Subgroups of direct products of tworight-angled Artin groups

First year report

University of Warwick

Jone Lopez de Gamiz Zearra

Supervisors:

Montserrat Casals Ruiz

Karen Vogtmann

Coventry, 7 April 2019

Contents

Academic activities v0.1 Papers studied . . . . . . . . . . . . . . . . . . . . . . . . . . v0.2 Books studied . . . . . . . . . . . . . . . . . . . . . . . . . . . v0.3 Courses for credit . . . . . . . . . . . . . . . . . . . . . . . . . v0.4 Courses not for credit . . . . . . . . . . . . . . . . . . . . . . vi0.5 Seminars followed . . . . . . . . . . . . . . . . . . . . . . . . . vi0.6 Talks given . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi0.7 Reading groups . . . . . . . . . . . . . . . . . . . . . . . . . . vi0.8 Conferences attended . . . . . . . . . . . . . . . . . . . . . . . vi

Notation vii

Introduction ix

1 Subgroups of right-angled Artin groups 1

2 Coherence of right-angled Artin groups 7

3 Basic properties of right-angled Artin groups 133.1 Decomposition of right-angled Artin groups as direct products 133.2 Marshall Hall’s theorem for right-angled Artin groups . . . . 143.3 K(G, 1)-complexes for right-angled Artin groups . . . . . . . 18

4 Subgroups of direct products of right-angled Artin groups 23

A Bass-Serre theory 27

Bibliography 33

iii

Academic activities

0.1 Papers studied

-A remark on generalised free products, Gilbert Baumslag, [2].

-Finiteness properties of automorphism groups of right-angled Artin groups,Ruth Charney and Karen Vogtmann.

-Automorphisms of two-dimensional right-angled Artin groups, Ruth Char-ney, John Crisp and Karen Vogtmann, [9].

-Moduli of graphs and automorphisms of free groups, Marc Culler and KarenVogtmann.

-Subgroups of Graph Groups, Carl Droms, [11].

-Graph Groups, Coherence, and Three-Manifolds, Carl Droms, [12].

-The Finite Basis Extension Property and Graph Groups, Carl Droms,Brigitte Servatius and Herman Servatius, [13]. -Subgroups of direct prod-ucts with a free group, Charles F. Miller III, [18].

-Automorphisms of Free Groups and Outer Space, Karen Vogtmann.

-Contractibility of Outer space: reprise, Karen Vogtmann.

0.2 Books studied

-Cohomology of Groups, Kenneth S. Brown, [7].

-Groups Acting on Graphs, Warren Dicks and Martin J. Dunwoody, [10].

0.3 Courses for credit

-Ring Theory.

-Cohomology and Poincare Duality.

-Geometric Group Theory.

v

vi 0.4. Courses not for credit

0.4 Courses not for credit

-Lie Groups.

-Hyperbolic Geometry.

-Course on Outer space and Automorphisms of free groups (TCC).

0.5 Seminars followed

-Junior Geometry and Topology seminar (GATEWAY).

-Geometry and Topology seminar.

0.6 Talks given

-‘Coherence of right-angled Artin groups’, Junior Geometry and Topologyseminar, University of Warwick.

-Geometry and Topology reading group, term 2, University of Warwick.

0.7 Reading groups

-Geometry and Topology reading group, terms 1 and 2.

-Ergodic Theory reading group, term 2.

0.8 Conferences attended

-Groups and Geometry in the South East, University College London (UCL),26/10/2018.

-Groups and Geometry in the South East, University of Southampton, 30/11/2018.

-Groups and Geometry, University of London, 12/12/2018.

-Geometric Analysis meets Geometric Topology, University of Heidelberg,25/02/2019–28/02/2019.

-Groups and Geometry in the South East, University of Warwick, 01/03/2019.

Notation

During this dissertation, X will be a finite graph. The vertex set and edgeset of X will be denoted by V (X) and E(X), respectively. The incidencefunctions of X will be ι : E(X)→ V (X) and τ : E(X)→ V (X). Let S be asubset of V (X). The subgraph X〈S〉 of X will be the graph obtained fromX by deleting the vertices in S and all edges which have an endpoint in S.The subgraph 〈S〉 of X will be the full subgraph induced by S. For example,if X is the following graph,

x y

z w

t

and S = {y, z}, then X〈S〉 and 〈S〉 are as follows, respectively:

x

w

t

y

z

If X1 and X2 are two graphs, the join graph, denoted by X1 ∨X2, willcorrespond to the graph obtained by joining the vertices of X1 with all thevertices of X2.

Let Y be a subgraph of X. A chord of Y is an edge e ∈ E(X) \ E(Y )connecting two vertices of Y . For example, the two possible chords of C4

are the following.

The center of a group G will be denoted by C(G), 1 or {1} will corre-spond to the identity group, and Fn, Zn and Zn will be the free group of

vii

viii

rank n, the cyclic group of order n and the free abelian group of rank n,respectively. Moreover, if H is a subgroup of G, the normal closure of H inG will be written as 〈〈H〉〉. If x is an element of G, the length of x, denotedby |x|, will be the length of the shortest word representing x. If G acts ona set Y , the orbit of an element y ∈ Y will be denoted by O(y) and thestabilizer by Gy.

Let X be a 2-dimensional cell complex, v a fixed 0-cell and k a non-negative constant. Then [X ]k will be the subcomplex of X consisting ofthose vertices which are joined to v by a cellular path of length smaller thank together with all 1 and 2-cells spanned by those vertices.

Finally, I will denote the unit interval [0, 1].

Remark 0.8.1. Instead of labels, colors are used in plenty of graphs withthe aim of facilitating the understanding.

Introduction

Let us start defining right-angled Artin groups. Their class includes freegroups of finite rank and finitely generated free abelian groups.

Definition 0.8.1. Given a graph X, the corresponding right-angled Artingroup (RAAG), denoted by GX, is given by the following presentation.

〈V (X) | if x, y ∈ V (X), xy = yx ⇐⇒ x and y are adjacent〉.

Example 0.8.1. If X is the following,

x y z

then GX = 〈x, y, z | ∅〉 ∼= F3. In general, the right-angled Artin groupcorresponding to a totally disconnected graph is a free group.

Example 0.8.2. If X is the following,

x y

z

then GX = 〈x, y, z | xy = yx, xz = zx, yz = zy〉 ∼= Z3. In general, the right-angled Artin group corresponding to a complete graph is a free abeliangroup.

Right-angled Artin groups have attracted much attention in GeometricGroup Theory. On the one hand, it has been shown that all hyperbolic3-manifold groups are virtually finitely presented subgroups of RAAGs. Onthe other hand, although the word, conjugacy and isomorphism problems aresolvable, there are subgroups of RAAGs where those algorithmic problemsare not solvable.

ix

x

That last property already reflects that not all subgroups of right-angledArtin groups are again RAAGs. Thus, Chapter 1 is devoted to understandunder which conditions does this hold.

Chapter 2 talks about their coherence; that is, it analyses when allfinitely generated subgroups of a RAAG are finitely presented.

The rest of the dissertation consits of explaining the necessary tools inorder to reach the final result, which will be summarised in the followinglines.

In the 80s many mathematicians started to study the structure of finitelypresented subgroups. In fact, the importance of Baumslag and Roseblade’spapar,[4], was the contrast between finitely presented and finitely generatedsubgroups. Finitely generated subgroups of the direct product of two freegroups are wild: there are uncountably many and the isomorphism problemis unsolvable, so there is no hope to classify them. However, they proved thatwhen restricted to finitely presented subgroups, they turn surprisingly mildand well-understood: they are free or virtually a direct product of two freegroups. This work motivated a systematic study on the structure of finitelypresented subgroups of direct products conducted by Bridson, Howie, Millerand Short in a series of papers.

In [5], the authors generalize Baumslag and Roseblade’s paper to thedirect product of finitely many limit groups over free groups (in particular,of finitely many free groups). Shortly speaking, they show that any FPn(Q)subgroup is virtually a direct product of limit groups over free groups.

In [6], it is shown that the classic algorithmic problems are solvablefor finitely presented subgroups of the direct product of finitely many limitgroups over free groups. This result further stresses the difference betweenfinitely presented and finitely generated subgroups of direct products of freegroups from the algorithmic point of view: conjugacy and membership prob-lems are not solvable for finitely generated subgroups but solvable for finitelypresented ones.

My PhD project consists of generalizing those three papers to the largestpossible class of right-angled Artin groups. In this report, Baumslag andRoseblade’s paper is generalized to the class of RAAGs such that all theirsubgroups are again RAAGs. It will be shown that in this case, any finitelypresented subgroup of the direct product of two RAAGs is again a right-angled Artin group or virtually a direct product of two RAAGs. In order toachieve this goal, some of their properties have been explained in Chapter 3.Finally, the main result is stated and proved in Chapter 4.

It has been necessary to learn about Bass-Serre theory and cohomologyof groups before starting with the main topic. Hence, a summary of the firstone is written in Appendix A. All the useful theory related to cohomologyof groups can be found in [7].

Chapter 1

Subgroups of right-angledArtin groups

This chapter will examine subgroups of right-angled Artin groups. It isknown that subgroups of free groups are again free, and the same prop-erty holds in the case of free abelian groups. Therefore, it is natural toask whether all subgroups of a right-angled Artin group are again right-angled Artin groups. Nevertheless, one can find easily a counterexample: Apresentation of F2 × F2 is

〈x, y, z, w | xy = yx, zw = wz〉,

so it can be realized as GX such that X is the following:

x y

z w

It is known that it has a finitely generated subgroup which is not finitelypresented (see [19]). In particular, that subgroup is not a right-angled Artingroup.

However, Carl Droms gave a sufficient and necessary condition for aright-angled Artin group to have all subgroups of this type ([11]). The mainpurpose of this chapter is to state and prove that theorem.

Let us start by defining some important notions.

Definition 1.0.1. A subset S of V (X) is said to be a separating set ofvertices of X if X〈S〉 is disconnected.

1

2

Definition 1.0.2. The graph C4 is defined to be

Definition 1.0.3. The graph L3 is defined to be

Let us state the main theorem.

Theorem 1.0.1. Every subgroup of GX is itself a right-angled Artin groupif and only if no full subgraph of X has either of the forms C4 or L3.

Example 1.0.1. Totally disconnected and complete graphs have that prop-erty, as expected.

Example 1.0.2. F2 × F2 = GC4, so there is a subgroup which is not aright-angled Artin group.

Let us start proving the sufficient condition. That is, suppose that nofull subgraph of X has either of the two forms C4 or L3 and let us do it byinduction on the number of vertices of X.

If X is not connected and X1, · · · , Xn are the connected componentsof X, then GX = GX1 ∗ · · · ∗ GXn. If X has two vertices, then GX isa free group. Now, suppose that the number of vertices is greater than 2and that the statement holds for graphs with fewer vertices. In particu-lar, it holds for GXi for all i ∈ {1, · · · , n}. By Kurosh subgroup theorem(see Theorem A.0.2) , any subgroup of GX is isomorphic to the free prod-uct of a free group and certain conjugates in GX of subgroups of thoseGXi, i ∈ {1, · · · , n}. By inductive hypothesis, those subgroups are againright-angled Artin groups, so are also their conjugates. In addition, the freeproduct of right-angled Artin groups is again a right-angled Artin group.

If X is connected, the next goal is to verify that there is a vertex in Xwhich is joined to every other vertex of X:

Lemma 1.0.2. Let X be a connected graph with no full subgraph isomorphicto either C4 or L3 and let S be a minimal separating set of vertices of X.Then, 〈S〉 is complete.

Proof. Let X1 be a component of X〈S〉. Then, every point of S is adjacentto a vertex of X1 (otherwise, S would not be minimal). Let u, v ∈ S and letX1 and X2 be two components of X〈S〉. For i ∈ {1, 2}, let Pi be a minimal

Chapter 1. Subgroups of right-angled Artin groups 3

path which connectes u and v through Xi. Since Pi is minimal, it has nochords except possibly the one joining u and v.

Set C = P1 ∪ P2. The length of C is greater than or equal to 4, so byassumption it has a chord. Observe that this chord cannot join a vertex ofP1 different to u and v, to a vertex of P2 different to u and v because theyare in distinct components of X〈S〉. Therefore, it must join two points of thesame path, but as argued, this is only possible if this edge joins u and v. Inconclusion, u and v are adjacent.

Lemma 1.0.3. Let X be a connected graph with no full subgraph isomorphicto either C4 or L3. Then, there is at least one vertex of X which is joinedto every other vertex of X.

Proof. If X is a complete graph, the statement is routine. If X is notcomplete, let us pick a minimal separating set S of vertices of X. Note thatS is not empty because X is connected and S is not V (X) because X is notcomplete. By Lemma 1.0.2, 〈S〉 is complete.

Let x be an element of S. Let us check that if y is a vertex of X〈S〉,then x and y are adjacent. Let Cy denote the component of X〈S〉 containingy, and let C be any other component of X〈S〉. By the same reasoning asbefore, there must be a vertex y′ in Cy adjacent to x and a vertex z in Cadjacent to x.

Suppose that x and y are not adjacent. Cy is connected, so it may beassumed that y and y′ are adjacent. However, the subgraph of X inducedby the vertices x, y, y′ and z is isomorphic to L3.

Let z be any vertex of X which is adjacent to all the other vertices of X(it exists by the previous lemma). Note that z is in the center of GX andX = X〈z〉 ∨ {z}. Hence,

GX = 〈z〉 ×GX〈z〉.

Let p : GX → GX〈z〉 denote the projection onto the second summand. Then,there is an exact sequence

1 〈z〉 GX GX〈z〉 1

Let us prove by induction on the number of vertices that the subgroupsof GX are right-angled Artin groups. If X has 1 or 2 vertices, it is routine.Now, suppose that the statement holds for graphs with fewer vertices andsuch that no full subgraph is isomorphic to C4 or L3.

Let H be a subgroup of GX. There is another exact sequence,

1 H ∩ 〈z〉 H p(H) 1α

The next aim is to show that the sequence right splits. The previousstatement is true for X〈z〉. Thus, p(H) is a right-angled Artin group, say

4

p(H) = GY for some graph Y .

Let h be an element of H. Then, h = znp(h) for some n ∈ Z. Then,p(h)h−1 ∈ 〈z〉. Now, let y be a vertex of Y and pick an element h of H suchthat p(h) = y. Then, yh−1 ∈ 〈z〉, so yh−1 = zn for some n ∈ Z. Hence,z−ny = h and since z is in the center of GX, yz−n ∈ H. Define ι : p(H)→ Hto be ι(y) = yzn.

Firstly, ι is well-defined, because if two vertices y and y′ commute, byusing that z is in the center, ι(y) and ι(y′) also commute. Secondly, ι is ahomomorphism. Let y and y′ be two vertices of Y and let us choose h, h′ ∈ Hwith p(h) = y and p(h′) = y′. Then, p(h) = zn1h and p(h′) = zn2h′ for somen1, n2 ∈ Z and ι(y) = yzn1 , ι(y′) = y′zn2 . Since p is a homomorphism,

p(hh′) = p(h)p(h′) = zn1hzn2h′ = zn1+n2hh′,

so ι(yy′) = yy′zn1+n2 = yzn1y′zn2 = ι(y)ι(y′). Finally, clearly p ◦ ι = 1p(H),so H = (H∩〈z〉)nϕp(H) where ϕ : p(H)→ Aut(H∩〈z〉) is a homomorphismdefined as follows: for each k ∈ p(H), if ϕ(k) is denoted by ϕk, for h anelement of H ∩ 〈z〉, ϕk(h) is the unique element of H ∩ 〈z〉 such that

ι(k)α(h)ι(k)−1 = α(ϕk(h)).

Since α(h) lies in the center of H, ϕk(h) = h. In conclusion, ϕ equals theidentity, so

H = (H ∩ 〈z〉)× pH.

Finally, since H ∩ 〈z〉 is a subgroup of 〈z〉 and right-angled Artin groupsare torsion-free, H ∩ 〈z〉 is trivial or infinite cyclic. In the first case, H isa right-angled Artin group because p(H) is. In the second case, H = GZwhere Z = Y ∨ {v} for some vertex v.

In order to prove the necessary condition, notice that it suffices to findsubgroups of GC4 and GL3 which are not right-angled Artin groups.

As pointed out before, GC4 is isomorphic to F2×F2 and it has a finitelygenerated subgroup which is not finitely presented. In particular, it is nota right-angled Artin group. In the case of

GL3 = 〈x, y, z, w | xy = yx, yz = zy, zw = wz〉,

let us consider a homomorphism s : GL3 → Z2 such that

s(x) = s(y) = s(z) = s(w) = 1,

and let K = kers. Let us show that K is not a right-angled Artin group.See Section 3.3 in order to check that

χ(GX) =

n∑i=0

(−1)idi,

Chapter 1. Subgroups of right-angled Artin groups 5

where di is the number of i-vertex complete subgraphs of X (and d0 = 1),and cd(GX) equals the number of vertices of the largest complete subgraphof X. If X is a tree, then di = 0 for all i > 2, so χ(GX) = 1− d1 + d2 = 0.In addition, the index of K in GX is 2 and GX is a torsion-free group.Therefore, χ(K) = 0 (see [7] 6.3).

It can be shown, by using the Reidemeister-Schreier theorem, that Khas presentation

〈a, b, c, d | ab = ba, bc = cb, bc2d = dbc2〉.

Suppose that K is a right-angled Artin group, and so that K = GY forsome graph Y . Note that

K�K ′ = 〈a, b, c, d | ab = ba, ac = ca, ad = da, bc = cb, bd = db, cd = dc〉,

where K ′ = [K,K], so it is a free abelian group of rank 4. Thus, Y musthave 4 vertices. In addition, again since the index of K in GX is 2 and GXis a torsion-free group, cd(K) equals cd(GX) = 2 (see [7] 3.1). Hence, Y hasno triangles. Then, 0 = χ(K) = 1 − d1 + d2, so Y is a tree. In conclusion,there are two possibilities for Y :

or

Let us rule out the first option. Observe that K is the amalgamatedproduct of 〈a, b, c | ab = ba, bc = cb〉 and 〈c′, d | c′d = dc′〉 via 〈bc2〉 and 〈c′〉.By Corollary 4.5 in [15], since no power of bc2 commutes with a, the centerof K is trivial. Nevertheless, by analysing the graph, the middle vertex isin the center of GY .

Finally, let us rule out the second option. Let K2 = K�[K,K ′] and

G2 = GL3�[GL3, GL′3]. By using that [xz, y] = [x, y]z[z, y],

1 = [bc2, d] = [b, d]c2[c2, d] = [b, d]c

2[c, d]c[c, d].

Let us compute [b, d]c2. Since xy = [x, y]yx,

[b, d]c2

= c−2[b, d]c2 = [c−2, [b, d]][b, d]c−2c2 = [c−2, [b, d]][b, d].

Thus, in K2, [b, d]c2 = [b, d]. In the same way, it can be shown that [c, d]c

equals [c, d]. To sum up, 1 = [b, d][c, d]2. Thus, modulo (K ′2)2, 1 = [b, d], sothe image of b lies in the center.

6

If K and GL3 are isomorphic, then K2∼= G2, so there is an element g in

G2 which is central modulo (G′2)2. Note that g can be written as xpyrzswtCfor some p, r, s, t ∈ Z and C ∈ G′2. Under these conditions,

[x, g] = [x, z]s[x,w]t,

and since modulo (G′2)2 g is central, [x, g] lies in (G′2)2. It can be shown thatG′2 is a free abelian group generated by [x,w], [x, z] and [y, w]. Hence, s andt must be even. In the same way, by considering [w, g], p and t must be even.

In conclusion, g is a square modulo G′2. Nevertheless, K2�K2′ = K�K ′ isfree abelian of rank 4 generated by the images of a, b, c and d, so b cannotbe a square modulo K ′2.

Chapter 2

Coherence of right-angledArtin groups

By definition of group presentations, finitely presented groups are finitelygenerated. However, in general it is a difficult problem to know whether afinitely generated group is finitely presented or not.

In the case of right-angled Artin groups, Carl Droms characterised theclass of RAAGs with that property ([12]). The main purpose of this chapteris to state and prove that theorem.

Let us start with basic algebraic and combinatorial definitions.

Definition 2.0.1. A group is called coherent if each of its finitely generatedsubgroups is finitely presented.

Free groups and free abelian groups are coherent. Therefore, it is naturalto ask whether all right-angled Artin groups are coherent. Nevertheless, thishope quickly escapes since F2 × F2 is a right-angled Artin group which isknown not to be coherent (see Chapter 1).

A restriction on the graph X is needed to ensure coherence:

Theorem 2.0.1. The group GX is coherent if and only if each circuit ofX of length greater than 3 has a chord.

Example 2.0.1. Free groups and free abelian groups correspond to totallydisconnected and complete graphs, respectively, and these graphs have thatproperty.

Example 2.0.2. Note that if X does not contain a L3 or a C4 (as in theprevious chapter), then GX is coherent.

7

8

Example 2.0.3. F2×F2 can be realized as a right-angled Artin group whereX is the following graph:

Therefore, it is not coherent as it was pointed out before.

The rest of the chapter will consist of proving this theorem.

Given an element g ∈ GX, suppose that g can be written as

xei1i1· · ·xeikik ,

where xij is a vertex of X and eij ∈ Z for all j ∈ {1, · · · , k}.Let us define s(g) to be ei1 + · · · + eik . For definition to make sense, it

is necessary to check that it is well-defined, so assume that

xei1i1· · ·xeikik = y

ai1i1· · ·xaitit ,

where xij , yis ∈ V (X), eij , ais ∈ Z for all j ∈ {1, · · · , k}, s ∈ {1, · · · , t}.Then,

xei1i1· · ·xeikik y

−aitit· · · y−ai1i1

=∏

gxixjx−1i x−1

j g−1,

for some vertices xi, xj that are adjacent in X and some g ∈ GX. Hence,ei1 + · · ·+ eik − ait − · · · − ai1 = 0 and so s(g) is well-defined.

Let KX be the subgroup of GX consisting of all the elements g ∈ GXwith s(g) = 0.

Let us convince ourselves that it suffices to prove the theorem for con-nected graphs. In fact, if X is not connected and X1, · · · , Xn are the con-nected components of X, then GX = GX1 ∗ · · · ∗GXn.

Proposition 2.0.2. If A and B are coherent groups, A ∗B is coherent.

Proof. Let H be a finitely generated subgroup of A∗B. By Kurosh subgrouptheorem (see Theorem A.0.2),

H = F ∗ ∗i∈IgiAig−1i ∗ ∗j∈JfjBjf−1

j ,

where F is a free group, Ai is a subgroup of A, Bj is a subgroup of B andgi, fj ∈ A ∗ B for all i ∈ I, j ∈ J . Since H is finitely generated, I andJ are finite and for i ∈ I, j ∈ J , Ai, Bj and F are finitely generated. Byassumption, A and B are coherent, so they are also finitely presented. Inconclusion, H is finitely presented.

This proposition implies that from now on, X can be assumed to beconnected.

The key part of the proof lies on the following proposition and corollary.

Chapter 2. Coherence of right-angled Artin groups 9

Proposition 2.0.3. Let X be a finite connected graph, and let U and V befull subgraphs of X with X = U ∪ V and W = U ∩ V . Then,

KX = KU ∗KW KV.

Proof. Note that in the above conditions, GX = GU ∗GW GV . By Bass-Serre theory, GX acts on a directed tree T. The vertices of T are the leftcosets of the subgroups GU and GV in GX and the edges are the left cosetsof GW in GX. Therefore, since KX is a subgroup of GX, KX also acts onT . In order to find a presentation of KX, a fundamental KX-transversalhas to be found.

KX acts transitively on the edges of T . To see this, let us fix w a vertexof W and let gGW be any edge of T . If s(g) = e1 + · · ·+ en, then

s(ws(g)

)= e1 + · · ·+ en,

so ws(g)g−1 ∈ KX and (ws(g)g−1)(gGW ) = ws(g)GW = GW . This meansthat there is a unique orbit of edges under the action of KX. The verticesGU and GV lie in different orbits of the KX-action on T . Thus, the fun-damental KX-transversal of T consists of the two vertices GU , GV and theedge GW . In conclusion, by Bass-Serre theory,

KX = KX ∩GU ∗KX∩GW KX ∩GV = KU ∗KW KV.

Corollary 2.0.4. Let T be a finite tree with n + 1 vertices, where n is anon-negative integer. Then, KT is a free group of rank n.

Proof. Let us prove it by induction on n. If n equals 0, the right-angled Artingroup GT is isomorphic to Z, so KT = {1}. If n is 1, GT is isomorphic toZ2 = {anbm | n,m ∈ Z}, and in this case KT = {anb−n | n ∈ Z} which isisomorphic to Z. Now, suppose that n ≥ 2 and that the statement is truefor finite trees with fewer than n vertices.

Let x be a vertex of T with valence 1 and let y be the unique vertex ofT adjacent to x. Let T ′ denote the tree obtained from T by removing thevertex x and the edge joining x and y. Then, by the previous proposition,

KT = KT ′ ∗K〈y〉 K〈x, y〉.

By inductive hypothesis, KT ′ is a free group of rank n− 1. As in the casesn ∈ {0, 1}, K〈x, y〉 ∼= Z and K〈y〉 ∼= {1}. To sum up, KT is a free group ofrank n.

Before proceeding with the proof, two facts related to amalgamated prod-ucts will be assumed. They can be found in [16] Theorem 8 and [2], respec-tively.

10

Proposition 2.0.5. Let G = A ∗U B, where A and B are coherent groupsand all the subgroups of U are finitely generated. Then, G is coherent.

Proposition 2.0.6. Let G = A ∗U B where A and B are finitely generatedgroups. Then, G is finitely presented if and only if U is finitely generated.

Recall that X is assumed to be connected and let us start proving thesufficient condition. That is, suppose that every circuit ofX of length greateror equal than 4 has a chord. The proof is done by induction on the numberof vertices of X, say n.

If n is 1 or 2, GX is free abelian, so it is coherent. Now, suppose thatthe statement holds for graphs where every circuit of length greater or equalthan 4 has a chord and with fewer than n vertices. If X is complete, GXis again free abelian, so it is coherent. Otherwise, by using Lemma 1.0.2, Xhas a separating set A of vertices such that 〈A〉 is a complete subgraph ofX.

Let us choose X1 and X2 full subgraphs of X such that X = X1 ∪ X2

and 〈A〉 = X1 ∩X2. In this case,

GX = GX1 ∗G〈A〉 GX2.

Note that the conditions of the theorem hold for X1 and X2. Thus, by in-ductive hypothesis, GX1 and GX2 are coherent. Moreover, 〈A〉 is a completesubgraph, so G〈A〉 is a finitely generated free abelian group. Therefore, byProposition 2.0.5, GX is coherent.

Let us check the necessary condition. Notice that it suffices to show thatif X is a circuit of length greater than 3 without a chord, then GX is notcoherent. Under this condition, let x and y be two nonadjacent vertices ofX and let us take X1 and X2 two trees such that

X = X1 ∪X2 and X1 ∩X2 = 〈x, y〉.

For example, if X = C4,

xX2

X1y

By Proposition 2.0.3,

KX = KX1 ∗K〈x,y〉 KX2,

and by Corollary 2.0.4, KX1 and KX2 are finitely generated free groups.Then, KX is finitely generated.

Chapter 2. Coherence of right-angled Artin groups 11

The last step is to prove that K〈x, y〉 equals 〈〈xy−1〉〉 in F2, which is notfinitely generated, so by Proposition 2.0.6, KX is not finitely presented.

Trivially, the normal closure of the element xy−1 is contained in K〈x, y〉.In order to show the other inclusion, let g ∈ K〈x, y〉 be a reduced word inthe alphabet {x, y}. Let us distinguish four different cases:

g = xa1ya2 · · ·xan ,g = ya1xa2 · · · yan ,g = xa1ya2 · · · yan or

g = ya1ya2 · · ·xan ,for ai ∈ Z \ {0}, i ∈ {1, · · · , n}.

If n = 0, g = 1 which is in 〈〈xy−1〉〉. If n = 1, it is not possible for g tobe in 〈〈xy−1〉〉. If n equals 2, g can be xny−n or ynx−n for some n ∈ Z\{0}.Let us check that xny−n ∈ 〈〈xy−1〉〉. The other option is proved similarly.

x2y−2 = x(xy−1)x−1(xy−1) ∈ 〈〈xy−1〉〉,

x3y−3 = xx(xy−1)x−1x−1(x2y−2) ∈ 〈〈xy−1〉〉.

In general,

xny−n = xn−1(xy−1)x1−n(xn−1y1−n) ∈ 〈〈xy−1〉〉.

Now, suppose that n is greater or equal than 3 and that the statement holdsfor reduced words in K〈x, y〉 with length smaller than n. If g is of the firsttype,

g = xa1ya2 · · · yan−1x−a2−···−an−1x−a1 ,

and by inductive hypothesis, ya2 · · · yan−1x−a2−···−an−1 ∈ 〈〈xy−1〉〉. The sameargument can be used if g is of the second type. If g is of the third type,

g = xa1ya2 · · ·xan−1y−a2−···−an−1y−a1 ,

and this equals xa1(ya2 · · ·xan−1y−a2−···−an−1)x−a1xa1y−a1 , which by induc-tive hypothesis lies in 〈〈xy−1〉〉.

Chapter 3

Basic properties ofright-angled Artin groups

In this chapter, X will not contain any full subgraph isomorphic to C4 orL3.

3.1 Decomposition of right-angled Artin groups asdirect products

It is known that if F is a free group and H is a finitely generated normalsubgroup of F , then H has finite index in F .

The goal of this section is to analyse to what extent does this hold inright-angled Artin groups.

The center of a right-angled Artin group GX is the subgroup generatedby the set of vertices S in X that are adjacent to every other vertex byan edge (see [9] Proposition 2.2). Let us check that X〈S〉 has at least twoconnected components. By contradiction, assume that it is connected. Ifthe number of vertices of X〈S〉 is 1 or 2, then GX = C(GX) and this is acontradiction. If the number of vertices is 3, since X〈S〉 is connected, there isa vertex which is connected to the other two, so it lies in the center. Finally,if the number of vertices is greater than or equal to 4, none of the vertices ofX〈S〉 can be adjacent to all the other vertices of X〈S〉, but this is not possibleby Lemma 1.0.3. In conclusion,

GX = C(GX)× (GX1 ∗GX2 ∗ · · · ∗GXn),

where X1, · · · , Xn are the connected components of X〈S〉.In [1] Section 6, Baumslag proved that any finitely generated normal

subgroup of a free product is of finite index. Therefore, if H is a finitelygenerated normal subgroup of GX,

H = (H ∩ C(GX))× p(H),

13

14 3.2. Marshall Hall’s theorem for right-angled Artin groups

where p : C(GX) × GX〈S〉 → GX〈S〉 is the projection map, and p(H) is afinitely generated normal subgroup of GX1 ∗GX2 ∗ · · · ∗GXn. Thus, p(H)is of finite index in GX1 ∗GX2 ∗ · · · ∗GXn.

3.2 Marshall Hall’s theorem for right-angled Artingroups

Marshall Hall proved that if F is a free group of finite rank and H is afinitely generated subgroup with basis B and {g1, · · · , gk} ⊆ F \ H, thenB can be extended to a basis of a finitely generated subgroup H∗ of finiteindex in F such that {g1, · · · , gk} ⊆ F \H∗ (see [14]).

The aim of this section is to show that the same holds for RAAGs. Fromnow on, it will be assumed that X is not complete. The following result willbe proved:

Proposition 3.2.1. Let H be a finitely generated subgroup of GX and letM ≥ 0. Then, H has a finite generating set which can be extended to a finitegenerating set for a subgroup H∗ with finite index in GX and such that ifx ∈ H∗ and |x| < M , then x ∈ H.

Note that this proposition implies that if {x1, · · · , xn} ⊆ GX \H, by taking

0 ≤M < min{|xi| | i ∈ {1, · · · , n}

},

then {x1, · · · , xn} ⊆ GX \H∗.Let us carry out the proof of the proposition.

Proof. Let us prove it by induction on the number of vertices. If n equals1 or 2, the result is already known. Now, suppose that the statement holdsfor graphs with fewer vertices and without full subgraphs isomorphic to C4

or L3. Let us distinguish two cases.

(1) Assume that X is connected. It has been proved in Lemma 1.0.3 thatthere is a vertex, say t, of X which is adjacent to all the other vertices ofX, GX = 〈t〉 ×GX〈t〉 and H = (H ∩ 〈t〉)× p(H) where p : GX → GX〈t〉 isthe projection map. Since p(H) is a finitely generated subgroup of GX〈t〉,by hyphotesis, there is a subgroup K of finite index in GX〈t〉 with a finitegenerating set Y such that some subset Z of Y is a finite generating set forp(H), and if k is an element of K and has length less than M in GX〈t〉, thenk lies in p(H). There are two possibilities for H ∩ 〈t〉.

Suppose that H ∩ 〈t〉 = {1} and that p(H) = GW for some graph Wwith V (W ) = {x1, · · · , xn}. As it has been shown,

1 H ∩ 〈t〉 H p(H) 1

Chapter 3. Basic properties of right-angled Artin groups 15

right splits with ι : p(H) → H such that ι(xi) = tnixi where ni ∈ Z for allxi ∈ V (W ). Let N = maxi∈{1,··· ,n}{|ni|}. Note that GX〈t〉 contains onlyfinitely many elements of length less than or equal to M , so T can be chosenso that any reduced product of T or more xi’s and their inverses has lengthgreater than M .

Let L = M+TN+1 and K∗ = 〈tL〉×K. K∗ is of finite index in GX anda generating set of p(H) (so of H) is contained in a generating set of K∗. Ifk is an element of K∗ and has length less than M , then p(k) ∈ K has lengthless than M , so p(k) lies in p(H) and k ∈ 〈tL, Z∗〉, where Z∗ = p−1(Z).Suppose that

k = (tL)s(tni1xi1)εi1 · · · (tnirxir)εir ,

for εij ∈ {1,−1}, ij ∈ {1, · · · , n}, j ∈ {1, · · · , r} and s 6= 0. Then,

|k| =∣∣∣Ls+

r∑j=1

nij

∣∣∣+ |xεi1i1 · · ·xεirir|.

Since |k| ≤M , by the definition of T , r has to be less than T . Therefore,

|k| ≥ |s|L−∣∣∣ r∑j=1

nij

∣∣∣ ≥ |s|L− r∑j=1

|nij | ≥ |s|L− TN ≥ L− TN = M + 1,

and this is a contradiction.

If H ∩ 〈t〉 = 〈tk〉 for some k ∈ N, then {tk} ∪ Z∗ is a generating set for〈tk〉 ×K, where Z∗ = p−1(Z). This is, in addition, of finite index in GX.Finally, if k ∈ K∗ has length less than M , then p(k) ∈ K has length lessthan M , so p(k) ∈ p(H) and so k ∈ H.

(2) Suppose that X is not connected. Then, GX = GX1 ∗ GX2, where X1

and X2 are graphs not containing C4 and L3 and with fewer vertices. ByKurosh subgroup theorem (see Theorem A.0.2),

H = F ∗ ∗i∈IgiCig−1i ∗ ∗j∈JfjDjf

−1j ,

where (Ci)i∈I is a family of subgroups of GX1, (Dj)j∈J is a family of sub-groups of GX2 and gi, fj are elements of GX for i ∈ I, j ∈ J . Since H isfinitely generated, I and J have to be finite.

Let A1 and A2 be the presentation complexes with based points p1 andp2 associated to GX1 and GX2, respectively. Then, GX is realized by join-ing p1 and p2 to a new vertex p by two edges e1 and e2 (see image). Let Fbe this complex.

A1 p1 pe1 e2 p2A2

16 3.2. Marshall Hall’s theorem for right-angled Artin groups

The subgroup H can be realized by a covering H of F . Let us fix p0 a vertex

of the fiber of p. Observe that Ai lifts to a disjoint union of covers A(j)i in

H, for i ∈ {1, 2}.The subgroup H is finitely presented. Therefore, there is H′ a finite

subcomplex of H such that

H = π1

(H, p0

)= π1

(H′, p0

).

Thus, there is a constant k so that π1

([H]k, p

0)

= H. Let us consider the

A(j)1 , A(j)

2 which intersect [H]k non-trivially. There are finitely many of thosebecause each of them realizes one of the giCig

−1 or fjDjf−1j of above. Let

U be the union of those A(j)1 , A(j)

2 and [H]k. Then,

H = π1

(U , p0

).

Let us construct a spanning tree T for U by first taking a spanning tree

T j1 , T j2 in each A(j)1 , A(j)

2 and then adding just enough of the lifts of theedges e1 and e2 to connect these trees into one. Let us choose pij a point

in A(j)i which is connected to p0 by the shortest reduced path in T , say

wij . Then, π1

(U , p0

)has a generating set consisting of wijA

(j)i w−1

ij , where

A(j)i = π1

(A(j)i , pij

)and a set E which corresponds to lifts of e1 and e2 not

living in T .

By inductive hypothesis, for i ∈ {1, 2}, since A(j)i is a finitely generated

subgroup of GXi, there is a subgroup B(j)i of finite index in GXi extending

the generating set of A(j)i . Again, let us realize the subgroup B

(j)i by a

covering B(j)i of Ai.

Let us pick xij in B(j)i the image of pij . By replacing each A(j)

i with B(j)i

along pij and xij , another finite complex U ′ is obtained. Thus, π1

(U ′, p0

)has a generating set consisting of wijB

jiw−1ij and the set E. Nevertheless,

note that U ′ is not a cover of F , because there may be vertices of B(j)i not

adjacent to a lift of ei. However, this can be solved by attaching a copy ofF \ Ai to such vertex to form a cover H∗ with the desired properties (seeexample).

Example 3.2.1. Suppose that the cover H is the following. So in this case,

GX = F ({a, b}) ∗ F ({c, d}), A(1)1 = 〈a3〉, A(2)

1 = 〈a, b〉, A(1)2 = 〈c, d2〉 and

A(2)2 = 〈dcd−1, d2cd−2, c〉.

Chapter 3. Basic properties of right-angled Artin groups 17

p

Then, H∗ is obtained as follows.

p

18 3.3. K(G, 1)-complexes for right-angled Artin groups

3.3 K(G, 1)-complexes for right-angled Artin groups

The goal of this section is to define the Salvetti complex associated to a right-angled Artin group and to check that it is an Eilenberg-Maclane complex oftype (G, 1).

It is known that if G is a finitely presented group with finite generatingset S = {s1, · · · , sn} and finite set of relators R, then there is a way ofobtaining a 2-dimensional cell complex, called the presentation complex ofG, with fundamental group G. The complex has a single vertex, one loopat the vertex for each element of S and there is one 2-cell for each elementof R with the boundary of the 2-cell attached along the appropriate word.

If G is a right-angled Artin group, there is another associated cell com-plex, called the Salvetti complex, constructed as follows: Let us begin witha wedge of circles attached to a vertex x0 and labeled by the generators.For each edge, say from si to sj in X, let us attach a 2-cell with boundarylabeled by sisjs

−1i s−1

j (a 2-torus). For each triangle in X connecting threevertices si, sj , sk, let us attach a 3-torus with faces corresponding to the torifor the three edges of the triangle. This process is continued by attaching ak-torus for each complete subgraph of X with k-vertices.

Example 3.3.1. Suppose that X is the following graph.

b a

c d

The presentation complex of GX is obtained in this way:

∪ ∪ ∪

∪ ∪

Chapter 3. Basic properties of right-angled Artin groups 19

The Salvetti complex of GX is the following:

∪ ∪

Let G = GX. Let us show that the Salvetti complex associated to G,which will be denoted by SG, is a K(G, 1)-complex.

Since the fundamental group of SG is isomorphic to the fundamentalgroup of its 2-skeleton, it is clear that it is isomorphic to G. Moreover, SGis connected, so it only remains to check that the universal cover of SG iscontractible.

In order to achieve that goal, two well-known theorems in GeometricGroup Theory will be used. Their proofs can be found, for example, in [20](Proposition 2.6, Theorem 1.4 and Section 2.5).

Theorem 3.3.1. Any CAT (0) space is contractible.

Theorem 3.3.2 (Cartan-Hadamard). If (X, d) is a complete, connected,non-positively curved length space, then its universal cover with the inducedmetric is CAT (0).

In conclusion, it suffices to check that the Salvetti complex is non-positively curved. For that, some properties of cube complexes and linksare needed.

Let us start recalling the basic definitions.

Definition 3.3.1. A cubical complex X is the quotient of a disjoint unionC = ∪λ∈AInλ by an equivalence relation ∼. The restriction maps pλ fromInλ to X of the natural projection p : C → X are required to satisfy thefollowing:

(1) For every λ ∈ A, the map pλ is injective.

(2) If pλ(Inλ)∩pλ′(Inλ′ ) 6= ∅, there is an isometry hλ,λ′ from a face Tλ of Inλ

onto a face Tλ′ ⊆ Inλ′ such that pλ(x) = pλ′(x′) if and only if x′ = hλλ′(x).

Note that Salvetti complexes are cube complexes. From now on, C andX are as above.

Definition 3.3.2. A local edge of C is a subinterval of length 13 on an edge

of a cube in C such that one of the endpoints is a vertex of a cube in C.A local edge of X is the image under p of a local edge in C.

20 3.3. K(G, 1)-complexes for right-angled Artin groups

Definition 3.3.3. Let v be a vertex in X. The link of v in X, lk(v), is asimplicial complex formed in the following way. The vertices of lk(v) are thelocal edges of X containing v. A collection of vertices {v0, · · · , vk} in lk(v)spans a k-dimensional simplex in lk(v) if and only if the corresponding localedges of X are images under p of a collection of local edges all contained inthe same cube in C and all which share a vertex.

Example 3.3.2. If C and the equivalence relation ∼ are represented in thefollowing picture,

v

the link of v is

Example 3.3.3. Suppose that C and the equivalence relation ∼ are asfollows.

Then, X can be realised as a cube,

Hence, the link of the vertex drawn in yellow is

Definition 3.3.4. A flag complex is a simplicial complex such that if ak-complete graph in the 1-skeleton of the complex is taken, it spans a k-simplex.

Finally, let us state a special case of Gromov’s link condition for cubecomplexes:

Theorem 3.3.3. Let X be a cube complex. If the links of all the verticesare simplicial flag complexes, X is non-positively curved.

Chapter 3. Basic properties of right-angled Artin groups 21

In conclusion, note that the Salvetti complex satisfies the conditions ofthe previous theorem. Thus, it is non-positively curved, and so by Theo-rem 3.3.1 and Theorem 3.3.2, its universal cover is contractible.

To sum up, SG is an n-dimensional K(G, 1)-space, where n is equal tothe number of vertices of the largest complete subgraph of X. Moreover,

Hn(G,Z) ∼= Hn(SG,Z) ∼= Zdn 6= 0,

where dn is the number of n-vertex complete subgraphs of X. Thus, bydefinition, the cohomological dimension of G, cdG, equals n (see [7] page184).

In addition, there is also an exact sequence of left ZG-modules,

1 Z(G)dn Z(G)dn−1 · · · Z(G)d0 1,

where di is the number of i-vertex complete subgraphs ofX for i ∈ {1, · · · , n}and d0 = 1 (see [7] 4.2). Thus, by [7] page 247,

χ(G) =

n∑i=0

(−1)idi.

Chapter 4

Subgroups of direct productsof right-angled Artin groups

Baumslag and Roseblade showed that the only finitely presented subgroupsof a direct product of two free groups are either free or are virtually a directproduct of two free groups (see [4]).

The aim of this chapter is to prove that in the case of right-angled Artingroups where the graph does not contain a C4 or a L3, a finitely presentedsubgroup is either a right-angled Artin group or, modulo the center, it isvirtually a direct product of two right-angled Artin groups. The proof isbased on Miller’s paper about this fact for free groups (see [18]).

As in the case of Proposition 2.0.5 and Proposition 2.0.6, a property ofHNN-extensions will be assumed (see [2]).

Proposition 4.0.1. Let

G = 〈D, t | t−1bt = φ(b), ∀b ∈ L〉,

be a HNN-extension of a finitely generated group D with associated sub-groups L and φ(L). Then, G is finitely presented if and only if L is finitelygenerated.

From now on, GX will be a right-angled Artin group such that X doesnot contain a full subgraph isomorphic to C4 or L3. Let us start showingthe key theorem.

Theorem 4.0.2. Let A×GX be the direct product of a group A with GX.Suppose that G is a subgroup of A×GX which intersects GX non-trivially.Then,

(1) If G is finitely generated, G has a subgroup G0 of finite index which isan HNN-extension of the form

〈D, t | t−1bt = b,∀b ∈ L〉,

23

24

where D is finitely generated and L = G ∩A is a subgroup of D.

(2) If G is finitely presented, L = G ∩A is finitely generated.

Proof. Let H be a subgroup of A × GX, p1 : H → A the first projectionmap and p2 : H → GX the second projection map. For i ∈ {1, 2}, let Hi

be the image of H under pi. Then, H2 is a subgroup of GX, so it is aright-angled Artin group. Moreover, H is a subgroup of H1 × H2. Thus,it can be assumed that if G is a subgroup of A×GX, the projection mapspA : G→ A and pGX : G→ GX are surjective.

By hypothesis, G intersects GX non-trivially. Then, there is an elementt in G∩GX such that t is not the identity. By the consequence of Proposi-tion 3.2.1, GX contains a subgroup M of finite index in GX which has thecyclic group 〈t〉 as a free factor. GX is finitely generated and M is of finiteindex in GX, so M is also finitely generated. Assume that {t, s1, · · · , ss} isa generating set for M . For i ∈ {1, · · · , n}, let si ∈ p−1

GX(si). Let L = A∩Gand G0 = p−1

GX(M).

Note that the index of G0 in G is finite. Otherwise, if

G =⋃

i∈IgiG0,

where I has infinite cardinality and gi ∈ G, then

GX = pGX(G) =⋃

i∈IpGX(giG0) =

⋃i∈IpGX(gi)M,

and this is a contradiction because M has finite index in GX.

Let us check that

G0 = 〈L, t, s1, · · · , sn | t−1bt = b, si−1bsi = φi(b), i ∈ {1, · · · , n}, ∀b ∈ L〉,

where φi is the automorphism of L induced by conjugation by si. Observethat there is an exact sequence

1 L G0 M 1α pGX

and that the sequence right splits by defining a map p′GX : M → G0 suchthat

p′GX(t) = t and p′GX(si) = si, for all i ∈ {1, · · · , n}.

Let M be 〈t, s1, · · · , sn〉 the subgroup of G0. Hence, G0 = L nφ M, where

φ : M → Aut(L) is a homomorphism defined as follows: for each k ∈ M , ifφ(k) is denoted by φk, φk(l) is the unique element of L such that

p′GX(k)α(l)p′GX−1

(k) = α(φk(l)).

Chapter 4. Subgroups of direct products of right-angled Artin groups 25

Observe that φt equals the identity because p′GX(t) = t lies in GX and α(l)in A. Therefore,

G0 = 〈L, t, s1, · · · , sn | t−1bt = b, si−1bsi = φi(b), i ∈ {1, · · · , n}, ∀b ∈ L〉,

where φi = φsi for i ∈ {1, · · · , n}.Now, suppose that G is finitely generated. Since the index of G0 in

G is finite, G0 is also finitely generated. Suppose that G0 is generated byelements a1, · · · , ak in L together with the elements t, s1, · · · , sn. Since tacts trivially by conjugation, L is generated by the ai’s and their conjugatesby words in {s1, · · · , sn}. Let D = 〈a1, · · · , ak, s1, · · · , sn〉. Note that D isfinitely generated and that L is a subgroup of D. Moreover,

G0 = 〈D, t | t−1bt = b,∀b ∈ L〉.

If G is finitely presented, G0 is also finitely presented. In conclusion, L hasto be finitely generated by Proposition 4.0.1.

Corollary 4.0.3. Let GX1 ×GX2 be the direct product of two right-angledArtin groups, G a subgroup of GX1 × GX2 and define Li = G ∩ GXi fori ∈ {1, 2}. Suppose that GXi = C(GXi) × GYi for some graphs Yi and letpi : GXi → GYi be the projection maps, for i ∈ {1, 2}.(1) If L1 = 1 or L2 = 1, then G is a right-angled Artin group.

(2) Suppose that L1 and L2 are non-trivial and one of them, say L1, isfinitely generated. Then, the right-angled Artin group p1(L1)×L2 has finiteindex in GY1 ×GX2.

(3) Otherwise, G is not finitely presented.

Proof. The projections pGX1 : G → GX1 and pGX2 : G → GX2 can be as-sumed to be surjective, as in the previous theorem.

If one of them is trivial, say L1, then pGX2 is injective: Let (g1, g2),(g′1, g

′2) be elements of G with g2 = g′2. Hence, (g′1g

−11 , 1) ∈ G, so (g′1g

−11 , 1) ∈

L1 = {1}. Thus, g1 = g′1. Therefore, G is isomorphic to GX2. In particular,it is a right-angled Artin group.

Suppose that L1 and L2 are non-trivial and one of them, say L1, is finitelygenerated. Note that L1 is a normal subgroup of G. Then, pGX1(L1) =L1 is a finitely generated normal subgroup of pGX1(G) = GX1, so p1(L1)is a finitely generated normal subgroup of p1(GX1) = GY1. Hence, bySection 3.1, p1(L1) has finite index in GY1. In conclusion, p−1

GX1(p1(L1)) =

p1(L1)× L2 has finite index in p−1GX1

(GY1) = (p1 × p2)(G).Otherwise, by the previous theorem, it is not possible for G to be finitely

presented.

Example 4.0.1. Note that if GX1 and GX2 are free groups, the alreadyknown result for subgroups of direct products of two free groups is achieved.

26

Stallings showed that

F2 × F2 = 〈a, b, c, d | ac = ca, ad = da, bc = cb, bd = db〉

is not coherent by checking that the kernel of F2 × F2 → Z sending a, b, cand d to 1 is finitely generated but not finitely presented.

Let K be that kernel. Proving that K is not finitely presented is aconsequence of the previous theorem: It is easy to check that L1 and L2 are〈〈ab−1〉〉 and 〈〈cd−1〉〉, respectively. These are not finitely generated, so byCorollary 4.0.3, the result is obtained.

Example 4.0.2. Let F1 = F ({a, b}) and F2 = F ({c, d}) be two free groupsof rank 2. Let φ1 : F1 → Z2 send a and b to 1, and let φ2 : F2 → Z2 send cand d to 1.

Let us define a subgroup of F1 × F2 as follows:

G = {(w, v) | w ∈ F1, v ∈ F2, φ1(w) = φ2(v)}.

Then, L1 equals {(w, 1) | w ∈ F1, φ1(w) = 0}, and this is isomorphic toker(φ1). Thus, L1 has index 2 in F1. Similarly, L2 has index 2 in F2.

On the one hand, note that G can be described as

{(w, v) | w ∈ F1, v ∈ F2 and l(w) ≡ l(v) mod 2}.

Then, sinceF1 × F2 = G ∪ (a, 1)G,

the index of G in F1 × F2 is 2. On the other hand, the index of L1 × L2 inF1 × F2 is 4.

If G = H1×H2 for some subgroups H1 and H2 of F1 and F2, respectively,for the index of G to be 2 in F1 × F2, H1 has to be F1, or H2 has to be F2.Suppose that H1 = F1. In addition, the index of L1 × L2 in F1 ×H2 has tobe 2, but the index of L1 in F2 is already 2. Thus, H2 = L2. Nevertheless,G is not equal to F1 × L2 because (a, c) ∈ G but (a, c) is not in F1 × L2.

In conclusion, G is virtually a direct product of two free groups but it isnot itself a direct product of two subgroups of F1 and F2.

Appendix A

Bass-Serre theory

From now on, G will be used to denote a group and T a tree.

It is known that every free group acts freely on a tree. In addition, theconverse also holds: if a group acts freely on a tree, then the group is free.

Therefore, it is a natural question to ask what happens if the conditionfreely is not required. Bass-Serre theory gives the answer to this question:G acts on T without global fixed point (that is, for all v ∈ V (T ) there issome g ∈ G such that g · v 6= v) if and only if G is the fundamental group ofa graph of groups.

The goal of this appendix is to define the necessary notions to understandthese two implications and to give explicit constructions of the trees and thegroups involved. Nevertheless, proofs will be omitted (see [3] Chapter 2 or[10] Chapter 1).

Let us start with the necessary condition. For that, the notion funda-mental group of a graph of groups has to be understood.

Definition A.0.1. A G-set is a set X with an action of G on X.

Definition A.0.2. A map α : X1 → X2 where X1 and X2 are G-sets is aG-map if α(g · x) = g · α(x) for all g ∈ G, x ∈ X1.

Definition A.0.3. A graph X is a G-graph if the vertex set and the edgeset are G-sets and the incidence and terminal functions are G-maps.

If X is a G-graph, the quotient graph will be denoted by X�G.

A property which will be used many times is the following: If X�G isconnected, there exist subsets Y0 ⊆ Y ⊆ X such that Y meets each G-orbitexactly once, Y0 is a subtree of X, V (Y0) = V (Y ) and for each e ∈ E(Y ),ι(e) ∈ V (Y ) = V (Y0). Such a set Y is called a G-transversal in X.

It is now possible to define what a graph of groups is.

Definition A.0.4. A graph of groups is a connected graph X together witha function G(−) which assigns to each v ∈ V (X) a group G(v), called vertex

27

28

group, and to each edge e ∈ E(X) a distinguished subgroup G(e) of G(ι(e)),called edge group, together with a monomorphism te from G(e) to G(τ(e)).The image of an element g in G(e) under te is denoted by gte .

Example A.0.1 (Important example). Suppose that X is a G-graph such

that X�G is connected. Let us choose a fundamental G-transversal Y in Xwith subtree Y0. A graph of groups associated can be defined as follows:

For each e ∈ E(Y ), by definition of Y , there are unique elements ι(e),τ(e) ∈ V (Y ) which lie in the same G-orbits as ι(e) and τ(e), respectively.In fact, ι(e) = ι(e). Then, it is easy to check that Y can be made into agraph with the functions ι and τ .

Since τ(e) and τ(e) lie in the same orbits, there is an element te in Gsuch that teτ(e) = τ(e). Moreover, if e lies in E(Y0), τ(e) equals τ(e), sote = 1.

Note that the stabilizer of an edge e is a subgroup of the stabilizers ofι(e) and τ(e). In addition, since ι(e) = ι(e), Gι(e) = Gι(e), and it is routineto show that

Gτ(e) = teGτ(e)t−1e .

Thus, there is a monomorphism te : Ge → Gτ(e) sending an element g tot−1e gte.

Example A.0.2. Suppose that G = 〈s | s4〉 and X is the following G-graph:

1 s

s3 s2

e

se

s2e

s3e

A fundamental G-transversal is drawn in purple. Note that τ(e) = s andτ(e) = 1, so te = s.

At this point, fundamental groups of graph of groups can be defined.

Definition A.0.5. Let (G(−), X) be a graph of groups and let Y0 be amaximal subtree of X. The associated fundamental group π(G(−), X, Y0) isgiven by the following presentation. The generating set consists of

{te | e ∈ E(X)} ∪⋃

v∈V (X)

G(v).

Moreover, there are three different type of relations:

(1) the relations for G(v), for each v ∈ V (X),

Appendix A. Bass-Serre theory 29

(2) t−1e gte = gte for all e ∈ E(X), g ∈ G(e),

(3) te = 1, for all e ∈ E(Y0).

Remark A.0.1. Let us explain better the second relation. The left handside of the equality corresponds to an operation on the generators in thegroup. If g is an element of G(e), gte is the image of g under the monomor-phism te, so it is an element of G(τ(e)). Hence, the right hand side corre-sponds to an element of the generating set.

Example A.0.3 (Important example). Suppose that X has one edge e andtwo vertices ι(e), τ(e). Let A = G(ι(e)), B = G(τ(e)) and C = G(e) suchthat C is a subgroup of A and there is a specified monomorphism t : C → B.

A BC

Here Y0 equalsX. The fundamental group associated to that graph of groupsis the free product A and B with amalgamated C and t(C); it is presentedon the generating set A ∪ B with relations saying that c = ct for all c ∈ C,together with the relations of A and B.

Example A.0.4 (Important example). Suppose that X has one edge e andone vertex v = ι(e) = τ(e). Let A = G(v) and C = G(e), so that C is asubgroup of A and there is specified a monomorphism t : C → A.

A

C

Here Y0 consists of a single vertex, and the fundamental group associatedto the graph of groups is the HNN-extension of A by t : C → A, A ∗C t; itis presented by A ∪ {t} with relations saying that t−1ct = ct for all c ∈ C,together with the relations of A.

If G is a fundamental group of a graph of groups (G(−), X), it actswithout global fixed point on a tree T which is constructed as follows:

V (T ) =⋃

v∈V (X)

G�G(v) and E(T ) =⋃

e∈E(X)

G�G(e).

Moreover, ι : E(T ) → V (T ) is defined such that ι(gG(e)) = gG(ι(e)), andτ : E(T )→ V (T ) sends gG(e) to gteG(τ(e)).

Example A.0.5. See the proof of Proposition 2.0.3.

30

Now, let us continue with the necessary condition. The aim is to showthat if a group acts on a tree, such a group is the fundamental group of agraph of groups.

Theorem A.0.1 (Structure Theorem for groups acting on trees). Supposethat G acts on T without global fixed point. Let us choose a fundamentalG-transversal Y and let us construct a graph of groups as in Example A.0.1.Then, G is isomorphic to the fundamental group associated to that graphof groups. Explicitly, G has the following presentation. The generating setconsists of

{te | e ∈ E(Y )} ∪⋃

v∈V (Y )

Gv.

Moreover, there are three different types of relation:

(1) the relations for Gv, for each v ∈ V (Y ),

(2) t−1e gte = gte for all e ∈ E(Y ), g ∈ Ge,

(3) te = 1, for all e ∈ E(Y0).

Remark A.0.2. Note that as a consequence of this theorem, it is obtainedthat if G acts freely on T , then G is free.

Example A.0.6. Let G = GL2(Z) and H = {z ∈ C | z > 0}. G acts in Has follows. Let (

a bc d

)∈ G and z ∈ H.

If ad− bc = 1, (a bc d

)· z =

az + b

cz + d,

and if ad− bc = −1, (a bc d

)· z =

az + b

cz + d.

Then, it is easy to check that by writting z = x+ iy for x, y ∈ R,(a bc d

)· (x+ iy) =

(ax+ b)(cx+ d) + acy2 + iy

(cx+ d)2 + c2y2.

Let Y = {cos θ + i sin θ | π/3 ≤ θ ≤ π/2} ⊆ H. Observe that Y can beviewed as the geometric realization of a graph with one edge, say e, havingι(e) = i, τ(e) = 1

2(1 + i√

3). Let T = O(Y ). Clearly, G acts on T . Let usshow the following properties:

(1) T is a tree,

(2) Y is a fundamental G-transversal in T ,

(3) Ge ∼= D2, Gι(e) ∼= D4 and Gτ(e)∼= D6, where Dn denotes the dihedral

Appendix A. Bass-Serre theory 31

group of order 2n.

If this is checked, by the previous theorem and by Example A.0.3,

G ∼= D4 ∗D2 D6.

It is routine to check property (2). Let us first prove property (3).

It is known that G is generated by t, s and r, where

t =

(0 11 0

), s =

(1 10 1

), r =

(1 00 −1

).

Moreover, by computing Gι(e), Gτ(e) and Ge, the following is obtained:

Gι(e) =

{±(

1 00 1

),±(

0 −11 0

),±(

1 00 −1

),±(

0 11 0

)},

Gτ(e) =

{±(

1 00 1

),±(

1 −11 0

),±(

0 1−1 1

),±(

0 11 0

),±(

1 01 −1

),±(−1 10 1

)},

Ge =

{±(

1 00 1

),±(

0 11 0

)}.

Therefore, property (3) is shown. Now, observe that t, r ∈ Gι(e) ⊆ 〈Gι(e), Gτ(e)〉.In addition,

srt =

(1 00 1

)∈ Gτ(e) ⊆ 〈Gι(e), Gτ(e)〉,

so s ∈ 〈Gι(e), Gτ(e)〉. In conclusion, G = 〈Gι(e), Gτ(e)〉.

Let us check that T is a tree. Firstly, T is connected: Let X be theconnected component of T containing Y . Observe that G permutes thecomponents of T . If g lies in Gι(e), g · e lies in X because

ι(g · e) = g · ι(e) = ι(e).

By the same reason, if g lies in Gτ(e), then g · e lies in X. Therefore, X isclosed under the action of Gι(e) ∪Gτ(e). Since G is generated by them, X isclosed under the action of G. To sum up, O(Y ) ⊆ X; that is, T = X, so Tis connected.

Secondly, T does not have a simple closed path: By doing some calcula-tions, it can be shown that

(1) i = ι(e) is the only vertex in the imaginary axis,

(2) If Re(g · z) > 0, then Re(g · 12(1 + i

√3)) > 0,

32

(3) If Re(g · z) < 0, then Re(g · 12(1 + i

√3)) < 0.

Hence, all the edges of T lie in

{z ∈ C | Re(z) > 0, Im(z) > 0}∪{z ∈ C | Re(z) < 0, Im(z) > 0},

and i = ι(e) is the only vertex in the imaginary axis. Now, suppose thatthere is a simple closed path in T . Then, if necessary, an element of G canbe applied to obtain a path in T from ι(e) to τ(e) which does not use theedge e. It is easy to show that the only edges incident to ι(e) are e and e′,where

e′ =

(0 −11 0

)e.

Note that e′ lies in {z ∈ C | Re(z) < 0, Im(z) > 0}. Thus, there is apath from τ(e′) to τ(e) which does not pass through ι(e) = i, but this is acontradiction.

Finally, Bass-Serre theory also gives the structure of subgroups of freeproducts:

Theorem A.0.2 (Kurosh Subgroup theorem). Let G = A ∗ B be the freeproduct of groups A and B and let H be a subgroup of G. Then,

H = F ∗ ∗i∈IgiAig−1i ∗ ∗j∈JfjBjf−1

j ,

where Ai is a subgroup of A, Bj is a subgroup of B and gi, fj ∈ G, for alli ∈ I, j ∈ J .

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