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[ ] ( ) We define representatives of our cohomology classes by constructing a low dimensional ap- proximation for a cellular chain complex of the universal cover of a K Sn, 1 -space. Furthermore, ( ) we show that the 2-chains of this chain complex are generated by the Sn orbits of three classes of elements: cij , dijkℓ, and eijk. Respectively, each of these three types of generators corresponds to the squaring, commuting, and braid relations in the presentation for the symmetric group. Therefore, the cocycle representing κ is determined as a function from the Sn module generated n ‰2Ž [ ] by these three 2-chains to Z2 . Using the presentation:

n ‰2Ž 2 Z2 = g¯ij 1 i j n g¯ = 1, gij ,gkℓ = 1 (1) ⟨{ } ≤ ≠ ≤ S ij [ ] ⟩ our second theorem defines the 2-cocycle classifying Zn.

Theorem 1.2. If n ≥ 4, then a representative for the cocycle classifying Zn as an extension of n ‰2Ž Sn by Z2 is given by:

κ cij = g¯ij ( ) ⎧g¯ik + g¯kj + g¯iℓ + g¯ℓj i < k < j < ℓ or k < i < ℓ < j κ dijkℓ = ⎪ ( ) ⎨0 otherwise ⎪ ⎩⎪ ⎧g¯ij + g¯jk i < k < j, j < i < k, or k < j < i κ eijk = ⎪ ( ) ⎨0 otherwise ⎪ ⎩⎪ In the construction of Theorem 1.2 we give a presentation for Zn. By considering Artin’s original presentation for the braid group given by:

bibi+1bi = bi+1bibi+1 for all i Bn = b1,...,bn (2) d W bibj = bj bi if i − j > 1 i S S our presentation for Zn yields a normal generating set for Bn 4 . [ ] Theorem 1.3. Let i < k. As a subgroup of the braid group, Bn 4 is normally generated by elements of the form: [ ] 1. b2,b2 [ i k] 4 2. bi

Outline This paper will begin with preliminaries which will define the conventions and def- initions for constructing group presentations, group cohomology, and braid/symmetric groups required. In section 3 we build an approximation for a low dimensional resolution for the univer- sal cover of a K Sn, 1 space and define the maps needed to construct the 2-cocycle using this ( ) 2 n Z‰2Ž resolution. Then in section 4 we construct the 2-cocycle φ ∈ Hθ Sn; and prove the order of this element. In section 5 we prove κ is determined by[φ], finishing( the proof) of Theorems 1.1 and 1.2. The proof of Theorem 1.3 is finished at the end of the paper with an explanation of the difficulties which arise in finding a finite generating set for Bn 4 . [ ]

2 1.1 Acknowledgments First I would like to give my deepest gratitude to my advisor Matthew Day for his continuous guidance, suggestions, and comments throughout this research and early drafts of this paper. I would also like to thank Dan Margalit and Matthew Clay for helpful conversations.

2 Preliminaries 2.1 Building a Group Presentation Given a group extension

1 KGQι π 1 and group presentations for K and Q, we describe the process to build a group presentation for G. This construction is previously known and included for completeness. Let SK RK and ⟨ S ⟩ S¯Q R¯Q be group presentations for K and Q respectively. For any set S, let F S denote the free⟨ groupS ⟩ generated by S. ( )

Generators: Since Q = G K, we may lift any elementg ¯ ∈ Q to an element g ∈ G by choosing ~ a g such that π g = g¯. Fix SQ ⊂ G as choice of lifts from elements in S¯Q such that π restricted ( ) to SQ is a bijection. Define S = SK ∪ SQ as the generators of G.

Relations: Since K ≈ SK RK , any element k ∈ K can be considered as a word k1⋯kn ∈ ⟨ S ⟩ −1 F SK where ki ∈ SK for all i. Furthermore since K is normal in G, sks ∈ K for any s ∈ SQ and( k )∈ K; therefore we can choose k′ ∈ K such that sks−1 = k′ and we get a relation in G. Note ′ that for any k ∈ SK, k is determined by the choice of each s ∈ SQ as a lift of an element in S¯Q. ′ Let R be the relations in G defined by conjugating elements of SK by elements of SQ and their inverses. Now, we can write any relation in R¯Q bys ¯1⋯s¯n = 1Q wheres ¯i ∈ S¯Q for all i, therefore s1⋯sn = k for some k ∈ K. Note that the choices for si as a lifts ofs ¯i determines k. We take s1⋯sn = k in G as the lift of the relations ¯1⋯s¯n = 1Q in Q. Let RQ denote the lift of relations in ′ ′ R¯Q determined by the choice of SQ. Define R = RK ∪ RQ ∪ R , then G = S R . ⟨ S ⟩ Lemma 2.1. Let K, G′,Q be groups with presentations be defined as above. Any element g ∈ G′ can be written as g = w1w2 where w1 ∈ F SQ and w2 ∈ F SK . ( ) ( ) ′ Proof. Let g ∈ G , then g = s1⋯sn where si ∈ S. By induction on n, suppose g = s1s2. In this ′ case the statement is trivial unless s1 ∈ Sk and s2 ∈ SQ. By the relations of R (conjugating SK −1 by elements of SQ and the inverses) there exists k ∈ K such that s2 s1s2 = k. Consider

−1 s1s2 = s2s2 s1s2 = s2k

Since K is generated by SK , k = w2 for some w2 ∈ F SK and hence the Lemma holds for n = 2. Now, suppose the Lemma is true for any element( of) G′ that is a reduced word of length n − 1 in F S . Suppose g = s1⋯sn, then by induction s1⋯sn 1 = w1w2 where w1 ∈ F SQ and ( ) − ( ) w2 ∈ F SK . If sn ∈ SK , then we are done. Therefore assume sn ∈ SQ, then we have g = w1w2sn. ( )

3 ′ In particular, w2 = k1⋯kℓ for k1,...,kℓ ∈ SK . By the relations of R there exists hi ∈ K such that −1 sn kisn = hi for every i such that 1 ≤ i ≤ ℓ. Thus we have

s1⋯sn = w1w2sn

= w1k1⋯kℓsn −1 = w1k1⋯kℓ−1snsn kℓsn −1 = w1k1⋯kℓ−2snsn kℓ−1snhℓ ⋮

= w1snh1⋯hℓ

Noting that each hi can be written as a word in F SK for all i proves the Lemma. ( ) Theorem 2.2. Let N, G′,Q with group presentations defined previously, then there exists maps ι′ and π′ such that ′ ′ 1 Nι G′ π Q 1 is a short exact sequence.

′ Proof. Recall SK RK , S¯Q R¯Q and S R are group presentations for K,Q and G ⟨ S ⟩ ⟨ S ⟩ ⟨ S ⟩ respectively. Since K ≈ SK RK , any element k ∈ K can be written as k1⋯kn where ki ∈ SK ′ ⟨ ′ S ′ ⟩ ′ for all i. Define ι ∶ K → G by ι k = k1⋯kn. Since RK ⊂ R it is clear that ι is a well defined ( ) ′ ′ ′ homomorphism. Furthermore, since k1⋯kn is unique, ι is injective. Now, define π ∶ G → Q ⎧s¯ s ∈ SQ by π′ s = ⎪ To show π′ is a well defined homomorphism it suffices to check the ( ) ⎨1 s ∈ S ⎪ K relations of ⎩⎪R are preserved in the image of π′. Since we can write any relation as r = 1 and by Lemma 2.1, any relation in R can be written as s1⋯snk1⋯kℓ = 1 where si ∈ SQ and ki ∈ F SK . ′ ′ ′ ( ) Therefore π s1⋯snk1⋯kℓ = s¯1⋯s¯n. This implies a relation in G is preserved under π ifs ¯1⋯s¯n ( ) ′ was a relation in Q. By the choice of lifting R¯Q to RQ,s ¯1⋯s¯n is a relation in Q and π is a well ′ defined homomorphism. Furthermore, since SQ is a lift of S¯Q, π is clearly surjective. It remains to show the sequence is exact. ′ ′ ′ First we show ι K ⊆ ker π . Since K ≈ SK RK and ι k = k for all k ∈ SK , it suffices ′ ( ) ⟨ S ⟩ ′ ( ) ′ ′ to show π k = 1 for all k ∈ SK . By the definition of π this is obvious. Hence ι K ⊆ ker π . ( ) ′ ( ) Now, let g ∈ ker π . By Lemma 2.1 g = w1w2 for some w1 ∈ F SQ and w2 ∈ F SK . Furthermore ′ ( ) ′ ( ) w1 = s1⋯sn for si ∈ SQ and by definition π g = s¯1⋯s¯n. Since g ∈ ker π ,s ¯1⋯s¯n = 1 in Q and ( ) thuss ¯1⋯s¯n is a relation in Q. Buts ¯1⋯s¯n lifts to a relation s1⋯sn = k by the definition of RQ as a lift of R¯Q. Furthermore k can be written uniquely as a word k1⋯kℓ ∈ F SK . Hence g can be ′ ′ ( ) written as a word in F SK and therefore g ∈ K. Hence ker π ⊆ ι K and the sequence ( ) ( ) ′ ′ 1 Nι G′ π Q 1 is exact. Theorem 2.3. Let 1 NGQι π 1 be a group extension. Given group presentations for K and Q defined above, there exists a group presentation for G.

4 Proof. Let SK RK and S¯Q R¯Q be group presentations for K and Q respectively. Define G′ = S R ⟨as inS the⟩ paragraphs⟨ S on⟩ generators and relations above. It suffices to show G ≈ G′. Notice⟨ thatS ⟩ by construction every element of S is an element of G, therefore define f ∶ G′ → G by f s = s for all s ∈ S. By Theorem 3.2 we have the following diagram: ( ) ′ ′ 1 Kι G′ π Q 1

≈ ≈ f ≈ ≈ 1 KGQι π 1

Furthermore, f clearly commutes with the diagram by the definition of ι′ and π′, therefore f is an isomorphism by the 5-Lemma.

2.2 Group cohomology In this section we will review the general group cohomology required to classify group extensions by 2-cocycles. We begin with constructing the normalized bar resolution, give the definition of equivalent group extensions, and describe the construction of a cocycle from an extension. The results of this section are known and can be found more thoroughly in Brown’s text [4].

Normalized standard resolution Let G be a groupand let Pt be the free Z-module generated by t + 1 tuples p0,...pt where pi ∈ G for all i. G acts on p0,...,pn by p ⋅ p0,...,pn = ( ) ( ) ( ) p ⋅ p0,...,p ⋅ pt for any p ∈ G. To construct a chain complex, we use the boundary operator ( ) t P ∶ P − i ∂t Pt → Pt−1 by ∂t = ∑ 1 di where i=0( )

di p0,...,pt = p0,...,pi 1,pi 1,...,pt ( ) ( − + ) As ZG modules, Pt is freely generated by elements 1,p1,...,pt which represent the G orbits ( ) of the t + 1 tuples where p0 = 1. We use the following bar notation to represent elements of Pt as G-orbits: p1 p2 ... pt = 1,p1,p1p2,...p1p2⋯pt . The change of basis results in the following [ S S S ] ( ) change to di in the boundary operator: ⋯ ⎧p1 p2 pt i = 0 ⎪ [ S S ] di p1 ⋯ pt = ⎪ p1 ⋯ pi 1 pipi 1 pi 2 ⋯ pt 0 < i < t [ S S ] ⎨[ S S − S + S + S S ] ⎪ 1 ⋯ 1 ⎪ p pt− i = t ⎩⎪[ S S ] P P Therefore ∂2 p1 p2 = p1 p2 − p1p2 + p1 and ∂1 p1 = p1 − 1 . The augmentation ([ S ]) [ ] [ ] [ ] ([ ]) ( )[ ] map εP ∶ P0 → Z defined by εP = 1 implies ([ ]) P P P ∂3 ∂2 ∂1 εP P ∶ ⋯ P2 P1 P0 Z 0 is a free resolution of Z over ZG modules. Now, let Dt be the subcomplex of Pt generated by elements p1 ... pt such that pi = 1 for some i. Then P¯t = Pt Dn with the maps of the chain complex P¯[ inducedS S from] P defines a free resolution of Z over ZG~ modules, called the normalized standard resolution. Furthermore, two projective resolutions of Z over ZG are homotopy equivalent. For a proof of this fact the interested reader is referred to chapter one of Brown’s text [4]. For a group G, changing the projective resolution of Z over ZG is equivalent to changing the choice of representative for the corresponding cohomology class.

5 Equivalent extensions Let K and Q be groups in which Q acts on K by the outer automor- phism θ ∶ Q → Aut K . An extension of Q by K giving rise to θ is a short exact sequence: ( ) 1 → K → E → Q → 1 where conjugation in E agrees with θ. Two group extensions E1 and E2 are equivalent if there exists an isomorphism ϕ ∶ E1 → E2 such that the diagram:

E2

1 KQϕ 1

E2 commutes. Let E Q; K denote the set of equivalence classes of these extensions. If K is abelian, ( ) 2 there is a one-to-one correspondence between E Q,K and Hθ Q,K (See Brown [4]). Suppose K is an abelian group and Q acts( on K) by the( action) of θ as in the preceding paragraph. Consider the extension:

0 KEQι π 1

A section, s, is a function s ∶ Q → E such that π ○ s = idQ; furthermore s is normalized if s 1Q = 1E. Since s is a set theoretical function, s p1 s p2 is not necessarily equivalent to ( ) ( ) ( ) s p1p2 in E. Therefore we can measure the failure of s to be a homomorphism by a cocycle ( ) 1 κ ∈ Hom P2,K such that s p1 s p2 = ι κ p1 p2 s p1p2 − . Hence we have a formula: ( ) ( ) ( ) ( ([ S ])) ( ) 1 κ p1 p2 = s p1 s p2 s p1p2 − ([ S ]) ( ) ( ) ( ) which determines a cocycle corresponding the to E as an extension of Q by K. The following theorem by Eilenberg and MacLane provides the classification of extensions by group cohomology which depends on the action of Q on K [6]. Theorem 2.4 (Eilenberg MacLane 1947). Suppose Q and K are groups such that Q acts on K. There exists a bijection between E Q,K and H2 Q; K . ( ) ( ) The choice of section determines the representative of the cohomology class in H2 Q; K . θ ( ) Furthermore, if the extension is split then E ≈ K ⋊θ Q corresponds to the cohomology class represented by the trivial cocycle.

2.3 Braid groups and symmetric groups Symmetric Group The standard presentation for the symmetric group generated by adjacent transpositions is:

2 σi = 1, σi, σj = 1 if i − j > 1 Sn = σ1,...,σn−1 [ ] S S (3) d W σiσi+1σi = σi+1σiσi+1 i

Let σi,j = represent the transposition which permutes i and j. As an element of (3) we take the convention: ⋯ −1 ⋯ −1 σi,j = σi σj−2σj−1σj−2 σi

6 where i < j. Throughout this paper, we will use the following presentation for the symmetric group: 2 σ = 1, σ , σ = 1 if i, j ∩ k,ℓ = 1 R i,j i,j k,ℓ Sn = σi,j 1≤i

Braid groups Let x1,...,xn be marked points in C. Elements of Bn can be represented by a collection of n non-colliding paths fi ∶ 0, 1 → C × 0, 1 such that fi 0 ,fi 1 ∈ x1,...,xn and [ ] [ ] ( ) ( ) { } fi t ∈ C× t . Then the collection fi determines a permutation on 1,...,n . Throughout this ( ) { } { } { } paper we will represent elements of Bn by strand diagrams representing the paths f1 t ,...,fn t . ( ) ( ) A positively oriented twist between strands i and i + 1, denoted by bi, corresponds to a clockwise twist in the strand diagram, where the i + 1 st strand passes over the ith strand. We define the half twist between the( i and) j strands by:

1 1 ⋯ 2 1 − ⋯ − bi,j = bi bj− bj− bj−2 bi

Under the strand diagrams, this is equivalent to pulling all strands between the ith and jth strands over the ith strand, half twist the ith and jth strands by pulling the jth strand over the th th i strand, then pull all strands between i and j over the j strand. This choice of bi,j yields the following presentation for Bn which is equivalent to the Birman-Ko-Lee presentation [2]:

bi,j ,bk,ℓ = 1 if j − k j − ℓ i − j i − ℓ > 0 R [ ] ( )( )( )( ) R −1 B ≈ b R bi,jbj,kb = bi,k if i < j < k, k < i < j, j < k < i (5) n i,j R i,j d{ }R 1 i R b− b b = b if j < i < k, i < k < j, k < j < i R i,j j,k i,j i,k R

Pure braids As a subgroup of the braid group elements of the pure braid group, PBn, are represented by strand diagrams where every strand begins and ends at the same point of C. In 2 terms of (5), PBn is generated by all bi,j . For any pure braid, we can define the winding number of the ith and jth strand to be the number of positively oriented full twists between them.

Level m braid group Let Zm denote Z mZ and Bn be the braid group. Evaluating the unreduced Burau representation (See Birman~ [1]), at t = −1, and reducing mod m (for any m ≥ 0) yields a map Bn → GLn Zm : ( ) ρ 1 −1 t=− Bn GLn Z t,t GLn Z GLn Zm ( [ ]) ( ) ( )

−1 The map ρ ∶ Bn → GLn Z t,t is the unreduced Burau representation of the braid group defined on the generators( of[ (2): ])

1 − t t ρ bi ↦ Ii 1 ⊕ ⊕ In i 1 ( ) − Œ 1 0‘ − −

A topological description of the Burau representation can be found in Brendle and Margalit’s paper [3]. For m ≥ 0, the level m braid group Bn m is defined as the kernel of this composition, [ ] 2 which is a finite index subgroup of Bn. Brendle and Margalit proved Bn 4 = PB , the subgroup [ ] n of PBn generated by squares of all elements [3].

7 Formalization of main result: Define Zn = Bn Bn 4 and PZn be the image of PBn in Zn. ~ [ ] The standard surjection of Bn onto the symmetric group Sn yields the following non split group extension [8] 1 →PZn →Zn → Sn → 1 n ‰2Ž Kordek and Margalit also proved PZn ≈ Z2 [8]. Therefore there exists a nontrivial cohomology n n 2 ‰2Ž ‰ Ž class κ ∈ H Sn; Z2 corresponding to this extension. Furthermore Z 2 ≈ H1 PBn; Z is the [ ] θ ( ) ( ) abelianization of PBn [7], therefore we can consider the group extension:

n ‰2Ž 0 → H1 PBn; Z → Gn → Sn → 1 ( ) where Gn is the quotient of Bn by the commutator subgroup of PBn. Thus there exists a 2 n Z‰2Ž cohomology class φ ∈ Hθ Sn; such that φ composed with the mod 2 reduction of integers n [ ] ( ) 2 ‰2Ž determines a representative for κ ∈ H Sn; Z2 corresponding to Zn. [ ] θ ( )

3 The universal cover of a K(Sn, 1)-space

In this section we construct a low dimensional approximation for a resolution of Z over ZSn- modules for the universal cover of a K Sn, 1 -space. We approximate the cellular chain complex in dimensions 0,1 and 2 for the universal( cover,) X˜, of the 2-skeleton for the Cayley complex of Sn. Let R be the chain complex of X˜ and recall P is the normalized bar resolution. Since each dimension of R is a free Sn module, for each i there exists γi ∶ Ri → Pi which commutes with the diagram: R R R ∂3 ∂2 ∂1 εR ⋯ R2 R1 R0 Z 0

γ2 γ1 γ0 id P P P ∂3 ∂2 ∂1 εP ⋯ P¯2 P¯1 P¯0 Z 0

The existence of γi yields the following theorem:

Theorem 3.1. Let 0 → K → E → Sn → 1 be a group extension which corresponds to κ ∈ 2 ′ [ ] Hθ Sn; K under the normalized standard resolution. There exists κ ∈ Hom R2,K such that ′ ( ) 2 ( ) κ = κ in H Sn; K defined by: [ ] [ ] ( ) ′ κ = κ ○ γ2

Cayley Complex To describe the universal cover of a K Sn, 1 -space, we use the group pre- sentation of the symmetric group generated by all possible( transp) ositions given by (4). Let X be the K Sn, 1 -space constructed from building the rose. As the 2-skeleton of a CW-complex ( ) X has one 0-cell, a 1-cell for each generator of Sn, and a 2-cell for each relation in Sn. Let x0 denote the 0-cell of X and xi,j denote each 1-cell of X. Note that we only use positive generators 2 to label 1-cells. Let ci,j be the two cell glued by the relation σi,j = 1 and di,j,k,ℓ be the two cell glued by the relation σi,j , σk,ℓ = 1 if i, j ∩ k,ℓ = ∅. Since we can rewrite the relation −1 [ −1 −1 ] { } { } σi,j σj,kσi,j = σi,k as σi,j σj,kσi,j σi,k = 1. Let ei,j,k be the two cell that glues along this relation in R X. It is clear that ∂2 ci,j = σi,j − 1 xi,j and by Figure 1 we have: ( ) ( ) R ∂2 di,j,k,ℓ = xi,j + σi,j xk,ℓ − σk,ℓxi,j − xk,ℓ ( ) R ∂2 ei,j,k = xi,j + σi,j xj,k − σi,kxi,j − xi,k ( )

8 Figure 1: Gluing of Cayley Complex

xi,j xi,j σk,ℓx0 σi,j σk,ℓx0 σi,kx0 σi,j σj,kx0

xk,ℓ di,j,k,ℓ xk,ℓ xi,k ei,j,k xj,k

x0 σi,j x0 x0 σi,j x0 xi,j xi,j

Every 0-cell of the Cayley Complex is of the form g ⋅ x0 for some g ∈ Sn and there exists a 1-cell joining g ⋅x0 to x0 for every g ∈ Sn. Therefore R0 is clearly generated by x0 as a ZSn module and by considering xi,j as the path from x0 to σi,j x0, the collection of all xi,j generate the 1-cells of R1 as a ZSn module. Now, every two cell in X˜ is glued in by ci,j , di,j,k,ℓ, or ei,k,j starting at the vertex gx0 for some g ∈ Sn. Therefore any 2-cell in X˜ can be considered as g ⋅ ci,j , g ⋅ di,j,k,ℓ, or g ⋅ ei,k,j for some g ∈ Sn and thus as a free Sn module R2 is generated by ci,j , di,j,k,ℓ, or ei,k,j . Therefore we have the following lemma.

Lemma 3.2. As a ZSn modules, R1 is generated by xi,j and R0 by x0. Furthermore R2 is { } generated by ci,j ∪ di,j,k,ℓ ∪ ei,j,k where i < j and k < ℓ. { } { } { } By Lemma 3.2 defining γ0 and γ1 by γ0 x0 = and γ2 xi,j = σi,j obvious commute with the chain complexes. ( ) [ ] ( ) [ ]

Theorem 3.3. Define γ2 ∶ R2 → P¯2 by:

γ2 ci,j = σi,j σi,j ( ) [ S ] γ2 di,j,k,ℓ = σi,j σk,ℓ − σk,ℓ σi,j ( ) [ S ] [ S ] γ2 ei,j,k = σi,j σj,k − σi,k σi,j ( ) [ S ] [ S ] Then γ2 commutes with the chain complexes.

Proof. γ2 is clearly a ZSn module homomorphism defined on generators. Therefore it suffices to R P show γ1 ○ ∂2 = ∂2 ○ γ2. Recall the generators of R1 are xi,j and γ1 xi,j = σi,j . By following the gluing maps in Figure 1 we have: ( ) [ ]

R γ1 ○ ∂2 ci,j = σi,j + σi,j σi,j ( ) [ ] [ ] R γ1 ○ ∂2 di,j,k,ℓ = σi,j + σi,j σk,ℓ − σk,ℓ σi,j − σk,ℓ ( ) [ ] [ ] [ ] [ ] R γ2 ○ ∂2 ei,k,j = σi,j + σi,j σj,k − σi,k σi,j − σi,k ( ) [ ] [ ] [ ] [ ]

Recall we are using the normalized bar resolution, therefore p1 ⋯ pt = 0 if pi is trivial for any R [ S S ] P i. Now, since ∂2 p1 p2 = p1 p2 − p1 p2 + p1 for any p1,p2 ∈ Sn and ∂2 is a ZSn module homomorphism, it([ is clear:S ]) [ ] [ S ] [ ]

P 2 ∂2 σi,j σi,j = σi,j σi,j − σ + σi,j + σi,j − σi,j + 1 ([ S ]) [ ] [ i,j ] [ ] [ ] [ ] [ ] = σi,j σi,j − 1 + σi,j [ ] [ ] [ ] = σi,j + σi,j σi,j [ ] [ ]

9 P R P Thus ∂2 ○γ2 ci,j = γ1 ○∂2 ci,j . To compute ∂2 di,j,k,ℓ , recall that i, j ∩ k,ℓ = ∅, therefore ( ) ( ) ( ) { } { } σi,j σk,ℓ = σk,ℓσi,j : P ∂2 σi,j σk,ℓ − σk,ℓ σi,j = σi,j σk,ℓ − σi,j σk,ℓ + σi,j − σk,ℓ σi,j + σk,ℓσi,j − σk,ℓ ([ S ] [ S ]) [ ] [ ] [ ] [ ] [ ] [ ] = σi,j σk,ℓ − σi,j σk,ℓ + σi,j − σk,ℓ σi,j + σi,j σk,ℓ − σk,ℓ [ ] [ ] [ ] [ ] [ ] [ ] = σi,j σk,ℓ + σi,j − σk,ℓ σi,j − σk,ℓ [ ] [ ] [ ] [ ]

P −1 Finally, to compute ∂2 ei,k,j recall that in Sn, σi,k = σi,j σj,kσi,j and therefore σi,kσi,j = σi,j σj,k. Hence we have: ( ) P ∂2 σi,j σj,k − σi,k σi,j = σi,j σj,k − σi,j σj,k + σi,j − σi,k σi,j + σi,kσi,j − σi,k ([ S ] [ S ]) [ ] [ ] [ ] [ ] [ ] [ ] = σi,j σj,k − σi,kσj,k + σi,j − σi,k σi,j + σi,j σj,k − σi,k [ ] [ ] [ ] [ ] [ ] [ ] = σi,j σj,k + σi,j − σi,k σi,j − σi,k [ ] [ ] [ ] [ ]

Therefore, by Theorems 2.1,2.2 and 2.4 we have a definition for κ . [ ] Theorem 3.4. Let E be the extension of Sn by K corresponding to the cohomology class 2 ′ H Sn; K represented by κ ∈ Hom P2,K . Define κ ∈ Hom R2,K by: θ ( ) ( ) ( ) ′ κ ci,j = s σi,j σi,j ( ) ( )( ) ′ −1 −1 κ di,j,k,ℓ = s σi,j s σk,ℓ s σi,j σk,ℓ − s σk,ℓ s σi,j s σk,ℓσi,j ( ) ( ) ( ) ( ) ( ) ( ) ( ) ′ −1 −1 κ ei,k,j = s σi,j s σj,k s σi,j σj,k − s σi,k s σi,j s σi,kσi,j ( ) ( ) ( ) ( ) ( ) ( ) ( ) Then κ = κ′ . [ ] [ ] Proof. By Theorem 2.1:

κ σi,j σi,j = s σi,j s σi,j ([ S ]) ( ) ( ) −1 −1 κ σi,j σk,ℓ − σk,ℓ σi,j = s σi,j s σk,ℓ s σi,j σk,ℓ − s σk,ℓ s σi,j s σk,ℓσi,j ([ S ] [ S ]) ( ) ( ) ( ) ( ) ( ) ( ) −1 −1 κ σi,j σk,ℓ − σi,k σi,j = s σi,j s σj,k s σi,j σj,k − s σi,k s σi,j s σi,kσi,j ([ S ] [ S ]) ( ) ( ) ( ) ( ) ( ) ( ) Therefore κ′ = κ ○ γ2 and by Section 2.2, κ′ = κ . [ ] [ ] n 4 Cohomology class of Sn with coefficients in Z‰2Ž

n ‰ Ž In this section we will build group presentations for the extension of Sn by Z 2 required to n evaluate the function φ ∶ R2 → Z‰2Ž described in Section 2.2.

n ‰ Ž Extension of Sn by Z 2 Let Rn be the commutator subgroup of PBn, since H1 PBn; Z ≈ n n ‰ Ž ‰ Ž ( ) Z 2 is the abelianization of PBn, we have Z 2 ≈ H1 PBn; Z ≈ PBn Rn. Furthermore Rn is a ( ) ~ characteristic subgroup of PBn and PBn is normal in Bn, therefore Rn is normal in Bn. By the third isomorphism theorem,

Bn Rn PBn Rn ≈ Bn PBn ≈ Sn ( ~ )~( ~ ) ~ Then the group Gn defined as the extension of Sn by H1 PBn; Z is isomorphic to Bn Rn. Therefore we have the group extension: ( ) ~

n ‰ Ž ι1 π1 0 Z 2 Gn Sn 1 where the action of Sn is induced by the conjugation of Bn on PBn.

10 Table 1: Relations of Gn

R1: gi,j ,gk,ℓ = 1 for all i,j,k,ℓ [ 2 ] R2:σ ˜i,j = gi,j for all i −1 R3:σ ˜i,j σ˜k,j σ˜i,j = σ˜i,k if k < i < j; i < j < k; or j < k < i

−1 R4:σ ˜i,j σ˜j,kσ˜i,j = σ˜i,k if i < k < j; j < i < k; or k < j < i −1 −1 i < k < j < ℓ ⎧gi,kgi,ℓgj,kgj,ℓ ⎪ −1 −1 R5: σ˜i,j , σ˜k,ℓ = ⎪g gk,j gi,ℓg k < i < ℓ < j [ ] ⎨ k,i ℓ,j ⎪1 otherwise ⎪ −1⎩ R6:σ ˜i,j gk,ℓσ˜i,j = gσi,j (k),σi,j (ℓ)

Normalized section Since Gn is a quotient of Bn, letσ ˜i,j represent the projection of the th th positively oriented half twist between the i and j strands, bi,j , in Gn. In particular, we need an algorithm which takes an element of Sn, and produces a normal form in terms of the generators σi,j . mp mp 1 ⋅ 1 Let p ∈ Sn, then there exists a product of 2-cycles in ∏i=1 σn+ −i,kn+1−i such that p ∏i=1 σn+ −i,kn+1−i = 1. Note this is taking the permutation and applying transpositions first fixing n, then n−1, contin- mp ∶ uing until the permutation is the identity on 1,...,n . Therefore p = ∏i=1 σi,ki . Let s Sn → Gn be the normalized section defined by: { }

mp

s p = M σ˜i,ki ( ) i=1

4.1 Presentation of Gn n ‰ Ž Generators of Gn Let gi,j 1 ≤ i < j ≤ n be the commuting generators of Z 2 . Since n ‰ Ž ( ) Z 2 ≈ PBn R, each gi,j represents a pure braid. In particular gi,j is the positively oriented full ~ th th −1 twist of the i and j strands while gi,j represents the negatively oriented full twist between th th −1 the i and j strands. Notice that gi,j = gj,i. By the choice of section, each σi,j (σi,j ) in the th generating set of Sn lifts to the positively (negatively) oriented half twist between the i and th −1 j strands, denotedσ ˜i,j (˜σi,j ). Thus by section 2.1 generating set for Gn is:

g1,2,...,gi,j ,...,gn 1,n, σ˜1,2,..., σ˜i,j ,..., σ˜n 1,n { − − } where i, j ∈ 1,...,n with i < j. { }

4.1.1 Relations of Gn

A full list of all the relations in Gn are given in Table 1. We will prove the relations of Gn in n ‰ Ž this section. Since Z 2 embeds into Gn, the relation gi,j ,gk,ℓ is preserved in Gn. Therefore n [ ] ‰2Ž R1 is obvious. We begin with a theorem describing how to determine elements of Z in Gn.

n Theorem 4.1. Let k ∈ ι1 Z‰2Ž . Then k is determined by the winding numbers of a strand diagram. ( )

11 −1 −1 Figure 2:σ ˜i,j σ˜j,kσ˜i,j σ˜i,k

k i j i j k j k i

σ˜i,j σ˜i,j

σ˜j,k σ˜j,i

−1 σ˜k,j σ˜i,j

σ˜j,k

−1 −1 σ˜i,j σ˜i,k

−1 −1 σ˜k,i σ˜j,i

k < i < j i < j < k

−1 σ˜k,i

j < k < i

n Proof. Let k ∈ ι1 Z‰2Ž , then k is a pure braid and the winding number between any two strands ( ) is well defined. Furthermore since gi,j ,gk,ℓ = 1, k can be written uniquely as [ ] n−1 n ǫi,j k = M M gi,j i=1 j=i+1 where ǫi,j ∈ Z is the number of positively oriented full twists between the i and j strands. Thus th th ǫi,j is the winding number for the i and j strands in the strand diagram. th th Sinceσ ˜i,j is the positively oriented half twist between the i and j strands, R2 is obvious as well. We begin by proving relations R3 and R4. Lemma 4.2. If k < i < j, i < j < k, or j < k < i, then

−1 −1 σ˜i,j σ˜j,kσ˜i,j σi,k = 1

−1 −1 −1 Proof. Consider the relation σi,j σj,kσi,j = σi,k in Sn. We can rewrite this relation as σi,j σj,kσi,j σi,k = n 1 1 Z‰2Ž − − 1 in Sn. Since ker π1 = ι1 we haveσ ˜i,j σ˜j,kσ˜i,j σ˜i,k is a pure braid. By Figure 2, if k < i < j, notice that strand j passes( over) both i and k at every crossing while strand i passes under k at −1 −1 every crossing, therefore every winding number is zero. Thusσ ˜i,j σ˜j,kσ˜i,j σ˜i,k = 1 if k < i < j. If i < j < k, notice strand i passes under every strand while strand j passes over every strand. −1 −1 Therefore both strands i and j must have winding number zero. Henceσ ˜i,j σ˜j,kσ˜i,j σ˜i,k = 1 for i < j < k

12 −1 −1 Figure 3:σ ˜i,j σ˜j,kσ˜i,j σ˜i,k

i k j j i k k j i

σ˜j,i σ˜j,i

σ˜i,j σ˜k,j

−1 σ˜j,k σ˜j,i

σ˜k,j

−1 −1 σ˜j,i σ˜k,i

−1 −1 σ˜i,j σ˜i,k

j < i < k k < j < i

−1 σ˜i,k

i < k < j

Now, suppose j < k < i. Strand j passes under every other strand, therefore the only possible nonzero winding number could be associated to gi,k. Strand i passes over strand k twice, then −1 −1 under twice, which is equivalent to the trivial braid and thereforeσ ˜i,j σ˜j,kσ˜i,j σ˜i,k = 1. Lemma 4.3. If i < k < j, j < i < k, or k < j < i, then

−1 −1 −1 σ˜i,j σ˜j,kσ˜i,j σ˜i,k = gi,j gk,j

−1 −1 −1 −1 Proof. By rewriting the relations as σi,j σk,ℓσi,j σi,k = 1 in Sn, we getσ ˜i,j σ˜k,ℓσ˜i,j σ˜i,k is in ker π. Suppose i < k < j, notice that by Figure 3 strands i and k have both a clockwise and counter- clockwise twist, therefore the winding number of gi,k is zero. Furthermore, strand j passes over strand i then under, so gi,j has a winding number of 1. Also, strand j passes under strand k then over it, thus gk,j has a winding number of −1. If j < i < k, by Figure 3, gj,i clearly has a winding number of 1. Now k passes over i twice and gi,k has a winding number of zero while k passes under j then over j, so gj,k has a winding number of −1. Suppose k < j < i, by Figure 3 strand k passes under i, over j, then under both i and j. Therefore gk,i has a winding number of zero while gk,j has a winding number of −1. Now, i passes over then under j, so gj,i has winding number 1 since i > j.

Relation 3 Notice that relation R3 is proven by the cases of Lemma 4.2. To prove relation −1 R4, we need to show Theorem 4.3 impliesσ ˜i,j σ˜k,ℓσ˜i,j = σ˜i,k. To prove this we need to prove the case of R5 when i, j ∩ k,ℓ = 1. S{ } { }S

13 −1 Figure 4:σ ˜i,j gj,kσ˜i,j

i j k i k j k i j

σ˜i,j σ˜i,j

σ˜i,j

gj,k

gk,j −1 σ˜i,j gk,j i < j < k

−1 σ˜i,j

σ˜i,j−1 k < i < j

i < k < j

Theorem 4.4. Suppose i, j ∩ k,ℓ = 1, then: S{ } { }S −1 σ˜i,j gk,ℓσ˜i,j = gα,β where α ∈ i, j ∖ k,ℓ and β ∈ k,ℓ ∖ i, j . { } { } { } { } n 1 n Z‰2Ž − Z‰2Ž ∩ Proof. Since is normal in Gn,σ ˜i,j gj,kσ˜i,j ∈ . First, assume i, j k,ℓ = j . Without loss of generality assume j = ℓ, then there are three cases: i < j < k,{i < k} < {j and} k{< }i < j. By Figure 4 it is clear that for both cases i < j < k and k < i < j that j has zero winding number with both i and k. Furthermore, in both cases there is a full clockwise twist between th th the i and j strands, so the resulting twist for both cases are gi,k = gk,i. If i < k < j, notice that the jth strand passes under the kth strand twice and over the ith strand twice. Therefore there is no twist between the jth strand and either of the other two. Furthermore, the kth strand passes over the ith strand, then under it, and over strand i twice −1 more. Therefore there is one full clockwise twist between strands i and k. Thusσ ˜i,j σ˜j,kσ˜i,j = gi,k. Now, suppose i, j ∩ k,ℓ = i . Assume i = ℓ, then there are the three cases as shown in Figure 5. In both{ cases} i <{ k <}j and{ } k < i < j the ith strand only passes under all other strands, therefore the winding number of any twist involving the ith strand is zero. If k < i < j then Figure 5 clearly shows a clockwise full twist between strands k and j. If i < k < j, notice the jth strand passes under the kth strand twice, then over it, then under again. Therefore strands k and j have the equivalent of a full clockwise twist. If i < j < k, Figure 5 shows the ith and jth strands have both a full clockwise and coun- terclockwise twist; therefore the winding number for gi,j is zero. Furthermore, Figure 5 clearly

14 −1 Figure 5:σ ˜i,j gi,kσ˜i,j

i j k i k j k i j

σ˜i,j σ˜i,j

σ˜i,j gk,i

gi,k −1 σ˜i,j gi,k k < i < j

−1 σ˜i,j

−1 i < j < k σ˜i,j

i < k < j

shows strands k and j have a positive full twist while strands i and k have a winding number of zero.

−1 −1 −1 Theorem 4.5. Supposeσ ˜i,j σ˜j,kσ˜i,j σ˜i,k = gi,jgk,j . Then

−1 σ˜i,j σ˜k,ℓσ˜i,j = σ˜i,k

−1 −1 −1 Proof. Supposeσ ˜i,j σ˜j,kσ˜i,j σ˜i,k = gi,j gk,j . By multiplying on the right byσ ˜i,k and on the right −1 by gi,j , by relation R1 we have

−1 −1 −1 σ˜i,j σ˜j,kσ˜i,j σ˜i,k = gi,j gk,j −1 −1 −1 gi,j σ˜i,j σ˜j,kσ˜i,j = gk,j σi,k

By rewriting Theorem 4.4 and R2 we have

−1 −1 −1 gi,j σ˜i,j σ˜j,kσ˜i,j = gk,j σi,k 1 −2 ˜ ˜ − −1 σ˜i,j σ˜i,j k, ji, j = σ˜i,kgi,j −1 −1 σ˜i,j σ˜k,j σ˜i,j gi,j = σ˜i,k

2 Since gi,j = σ˜i,j the theorem and R4 is proven.

15 Theorem 4.6. Suppose i, j ∩ k,ℓ ≠ 1. Then S{ } { }S −1 −1 ⎧gi,kgi,ℓgj,kgj,ℓ i < k < j < ℓ ⎪ −1 −1 σ˜i,j , σ˜k,ℓ = ⎪g gk,j gi,ℓg k < i < ℓ < j [ ] ⎨ k,i ℓ,j ⎪1 otherwise ⎩⎪ Proof. Suppose i, j ∩ k,ℓ = ∅. By assumption i < j and k < ℓ, furthermore if j < k or ℓ < i then { } { } −1 it is clear σ˜i,j , σ˜k,ℓ = 1. Since σ˜i,j , σ˜k,ℓ = σ˜k,ℓ, σ˜i,j , it suffices to show the cases i < k < j < ℓ [ ] [ ] [ ] and i < k < ℓ < j. If i < k < ℓ < j, then i−k i−ℓ j −k j −ℓ > 0. Therefore bi,j ,bk,ℓ = 1 in (5). ( )( )( )( ) [ ] Since Gn is a quotient of Bn, σ˜i,j , σ˜k,ℓ = 1. Therefore it remains to show the case i < k < j < ℓ. Now, suppose i < k < j <[ℓ. By Figure] 6 the ith strand passes under the kth strand then th over, therefore gi,k has a winding number of 1. Furthermore strand i passes under the j strand twice; also the ith strand passes over the ℓth strand once then under it and therefore has a counterclockwise full twist. Therefore gi,j has a winding number of zero and gi,ℓ has a winding th number −1. Now, the j strand passes under the strand k then over it, therefore gk,j has a winding number of −1. Strand j passes under strand ℓ then under it and thus gj,ℓ has a winding number of 1. Finally strand ℓ passes over strand k twice, therefore gk,ℓ has a winding number of zero. Notice that by Figure 6 the crossings in the braid diagram for k < i < ℓ < j have the reverse orientation to those in i < k < j < ℓ. Thus we have

−1 −1 ⎧gi,kgi,ℓgk,j gj,ℓ i < k < j < ℓ ⎪ −1 −1 σ˜i,j , σ˜k,ℓ = ⎪g gi,ℓgk,j g k < i < ℓ < j [ ] ⎨ i,k j,ℓ ⎪ ⎪1 otherwise ⎩⎪

To finish the proof this presentation for Gn is correct, it remains to show the second part of R5. Theorem 4.7. If i, j ∩ k,ℓ ≠ 1, then S{ } { }S −1 σ˜i,j gk,ℓσ˜i,j = gk,ℓ

Proof. First suppose i, j ∩ k,ℓ = i, j . Then by relation R3 we have { } { } { } −1 2 −1 σ˜i,j gk,ℓσ˜i,j = σ˜i,j σ˜i,j σ˜i,j 2 = σ˜i,j

= gi,j

Now, suppose i, j ∩ k,ℓ = ∅. If i < k < j < ℓ, then by Figure 7, strands i and k have one full { } { } clockwise twist followed by a full counterclockwise twist, therefore the winding number for gi,k is zero. Also, the ℓth strand has a clockwise twist with the kth strand and does not interact with either of the other strands. Therefore giℓ and gj,ℓ have winding numbers of zero while gk,ℓ has a winding number of 1. Furthermore, strand j passes under strand k twice, so gk,j has winding number zero as well. Thus, if i < k < j < ℓ then the theorem holds. Now, suppose k < i < ℓ < j. By Figure 7 the ith strand always passes under the other strands, th therefore gk,i, gi,ℓ, and gi,j all have winding number zero. Furthermore, the j strand passes under the strand ℓ twice and over strand k twice, therefore gk,j and gℓ,j both have winding

16 Figure 6: σ˜i,j , σ˜k,ℓ [ ] i k j ℓ k i ℓ j

σ˜i,j σ˜i,j

σ˜k,ℓ σ˜k,ℓ

−1 −1 σ˜i,j σ˜i,j

−1 −1 σ˜k,ℓ σ˜k,ℓ

i < k < j < ℓ k < i < ℓ < j

17 −1 Figure 7:σ ˜i,j gk,ℓσ˜i,j

i k j ℓ k i ℓ j

σ˜i,j σ˜i,j

gk,ℓ gk,ℓ

−1 −1 σ˜i,j σ˜i,j

18 number zero. Finally, Figure 7 shows one full clockwise twist between the kth and ℓth strands. −1 Thereforeσ ˜i,j gk,ℓσ˜i,j = gk,ℓ if k < i < ℓ < j. Otherwise, by R3 and R5,

−1 2 −1 σ˜i,j gk,ℓσ˜i,j = σ˜i,j σ˜k,ℓσ˜i,j −1 −1 = σ˜i,j σ˜k,ℓσ˜i,j σ˜i,j σ˜k,ℓσ˜i,j −1 −1 −1 = σ˜i,j σ˜k,ℓσ˜i,j σ˜k,ℓσ˜k,ℓσ˜i,j σ˜k,ℓσ˜i,j −1 −1 −1 −1 = σ˜i,j σ˜k,ℓσ˜i,j σ˜k,ℓσ˜k,ℓσ˜i,j σ˜k,ℓσ˜i,j σ˜k,ℓσ˜k,ℓ

= 1 ⋅ σ˜k,ℓ ⋅ 1 ⋅ σ˜k,ℓ 2 = σ˜k,ℓ

= gk,ℓ

2 n 4.2 Cohomology class in H (Sn; Z‰2Ž)

Given the group presentation for Gn with relations in Table 1, by Theorem 3.4 we can determine n ‰ Ž the cohomology class representing the group extension of Sn by Z 2 . Recall the normalized section chosen for Sn at the beginning of Section 4 changes a permutation to a product of 2- cycles which permutes integers in ascending order. This normalized section s ∶ Sn → Gn applied to generators implies s σi,j = σ˜i,j , therefore relations in Table 1 determine the image of a ( ) representative for the corresponding cocylce by Theorem 3.4. Since we lift any element of Sn by the normal form, by Theorem 3.4 we first need to determine the normal forms for products of two transpositions. Lemma 4.8. Suppose i, j ∩ k,ℓ = ∅, then { } { }

⎧σ˜i,j σ˜k,ℓ j < ℓ s σk,ℓσi,j = s σi,j σk,ℓ = ⎪ ( ) ( ) ⎨σ˜ σ˜ ℓ < j ⎪ k,ℓ i,j ⎩⎪ Proof. Both σk,ℓσi,j and σi,j σk,ℓ have the same normal form by Lemma ??. If ℓ < j then we take σi,j σk,ℓσi,j σk,ℓ = 1. Therefore σi,j σk,ℓ = σk,ℓσi,j and s σi,j σk,ℓ = σ˜k,ℓσ˜i,j . Now, suppose ( ) j < k. Then we take the transposition fixing k first, yielding σi,j σk,ℓσk,ℓσi,j and thus s σi,j σk,ℓ = ( ) σ˜k,ℓσ˜i,j . Lemma 4.9. The normal form for products of transpositions with intersection are:

⎧σ˜i,kσ˜i,j max i,k,j = j ⎪ { } s σi,kσi,j = s σi,j σj,k = ⎪σ˜i,j σ˜j,k max i,k,j = k ( ) ( ) ⎨ { } ⎪ ⎪σ˜k,j σ˜k,i max i,k,j = i ⎩⎪ { } The proof of Lemma 4.9 is a straightforward computation similar to the proof of Lemma 4.8 Now that we have the image of the normal forms in the definition of the cohomology class n ‰ Ž describing this extension of Sn by Z 2 . Theorem 4.10. Let n ‰ Ž ι1 π1 0 Z 2 Gn Sn 1

19 n ‰ Ž be the group extension where the action of Sn on Z 2 is determined by the conjugation of pure braids by half twists in Bn. The homomorphism:

φ ci,j = gi,j ( ) − − + ⎧gi,k gi,ℓ gk,j gj,ℓ i < k < j < ℓ ⎪ φ di,j,k,ℓ = ⎪−gi,k + gk,j + gi,ℓ − gℓ,j k < i < ℓ < j ( ) ⎨ ⎪ ⎪0 otherwise ⎩⎪ ⎧gi,j − gk,j i < k < j, j < i < k, k < j < i φ ei,k,j = ⎪ ( ) ⎨0 otherwise ⎪ ⎩⎪ n n ‰ Ž 2 ‰ Ž is a representative for this extension of Sn by Z 2 in H Sn; Z 2 . θ ( ) n Before we prove Theorem 4.10, note that the image of our cocycle is the abelian group Z‰2Ž n with additive notation. However the computations to determine the elements of Z‰2Ž in the image is done within the extension Gn with multiplicative notation. Proof. First, by Theorem 3.4 we have

φ ci,j = s σi,j s σi,j ( ) ( ) ( ) = σ˜i,j σ˜i,j

= gi,j

Now, consider di,j,k,ℓ, without loss of generality assume i < j and k < ℓ. By Lemma 4.8 s σk,ℓσi,j = s σi,j σk,ℓ depends on max j, ℓ . Suppose j < ℓ, then by Lemma 4.8, s σk,ℓσi,j = ( ) ( ) { } ( ) σ˜i,j σ˜k,ℓ. Therefore by Theorem 3.4 and relation R5

−1 −1 φ di,j,k,ℓ = σ˜i,j σ˜k,ℓ σ˜i,j σ˜k,ℓ − σ˜k,ℓσ˜i,j σ˜i,j σ˜k,ℓ ( ) ( ) ( ) −1 −1 = −σ˜k,ℓσ˜i,j σ˜k,ℓσ˜i,j −1 −1 ⎧gi,kgi,ℓgk,j gj,ℓ i < k < j < ℓ = − ⎪ ⎨0 otherwise ⎪ ⎩⎪ ⎧−gi,k + gi,ℓ + gk,j − gj,ℓ i < k < j < ℓ = − ⎪ ⎨0 otherwise ⎪ ⎩⎪ ⎧gi,k − gi,ℓ − gk,j + gj,ℓ i < k < j < ℓ = ⎪ ⎨0 otherwise ⎪ ⎩⎪ Thus for φ di,j,k,ℓ it remains to show the case ℓ < j. By Lemma 4.8, Theorem 3.4, and R5 ( ) −1 −1 φ di,j,k,ℓ = σ˜i,j σ˜k,ℓ σ˜k,ℓσ˜i,j − σ˜k,ℓσ˜i,j σ˜k,ℓσ˜i,j ( ) ( ) ( ) −1 −1 = σ˜i,j σ˜k,ℓσ˜i,j σ˜k,ℓ −1 −1 ⎧gk,igk,j gi,ℓgℓ,j k < i < ℓ < j = ⎪ ⎨0 otherwise ⎪ ⎩⎪ ⎧−gk,i + gk,j + gi,ℓ − gℓ,j k < i < ℓ < j = ⎪ ⎨0 otherwise ⎪ ⎩⎪

20 Thus it only remains to prove the image of ei,k,j . By Lemma 4.9 there are three cases for the normal form of the permutation σi,kσi,j = σi,j σj,k depending on max i,k,j . Suppose max i,k,j = j, then by Theorem 3.4 and Lemma 4.9 we have: { } { } −1 −1 φ ei,k,j = s σi,j s σj,k s σi,j σk,j − s σi,k s σi,j s σi,kσi,j ( ) ( ) ( ) ( ) ( ) ( ) ( ) −1 −1 = σ˜i,j σ˜k,j σ˜i,kσ˜i,j − σ˜i,kσ˜i,j σ˜i,kσ˜i,j ( ) ( ) −1 −1 = σ˜i,j σ˜k,j σ˜i,j σ˜i,k By relations R3 and R4 together with Theorem 4.4 we get:

−1 −1 φ ei,k,j = σ˜i,j σ˜k,j σ˜ σ˜ ( ) i,j i,k −1 ⎧σ˜i,kσ˜i,k k < i < j = ⎪ 1 1 1 ⎨g σ˜− σ˜ σ˜ g− σ˜− i < k < j ⎪ i,j i,j k,j i,j i,j i,k ⎩⎪ ⎧0 k < i < j = ⎪ 1 1 ⎨g σ˜ g− σ˜− i < k < j ⎪ i,j i,k i,j i,k ⎩⎪ ⎧0 k < i < j = ⎪ 1 ⎨g g− i < k < j ⎪ i,j k,j ⎩⎪ ⎧0 k < i < j = ⎪ ⎨g − g i < k < j ⎪ i,j k,j ⎩⎪ Hence if max i,k,j = j then we are done. { } Now, let max i,k,j = k, then by Lemma 4.9 s σi,kσi,j = σ˜i,j σ˜j,k. Therefore, as above we have { } ( )

−1 −1 φ ei,k,j = σ˜i,j σ˜j,k σ˜i,j σ˜j,k − σ˜i,kσ˜i,j σ˜i,j σ˜j,k ( ) ( ) ( ) −1 −1 = −σ˜i,kσ˜i,j σ˜j,kσ˜i,j

−1 −1 Now, if i < j < k then R3 applies toσ ˜i,j σ˜j,kσ˜i,j and if j < i < k then R4 applies. Thus

−1 ⎧σ˜i,kσ˜i,k i < j < k φ ei,k,j = − ⎪ 1 1 1 ( ) ⎨σ˜ g σ˜− σ˜− σ˜ g− j < i < k ⎪ i,k i,j i,j j,k i,j i,j ⎩⎪ ⎧0 i < j < k = − ⎪ 1 1 ⎨σ˜ g σ˜− g− j < i < k ⎪ i,k i,j i,k i,j ⎩⎪ ⎧0 i < j < k = − ⎪ 1 ⎨g− g j < i < k ⎪ i,j k,j ⎩⎪ ⎧0 i < j < k = ⎪ ⎨g − g j < i < k ⎪ i,j k,j ⎩⎪ It remains to prove the result if max i,k,j = i. By Lemma 4.9, if max i,k,j = i then { } { } s σi,j σi,j = σ˜k,j σ˜k,i. Thus ( ) −1 −1 φ ei,k,j = σ˜j,iσ˜j,k σ˜j,kσ˜k,i − σ˜k,iσ˜j,i σ˜j,kσ˜k,i ( ) ( ) ( ) −1 −1 − −1 −1 = σ˜j,iσ˜j,kσ˜k,iσ˜j,k σ˜k,iσ˜i,j σ˜k,iσ˜j,k

21 −1 −1 If j < k < i, then relation R3 applies toσ ˜j,kσ˜k,iσ˜j,k while R4 applies if k < j < i. Therefore,

−1 −1 −1 ⎧σ˜j,iσ˜j,i j < k < i σ˜j,iσ˜j,kσ˜ σ˜ = ⎪ 1 1 1 k,i j,k ⎨σ˜ g σ˜− σ˜− σ˜ g− k < j < i ⎪ j,i k,j k,j k,i k,j k,j ⎩⎪ ⎧0 j < k < i = ⎪ 1 1 ⎨σ˜ g σ˜− g− k < j < i ⎪ j,i k,j j,i k,j ⎩⎪ ⎧0 j < k < i = ⎪ 1 ⎨g g− k < j < i ⎪ k,i k,j ⎩⎪ −1 Furthermore, if j < k < i then R3 applies toσ ˜k,iσ˜j,iσ˜k,i while R4 applies if k < j < i. Hence

−1 −1 −1 ⎧σ˜j,kσ˜j,k j < k < i σ˜k,iσ˜i,j σ˜k,iσ˜j,k = ⎪ 1 1 1 ⎨g σ˜− σ˜ σ˜ g− σ˜− k < j < i ⎪ k,i k,i j,i k,i k,i k,j ⎩⎪ ⎧0 j < k < i = ⎪ 1 1 ⎨g σ˜ g− σ˜− k < j < i ⎪ k,i k,j k,i k,j ⎩⎪ ⎧0 j < k < i = ⎪ 1 ⎨g g− k < j < i ⎪ k,i j,i ⎩⎪ Thus we have

⎧0 − 0 j < k < i φ ei,k,j = ⎪ 1 1 ( ) ⎨g g− − g g− k < j < i ⎪ k,i k,j k,i j,i ⎩⎪ ⎧0 j < k < i = ⎪ ⎨gk,i − gk,j − gk,i − gj,i k < j < i ⎪ ( ) ⎩⎪ ⎧0 j < k < i = ⎪ ⎨g − g k < j < i ⎪ j,i k,j ⎩⎪

4.3 Order of [φ]

n t Z‰2Ž To determine the order of φ we build the extensions Gn of Sn by which correspond to ⋅ [ ] Z t the class of t φ and show that if t ∈ 2 then Gn is a split extension. To construct extensions corresponding[ to] multiples of φ we observe the relation between the lifting relations in the group presentation and the image[ ] of the cocycle. The formal process would be to construct new extensions from quotients of direct products Gn × Gn using a technique similar to Baer sums. Note that Baer sums do not generally result in new extension when the cokernel of an extension is nonabelian, however no obstructions arise when taking a Baer sum of two equivalent extensions and our construction yields the same results. n t Z‰2Ž Lemma 4.11. Let Gn be the group extension of Sn by corresponding to the cohomology n 2 ‰ Ž t class of t ⋅ φ ∈ H Sn, Z 2 . Then G is generated by σ˜i,j ∪ gk,ℓ with relations in Table 2. [ ] ( ) n { } { } The proof follows from the proof of Theorem 4.10 noting that multiplicative powers of the group elements in the image of the cohomology class becomes additive powers of the cohomology class.

22 t Table 2: Relations of Gn

t R 1: gi,j ,gk,ℓ = 1 for all i,j,k,ℓ [ ] t 2 t R 2:σ ˜i,j = gi,j for all i t −1 R 3:σ ˜i,j σ˜k,j σ˜i,j = σ˜i,k if k < i < j; i < j < k; or j < k < i

t −1 R 4:σ ˜i,j σ˜j,kσ˜i,j = σ˜i,k if i < k < j; j < i < k; or k < j < i t −t −t t i < k < j < ℓ ⎧gi,kgi,ℓgj,kgj,ℓ t ⎪ −t t t −t R 5: σ˜i,j , σ˜k,ℓ = ⎪g g g g k < i < ℓ < j [ ] ⎨ k,i k,j i,ℓ ℓ,j ⎪1 otherwise ⎩⎪ g i, j ∩ k,ℓ = 1 and j = ℓ t −1 ⎧ i,k R 6:σ ˜i,j gk,ℓσ˜ = ⎪ S{ } { }S i,j ⎨g otherwise ⎪ k,ℓ ⎩⎪

Theorem 4.12. If t ∈ 2Z, then the group extension:

n Z‰2Ž t 0 → → Gn → Sn → 1 is split.

Proof. First, note that the existence of a splitting is independent of group presentation for Sn. t Therefore it suffices to define a homomorphism ω ∶ Sn → Gn using the presentation (3) of Sn. Set: −t~2 ω σi = σ˜i,i 1g ( ) + First note that:

2 −t~2 −t~2 ω σi = σ˜i,i 1g 1σ˜i,i 1g 1 ( ) + i,i+ + i,i+ −1 2 = σ˜i,i+1gi,i+1σ˜i,i+1σ˜i,i+1gi,i+1 −t~2 t −t~2 = gi,i+1gi,i+1gi,i+1 = 1

Now, if i − j > 1, then by the relations in Table 2 it is clear: S S −1 −1 ω σi ω σj ω σi ω σj = 1 ( ) ( ) ( ) ( ) Thus it only remains to show the braid relation is preserved under ω. Applying relations Rt1,

23 Table 3: Relations of Zn R 2 1:g ¯i,j = 1 for all i, j

R2: g¯i,j , g¯k,ℓ = 1 for all i,j,k,ℓ [ ] R 2 3:σ ¯i,j = g¯i,j for all i R −1 4:σ ¯i,j σ¯k,j σ¯i,j = σ¯i,k if k < i < j; i < j < k; or j < k < i

R −1 5σ ¯i,j σ¯j,kσ¯i,j = σ¯i,k if i < k < j; j < i < k; or k < j < i

⎧g¯i,kg¯i,ℓg¯j,kg¯j,ℓ i < k < j < ℓ or k < i < ℓ < j R6 σ¯i,j , σ¯k,ℓ = ⎪ [ ] ⎨1 otherwise ⎪ ⎩⎪ if j = ℓ −1 ⎧g¯i,k R7:σ ¯i,j g¯k,ℓσ¯ = ⎪ i,j ⎨g¯ otherwise ⎪ k,ℓ ⎩⎪

Rt2, and Rt6 together with Rt3 on the image we get:

−t~2 −t~2 −t~2 t~2 −1 t~2 −1 t~2 −1 σ˜i,i+1gi,i+1σ˜i+1,i+2gi+1,i+2σ˜i,i+1gi,i+1gi+1,i+2σ˜i+1,i+2gi,i+1σ˜i,i+1gi+1,i+2σ˜i+1,i+2 −t~2 −1 −t~2 −1 2 −t~2 t~2 −1 t~2 −1 −1 −1 = gi,i+1σ˜i,i+1σ˜i+1,i+2σ˜i,i+1σ˜i,i+1gi+1,i+2σ˜i,i+1σ˜i,i+1gi,i+1gi+1,i+2σ˜i+1,i+2gi,i+1σ˜i,i+1σ˜i+1,i+2gi+1,i+2 −t~2 −t~2 t −t~2 t~2 −2 t~2 −1 −1 −1 t~2 = gi,i+1σ˜i,i+2gi,i+2gi,i+1gi,i+1gi+1,i+2σ˜i+1,i+2σ˜i+1,i+2gi,i+1σ˜i+1,i+2σ˜i+1,i+2σ˜i,i+1σ˜i+1,i+2gi+1,i+2 −t~2 −t~2 t~2 t~2 −t t~2 −1 −1 t~2 = gi,i+1σ˜i,i+2gi,i+2gi,i+1gi+1,i+2gi+1,i+2gi,i+2σ˜i+1,i+2σ˜i,i+1σ˜i+1,i+2gi+1,i+2 −t~2 t~2 −t~2 2 1 1 2 t~2 2 − − − = gi,i+1σ˜i,i+ gi,i+1gi+1,i+2σ˜i+1,i+2σ˜i+1,i+2σ˜i,i+1σ˜i+1,i+2σ˜i+1,i+2gi+1,i+2 −t~2 t~2 −t~2 t −1 −t −t~2 = gi,i+1σ˜i,i+2gi,i+1gi+1,i+2gi+1,i+2σ˜i,i+2gi+1,i+2gi+1,i+2 −t~2 t~2 −1 t~2 −1 −t~2 = gi,i+1σ˜i,i+2gi,i+1σ˜i,i+2σ˜i,i+2gi+1,i+2σ˜i,i+2gi+1,i+2 −t~2 t~2 t~2 −t~2 = gi,i+1gi+1,i+2gi,i+1gi+1,i+2 = 1

Thus s is well defined and the short exact sequence is split.

5 Zn and Bn[4]

5.1 Cohomology class corresponding to Zn Recall the group extension

n ‰2Ž ι π 0 Z2 Zn Sn 1

n n ‰2Ž ‰2Ž Since Z2 ≈ PZn and Zn ≈ Bn Bn 4 , the action of Sn on Z2 is induced by the conjugation n ‰2Ž~ [ ] action of Bn on PBn. Since Z2 ≈ PZn, eachg ¯ij corresponds to the pure braid between strands i and j.

24 2 Theorem 5.1. Let An be the subgroup of Gn normally generated by g 1 i j n. Then { i,j} ≤ < ≤ Gn An ≈ Zn. ~ 2 Proof. Since An is generated by gi,j for 1 ≤ i < j ≤ n and gi,j corresponds to pure braids, An is the subgroup of Bn generated by squares of pure braids and commutators of pure braids. Therefore An is the image of Bn 4 in Gn. Thus Gn An ≈ Zn. [ ] ~

′ Presentation of Zn Let s be the normalized section lifting σi,j toσ ¯i,j , the image of a clockwise th th half twist between the i and j strands in Zn. Since An is the image of Bn 4 in Gn, by [ ] Theorem 2.3, a group presentation for Zn is the presentation for Gn (replacing gi,j withg ¯i,j) with 2 the extra relationg ¯i,j = 1 and the winding numbers of eachg ¯i,j taken mod 2 for all 1 ≤ i < j ≤ n. Thus g¯1,2,..., g¯i,j ,..., g¯n 1,n, σ¯1,2,..., σ¯i,j ,..., σ¯n 1,n { − − } is a generating set for Zn. A full list of relations is given in Table 3. The following theorem completes the proofs of both Theorem 1.1 and Theorem 1.2.

n n ‰2Ž 2 ‰2Ž Theorem 5.2. Let κ ∈ Hom R2, Z2 such that κ ∈ H Sn; Z2 corresponds to the group extension ( ) [ ] ( ) n ‰2Ž ι π 0 Z2 Zn Sn 1

n n ‰ Ž ‰2Ž Let η ∶ Z 2 → Z2 be the mod 2 reduction map. Then κ = η ○ φ.

Proof. Define f ∶ Gn → Zn by f σ˜i,j = σ¯i,j and f gi,j = g¯i,j for all i, j. By Theorem 5.1 the following diagram commutes: ( ) ( )

n ‰ Ž ι1 π1 0 Z 2 Gn Sn 1 η f id

n ‰2Ž ι 0 Z2 Zn Sn 1

Furthermore, by Theorem 3.4

′ ′ κ ci,j = s σi,j s σi,j ( ) ′ ( ) ′ ( ) ′ ′ ′ ′ −1 −1 κ di,j,k,ℓ = = s σi,j s σk,ℓ s σi,j σk,ℓ − s σk,ℓ s σi,j s σk,ℓσi,j ( ) ′ ( ) ′ ( ) ′ ( ) ′ ( ) ′ ( ) ′ ( ) ′ −1 −1 κ ei,k,j = s σi,j s σj,k s σi,j σj,k − s σi,k s σi,j s σi,kσi,j ( ) ( ) ( ) ( ) ( ) ( ) ( ) ′ ′ By the choices of s and s with the definition of f, we have s σi,j = f ○s σi,j . Thus κ = φ○f. ( ) ( )

5.2 Generating sets of Bn[4] In this section we begin with a well known proposition which is a consequence of Schreier’s formula which guarantees the existence of a finite generating set for Bn 4 . Then we prove Theorem 1.3 and give a summary of the difficulties in refining our normal[ generating] set to a finite generating set. Proposition 5.3. If G is a group with generating set of size j and H is a subgroup of finite index k, then there exists a finite generating set of size at most j k − 1 + 1. ( )

25 This proposition is a well know and a proof can be found in Lyndon and Schupp [9]. Since n n ‰2Ž ‰ Ž Zn is an extension of Sn by Z2 , therefore Zn is finite of order n! ⋅ 2 2 . Furthermore, Bn is finitely generated with a generating set of size n − 1, therefore there exists a generating set for Bn 4 of size at most: [ ] n n − 1 ⋅ n! ⋅ 2‰2Ž − 1 + 1 ( ) Š 

Proof of Theorem 1.3 Consider the quotient of the Artin presentation of Bn given by (2) by the normal subgroup generated by b4 and b2,b2 . Then this quotient has a group presentation: i [ i j ] 4 2 2 R bi = 1, bi ,bj = 1, bibi+1bi = bi+1bibi+1 b1,...,b R [ ] nR − d R bibj = bjbi if i j > 1 i R S S R R Now, consider the subgroup of Zn generated byσ ¯i,i+1. By relations 3 and 4 this is a generating 2 −1 −1 set for Zn. Furthermore, note thatg ¯i,j = σ¯i⋯σ¯j 2σ¯ 1σ¯ 2⋯σ¯ . Notice also that g¯i,j, g¯k,ℓ − j− j− i {[ ]} is normally generated by g¯i,i 1, g¯k,k 1 since the conjugation action by anyσ ¯α,β permutes the [ + + ] indices. Replacingσ ¯i with bi gives the presentation above. Thus Bn 4 is normally generated by b4 and b2,b2 . [ ] i [ i j ]

Finite generating set for Bn 4 Proposition 5.3 yields a finite generating set for Bn 4 , however the size of this generating[ ] set grows faster than exponentially in n. We would hope[ for] a generating set which grows polynomial in n and the natural method of finding this would be to add necessary generates from our normal generating set. The following theorem by Cohen, Falk, and Randell provides the difficulty in this method [5].

Theorem 5.4. If n ≥ 3, then there is a single class of epimorphisms from PBn to the free group of rank 2.

The existence of an epimorphism from PBn to the free group of rank 2 implies the commutator subgroup of PBn is not finitely generated since the commutator subgroup of the free group of rank 2 is not finitely generated. Under the inclusion map PBn → Bn, the image of the commutator subgroup of PBn is contained in Bn 4 . Furthermore, commutators of pure braids are elements of our normal generating set, which implies[ ] we must determine how elements of the commutator subgroup of the pure braid group (as a subgroup of Bn) can be determined as elements of the normal subgroup (of Bn) generated by squares of pure braids.

References

[1] Joan S. Birman. Braids, links, and mapping class groups. Princeton University Press, Prince- ton, N.J.; University of Tokyo Press, Tokyo, 1974. Annals of Mathematica Studies, No. 82. [2] Joan Birman, Ki Hyoung Ko, and Sang Jin Lee. A new approach to the word and conjugacy problems in the braid groups. Adv. Math, 139(2):322-353, 1998 [3] Tara E. Brendle and Dan Margalit. The level 4 braid group. J. Reine Angew. Math., 735:249- 264, 2018. [4] Kenneth S. Brown, Cohomology of Groups, volume 87 of Graduate Texts in Mathematics. Springer-Verlag, New York-Berlin, 1982.

26 [5] Daniel C. Cohen, Michael Falk, and Richard Randell. “Pure Braid Groups Are Not Residually Free.” Configuration Spaces (2012): 213–230. [6] Samuel Eilenberg and Saunders Maclane, “Cohomology Theory in Abstract Groups. II: Group Extensions with a Non-Abelian Kernel.” Annals of Mathematics, vol. 48, no. 2, 1947, pp. 326–341. [7] Benson Farb and Dan Margalit, A primer on mapping class groups, volume 49 of Princeton Mathematical Series. Princeton University Press, Princeton NJ, 2012 [8] Kevin Kordek and Dan Margalit. Representation stability in the level 4 braid group. [9] Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory, volume 89 of Ergebnisse der Mathematik und ihrer Grenzgebiete. Springer-Verlag, Berlin-Heidelberg-New York 1977.

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