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Preprint submitted on 28 May 2013
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Conjugation-free groups, lower central series and line arrangements
Michael Friedman
To cite this version:
Michael Friedman. Conjugation-free groups, lower central series and line arrangements. 2013. �hal-
00826941�
ARRANGEMENTS
MICHAEL FRIEDMAN
Abstract. The quotients G
k/G
k+1of the lower central series of a finitely presented group G are an important invariant of this group. In this work we investigate the ranks of these quotients in the case of a certain class of conjugation-free groups, which are groups generated by x
1, . . . , x
nand having only cyclic relations:
x
itx
it−1· . . . · x
i1= x
it−1· . . . · x
i1x
it= · · · = x
i1x
it· . . . · x
i2.
Using tools from group theory and from the theory of line arrangements we explicitly find these ranks, which depend only at the number and length of these cyclic relations. It follows that for these groups the associated graded Lie algebra gr(G) decomposes, in any degree, as a direct product of local components.
1. Introduction
Let G be a group and consider its lower central series defined by G 1 = G, G k = [G k− 1 , G], k ≥ 2.
The graded sum
gr(G) .
= M
k≥1
G k /G k+1
has a graded Lie algebra structure induced by the commutator bracket on G. In general, gr(G) reflects many properties of the group G. For example, if G is finitely generated, then the abelian groups G k /G k+1 are also finitely generated. It is, however, difficult to find out the structure of G k /G k+1 even in the case when G is finitely generated. Thus, determining this structure, or even the ranks of the quotients φ k (G) .
= rank(G k /G k+1 ), for a certain class of finitely generated groups is a significant problem.
In the case of G = F n , the free group of rank n, Magnus (see e.g. [16]) showed that gr(F n ) is the free Lie algebra on n generators, whose ranks were computed by Witt [24]. Hall [12] introduced the basic commutators of F n , showing that the coset classes of weight k form a basis of G k /G k+1 . Therefore, for the free group, the structure of gr( F n ) is completely known. Another well-known example is the work of Chen [1], computing the structure of gr(F n /(F n ) ′′ ).
In this paper we concentrate on a particular class of finitely presented groups: groups having a presentation where all the relations are cyclic relations; that is, of the form:
x s i
tt
x s i
t−1t−1· . . . · x s i
11= x s i
t−1t−1· . . . · x s i
11x s i
tt
= · · · = x s i
11x s i
tt
· . . . · x s i
22,
where {x 1 , . . . , x n } are the generators of the group, 2 ≤ t, {i 1 , i 2 , . . . , i t } ⊆ {1, . . . , n} is an in- creasing subsequence of indices and s j ∈ hx 1 , . . . , x n i. We call such groups cyclic-related groups (see Definition 2.1) and the above relation a cyclic relation of length t. When all the conjugating elements are equal to the identity, i.e. s j = e, then we call such a group conjugation-free (see Defi- nition 2.6). The simplest example of such a group is the group F n /hR n i generated by n generators with one cyclic relation of length n:
R n : x n x n− 1 · . . . · x 1 = x n− 1 x n− 2 · . . . · x 1 x n = · · · = x 1 x n · . . . · x 2 . Obviously, F n /hR n i ≃ F n−1 ⊕ Z, and thus the structure of gr(F n /hR n i) is known.
A less-known example of a cyclic-related group is the pure braid group P B n (see [17, Theorem 2.3] for a cyclic-related presentation of this group). Note that given an arrangement of hyperplanes
1
A = {H 1 , . . . , H n } ⊂ C m , the fundamental group π 1 (C m − A) is a cyclic-related group (see Remark 4.9); thus, from this perspective, it is clear why the pure braid group is a cyclic-related group, as it is the fundamental group of the complement of a hyperplane arrangement known as the braid arrangement. In general, if W is a real reflection group, A W ⊂ C m the associated hyperplane arrangement and B W the Artin group associated to W , then π 1 ( C m − A W ) ≃ ker(B W ։ W ) is a cyclic-related group. Note that the study of the lower central series quotients of the pure braid group was initiated by Kohno [14].
As an arrangement of lines in C 2 is an example of a hyperplane arrangement, for any line arrangement L the associated fundamental group G = π 1 (C 2 − L) is a cyclic-related group. Kohno [13] proved, using the mixed Hodge structure on H 1 (C 2 − L, Q), that gr(G) ⊗ Q is isomorphic to the nilpotent completion of the holonomy Lie algebra of C 2 − L. Falk proved [6], using Sullivan’s 1- minimal models, that the lower central series ranks φ k (G) are determined only by the combinatorics of L. However, a precise combinatorial formula for φ k (G), and even for φ 3 (G), is not known (for the general formula for φ 3 , see e.g. [21, Corollary 3.6] or Remark 3.12 below).
These motivations lead us to investigate these ranks for cyclic-related and conjugation-free groups. While it is fairly easy to find out the rank φ 2 for any cyclic-related group (see Section 3.1.1), it is a harder task when considering the rank φ 3 . We find that for a certain class of conjugation-free groups, there is an upper bound on φ 3 (see Proposition 3.9). Explicitly, for such a group G, using only group-theoretic arguments, we prove that φ 3 (G) ≤ P
i≥ 3 n i φ 3 ( F i− 1 ), where n i is the number of cyclic relations of length i of G.
We then associate to this group G a line arrangement L(G), such that π 1 (C 2 − L(G)) ≃ G.
By [7], given any line arrangement L, φ 3 (π 1 (C 2 − L)) is bounded from below by the above upper bound. Thus, for this class of conjugation-free groups, one can calculate φ 3 directly, i.e. the third quotient of the lower central series behaves as if the group is a direct product of free groups. From this it follows, by [20], that these conjugation-free groups are decomposable (see Definition 4.19), that is, the quotient G k /G k+1 decomposes, for k ≥ 2, as a direct product of local components and that φ k (G) = P
i≥ 3 n i φ k (F i−1 ) for every k ≥ 2 (see Theorem 4.18). Hence, for this class of groups, we have found the complete structure of gr(G).
The paper is organized as follows. In Section 2 we define the main object of our research: cyclic- related and conjugation-free groups. We prove in Section 3 that for a certain class of conjugation- free groups, there is an upper bound on the third rank φ 3 . Associating to such a group a line arrangement, we show in Section 4 that this upper bound is obtained. This leads us to find an explicit formula on the ranks φ k for k ≥ 3 and to find a new series of examples of decomposable groups.
Acknowledgements: The author would like to thank Michael Falk, Uzi Vishne, Roland Bacher and Mikhail Zaidenberg for stimulating talks, and the Fourier Institut in Grenoble for the warm hospitality and support.
2. Cyclic-related and Conjugation-free groups
In this section, we define the main object of our research: cyclic-related groups and conjugation- free groups. Let G be a group, and denote [a, b] .
= a −1 b −1 ab, a b = b −1 ab for a, b ∈ G.
Definition 2.1. Let G be a finitely presented group. We say that G is cyclic-related if G has a presentation such that it is generated by x 1 , . . . , x n with relations R 1 , . . . , R w and the following requirements hold:
(1) All the relations are of the following form:
R p = R p,(i
t,...,i
1) : x s i
p,tt
x s i
t−1p,t−1· . . . · x s i
1p,1= x s i
t−1p,t−1· . . . · x s i
1p,1x s i
p,tt
= · · · = x s i
1p,1x s i
p,tt
· . . . · x s i
2p,2,
where 1 ≤ p ≤ w, 2 ≤ t, {i 1 , i 2 , . . . , i t } ⊆ {1, . . . , n} is an increasing subsequence of indices, s p,j ∈ hx 1 , . . . , x n i for 1 ≤ j ≤ t. These relations are called cyclic relations of length t. Note that when t = 2 we get the commutator [x s i
1, x s i
2′] = e.
(2) For every pair of indices j 1 , j 2 ∈ {1, . . . , n}, j 1 6= j 2 , there is a unique relation R p,(i
t,...,i
1) such that {j 1 , j 2 } ⊆ {i 1 , i 2 , . . . , i t }.
(3) For every two relations R p,(i
t,...,i
1) , R p
′,(j
s,...,j
1) , p 6= p ′ we have that:
|{i 1 , i 2 , . . . , i t } ∩ {j 1 , j 2 , . . . , j s }| ≤ 1.
Example 2.2. (1) The pure braid group P B n is an example of a cyclic-related group; see [17, Theorem 2.3] for a presentation of P B n with generators and relations satisfying the requirements above.
(2) For any hyperplane arrangement A ⊂ C k , π 1 ( C k − A) is a cyclic-related group (see Remark 4.9(1)).
Remark 2.3. Note that R p,(i
t,...,i
1) a cyclic relation of length t can be written as a list of t − 1 commutators:
[x s i
p,kk
, x s i
k−1p,k−1· . . . · x s i
1p,1· x s i
tp,t· . . . · x s i
k+1p,k+1] = e, where 1 ≤ k ≤ t.
Definition 2.4. We say that a cyclic relation of length t is multiple if t ≥ 3.
Definition 2.5. Given a cyclic-related group G, we define its associated graph Gr(G) in the following way:
• Vertices: for every multiple cyclic relation R p = R p,(i
t,...,i
1) (where t ≥ 3) associate a vertex v p = v p,(i
t,...,i
1) . When it is clear what are the corresponding generators to v p,(i
t,...,i
1) we write only v p .
• Edges: two vertices v p = v p,(i
t,...,i
1) , v p
′= v p
′,(j
s,...,j
1) are connected by a edge e i
vif {i 1 , i 2 , . . . , i t } ∩ {j 1 , j 2 , . . . , j s } = {i v } = {j u },
where 1 ≤ v ≤ t, 1 ≤ u ≤ s.
Definition 2.6. Let G be a cyclic-related group. If every cyclic relation is of the form R p,(i
t,...,i
1) : x i
tx i
t−1· . . . · x i
1= x i
t−1· . . . · x i
1x i
t= · · · = x i
1x i
t· . . . · x i
2,
(that is, there are no conjugations on the generators appearing in the relation) then we say that G is conjugation-free and that its relations are conjugation-free.
Example 2.7. (1) If the graph Gr(G) of a conjugation-free group G is empty, then all the relations of G are commutators between all the generators and hence G is a free abelian group, i.e. if G is generated by n generators then G ≃ Z n . This is since, by requirement (2) of Definition 2.1, every generator commutes with all other generators. Note that this is not true any more if G is cyclic-related, but not conjugation-free. For example, for the cyclic-related group
G 1 .
= hx 1 , x 2 , x 3 : [x 1 , x x 2
3] = [x 1 , x x 3
2] = [x 2 , x x 3
1] = ei,
the graph Gr(G 1 ) is empty (as there are no multiple relations) but G 1 is not abelian (one can check that there exists an epimorphism to Sym 3 ).
(2) Let G be a conjugation-free group with n generators and one cyclic relation of length n, i.e.
G ≃ hx 1 , . . . , x n : x n x n− 1 · . . . · x 1 = · · · = x 1 x n · . . . · x 2 i.
Then G ≃ F n− 1 ⊕ Z and the graph consists of only one vertex.
(3) Let G be a conjugation-free group such that Gr(G) is a disjoint union of two graphs. Then
G is a direct sum of two conjugation-free groups, whose graphs correspond to the two graphs which
are the components of Gr(G).
(4) If the graph of a conjugation-free group consists of only vertices (i.e. there are no edges), then, by the previous examples, this group is isomorphic to a direct sum of a free abelian group and free groups. See remark 4.17 for a generalization of this when Gr(G) does not have cycles.
(5) Let G 2 be a cyclic-related group generated by x 1 , . . . , x 6 with the following relations:
• cyclic relations of length 3:
R 1 : x x 3
4x 2 x 1 = x 2 x 1 x x 3
4= x 1 x x 3
4x 2 , R 2 : x 6 x 5 x 1 = x 5 x 1 x 6 = x 1 x 6 x 5 , R 3 : x 5 x 4 x 3 = x 4 x 3 x 5 = x 3 x 5 x 4 .
• cyclic relations of length 2:
R 4 : x 4 x 2 = x 2 x 4 , R 5 : x 5 x x 2
3= x x 2
3x 5 , R 6 : x 6 x 2 = x 2 x 6 , R 7 : x 6 x 4 = x 4 x 6
Then, the vertices of Gr(G 2 ) are v 1 , v 2 , v 3 (associated to the relations R 1 , R 2 , R 3 ) and the edges are e 1 , e 3 and e 5 (see Figure 1). Note that with this presentation G 2 is not conjugation-free.
v
1v
2v
3e
1e
3e
5Figure 1. The graph Gr(G 2 ) associated to the cyclic-related group G 2 .
Remark 2.8. Let G be a cyclic-related group. By requirement (3) in Definition 2.1, every two vertices of Gr(G) are connected with at most one edge.
3. Lower central series and conjugation-free groups
In this section we prove the main result: that for a certain class of conjugation-free groups, the rank of the third quotient of the lower central series is bounded from above. We will see in Section 4.3 that this inequality is in fact an equality, using tools arising from line arrangements.
First, we give in Section 3.1 a short survey of some of the known results concerning the quotients of the lower central series, afterwards proving the main theorem in Section 3.2.
3.1. The lower central series. We begin with some notations. Let G be a group generated by x 1 , . . . , x n and consider its lower central series G k , where G 1 = G and G k = [G k−1 , G], k ≥ 2.
Denote φ k (G) = rank(G k /G k+1 ). If G = F n the free group with n generators, then we denote ω k (n) .
= φ k (F n ).
Notation 3.1. (1) x i,j .
= [x i , x j ] ∈ G 2 for i, j ∈ {1, ..., n}.
(2) x i,jk .
= [x i , x j x k ], x ij,k .
= [x i x j , x k ] ∈ G 2 for i, j, k ∈ {1, ..., n}.
(3) x i,j,k .
= [x i,j , x k ] = [[x i , x j ], x k ] ∈ G 3 for i, j, k ∈ {1, ..., n}.
(4) x i,j,kl .
= [x i,j , x k x l ] = [[x i , x j ], x k x l ] ∈ G 3 for i, j, k, l ∈ {1, ..., n}.
(5) More generally, if α ∈ G, then we denote x i,α = [x i , α] and x i,α,k = [x i,α , x k ] ∈ G 3 and in the same way for expressions of the form x α,j , x α,j,k or x i,j,α .
Remark 3.2. If G is a group generated by x 1 , . . . , x n , then it is well-known that G 2 /G 3 is generated by x j,i where j > i and G 3 /G 4 is generated by x j,i,k where j > i and k ≥ i (see [12]).
Concentrating on the groups G 2 /G 3 and G 3 /G 4 , we recall a few of their well-known properties.
Proposition 3.3. (I) For every i, j, k ∈ {1, . . . , n}, the following equivalences hold in G 2 /G 3 : [x i , x x j
k] ≡ [x i , x j ], x i,jk ≡ x i,k x i,j .
(II) The following equivalences hold in G 3 /G 4 :
(1) x i,j,kl ≡ x i,j,l x i,j,k , x i,jk,l ≡ x i,k,l x i,j,l , x ij,k,l ≡ x j,k,l x i,k,l .
(2) x i,j,k ≡ x −1 j,i,k ,
where i, j, k, l are either words in G or indices in {1, . . . , n}.
Proof. (I) Obvious.
(II)(1) We give the proof only for the expression x i,jk,l . The proof for the other expressions is similar.
x i,jk,l = [[x i , x j x k ], x l ] = [[x i , x k ][x i , x j ] x
k, x l ] = [[x i , x k ][x k , [x i , x j ] −1 ][x i , x j ], x l ].
When expanding the commutator brackets of the right hand side, we see that both the expression [x k , [x i , x j ] −1 ] and its inverse appear. As [x l , [x i , x j ] −1 ] ∈ G 3 , it commutes in the group G 3 /G 4 with any other element. Therefore,
x i,jk,l ≡ [[x i , x k ][x i , x j ], x l ] ≡ [[x i , x k ], x l ] · [[x i , x j ], x l ](mod G 4 ),
where in the last equivalence we used that [ab, c] = [a, c] b [b, c] = [b, [a, c] − 1 ][a, c][b, c], and that [b, [a, c] −1 ] ∈ G 4 if a ∈ G 2 .
(2) Let a = x i,j ; thus a −1 = x j,i . Therefore x i,j,k = [a, x k ] = a −1 x −1 k ax k and x j,i,k = [a −1 , x k ] = ax − k 1 a − 1 x k . Hence, a − 1 x j,i,k a = x − k 1 a − 1 x k a = x − i,j,k 1 . Thus, in G 3 /G 4 , we get that x i,j,k ≡ x − j,i,k 1 . 3.1.1. The rank of G 2 /G 3 . From now on, let G be a cyclic-related group with relations R 1 , . . . , R w . In this subsection we give a combinatorial description of φ 2 (G) (see also Remark 4.21(1) for a description of φ 2 (G) via topological invariants). Recall again that G 2 /G 3 is generated by x j,i when j > i.
Note that by Remark 2.3 and Proposition 3.3(I), every cyclic relation of length t in G is equivalent, in G 2 /G 3 , to a list of t − 1 commutators, where the generators appear without conjugations. That is, the relation R p of G:
R p = R p,(i
t,...,i
1) : x s i
tp,tx s i
t−p,t−11· . . . · x s i
1p,1= x s i
t−p,t−11· . . . · x s i
1p,1x s i
tp,t= · · · = x s i
1p,1x s i
tp,t· . . . · x s i
2p,2, is equivalent in G 2 /G 3 to the following list of commutators:
R p,k : [x i
k, x i
k−1· . . . · x i
1· x i
t· . . . · x i
k+1] = e,
where 1 ≤ k ≤ t. Moreover, by Proposition 3.3(I) we see that every relation R p,k , where 1 ≤ k ≤ t, is equivalent to
(3) x i
k,i
k+1· . . . · x i
k,i
t· x i
k,i
1· . . . · x i
k,i
k−1= e.
By requirement (2) of Definition 2.1, for every pair of indices i, j, the generator x j,i of G 2 /G 3 appears as a term in the relations of G 2 /G 3 only once. Therefore, for every two different relations R p , R p
′of G, when considering these relations in G 2 /G 3 (as in Equation (3)), the generators that appear in these relations are different.
For example, every cyclic relation of length 3: x α k x β j x γ i = x β j x γ i x α k = x γ i x α k x β j (where α, β, γ ∈ G and k > j > i) is equivalent to equalities of the form x j,i ≡ x k,j ≡ x − k,i 1 in G 2 /G 3 . Hence, while in every such cyclic relation appear three generators of G 2 /G 3 , two of them can be expressed as the third (or as its inverse).
In the same way, while in a cyclic relation R p (in G 2 /G 3 ) of length m appear m 2
generators
of G 2 /G 3 , m − 1 of them can be expressed as a product of the others; thus a cyclic relation
R p of length m contributes v(p) .
= m 2
− m + 1 independent generators to G 2 /G 3 . Note that v(p) = m−1 2
= ω 2 (m − 1).
Hence P w
p=1 v(p) = P
i≥3 n i ω 2 (i − 1), where n i is the number of cyclic relations of length i. We therefore see that
rank(G 2 /G 3 ) = φ 2 (G) = X
i≥3
n i ω 2 (i − 1).
3.2. The group G 3 /G 4 . Here we would like to generalize the above equality, proved for the rank of G 2 /G 3 , for the group G 3 /G 4 , for a special class of conjugation-free groups. We first introduce some notations.
Notation 3.4. Let G be a cyclic-related group.
(1) Let n m be the number of cyclic relations of length m.
(2) Let β(Gr(G)) be the first Betti number of the graph Gr(G).
Thus, we want to prove the following theorem:
Theorem 3.5. Let G be a conjugation-free group such that β(Gr(G)) ≤ 1. Then:
φ 3 (G) = X
i≥ 3
n i ω 3 (i − 1).
The proof of this theorem is divided into two parts. The first part, given in Proposition 3.9, proves that φ 3 (G) ≤ P
i≥3 n i ω 3 (i − 1), using only arguments from group-theory. The second part, given in section 4.3, proves that φ 3 (G) ≥ P
i≥ 3 n i ω 3 (i − 1) and uses arguments from the theory of line arrangements. In fact, as will be seen in Remark 3.13, the condition that β(Gr(G)) ≤ 1 can be relaxed. Before proving these propositions, we need to examine closely the generators of G 3 /G 4 . Example 3.6. Let G ≃ hx 1 , . . . , x n : x n x n−1 · . . . · x 1 = · · · = x 1 x n · . . . · x 2 i. As G ∼ = F n−1 ⊕ Z, we get that φ k (G) = ω k (n − 1).
Definition 3.7. Let G be a conjugation-free group generated by x 1 , . . . , x n , with the relations R 1 , . . . , R w .
(1) A generator of G 3 /G 4 of the form x j,i,k = [[x j , x i ], x k ] is called local if there is a multiple relation R p = R p,(u
t,...,u
1) (2 < t, 1 ≤ p ≤ w) such that {i, j, k} ⊆ {u r } t r=1 .
(2) We say that a generator x k participates in a cyclic relation R p = R p,(u
t,...,u
1) if k ∈ {u r } t r=1 . Remark 3.8. Every conjugation-free cyclic relation of length t induces t 3 − t local generators of G 3 /G 4 , which are not induced from other relations (since we subtract the number of generators of the form x i,i,i . Note also that x i,i,i = e). However, by Example 3.6, from such relation there are only ω 3 (t − 1) non-trivial independent local generators, that is, the other local generators can be expressed as a multiplication of powers of the other basic local generators.
We are now ready to prove the first part of Theorem 3.5:
Proposition 3.9. Let G be a conjugation-free group such that β(Gr(G)) ≤ 1. Then:
φ 3 (G) ≤ X
i≥3
n i ω 3 (i − 1).
Proof. Assume that G is generated by n generators x 1 , . . . , x n . In order to prove that φ 3 (G) ≤ P
i≥3 n i ω 3 (i − 1), we prove that every non-local generator of the form x j,i,k = [[x j , x i ], x k ]
(where j > i and k ≥ i) is equivalent (modulo G 4 ) to the identity element e. From now on, we
denote by ≡ equalities that take place in G 3 , modulo the group G 4 .
We prove this claim by splitting our proof into cases. If the generators x i , x j commute, then x j,i = e and thus x j,i,k = e. Assume thus that the generators x j , x i participate in the same multiple relation R 1 , and let v 1 be the corresponding vertex to this relation in Gr(G). Explicitly, as j > i,
R 1 = R 1,(y
′′k′′
,...,y
0′′,j,y
′k′
,...,y
0′,i,y
k,...,y
0) .
Remark 3.10. Note that some (but not all) of the sets {y 0 , . . . , y k }, {y 0 ′ , . . . , y ′ k
′}, {y 0 ′′ , . . . , y ′′ k
′′} may be empty.
If x i,k = x j,k = e, that is, the generator x k commutes with the generators x i and x j , then x j,i,k = e. We therefore examine the cases when the generator x k participates with at least one of the generators, x i or x j , in the same multiple relation. We consider three cases: when x i,k = e but x j,k 6= e; when x j,k = e but x i,k 6= e; and when x i,k 6= e and x j,k 6= e.
First case: In this case x k , x i commute, i.e. x k,i = e, and x k , x j participate in the same multiple relation R 2 ; let v 2 be the corresponding vertex to this relation in Gr(G) (see Figure 2).
v
1 aR
1=R
1,(..,j,..,yu,..,i,..)v
2 aR
2=R
2,(..,j,..,k,..)Figure 2. Part of the graph Gr(G): The generators x i , x j participate in a multiple relation R 1 (to whom the vertex v 1 is associated) and the generators x k , x j partici- pate in a multiple relation R 2 (to whom the vertex v 2 is associated when we assume that k > j). The generator x y
′uis a another generator participating in R 1 .
Let us prove that x j,i,k ≡ e.
(4) x − j,i,k 1 Eqn.(2) ≡ x i,j,k = x − j 1 x − i 1 x j x i x − k 1 x − i 1 x − j 1 x i x j x k x
k,i= =e x − j 1 x − i 1 x j x − k 1 x − j 1 x i x j x k . Denote Y = x y
k· . . . · x y
0, Y ′ = x y
′k′
· . . . · x y
′0
, Y ′′ = x y
′′k′′
· . . . · x y
′′0
. Therefore, since G is conjugation-free, by Definition 2.6 part of the relations of R 1 are the following:
(5) Y ′′ x j Y ′ x i Y = x j Y ′ x i Y Y ′′ = Y ′ x i Y Y ′′ x j = x i Y Y ′′ x j Y ′ = Y Y ′′ x j Y ′ x i .
Remark 3.11. Let us note that if a ∈ G 3 then for every b ∈ G, a ≡ b −1 ab (mod(G 4 )), that is, we can conjugate an element in G 3 by any element in G and remain in the same conjugacy class.
We now consider two cases: either that [x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e or that at least one of these equalities does not hold.
Case (1): Assume that [x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e. This happens when all the generators participating in R 1 , except the generator x j , commute with x k (for example, this situation occurs when the vertex v 1 is not a part of the cycle of the graph Gr(G)).
By Remark 3.11, we can conjugate x − j,i,k 1 by any element. Thus, following Equation (4):
(6) x −1 j,i,k Conj. by Y Y
′′Y
′≡ (Y Y ′′ Y ′ ) −1 · x −1 j x −1 i x j x −1 k x −1 j x i x j x k · Y Y ′′ Y ′
[x
k,Y Y
′′Y
′]=e
= (Y Y ′′ Y ′ ) −1 · x −1 j x −1 i x j x −1 k x −1 j x i x j · Y Y ′′ Y ′ · x k . We now examine the following expression: x i x j · Y Y ′′ Y ′ .
x i x j · Y Y ′′ Y ′ = x i Y Y ′′ x j [x j , Y Y ′′ ]Y ′ = x i Y Y ′′ x j Y ′ [x j , Y Y ′′ ][[x j , Y Y ′′ ], Y ′ ]
Rel.(5)
= x j Y ′ x i Y Y ′′ [x j , Y Y ′′ ][[x j , Y Y ′′ ], Y ′ ] = x j Y ′ x i Y Y ′′ [x j , Y Y ′′ ]m,
where we denote m = [[x j , Y Y ′′ ], Y ′ ]. Note that m ∈ G 3 and therefore commutes in G 3 /G 4 with any other element that belongs to G. Thus, Equation (6) turns to:
(7) x − j,i,k 1 ≡ m −1 [x j , Y Y ′′ ] −1 Y ′′−1 Y −1 x − i 1 Y ′−1 x − j 1 · x j x − k 1 x − j 1 · x j Y ′ x i Y Y ′′ [x j , Y Y ′′ ]mx k
≡ [x j , Y Y ′′ ] −1 Y ′′−1 Y −1 x −1 i Y ′−1 x −1 k Y ′ x i Y Y ′′ [x j , Y Y ′′ ]x k . Since [x k , x i ] = [x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e, then Equation (7) becomes:
x −1 j,i,k ≡ [x j , Y Y ′′ ] −1 x −1 k [x j , Y Y ′′ ]x k . Denoting y = Y Y ′′ , we get:
x −1 j,i,k ≡ x j,y,k Eqn.(2)
≡ x −1 y,j,k . Thus:
(8) x j,i,k ≡ x y,j,k = x − j 1 y − 1 x j yx − k 1 y − 1 x − j 1 yx j x k x
k,y=e, Conj. by Y
′x
i≡
(Y ′ x i ) −1 · x − j 1 y −1 x j x − k 1 x − j 1 yx j x k · Y ′ x i x
k,Y′=x =
k,i=e (Y ′ x i ) −1 x − j 1 y −1 x j x − k 1 x − j 1 yx j Y ′ x i x k . Now, yx j Y ′ x i
Rel.(5)
= x j Y ′ x i y. Therefore, Equation (8) becomes:
x j,i,k ≡ (x j Y ′ x i y) −1 · x j x −1 k x −1 j · x j Y ′ x i yx k = y −1 x −1 i Y ′−1 x −1 k Y ′ x i yx k = e,
when in the last equality we use the fact that x k commutes with all the other terms in the product.
Case (2): Let us now check when [x k , y 0 ] 6= e, when y 0 ∈ {Y, Y ′ , Y ′′ }. This happens when the generator x k participates in a multiple relation R 3 , where R 3 6= R 2 , and one of the generators in the set {x v } v∈Y , where Y = {y 0 , . . . , y k } ∪ {y 0 ′ , . . . , y ′ k
′} ∪ {y 0 ′′ , . . . , y ′′ k
′′}, also participates in R 3 (see Figure 3(a)). Note that x k participates in (at least) two multiple relations: R 2 and R 3 .
v
1 aR
1=R
1,(..,j,..,yu,..,i,..)v
2 aR
2=R
2,(..,k,..,cu,..,j,..)v
3 aR
3=R
3,(..,k,..,yu,..)(a)
v
1 aR
1=R
1,(..,j,..,yu,..,i,..)v
2 aR
2= R
2,(..,k,..,cu,..,j,..)v
3 aR
3(b)
v
aR=R
(..,cu,..,i,..)Figure 3. Part (a): a part of the graph Gr(G): The generators x i , x j participate in a multiple relation R 1 (to whom the vertex v 1 is associated) and the generator x k participates in two multiple relations R 2 , R 3 (to whom the vertices v 2 , v 3 are associated, when we assume that k > j). The generator x y
′uis another generator participating in R 1 and R 3 , and x c
′u′
is another generator participating in R 2 . Part (b): a forbidden case when x c
′u′
and x i participate in the same multiple relation R ′ . In this case, β(Gr(G)) > 1.
Assume that k > j (the proof for k < j is similar and we leave it to the reader; note that if k = j then x j,i,j is a local generator). Examining closely R 2 , we see that:
R 2 = R 2,(c
′′k′′
,...,c
′′0,k,c
′k′,...,c
′0,j,c
k,...,c
0) .
(when we take into consideration Remark 3.10 with respect to the sets {c i }, {c ′ i
′}, {c ′′ i
′′}). Denote C = x c
k· . . . · x c
0, C ′ = x c
′k′
· . . . · x c
′0
, C ′′ = x c
′′k′′
· . . . · x c
′′0
. By Definition 2.6, part of the relations of
R 2 are the following:
(9) C ′′ x k C ′ x j C = x k C ′ x j CC ′′ = C ′ x j CC ′′ x k = x j CC ′′ x k C ′ = CC ′′ x k C ′ x j .
By the Hall-Witt identity in G 3 /G 4 , we get that [[x j , x i ], x k ] · [[x − i 1 , x − k 1 ], x j ] ·[[x k , x − j 1 ], x − i 1 ] ≡ e.
As [x i , x k ] = e, we get that [[x j , x i ], x k ] · [[x k , x − j 1 ], x − i 1 ] ≡ e. It is easy to see that in G 3 /G 4 , [[x k , x −1 j ], x −1 i ] ≡ [[x j , x k ], x i ] −1 , and thus x j,i,k = [[x j , x i ], x k ] ≡ [[x j , x k ], x i ] = x j,k,i . Since x j,k,i Eqn.(2) ≡ x − k,j,i 1 , we prove that x k,j,i ≡ e.
(10) x k,j,i = x − j 1 x − k 1 x j x k x − i 1 x − k 1 x − j 1 x k x j x i x
k,i=e, Conj. by C
′CC
′′≡ (C ′ CC ′′ ) −1 · x −1 j x −1 k x j x −1 i x −1 j x k x j x i · C ′ CC ′′ .
Let C = {c 0 , . . . , c k } ∪ {c ′ 0 , . . . , c ′ k
′} ∪ {c ′′ 0 , . . . , c ′′ k
′′}. Note that x i commutes with every generator in the set {x v } v∈C , since otherwise β(Gr(G)) would be greater than 1. Indeed, if both x i and x c
participated in the same multiple relation R ′ for an index c ∈ C, then Gr(G) would have (at least) two cycles (see Figure 3(b)). Therefore [x i , C] = [x i , C ′ ] = [x i , C ′′ ] = e and
(11) x k,j,i ≡ (C ′ CC ′′ ) − 1 · x − j 1 x − k 1 x j x − i 1 x − j 1 x k x j · C ′ CC ′′ · x i . Let us examine the expression x k x j C ′ CC ′′ :
x k x j C ′ CC ′′ = x k C ′ x j [x j , C ′ ]CC ′′ t
=[[x .
j,C
′],CC
′′]
= x k C ′ x j CC ′′ [x j , C ′ ]t Rel.(9) = x j CC ′′ x k C ′ [x j , C ′ ]t.
Note that t ∈ G 3 and thus, in G 3 /G 4 , it commutes with any other element. Therefore, Equation (11) becomes:
(12) x k,j,i ≡ t − 1 [x j , C ′ ] − 1 C ′− 1 x − k 1 C ′′− 1 C − 1 x − j 1 · x j x − i 1 x − j 1 · x j CC ′′ x k C ′ [x j , C ′ ]tx i
≡ [x j , C ′ ] − 1 C ′− 1 x − k 1 C ′′− 1 C − 1 x − i 1 CC ′′ x k C ′ [x j , C ′ ]x i = [x j , C ′ ] − 1 x − i 1 [x j , C ′ ]x i , when in the last equality we used that [x i , x k ] = [x i , C] = [x i , C ′ ] = [x i , C ′′ ] = e. Thus:
x k,j,i
Eqn. (12)
≡ x j,C
′,i Eqn. (2)
≡ x −1 C
′,j,i ⇒ x C
′,j,i = x −1 j C ′− 1 x j C ′ x −1 i C ′− 1 x −1 j C ′ x j x i
[x
i,C
′]=e, Conj. by CC
′′x
k≡ (CC ′′ x k ) −1 · x −1 j C ′−1 x j x −1 i x −1 j C ′ x j x i · CC ′′ x k =
[x
i,CC
′′]=[x
i,x
k]=e
= (x −1 k C ′′−1 C −1 x −1 j C ′−1 )x j x −1 i x −1 j (C ′ x j CC ′′ x k )x i = e, where in the last equality we used the relation:
C ′ x j CC ′′ x i x k Rel.(9) = x j CC ′′ kC ′ and then the fact that x i commutes with C, C ′ , C ′′ and x k .
Second case: In this case x k , x j commute, i.e. x j,k = e, and x k , x i participate in the same multiple
relation R 4 , whose associated vertex in Gr(G) is denoted by v 4 . The proof that x j,i,k ≡ e is similar
to the proof presented in the First case. Thus, we only outline the main steps and leave the
detailed checks to the reader. We use the same notations, regarding the relation R 1 and generators
participating in it, that appeared in the First case. Again, we consider two cases: either that
[x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e or that at least one of these equalities is not true.
Case (1): Assuming that [x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e, we conjugate x j,i,k by Y ′ Y Y ′′ and get, using Relation (5), that x j,i,k ≡ x i,Y
′,k . Thus x j,i,k ≡ x − Y 1
′,i,k . Conjugating x Y
′,i,k by Y Y ′′ x j and again using Relation (5), we see that x Y
′,i,k ≡ e.
Case (2): This case happens when the generator x k participates in another multiple relation R 3 , where R 4 6= R 3 , and in R 3 participate also one of the generators in the set {x v } v∈Y , where Y = {y 0 , . . . , y k } ∪ {y 0 ′ , . . . , y ′ k
′} ∪ {y ′′ 0 , . . . , y ′′ k
′′} (see Figure 4).
v 1 a R 1 =R 1,(..,j,..,y
u,..,i,..)
v 4 a R 4 =R 4,(..,k,..,i,..)
v 3 a R 3 =R 3,(..,y
u,..,k,..)
Figure 4. a portion of the graph Gr(G): The generators x i , x j participate in a multiple relation R 1 (to whom the vertex v 1 is associated) and the generator x k par- ticipates in two multiple relations R 3 , R 4 (to whom the vertices v 3 , v 4 are associated).
The generator x y
′uis another generator participating in R 1 and R 3 . Note that it is known that i < k. Examining closely R 3 , we see that:
R 3 = R 3,(z
′′k′′
,...,z
′′0,k,z
′k′
,...,z
0′,i,z
k,...,z
0) .
(when we take into consideration Remark 3.10 with respect to the sets {z i }, {z i ′
′}, {z ′′ i
′′}). Denote Z = x z
k· . . . · x z
0, Z ′ = x z
′k′
· . . . · x z
′0, Z ′′ = x z
′′k′′
· . . . · x z
′′0. Therefore, by Remark 4.9, part of the relations of R 3 are the following:
(13) Z ′′ x k Z ′ x i Z = x k Z ′ x i ZZ ′′ = Z ′ x i ZZ ′′ x k = x i ZZ ′′ x k Z ′ = ZZ ′′ x k Z ′ x i .
Again, by the Hall-Witt identity in G 3 /G 4 , we get that x j,i,k ≡ x − i,k,j 1 and therefore x j,i,k ≡ x k,i,j . Noting that [x j , Z] = [x j , Z ′ ] = [x j , Z ′′ ] = e (otherwise β(Gr(G)) > 1), we conjugate x k,i,j by Z ′ ZZ ′′ , and using Relation (13), we get that x k,i,j ≡ x i,Z
′,j . Thus x k,i,j ≡ x −1 Z
′,i,j . Conjugating x Z
′,i,j by ZZ ′′ x k and again using Relation (13), we get that x Z
′,i,j ≡ e.
Third case: In this case x i,k 6= e and x j,k 6= e. There are two cases when this may occur: the first is when all of the three generators x i , x j , x k participate in the same multiple relation. In this case, the generator x j,i,k ∈ G 3 /G 4 is a local generator, so we are done. The second case, of a cycle of length 3 in the graph Gr(G), occurs when the vertices of the cycle correspond to the different multiple relations R 1 , R 2 and R 4 , where x i , x j participate in R 1 , x k , x j participate in R 2 and x i , x k
participate in R 4 (see Figure 5(a)).
We use the notations of the First case, i.e. that R 1 = R 1,(y
′′k′′
,...,y
0′′,j,y
′k′
,...,y
0′,i,y
k,...,y
0) .
We also use the notations Y, Y ′ , Y ′′ . Note that [x k , Y ] = [x k , Y ′ ] = [x k , Y ′′ ] = e, as otherwise x k
would participate in (at least) three multiple relations and β (Gr(G)) > 1 (see Figure 5(b)). Thus:
(14) x j,i,k = x −1 i x −1 j x i x j x −1 k x −1 j x −1 i x j x i x k Conj. by Y
′
Y Y
′′, [x
k,Y
′Y Y
′′]=e
≡
v
4 aR
4=R
4,(..,k,..,i,..)v
2 aR
2=R
2,(..,k,..,j,..)v
1 aR
1=R
1,(..,j,..,yu,..,i,..)(a)
v
4 aR
4=R
4,(..,k,..,i,..)v
2 aR
2=R
2,(..,k,..,j,..)v
1 aR
1=R
1,(..,j,..,yu,..,i,..)(b) v
aR=R
(..,k,..,yu,..)Figure 5. Part (a): A cycle of length three: all pairs of generators participate in different multiple relations. The generator x y
′uparticipates in R 1 . Part (b): a forbidden case when x y
′uand x k participate in the same multiple relation R ′ . In this case, β(Gr(G)) > 1.
≡ (Y ′ Y Y ′′ ) −1 x − i 1 x − j 1 x i x j x − k 1 x − j 1 x − i 1 x j x i Y ′ Y Y ′′ x k . Let us examine the expression x j x i Y ′ Y Y ′′ .
x j x i Y ′ Y Y ′′ = x j Y ′ x i [x i , Y ′ ]Y Y ′′ = x j Y ′ x i Y Y ′′ [x i , Y ′ ][[x i , Y ′ ], Y Y ′′ ] Rel.(5) = x i Y Y ′′ x j Y ′ [x i , Y ′ ]r, where we set r = [[x i , Y ′ ], Y Y ′′ ] ∈ G 3 , which commutes in G 3 /G 4 with every other element. Thus Equation (14) becomes:
(15) x j,i,k ≡ r −1 [x i , Y ′ ] −1 Y ′−1 x −1 j Y ′′−1 Y −1 x −1 i · x i x j x −1 k x −1 j x −1 i · x i Y Y ′′ x j Y ′ [x i , Y ′ ]rx k ≡ [x i , Y ′ ] − 1 Y ′− 1 x − j 1 Y ′′− 1 Y − 1 x j x − k 1 x − j 1 Y Y ′′ x j Y ′ [x i , Y ′ ]x k .
Noting that [x i , Y ′ ]x k = x k [x i , Y ′ ][[x i , Y ′ ], x k ] x
i,Y′,k
∈G
3≡ x k [[x i , Y ′ ], x k ][x i , Y ′ ], we denote f = Y ′− 1 x −1 j Y ′′− 1 Y − 1 x j x −1 k x −1 j Y Y ′′ x j Y ′ x k ,
and get from Equation (15) that:
(16) x j,i,k ≡ (f · [[x i , Y ′ ], x k ]) [x
i,Y
′] x
j,i,k∈G3≡
, Rem.3.11f · [[x i , Y ′ ], x k ].
Since x j,i,k , [[x i , Y ′ ], x k ] ∈ G 3 /G 4 , then also f ∈ G 3 /G 4 . Denoting y = Y Y ′′ , we examine the expression f :
f Conj. by x ≡
ix −1 i Y ′−1 x −1 j y −1 x j x −1 k x −1 j yx j Y ′ x k x i = x −1 i Y ′−1 x −1 j y −1 x j x −1 k x −1 j · yx j Y ′ x i · x k [x k , x i ].
Now,
yx j Y ′ x i Rel.(5)
= x j Y ′ x i y.
Thus,
f ≡ y − 1 x − i 1 Y ′− 1 x − j 1 · x j x − k 1 x − j 1 · x j Y ′ x i yx k [x k , x i ] = y − 1 x − i 1 Y ′− 1 x − k 1 Y ′ x i yx k [x k , x i ] [x
k,Y
′