.

#### Acknowledgments

The author would like to thank several colleagues for advice and feedback: Jon Hall, Ian Leary, Ulrich Meierfrankenfeld, and especially Mike Davis and Tadeusz Januskiewicz. Furthermore, the author owes many of the ideas herein to Kim, Crisp et al., as this paper is the product of his attempts to better understand their body of work on this problem.