Cocyclic Hadamard Matrices Kshitija Vaidya

Cocyclic Hadamard Matrices

Kshitija Vaidya Supervised by Dr Santiago Barrera Acevedo Monash University

Vacation Research Scholarships are funded jointly by the Department of Education and

Training and the Australian Mathematical Sciences Institute. T Abstract: A square {±1}-matrix H is Hadamard if HH = nIn. The existence of a Hadamard matrix of order

4n for each n ∈ N is conjectured. A recent algebraic approach to constructing Hadamard matrices connects cohomology with combinatorics, giving rise to cocyclic Hadamard matrices. Flannery’s paper ’Calculation of

Cocyclic Matrices’ [4], which is the focus of this report addresses the problem of computing cocyclic martrices.

1 Introduction

T A Hadamard matrix H is an n × n {±1}-array such that HH = nIn, where In is the n × n identity matrix and HT is the transpose of H. Hadamard proved that all Hadamard matrices have order 1, 2 or 4n for some n ∈ N [4]. The following conjecture, known as the Hadamard conjecture, remains an open problem:

Conjecture 1.1. There exists a Hadamard matrix of order 4n for each n ∈ N.

Central to the study of Hadamard matrices is the notion of Hadamard equivalence.

Deﬁnition 1.2. Two {±1}-matrices M1 and M2 are said to be Hadamard equivalent (written M1 ∼

M2) if M1 = PM2Q where P and Q are signed permutation matrices.

This deﬁnes an equivalence relation on the set of Hadamard matrices of order n. Some classical constructions of Hadamard matrices include the Sylvester, Williamson and Ito constructions [7]. Cocyclic Hadamard matrices constitute another family of Hadamard matrices. The connection between 2-cocycles in cohomology and Hadamard matrices was established by Horadam and de Launey [6]. We now know that many classical constructions of Hadamard matrices are cocyclic. As such, cocyclic Hadamard matrices may provide a uniform approach to Conjecture 1.1. This makes the calculation of cocyclic matrices an important problem. In response to this, Flannery developed a new method of calculating cocyclic matrices [4]. Section 2 of this report covers background on cohomology, cocycles and cocyclic matrices. Sections 3, 4 and 5 detail the results of Flannery’s paper [4]. In Section 5 we apply Flannery’s theory to construct the cocycles from the Klein four-group to C2. The list we obtain matches earlier computations. The ideas presented in this report are Flannery’s; we simply expand on their explanations [4].

2 Preliminaries

Knowledge of basic group theory is assumed. The following concepts are deﬁned in Appendix A: the centre of a group, torsion subgroups, group actions, G-modules, exact sequences of groups, group

1 extensions, lifts, free groups, commutators and commutator subgroups. In the following two deﬁnitions G is a group and U a G-module under some action ?.

Deﬁnition 2.1. [7, p.114 Deﬁnition 6.1] For n ≥ 0, the abelian group of n-cochains is the set

n n C (G, U) = {f : G → U|f(x1, ..., xn) = 0 whenever xi = 1 for some i} with group action given by pointwise addition. By definition, |G0| = 1, and so C0(G, U) =∼ {0}. The n n+1 degree n differential is the homomorphism ∂n : C (G, U) → C (G, U) defined by n X i ∂n(f)(x1, ..., xn+1) = x1 ? f(x2, ..., xn+1) + (−1) f(x1, ..., xi−1, xixi+1, ..., xn+1) i=1 n+1 + (−1) f(x1, ..., xn),

n which satisfies ∂n ◦ ∂n−1 = 0. The group of n-cocycles from G to U is denoted Z (G, U) and de- n fined as ker(∂n). The group of n-coboundaries from G to U is denoted B (G, U) and defined as 2 Im(∂n−1). The nth-cohomology group of G with coefficients in U is denoted H (G, U) and defined as Zn(G, U)/Bn(G, U). Each coset f + B2(G, U) is called a cohomology class and is denoted [f].

The following is an equivalent deﬁnition of 2-cocycles and 2-coboundaries, or simply (normalised) cocycles and coboundaries. Multiplicative notation is from now on unless otherwise stated.

Deﬁnition 2.2. [7, p.115, Deﬁnition 6.2] A (normalised) 2-cocycle (or simply cocycle) is a map ψ : G × G → U such that

ψ(g, h)ψ(gh, k) = (g ? ψ(h, k))ψ(g, hk), ψ(g, 1) = 1 = ψ(1, g), for all g, h, k ∈ G.A coboundary ∂φ is a cocycle of the form ∂φ(g, h) = φ(g)−1(g ? φ(h))−1φ(gh) for some 1-cochain φ : G → U.

Deﬁnition 2.3. Let G = {g1 = 1, g2, ..., gn} be a ﬁnite group and U a G-module. A v × v matrix M is G-cocylic over U if M ∼ [ψ(gi, gj)]1≤i,j≤v. A cocyclic Hadamard matrix is a Hadamard matrix that is G-cocyclic over C2.

3 Inﬂation, transgression and a ﬁve-term exact sequence

Here, we introduce the inﬂation and transgression homomorphisms which will be used in Section 4 to construct embeddings into the second cohomology group. These homomorphisms appear in an exact sequence related to the Lyndon-Hoschild-Serre spectral sequence.

2 3.1 Inﬂation

Let G be a finite group and U a G-module under the group action (g, u) 7→ g ? u. Let N E G. Define U N := {u ∈ U such that n ? u = u for all n ∈ N}. The action of G on U is used to define an induced action of G/N on U N where (gN, u) 7→ g ? u for all u ∈ U N and gN ∈ G/N. Now define a map

inf : Zn(G/N, U N ) → Zn(G, U N ), where ψ 7→ infψ, and infψ(g1, g2, ...., gn) := ψ(g1N, g2N, ...., gnN). This is the inflation map on cocycles. We define an inflation map on cohomology classes

inf : Hn(G/N, U N ) → Hn(G, U N ), where [ψ] 7→ [infψ].

Inﬂation is well-deﬁned (see Appendix B). It is straightforward to prove that inf is a homomorphism.

3.2 Transgression

For the rest of Section 3 suppose N ≤ Z(G) and G acts trivially on U. This induces a trivial action of G/N on U. Let proj. 1 → N −−→inc. G −−−→ G/N → 1 be a central extension of N by G/N (i.e. an extension such that N ≤ Z(G)). For convenience we identify i(N) with N, treating N as a subgroup of G. Let σ : G/N → G be a normalised lift. Deﬁne

µσ to be the cocycle associated with this extension, constructed in the usual way:

−1 µσ(gN, hN) := σ(gN)σ(hN)σ(gN · hN) ,

−1 where gN, hN ∈ G/N. Clearly µσ is well-deﬁned. Since π(σ(gN)σ(hN)σ(gN · hN) ) = 1, µσ maps elements from G/N into N. The construction of µσ is independent of the choice of normalised lift, 2 and µσ ∈ Z (G/N, N). A more general proof of this is in Appendix E (see the proof of Theorem E.1). We deﬁne transgresssion:

2 τ : Hom(N,U) → H (G/N, U), where φ 7→ [φ ◦ µσ].

Again, it is straightforward to show that τ is a homomorphism.

3.3 A ﬁve-term exact sequence

The inﬂation and transgression homomorphisms arise in the following ﬁve-term exact sequence which comes from the Lyndon-Hoschild-Serre spectral sequence (for a proof of exactness see Appendix C):

inf 0 → Hom(G/N, U) −−→infa Hom(G, U) −−→res Hom(N,U) −→τ H2(G/N, U) −−−→b H2(G, U). (1)

3 The homomorphism res is induced by restricting the domain of elements in Hom(G, U) to N. The exactness of (1) is used to show that the trangsression map is injective.

Lemma 3.1. [4, p.183, Lemma 2.1] Suppose that G, N, U are deﬁned as above, and N ≤ G0. Then τ : Hom(N,U) → H2(G/N, U) is injective.

Proof. Let ψ ∈ Hom(G, U) and n ∈ N, then n = [g, h] := g−1h−1gh for some g, h ∈ G. So,

ψ(n) = ψ(g−1h−1gh) = ψ(g−1)ψ(h−1)ψ(gh) = ψ(h−1)ψ(g−1)ψ(gh) = ψ(h−1g−1gh) = ψ(1) = 1

The above implies that Im(res) = {1} where res : Hom(G, U) → Hom(N,U). Since (1) is exact, we have Ker(τ) = {1} and thus τ is injective.

4 The Universal Coeﬃcient Theorem and a decomposition of H2(G, U)

In this section we introduce the Universal Coeﬃcient Theorem which expresses H2(G, U) as an external direct sum. Via embeddings, we obtain this decomposition as an internal direct sum. This ultimately provides a method of calculating cocyclic matrices associated with the elements of Z2(G, U).

4.1 The Universal Coeﬃcient Theorem

From now on, let G be finite and U a finitely generated, trivial G-module. A cocylce ψ ∈ Z2(G, U) is symmetric if ψ(g, h) = ψ(h, g) for all g, h ∈ G. If A is abelian then Ext(A, U) is defined to be the set of all 2-cocycle classes in H2(A, U) containing at least one symmetric cocycle.

Theorem 4.1. [8, p.179, Theorem 3.3] (Universal Coeﬃcient Theorem). For a ﬁnite group G and a trivial G module, U, 2 ∼ 0 H (G, U) = Ext(G/G ,U) × Hom(H2(G),U) (2)

2 ∗ where H2(G) denotes the Schur multiplier of G which we deﬁne in the usual way as H (G, C ).

Proof. See [8, Theorem 3.3].

Note that G0 is the commutator subgroup of G which is deﬁned in Appendix A.

0 0 Lemma 4.2. Let G be any group. Let A, B ≤ G, N,H E G with H ≤ G and G ≤ N. Then,

0 (i) G = {[x1, y1]...[xn, yn]|xi, yi ∈ G, n ∈ N}.

(ii) [A, B] E G whenever A, B E G ( where the subgroup [A, B] is deﬁned in Appendix A),

4 (iii) G/N is abelian,

Proof. See Appendix D.

Part (iii) of Lemma 4.2 implies that G/G0 is abelian. Thus the ﬁrst summand of decomposition (2) is the group of all 2-cocycle classes in H2(G/G0,U) containing at least one symmetric cocycle. The aim of the remainder of this section is to express (2) as an internal direct sum by ﬁnding 0 2 embeddings of Ext(G/G ,U) and Hom(H2(G),U) in H (G, U).

4.2 The ﬁrst embedding: Inﬂation

Consider the restriction to Ext(G/G0,U) of the inﬂation map deﬁned in Section 3:

0 2 0 2 inf|Ext(G/G0,U) : Ext(G/G ,U) ⊆ H (G/G ,U) → H (G, U) with φ 7→ [infφ].

For convenience we denote inf|Ext(G/G0,U) by inf. We have already shown that inf deﬁnes a homomorphism. Thus, to prove that inf is an embedding of Ext(G/G0,U) in H2(G, U), it suﬃces to show that inf is injective. To do this we use the following proposition:

Proposition 4.3. [4, p.184, Proposition 3.2] Let K be a group acting trivially on U, and N a normal ∼ 0 0 subgroup of K such that K/N = G and N ≤ K . Denote by inf1 the restriction to Ext(G/G ,U) of 2 0 2 2 2 the inﬂation H (G/G ,U) → H (G, U), and by inf2 the inﬂation H (G, U) → H (K,U). Then the composite inf2inf1 is injective.

0 2 Proof. Let [ψ] ∈ Ext(G/G ,U) such that inf2inf1[ψ] = 0 (i.e. inf2inf1ψ ∈ B (K,U)). For the remainder of the proof, we assume, without loss of generality, that the representative cocycle ψ is symmetric. 2 Since inf2inf1ψ ∈ B (K,U), for all x, y ∈ K we have the following:

0 0 −1 −1 (inf2inf1ψ)(x, y) = (inf1ψ)(xN, yN) = ψ((xN)·(K/N) , (yN)·(K/N) ) = φ(x) φ(y) φ(xy) (3) where φ : K → U is a 1-cochain. We note the following about φ: 0 Firstly, let t ∈ K , then t = [x1, y1]·...·[xn, yn] for some n ∈ N where xi, yi ∈ K. Thus, 0 0 tN = [x1N, y1N]·...·[xnN, ynN] ∈ (K/N) . Thus, (tN)·(K/N) = 1, implying that ψ((tN)·(K/N)0, (kN)·(K/N)0) = φ(t)−1φ(k)−1φ(tk) = 1. Hence,

φ(tk) = φ(k)φ(t) = φ(t)φ(k). (4)

Secondly, (4) implies that for all [x, y] ∈ K0 with x, y ∈ K,

φ(yx)φ([x, y]) = φ(yxx−1y−1xy) = φ(xy). (5)

5 Thirdly, (5) and the symmetry of ψ imply that for all [x, y] ∈ K0 where x, y ∈ K,

φ([x, y]) = 1 (6)

We now deﬁne φ0 : G/G0 → U, with φ0((kN)·(K/N)0) := φ(k) for all x ∈ K. We prove that φ0 is 0 0 −1 0 well-deﬁned: assume (k1N)·(K/N) = (k2N)(K/N) . This implies that (k1 k2)N ∈ (K/N) . So,

−1 (k1 k2)N = [x1N, y1N]·...·[xnN, ynN] = ([x1, y1]·...·[xn, yn])N, where xi, yi ∈ K for all i and n ∈ N. Thus, k2N = (k1[x1, y1]·...·[xn, yn])N. We then have the following:

0 0 φ ((k2N)·(K/N) ) = φ(k1[x1, y1]·...·[xn, yn])

= φ(k1)·φ([x1, y1])·...·φ([xn, yn]) from the repeated application of (4)

= φ(k1) from (6)

0 0 = φ ((k1N)·(K/N) )

Thus, φ0 is well-deﬁned. Moreover, φ0((1N)·(K/N)0) = φ(1) = 1 since φ is a 1-cochain. This shows 0 0 that φ is a 1-cochain from G/G to U. Using (3), for all k1, k2 ∈ K, we can express

0 0 ψ((k1N)·(K/N) , (k2N)·(K/N) ) =

0 0 −1 0 0 −1 0 0 0 φ ((k1N)·(K/N) ) φ ((k2N)·(K/N) ) φ (((k1N)·(K/N) )((k2N)·(K/N) ))

2 0 Thus, ψ ∈ B (G/G ,U) which shows that the composite map inf2inf1 is injective.

Corollary 4.4. [4, p.184, Corollary 3.2] Let G be a ﬁnite group that acts trivially on U. Then inf : Ext(G/G0,U) → H2(G, U) is injective.

Proof. In Proposition 4.3, take K = G and N = {1}.

It follows that inf : Ext(G/G0,U) → H2(G, U) is an embedding. We now describe a process for explicitly determining a full list of cocycle representatives for the elements of Ext(G/G0,U). By Lemma 4.2, G/G0 is abelian. Since G/G0 is also ﬁnite, we can write 0 ∼ G/G = Z e1 × Z e2 × ... × Z en where n ∈ , the ei’s range over a ﬁnite set of positive integers and p1 p2 pn N the pi’s are primes. By a property of the bifunctor Ext, we have the following [4, p.185]:

0 ∼ ∼ Ext(G/G ,U) = Ext(Z e1 × Z e2 × ... × Z en ,U) = Ext( e1 ,U) × Ext( e2 ,U) × ... × Ext( en ,U). p1 p2 pn Zp1 Zp2 Zpn

By computing a full set of representative cocyclic matrices for the elements of Ext( ei ,U) and forming Zpi all possible Kronecker products of these, we obtain the full set of representative cocyclic matrices for the

6 0 elements of Ext(G/G ,U). To obtain a representative cocyclic matrix for each element of Ext( ei ,U), Zpi 2 we note that Ext( ei ,U) = H ( ei ,U) since ei is cyclic. The proof of this result makes use of Zpi Zpi Zpi the connection between the second cohomology group and group extensions. A description of this connection as well as a proof of the stated equality is provided in Appendix E . 2 The equality between Ext( ei ,U) and H ( ei ,U) enables us to calculate the elements of Zpi Zpi

Ext( ei ,U) due to the following isomorphism: Zpi

Theorem 4.5. [9, p.52, Theorem 3.1] If C = {r, r2, ...., rn} is a cyclic group of order n, then H2(C,A) =∼ A/An where A is a trivial C-module. The map below deﬁnes the isomorphism:

2 ϕ : A → H (C,A) with a 7→ [fa],

i j i j where fa(r , r ) = 1 if i + j < n and fa(r , r ) = a if i + j ≥ n.

Proof. See [9, p.52, Theorem 3.1].

4.3 The second embedding: Transgression

∼ Below, G is a ﬁnite group with presentation G = F/R where F is free of ﬁnite rank and R E F .

Lemma 4.6.

(i) [F,R] E R,

0 (ii) [F,R] E (R ∩ F ),

0 (iii) (R ∩ F ) E R,

0 0 (iv) F E RF .

Proof. See Appendix F.

∼ 0 Theorem 4.7. (Hopf’s Formula) H2(G) = (R ∩ F )/[F,R].

Proof. For a proof of Hopf’s formula, see [10, p.50, Theorem 2.4.6].

Theorem 4.8. [10, p.50, Theorem 2.4.6] (i) R/[F,R] is a ﬁnitely generated abelian group,

(ii) (R ∩ F 0)/[F,R] is the torsion subgroup of R/[F,R] [3, p.248, Lemma 20.7.4],

(iii) R/[F,R] = (R ∩ F 0)/[F,R] × S/[F,R] for some S/[F,R] which is free abelian and has the same rank as F [3, p.249, Lemma 20.8.1].

7 Proof. (i) R/[F,R] is well-defined by Lemma 4.6. Since R E F , by the third isomorphism theorem, F/[F,R] ∼ ∼ R/[F,R] = F/R. But F/R = G and so R/[F,R] has finite index in F/[F,R] which is finitely generated since F has finite rank. Subgroups of finite index in finitely generated groups are finitely generated [11, p.376, Corollary 1.2]. Thus, R/[F,R] is finitely generated. To see that −1 R/[F,R] is abelian, note that r1r2(r2r1) ∈ [F,R] for all r1, r2 ∈ R.

(ii) F/F 0 is the group with presentation hX|f −1g−1fg = 1 for all f, g ∈ F i where F is free on the set 0 0 X. Thus, F/F is a free abelian group on X and therefore torsion-free. We have (R ∩ F ) E R R/[F,R] ∼ 0 by Lemma 4.6. Thus, by the third isomorphism theorem, (R∩F 0)/[F,R] = R/(R ∩ F ). Define a surjective map ϕ : R 7→ RF 0/F 0 with r 7→ r·F 0, then ϕ is a homomorphism with kernel R ∩ F 0. Hence, by the first isomorphism theorem, R/(R ∩ F 0) =∼ RF 0/F 0. From all of this, we can conclude that, R/[F,R] =∼ R/(R ∩ F 0) =∼ RF 0/F 0 ≤ F/F 0. (R ∩ F 0)/[F,R] Since F/F 0 is torsion-free, the torsion subgroup of R/[F,R] is contained in (R ∩ F 0)/[F,R]. The result follows because (R ∩ F 0)/[F,R] is finite [11, p.278, Remark 1].

(iii) Since R/[F,R] is finitely generated and abelian, it is the direct sum of its torsion subgroup and some free abelian subgroup [12, p.100, 4.2.10]. i.e. R/[F,R] = (R ∩ F 0)/[F,R] × S/[F,R] for some free abelian subgroup S/[F,R] ≤ R/[F,R]. We prove that S/[F,R] has the same rank as F . Since S/[F,R] =∼ (R/[F,R])/((R ∩ F 0)/[F,R]), the proof of the previous part implies that ∼ 0 0 0 0 S/[F,R] = RF /F . Since R,F E F , we have RF E F . So, by the third isomorphism theorem, F/F 0 ∼ 0 0 0 0 0 RF 0/F 0 = F/RF . Since RF ⊇ R and F/R is finite, F/RF is also finite. Thus, RF /F is a finite index subgroup of the free abelian group F/F 0, implying S/[F,R] has the same rank as F .

The term S/[F,R] introduced in Theorem 4.8 is a Schur complement of (R∩F 0)/[F,R] in R/[F,R].

Lemma 4.9.

(i) S E F

(ii) S E R

Proof. (i) For all s ∈ S and f ∈ F , we have fs = ((fsf −1s−1)s)f. Since [F,R] ≤ [R,F ] ≤ S, it

follows that S E F .

8 (ii) From Theorem 4.8, S/[F,R] E R/[F,R]. The result follows.

Lemma 4.9 allows us to construct the extension of R/S by F/R presented in the theorem below ∼ F/S where F/R = R/S by the third isomorphism theorem since R E F .

Theorem 4.10. 1 → R/S −→i F/S −→π F/R → 1

∼ ∼ is a central extension of R/S = H2(G) by F/R = G.

−1 Proof. Let fS ∈ F/S and rS ∈ R/S. Then since [F,R] E S, we have (rf)(fr) ∈ S, implying that R/S ≤ Z(F/S). ∼ R/[F,R] ∼ Since S E R, we have R/S = S/[F,R] by the third isomorphism theorem. This implies R/S = 0 0 ∼ ∼ (R ∩ F )/[F,R] by Theorem 4.8. By Hopf’s formula (R ∩ F )/[F,R] = H2(G). It follows that R/S =

H2(G).

∼ ∼ Taking R/S = H2(G) and F/R = G, we use the extension shown in Theorem 4.10 to deﬁne the 2 transgression map τs : Hom(R/S, U) → H (F/R, U) in the way described in Section 3. We show that

τs is injective:

Proposition 4.11. [4, p.185, Proposition 3.4] τs is injective.

Proof. We ﬁrst show that F 0R/S = F 0S/S. Since S ⊆ R, clearly F 0S/S ⊆ F 0R/S. 0 0 0 0 0 Let f 1r ∈ F R, where f 1 ∈ F and r ∈ R. Since R/[F,R] = (R ∩ F )/[F,R] ⊕ S/[F,R], we have 0 0 0 0 r[F,R] = f 2[F,R]s[F,R] or equivalently r[F,R] = f 2s[F,R] for some f 2 ∈ R ∩ F and s ∈ S. Thus, 0 0 0 0 0 0 r = f 2sg 1 for some g 1 ∈ [F,R]. Since [F,R] E F (by Lemma 4.2), sg 1 = g 2s where g 2 ∈ [F,R]. So 0 0 0 0 0 0 0 we have that f 1r = f 1f 2sg 1 = f 1f 2g 2s. 0 0 0 0 0 0 0 0 0 Clearly [F,R] ≤ F , implying that g 2 ∈ F and therefore that f 1f 2g 2 ∈ F . Hence, f 1r ∈ F S. So we have shown that F 0R/S ⊆ F 0S/S. Thus, F 0R/S = F 0S/S. It follows that

R/S ≤ F 0R/S = F 0S/S = (F/S)0

Taking N = R/S and G = F/S in Lemma 3.1, we ﬁnd that the transgression is injective.

Since we already know τs is a homomorphism, this implies that τs deﬁnes an embedding of 2 Hom(H2(G),U) in H (G, U).

9 Theorem 4.12. [4, p.185, Proposition 3.5] Take G =∼ F/R ﬁnite and S as deﬁned above. Then, im inf ∩ imτs = {0}.

Proof. Let K = F/S and N = R/S in Proposition 4.3. Since R/S ≤ (F/S)0 with (F/S)/(R/S) =∼ ∼ F/R = G, the hypotheses of Proposition 4.3 are satisﬁed. Then, inf1 from Proposition 4.3 represents inf. By (1), inf2 has kernel imτs. Let [ψ] ∈ im inf∩imτ. Then, [ψ] ∈ im inf1 ∩ker inf2. If there is some non-zero element in im inf1 ∩ ker inf2, then inf2inf1 would not be injective. Since inf2inf1 is injective by Proposition 4.3, [ψ] = 0.

Theorem 4.13. [4, p.186, Theorem 3.6] Again, take G =∼ F/R finite and S as defined above and let 2 U be finitely generated. Then, H (G, U) =im inf × imτs.

Proof. Since H2(G, U) is finite under these conditions, the result follows from the injectivity of the embeddings defined earlier, the Universal Coefficient Theorem and Theorem 4.12.

2 This shows that for all [ψ] ∈ H (G, U), we have [ψ] = [ψI ][ψT ], where ψI is a cocycle representative of some element in im inf and ψT is a cocycle representative of some element in imτs. We refer to a cocycle representative of an element in im inf as an inflation cocycle and to a cocycle representative of an element in imτs as a transgression cocycle. Likewise, we refer to the associated matrices as an inflation matrix and a transgression matrix. It follows from the above decomposition of [ψ] that for 2 each cocycle ψ ∈ Z (G, U), we have ψ = ψI ψT ∂φ where ∂φ is a coboundary [4]. Similarly, we can decompose the associated matrix Mψ = MψI •MψT •M∂φ, where • denotes elementwise multiplication [4]. The elements of Z2(G, U) can thus be computed by individually computing inflation cocycles, transgression cocycles and coboundaries. Note that the decomposition of H2(G, U) presented in this section is not canonical since the transgression map depends on a choice of presentation for G and a choice of Schur complement.

5 Symmetry in the decomposition of cocycles

ψ ∈ Z2(G, U) is almost symmetric if ψ(g, h) = ψ(h, g) for all g, h ∈ G such that gh = hg. It is straightforward to show that every coboundary is almost symmetric. A presentation of G =∼ F/R is 0 said to satisfy (P) if (R ∩ F )/[F,R] is generated by elements in {[f1, f2][R,F ]|fi ∈ F }, which is a set of generators of F 0/[F,R].

10 Lemma 5.1. [4, p.187, Lemma 4.1] Suppose G has a presentation F/R satisfying (P) and that

H2(G) 6= {0}. Choose a Schur complement S/[F,R]. Then for all nonzero [ψ] ∈ imτs, ψ is not almost symmetric.

Proof. We choose a normalised lift v : F/R → F . We then deﬁne a function σ : F/R → F/S with fR 7→ v(fR)S for all fR ∈ F/R. • Assume fR = gR for some f, g ∈ F . Then σ(fR) = v(fR)S = v(gR)S = σ(gR). Thus, σ is well-deﬁned. • σ(R) = v(R)S = S since v is normalised. Thus, σ is normalised. F/S ∼ • Note that R/S = F/R via the natural isomorphism (fS)(R/S) 7→ fR. Thus, we can consider F/S F/S σ to be a map from R/S → F/S with (fS)(R/S) 7→ v(fR)S for all (fS)(R/S) ∈ R/S . We also have the following:

π(σ((fS)(R/S))) = π(v(fR)S) = (v(fR)S)(R/S) = ((frf )S)(R/S) = (fS)(R/S),

where rf ∈ R. This shows that σ deﬁnes a normalised transversal function. Since we have assumed that (P) holds for our choice of presentation, we can conclude using the natural ismorphism between (R ∩ F 0)/[F,R] and R/S, that R/S can be generated by some subset of

{[f1, f2]S|fi ∈ F }.

Let [ψ] ∈ imτs be non-zero. Thus, [ψ] = [φ ◦ µσ] 6= 0 for some φ ∈ Hom(R/S, U). If φ were trivial on every generator of R/S, we would have [ψ] = [φ◦µσ] = 0. Thus φ is non-trivial on some generating −1 −1 −1 −1 0 element [f1 , f2 ]S where f1, f2 ∈ F . Now, [f1 , f2 ][F,R] ∈ (R ∩ F )/[F,R] and we assume without −1 −1 0 −1 −1 −1 loss of generality that [f1 , f2 ] ∈ R ∩ F ⊆ R. Thus, f1f2f1 f2 = (f1f2)(f2f1) ∈ R, implying that f1R and f2R commute. Assume for contradiction that ψ is almost symmetric. Since every coboundary is almost symmetric, this implies that φ ◦ µσ is almost symmetric. From this, and what −1 we have shown above, φ(µσ(f1, f2)) = φ(µσ(f2, f1)). Equivalently, φ(v(f1R)v(f2R)v(f1f2R) S) = −1 −1 −1 −1 φ(v(f2R)v(f1R)v(f2f1R) S) implying φ(v(f1R)v(f2R)v(f1f2R) v(f2f1R)v(f1R) v(f2R) S) = −1 −1 1. By the commutativity of f1R and f2R, it follows that φ(v(f1R)v(f2R)v(f1R) v(f2R) S) = 1. By the centrality of R/S in F/S, we have the following:

−1 −1 −1 −1 −1 −1 v(f1R)v(f2R)v(f1R) v(f2R) S = f1r1f2r2r1 f1 r2 f2 S −1 −1 −1 −1 = (r1r1 r2r2 S)([f1 , f2 ]S) −1 −1 = [f1 , f2 ]S.

−1 −1 −1 Thus, φ([f1 , f2 ] S) = 1 which gives the required contradiction.

11 Corollary 5.2. [4, p.185, Proposition 3.5] Suppose G has a presentation F/R satisfying (P). Then,

im inf = {[ψ] ∈ H2(G, U)|ψ is almost symmetric}.

Proof. ”⊆” Let [ψ] ∈ im inf. Then [ψ] = [infϕ] where [ϕ] ∈ Ext(G/G0,U). Since every cocycle class in Ext(G/G0,U) contains a symmetric cocycle, we can assume ϕ is symmetric. We can also assume without loss of generality that ψ = infϕ, implying that ψ is almost symmetric. The inclusion follows. ”⊇” Let [ψ] ∈ H2(G, U) with ψ almost symmetric. By what is discussed in the previous section, we can write ψ = ψI ψT ∂φ where ψI ∈ im inf, ψT ∈ imτs and ∂φ is a coboundary. Since im inf 0 contains the inﬂated cocycle classes of Ext(G/G ,U), we can assume without loss of generality that ψI is symmetric. ∂φ is almost symmetric. It follows that ψT must be almost symmetric. However, since our presentation satisﬁes the condition (P), this implies that ψT is trivial. The result follows.

This shows that we can write each ψ ∈ Z2(G, U) as the product of a symmetric inﬂation cocycle, an almost symmetric coboundary and an asymmetric transgression cocycle.

6 Applying Flannery’s Method

In this section, we apply Flannery’s method to calculate the matrix representations of all the cocycles 2 2 2 2 ∼ 2 in Z (G, C2) where G = ha, b : a = b = (ab) = 1i. Note that G = C2 via the isomorphism 1 7→ (1, 1), a 7→ (1, −1), b 7→ (−1, 1), ab 7→ (−1, −1). The indexing used in the following matrix representations of cocycles from G to C2 is: {g1 = 1, g2 = a, g3 = b, g4 = ab}. 2 Coboundary Matrices: The coboundaries in Z (G, C2) are exactly

1 a b ab   1 1 1 1 1     a  1 1 γ γ  M∂φ =     b  1 γ 1 γ    ab 1 γ γ 1 where γ ∈ {±1} (see Appendix G). 2 Inflation Matrices: To compute the matrices corresponding to inflation cocyles in Z (G, C2), we 0 first determine representative cocyclic matrices for the elements of Ext(G/G ,C2) via the method de- 0 ∼ ∼ scribed in Section 3. Since G is abelian and by the isomorphism mentioned ealier, G/G = G = C2×C2. 2 We begin by selecting a representative cocyclic matrix for each element in Ext(C2,C2) = H (C2,C2).

12 By the isomorphism described in Theorem 4.5, we ﬁnd that the following are representative cocyclic 2 matrices for each element in H (C2,C2), where the indexing used is {1, −1}:

1 −1 1 −1     1 1 1 1 1 1  ,   −1 1 −1 −1 1 1

The set of all possible Kronecker products of the above two matrices gives a full set of representative cocyclic matrices for the elements of Ext(C2×C2,C2), under the indexing {(1, 1), (1, −1), (−1, 1), (−1, −1)}. Since G/G0 =∼ G, the inflation is trivial and so the matrices corresponding to inflation cocycles in 2 Z (G, C2) are exactly the full set of all Kronecker products of the matrices shown above. Rewriting 2 the indexing via the isomorphism between G and C2 , we find that the inflation matrices are exactly:

1 a b ab   1 1 1 1 1     a  1 α 1 α      b  1 1 β β    ab 1 α β αβ where again α, β ∈ {±1}. Transgression Matrices: Recall that G = ha, b : a2 = b2 = (ab)2 = 1i. If we let F be free on 2 2 a, b and let R E F be the normal closure in F of {a , b , abab}, then F/R gives a presentation of G. ∼ It is known that H2(G) = C2 [10, p.98, Theorem 2.11.3]. We note that since F/R is commutative, R must contain F 0.

Lemma 6.1. (R ∩ F 0)/[F,R] = h[b−1, a][F,R]i.

Proof. We note that [b−1, a] = b2(abab)−1a2, implying [b−1, a][F,R] ∈ (R ∩ F 0)/[F,R]. We now prove that [b−1, a][F,R] has order 2 in (R ∩ F 0)/[F,R]. Assume for contradiction that [b−1, a] ∈ [F,R]. This −2 −2 0 ∼ implies that {b [F,R], a [F,R]} forms a generating set of R/[F,R]. Since (R ∩ F )/[F,R] = H2(G) has a complement S/[F,R] in R/[F,R] which is free abelian and of the same rank as F , we ﬁnd that ∼ R/[F,R] = C2 × Z × Z. Thus, R/[F,R] can be generated by no fewer than 3 of its elements, giving our contradiction. Thus, [b−1, a] ∈/ [F,R] and therefore [b−1, a][F,R] does not have order 1 in R∩F 0/[F,R]. Since R/[F,R] is central in F/[F,R] and since F 0 ≤ R, we can show that ([b−1, a][F,R])2 = [F,R]. −1 0 0 ∼ Thus, [b , a][F,R] has order two in (R ∩ F )/[F,R]. However, since (R ∩ F )/[F,R] = C2, it is a group of order 2. It follows that (R ∩ F 0)/[F,R] = h[b−1, a][F,R]i.

13 By the above lemma, the natural isomorphism between (R ∩ F 0)/[F,R] and R/S implies that R/S has two elements, namely [b−1, a]S and S.

Lemma 6.2. R/[F,R] = h[b−1, a][F,R]i×hb−2[F,R], a−2[F,R]i, where (R∩F 0)/[F,R] = h[b−1, a][F,R]i and the second factor is a Schur complement S/[F,R].

Proof. Since R is generated by a2, b2, abab and R/[F,R] is abelian, each element of R/[F,R] can be written as the product of an element from h[b−1, a][F,R]i and an element from hb−2[F,R], a−2[F,R]i. For contradiction assume this decomposition is not unique, then h[b−1, a][F,R]i∩hb−2[F,R], a−2[F,R]i must contain a non-identity element. In particular, it must contain [b−1, a][F,R]. However, this would imply that R/[F,R] can be generated by fewer than 3 elements which is a contradiction by the argument described in the proof of the previous lemma. Thus, R/[F,R] = h[b−1, a][F,R]i×hb−2[F,R], a−2[F,R]i where (R ∩ F 0)/[F,R] = h[b−1, a][F,R]i (by Lemma 6.1) and the second factor is a Schur complement S/[F,R].

We now define σ : F/R → F/S by σ(1) := S, σ(a) := aS, σ(b) := bS, σ(ab) := abS, where 1, a, b, ab denote the coset representatives of the distinct classes R, aR, bR, abR. The map is clearly well-defined and normalised. We prove that σ defines a lift. (F/R) =∼ (F/S)/(R/S) via the isomorphism fR 7→ (fS)(R/S). Thus, the projection map π : F/S → F/R maps fS 7→ fR for all fS ∈ F/S. From here, it is clear that π ◦ σ gives the identity map.

Thus, σ is a normalised lift. We can now define the associated cocycle µσ in the usual way with, −1 µσ(f1, f2) = σ(f1)σ(f2)σ(f1f2) , where f1, f2 ∈ {1, a, b, ab}. ∼ Since R/S = C2, the only elements of Hom(R/S, C2) are the trivial homomorphism and the −1 homomorphism that maps S to 1 and [b , a]S to −1. We label these as φ1 and φ2 respectively. The transgression cocycles that we want to explicitly find are then φ1 ◦ µσ and φ2 ◦ µσ. Note φ1 ◦ µσ just gives rise to the matrix of all 1s. To construct the matrix representation of φ2 ◦ µσ, we find the values of µσ at all possible arguments (see Appendix H):

• Let f = 1, a, b, then, µσ(f, f) = S. Let f = 1, a, b, ab, then, µσ(1, f) = S. −1 • µσ(a, b) = µσ(a, ab) = µσ(ab, b) = S, µσ(b, ab) = µσ(ab, ab) = µσ(b, a) = µσ(ab, a) = [b , a]S.

Given this and given that φ1 ◦ µσ is the trivial cocycle, the transgression matrices are exactly:

1 a b ab   1 1 1 1 1     a  1 1 1 1      b  1 κ 1 κ    ab 1 κ 1 κ

14 where κ ∈ {±1}. 2 Since the elements of Z (G, C2) are simply all possible products of the transgression, inﬂation and coboundary cocycles, we have the following:

2 2 2 2 Example 6.3. Let ψ ∈ Z (G, C2) where G = ha, b : a = b = (ab) = 1i, then the associated matrices

Mψ are of the form: 1 a b ab   1 1 1 1 1     a  1 α γ αγ  Mψ =     b  1 γκ β βγκ    ab 1 αγκ βγ αβκ where α, β, γ, κ ∈ {±1}.

It should be noted that Example 6.3 was ﬁrst derived via a diﬀerent method in [2].

7 Conclusion

Cocyclic Hadamard matrices may provide a uniform approach to the Hadamard Conjecture. Flan- nery’s work uses Homological algebra to shed light on the structure of coyclic matrices. His method is already used as a basis for algorithms that compute cocycles. Developing new and eﬃcient algorithms that apply Flannerys theory could be a focus of further research. To conclude, the cocyclic approach to Hadamard matrices has evolved into a rich, interesting area of study that is likely to grow.

References

[1] Bhattacharya, P. B., Jain, S. K., Nagpaul, S. R. (1994). Basic abstract algebra. Cambridge University Press. [2] De Launey, W. (1990). On the construction of n-dimensional designs from 2-dimensional designs. Australasian J. Combinatorics, 1, 67-82. [3] De Launey, W., Flannery, D. L. (2011). Algebraic design theory (No. 175). American Mathematical Soc.. [4] Flannery, D. L. (1996). Calculation of cocyclic matrices. Journal of Pure and Applied Algebra, 112(2), 181-190. [5] Hadamard, J. (1893). Resolution d’une question relative aux determinants. Bull. des sciences math., 2, 240-246. [6] Horadam, K. J., de Launey, W. (1993). Cocyclic development of designs. Journal of Algebraic Combinatorics, 2(3), 267-290. [7] Horadam, K. J. (2012). Hadamard matrices and their applications. Princeton university press. [8] Hilton, P. J., Stammbach, U. (2012). A course in homological algebra (Vol. 4). Springer Science Business Media. [9] Karpilovsky, G. (1985). Projective representations of ﬁnite groups. New York-Basel. [10] Karpilovsky, G. (1987). The schur multiplier. Oxford University Press, Inc.. [11] Karpilovsky, G., 1994. Group representations (Vol. 2). Elsevier. [12] Pepper, T. J. (2015). Structure of Finitely Generated Abelian Groups.

[13] Vermani, L. R. (2003). An elementary approach to homological algebra. CRC Press.

15 Appendix A

Deﬁnition A.1. The centre of a group G is {a ∈ G | g · a = a · g}, denoted Z(G).

Deﬁnition A.2. Let G be a group. The torsion subgroup of G, denoted T (G), is the subgroup of G consisting of all elements of ﬁnite order.

Deﬁnition A.3. [1, p.107] The group (G, ·) acts on a set S if ∃ a map ? : G×S → S, with (g, s) 7→ g?s such that for all a, b ∈ G and s ∈ S:

(i) a · b ? s = a ? (b ? s),

(ii) eG ? s = s where eG denotes the identity element in G.

The map ? is called the action of G on S.

Deﬁnition A.4. Let G be a group. Then a G-module is an abelian group (M, +) combined with a group action ? : G × M → M such that g ? (m + n) = g ? m + g ? n for all m, n ∈ M.

Deﬁnition A.5. An n-term exact sequence of groups consists of n groups, G1, ..., Gn and n − 1 homomorphisms α1, ..., αn−1 such that αi : Gi → Gi+1 with imαi = kerαi+1 for all 1 ≤ i ≤ n − 2.

Deﬁnition A.6. The group G is an extension of the group N by the group H if N E G up to isomorphism and H =∼ G/N. The extension can be represented by the following exact sequence:

1 → N −→i G −→π H → 1, where i denotes inclusion and π denotes projection.

Deﬁnition A.7. Let G be a group, N E G and π : G → G/N the projection map. A map σ : G/N → G is called a lift if π ◦ σ gives the identity map. σ is a normalised lift if σ(N) = 1.

Definition A.8. The group F is free on a generating set X if for all maps f : X → G, there exists a unique homomorphism ϕ : F → G such that ϕ ◦ i = f, where i : X → F is the inclusion map. In the above, G is an arbitrary group. The group F is free abelian if F satisfies the above condition given that G is an arbitrary abelian group. A free, or free abelian, group F is finitely generated if X is finite. In this case, the rank of F is |X|.

Deﬁnition A.9. Let G be a group. For g, h ∈ G, the commutator of g and h is g−1h−1gh, denoted 0 [g, h]. For G1,G2 ≤ G, we deﬁne [G1,G2] := h[x, y]|x ∈ G1, y ∈ G2i. And, [G, G] is denoted G and is called the commutator subgroup of G.

16 Appendix B

The inﬂation map on cohomology classes is shown below:

inf : Hn(G/N, U N ) → Hn(G, U N )

[ψ] 7→ [infψ].

We prove that the inﬂation of cohomology classes is well deﬁned. Note that additive notation is used. n N Let ψ1, ψ2 ∈ Z (G/N, U ) such that [ψ1] = [ψ2].

(infψ1 − infψ2)(g1, ..., gn) = infψ1(g1, ..., gn) − infψ2(g1, ..., gn)

= ψ1(g1N, ..., gnN) − ψ2(g1N, ..., gnN)

= (ψ1 − ψ2)(g1N, ..., gnN)

n N Since [ψ1] = [ψ2], we have ψ1 − ψ2 ∈ B (G/N, U ). Thus, (ψ1 − ψ2)(g1N, ..., gnN) = ∂n−1f for some f ∈ Cn−1(G/N, U N ). This implies,

n−1 X i (infψ1 − infψ2)(g1, ..., gn) = g1 ? f(g2N, ..., gnN) + (−1) f(g1N, ..., gi−1N, gigi+1N, ..., gnN) i=1 n + (−1) f(g1N, ..., gn−1N).

n−1 N n−1 Deﬁne f˜ : G → U such that f˜(x1, ..., xn−1) := f(x1N, ..., xn−1N) for all (x1, ..., xn−1) ∈ G .

If xi = 1 for some i ∈ {1, ..., n − 1}, then f˜(x1, ..., xi, ..., xn−1) = f(x1N, ..., N, ..., xn−1N) = 1 since f ∈ Cn−1(G/N, U N ). Thus, f˜ ∈ Cn−1(G, U N ) and

n−1 X i (infψ1 − infψ2)(g1, ..., gn) = g1 ? f˜(g2, ..., gn) + (−1) f˜(g1, ..., gi−1, gigi+1, ..., gn) i=1 n + (−1) f˜(g1, ..., gn−1).

n N This implies that infψ1 − infψ2 ∈ B (G, U ) and so [infψ1] = [infψ2].

Appendix C

The material in this appendix is from [13, Section 11.9]. We prove that (1) is exact:

• infa is injective:

17 Assume infaψ1 = infaψ2. Let gN ∈ G/N, then ψ1(gN) = infaψ1(g) = infaψ2(g) = ψ2(gN). Thus,

ψ1 = ψ2.

• im(infa) = ker(res):

“⊆”: Let infaψ ∈ im(infa) for some ψ ∈ Hom(G/N, U). Recall that res(infaψ) ∈ Hom(N,U) is the homomorphism induced by restricting the domain of infaψ to N.

For all n ∈ N, (res(infaψ))(n) = (infaψ)(n) = ψ(nN) = 1U since ψ ∈ Hom(G/N, U). Thus, im(infa) ⊆ ker(res). “⊇”: Now, let ψ0 ∈ ker(res). i.e. ψ0 ∈ Hom(G, U) such that ψ0(n) = 1 for all n ∈ N. Deﬁne ψ : G/N → U by ψ(gN) := ψ0(g). We show as follows that ψ is well-deﬁned: Let gN, hN ∈ G/N with gN = hN, then g = h · n for some n ∈ N. Hence,

ψ(gN) = ψ0(g) = ψ0(h · n) = ψ0(h) · ψ0(n) = ψ0(h) = ψ(hN).

Thus, ψ is well-deﬁned. We now show that ψ is a homomorphism. Let gN, hN ∈ G/N, then

ψ(gN · hN) = ψ((gh)N) = ψ0(gh) = ψ0(g) · ψ0(h) = ψ(gN) · ψ(hN).

0 0 Thus, ψ ∈ Hom(G/N, U). Moreover, infaψ(g) = ψ(gN) = ψ (g) for all g ∈ G. Hence, infaψ = ψ , 0 which implies that ψ ∈ im(infa). So we have ker(res) ⊆ im(infa).

• im(res) = ker(τ):

“⊆”: Let ψ¯ ∈ im(res), then there exists ψ ∈ Hom(G, U) such that ψ restricted to N gives ψ¯. We 2 show that τ(ψ¯) := [ψ¯◦µσ] = 1 ∈ H (G/N, U) where µσ is deﬁned as described in Subsection 3.2. Let gN, hN ∈ G/N, then

−1 ψ¯◦µσ(gN, hN) = ψ¯(σ(gN)σ(hN)σ(ghN) )

= ψ(σ(gN)σ(hN)σ(ghN)−1)

= ψ(σ(gN))ψ(σ(hN))ψ(σ(ghN)−1)

= ψ(σ(gN))ψ(σ(hN))ψ(σ(ghN))−1

= ψ◦σ(gN)ψ◦σ(hN)ψ◦σ(ghN)−1

= φ(gN)−1φ(hN)−1φ(ghN), where φ : G/N → U with φ(gN) := ψ◦σ(gN)−1 for all gN ∈ G/N. Since φ : G/N → U satisﬁes ¯ 2 φ(N) = ψ(σ(N)) = ψ(1G) = 1, φ is a 1-cochain. Thus, ψ◦µσ ∈ B (G/N, U). This shows that

18 im(res) ⊆ ker(τ). 0 0 0 2 0 2 “⊇”: Let ψ ∈ Hom(N,U) such that τ(ψ ) := [ψ ◦µσ] = 1 ∈ H (G/N, U), then ψ ◦µσ ∈ B (G/N, U). So there exists a 1-cochain φ : G/N → U with

0 −1 −1 ψ ◦µσ(gN, hN) = φ(gN) φ(hN) φ(ghN) (7) for all gN, hN ∈ G/N. Now, each g ∈ G can be expressed uniquely as ngσ(gN) for some ng ∈ N. 0 −1 Deﬁne ψ : G → U where ψ(g) := ψ (ng)φ(gN) . We show as follows that ψ is a homomorphism. Let g, h ∈ G, then

ψ(gh) = ψ(ngσ(gN)nhσ(hN))

= ψ(ngnhσ(gN)σ(hN))

= ψ(ngnhµσ(gN, hN)σ(ghN))

0 −1 = ψ (ngnhµσ(gN, hN))φ(ghN)

0 0 0 −1 = ψ (ng)ψ (nh)ψ ◦µσ(gN, hN)φ(ghN)

0 0 −1 −1 = ψ (ng)ψ (nh)φ(gN) φ(hN) by (7)

0 −1 0 −1 = ψ (ng)φ(gN) ψ (nh)φ(hN)

= ψ(g)ψ(h)

Thus, ψ ∈ Hom(G, U). We show that res(ψ) = ψ0. Let n ∈ N, then

ψ(n) = ψ(nσ(nN)) = ψ0(n)φ(N)−1 = ψ0(n) as φ is a cochain.

Thus, ψ0 ∈ im(res). This shows ker(τ) ⊆ im(Res).

• im(τ) = ker(infb):

“⊆”: Consider some τ(ψ) := ψ◦µσ ∈ Im(τ) where ψ ∈ Hom(N,U). We prove that infb(τ(ψ)) = 1 ∈ H2(G, U): For all g, h ∈ G,

−1 infbψ◦µσ(g, h) = ψ◦µσ(gN, hN) = ψ(σ(gN)σ(hN)σ(ghN) ) (8)

−1 Since each g ∈ G can be expressed uniquely as g = ngσ(gN) for some ng ∈ N, we have σ(gN) = ng g. By (8),

−1 −1 −1 −1 infbψ◦µσ(g, h) = ψ(ng gnh h(ngh gh) ) −1 −1 = ψ(ng nh ngh) −1 −1 = ψ(ng) ψ(nh) ψ(ngh)

= φ(g)−1φ(h)−1φ(gh),

19 where φ : G → U such that φ(g) := ψ(ng) for all g ∈ G. Since 1 = 1σ(N), we have φ(1) = 1. Thus, 2 φ : G → U is a 1-cochain. Thus infbψ◦µσ ∈ B (G, U). This shows that im(τ) ⊆ ker(infb). 2 2 “⊇”: Let [ψ] ∈ ker(infb) for some ψ ∈ Z (G/N, U), then infbψ ∈ B (G, U). i.e. infbψ(g, h) = φ(g)−1φ(h)−1φ(gh) for all g, h ∈ G for some 1-cochain φ. Deﬁne φ¯ : N → U to be the restriction of φ to N. We note the following: if n ∈ N and g ∈ G, then,

−1 −1 1 = ψ(nN, gN) = infbψ(n, g) = φ(n) φ(g) φ(ng).

This implies that φ(ng) = φ(n)φ(g). (9)

A consequence of the above argument is that φ¯ ∈ Hom(N,U). We now show as follows that [ψ] =

[φ¯◦µσ] (below we are swapping to additive notation):

(ψ − φ¯◦µσ)(gN, hN) = (ψ − φ¯◦µσ)(gN, hN)

= ψ(gN, hN) − (φ¯◦µσ)(gN, hN) ¯ = infbψ(σ(gN), σ(hN)) − (φ◦µσ)(gN, hN)

−1 = −φ(σ(gN)) − φ(σ(hN)) + φ(σ(gN)σ(hN)) + φ¯(µσ(gN, hN) )

−1 = −φ(σ(gN)) − φ(σ(hN)) + φ(σ(gN)σ(hN)) + φ(µσ(gN, hN) )

−1 = −φ(σ(gN)) − φ(σ(hN)) + φ(σ(gN)σ(hN)µσ(gN, hN) ) by (9)

= −φ◦σ(gN) − φ◦σ(hN) + φ◦σ(ghN).

Since −φ◦σ(N) = −φ(1) = 0, −φ◦σ : G/N → U is a 1-cochain. Thus, [ψ] = [φ¯◦µσ] and so ker(infb) ⊆ im(τ). This completes our proof of the exactness of (1).

Appendix D

We prove Lemma 4.2.

0 e en Proof. (i) G := h[x, y]|x, y ∈ Gi = {[x1, y1] 1 ...[xn, yn] |xi, yi ∈ G, ei = ±1, n ∈ N}. However, since [x, y]−1 = [y, x] where x, y ∈ G0, the above set can be rewritten as

{[x1, y1]...[xn, yn]|xi, yi ∈ G, n ∈ N}.

20 e en (ii) Since [A, B] := h{[a, b]|a ∈ A, b ∈ B}i, we have [A, B] ≤ G. Let [a1, b1] 1 ...[an, bn] ∈ [A, B] e where ai ∈ A, bi ∈ B, ei = ±1, n ∈ N. We can show that for all g ∈ G and for each [ai, bi] i , we have the following:

ei −1 −1 ei g[ai, bi] = [gaig , gbig ] g.

Thus,

e1 en −1 −1 e1 −1 −1 en g[a1, b1] ·....·[an, bn] = [ga1g , gb1g ] ·...·[gang , gbng ] g,

Hence, [A, B] E G given that A, B E G.

(iii) Let g1N, g2N ∈ G/N. Then we have:

g1NggN = g1g2N

0 = g1g2[g2, g1]N since G ≤ N

−1 −1 = g1g2g2 g1 g2g1N

= g2g1N

= g2Ng1N

Thus, G/N is abelian.

Appendix E

We provide some background on group extensions and their connection to group cohomology. Ulti- mately we use this to prove that Ext(C,U) = H2(C,U) for any cyclic group C and trivial C-module U. The material in this appendix is from [13, Section 11.7].

Let A be an abelian group and G any group. Then, a group extension of A by G is an exact sequence

E : 1 A i M π G 1.

For convenience, we can identify i(A) with A and therefore treat A as a subgroup of M.

Two extensions E1 and E2 are equivalent if ∃ a homomorphism α making the following diagram commute:

21 π1 E1 : 1 A M1 G 1 α

π2 E2 : 1 A M2 G 1 Now, consider the following extension (where A is regarded as a subgroup of M):

E : 1 A M π G 1.

Let σ : G → M be a lift. We note that since σ is a lift, for all x, y ∈ G,

σ(x)σ(y) = µσ(x, y)σ(xy), where µσ(x, y) ∈ A. In this sense, every lift σ : G → M induces a map µσ : G × G → A such that

−1 µσ(x, y) = σ(x)σ(y)σ(xy) . (10)

Also, for all x ∈ G and a ∈ A, we have the following:

π(σ(x)aσ(x)−1) = π(σ(x))π(a)π(σ(x))−1

= π(σ(x))π(σ(x))−1

= 1

This implies that σ(x)aσ(x)−1 ∈ A for all x ∈ G. Thus, we can deﬁne a binary operation ∗ : G×A → A with x ∗ a := σ(x)aσ(x)−1. We now prove that A forms a G-module under the action deﬁned by ∗. For all x, y ∈ G and a, b ∈ A we have the following:

(i) x ∗ 1 = σ(x)1σ(x)−1 = 1

(ii) x ∗ (y ∗ a) = x ∗ (σ(y)aσ(y)−1)

= σ(x)σ(y)aσ(y)−1σ(x)−1

= σ(x)σ(y)a(σ(x)σ(y))−1

−1 = µσ(x, y)σ(xy)a(µσ(x, y)σ(xy))

−1 −1 = µσ(x, y)σ(xy)aσ(xy) µσ(x, y)

−1 −1 −1 = σ(xy)aσ(xy) since σ(xy)aσ(xy) , µσ(x, y) and µσ(x, y) ∈ A

= xy ∗ a

(iii) x ∗ (ab) = σ(x)abσ(x)−1

= (σ(x)aσ(x)−1)(σ(x)bσ(x)−1)

= (x ∗ a)(x ∗ b)

22 Thus, A is a G-module under the action ∗. We now show the map ∗ is independent of our choice of lift from G to M. Let σ : G → M and τ : G → M both be lifts. Then, for all x ∈ G, we have

τ(x) = ax·σ(x) where ax ∈ A. This implies that for all a ∈ A and x ∈ G,

−1 −1 τ(x)aτ(x) = axσ(x)a(axσ(x))

−1 −1 = axσ(x)aσ(x) ax

−1 −1 −1 = σ(x)aσ(x) since ax, σ(x)aσ(x) and ax ∈ A

Thus, our deﬁnition of ∗ is independent of our choice of lift. So the G-module structure endowed on A by ∗ is said to be the module structure induced by the extension E.

We now prove that equivalent extensions induce the same G-module structure on A. Let E (from above) and E2 be equivalent extensions: E : 1 A M π G 1 α

π2 E2 : 1 A M2 G 1 where α is a homomorphism between M and M2 such that the above diagram commutes. Let ∗ and

∗2 denote the actions of G on A induced by E and E2 respectively. Again, let σ : G → M be a lift.

Then we deﬁne the map σ2 : G → M2 with σ2(x) = α(σ(x)). Now, since π2(σ2(x)) = π2(α(σ(x))) =

π(σ(x)) = x, we have that σ2 is a lift. Moreover,

−1 x∗2a = σ2(x)aσ2(x)

= α(σ(x))aα(σ(x))−1

= α(σ(x))α(a)α(σ(x))−1

= α(σ(x)aσ(x)−1)

= σ(x)aσ(x)−1 since σ(x)aσ(x)−1 ∈ A

= x∗a.

Thus E and E2 induce the same G-module structure on A. So the G-module structure endowed by ∗ on A can be considered as induced by the whole equivalence class represented by E.

We now prove the following Theorem on group extensions:

23 Theorem E.1. (Classiﬁcation Theorem on Group Extensions) Let A be a G-module under some action ∗. Let opext(G, A) denote the set of all equivalence classes of extensions of A by G whose elements induce the G-module structure given by ∗. There is a one to one correspondence between the elements of opext(G, A) and those of H2(G, A).

Proof. We deﬁne a map, θ : opext(G, A) → H2(G, A) with

θ([E]) := [µσ] where E is a represenatative of the equivalence class of group extensions denoted by [E]:

E : 1 A M π G 1.

σ is some normalised lift from G to M and µσ is deﬁned as done above in (10). We proceed by showing that the map θ is well-deﬁned:

2 −1 (i) We ﬁrst prove that µσ ∈ Z (G, A). Let x, y, z ∈ G. Then µσ(1, x) = σ(1)σ(x)σ(x) = 1 = −1 σ(x)σ(1)σ(x) = µσ(x, 1). Moreover,

−1 −1 (x ∗ µσ(y, z))µσ(x, yz) = (x ∗ (σ(y)σ(z)σ(yz) )(σ(x)σ(yz)σ(xyz) )

= σ(x)σ(y)σ(z)σ(yz)−1σ(x)−1σ(x)σ(yz)σ(xyz)−1

= σ(x)σ(y)σ(z)σ(xyz)−1

= (σ(x)σ(y)σ(xy)−1)(σ(xy)σ(z)σ(xyz)−1)

= µσ(x, y)µσ(xy, z)

2 Thus µσ ∈ Z (G, A).

(ii) We now prove that θ([E]) is independent of our choice of the normalised lift from G to M. Let

24 σ and τ be normalised lifts from G to M. Then for all x, y ∈ G,

−1 µτ (x, y) = τ(x)τ(y)τ(xy)

−1 = axσ(x)ayσ(y)(axyσ(xy)) where ax, ay, axy ∈ A

−1 −1 = axσ(x)ayσ(y)σ(xy) axy

−1 −1 −1 = axσ(x)ayσ(y)σ(y) σ(x) µσ(x, y)axy

−1 −1 = axσ(x)ayσ(x) µσ(x, y)axy

−1 = ax(x ∗ ay)µσ(x, y)axy

−1 −1 = ax(x ∗ ay)axy µσ(x, y) since both µσ(x, y) and axy ∈ A

−1 We deﬁne φ : G → A such that φ(x) := ax for all x ∈ G. Since σ and τ are both normalised, φ(1) = 1 and so φ is a 1-cochain. From above, we have

−1 −1 −1 −1 µσ(x, y) = φ(x) (x ∗ φ(y) )φ(xy)µσ(x, y) = φ(x) (x ∗ φ(y)) φ(xy)µσ(x, y).

And hence, [µσ] = [µτ ], making θ([E]) independent of our choice of the normalised lift.

(iii) We now prove that θ([E]) is independent of our choice of representatives of [E]. Let E1 and E2 be the following equivalent extensions:

π1 E1 : 1 A M1 G 1 α

π2 E2 : 1 A M2 G 1

Let σ1 be a normalised lift from G to M1. Then, as shown earlier, σ2 := α◦σ1 deﬁnes a lift from

G to M2. Moreover, since α(σ1(1)) = α(1) = 1, σ2 is also normalised. And, for all x, y ∈ G,

−1 µσ2 (x, y) = σ2(x)σ2(y)σ2(xy)

−1 = α(σ1(x))α(σ1(y))α(σ1(xy))

−1 = α(σ1(x)σ1(y)σ1(xy) )

−1 −1 = σ1(x)σ1(y)σ1(xy) since σ1(x)σ1(y)σ1(xy) ∈ A

= µσ1 (x, y)

Thus µσ2 = µσ1 implying that [µσ2 ] = [µσ1 ].

From (i)-(iii) we se that the map θ is well-deﬁned.

We now show that θ is one-to-one. Let [E], [E0] ∈ opext(G, A) where

25 E : 1 A M π G 1

0 E0 : 1 A M 0 π G 1 0 0 0 Let σ : G → M and σ : G → M be normalised lifts. Assume [µσ] = [µσ]. Then,

−1 −1 µσ0 (x, y) = µσ(x, y)φ(x) (x ∗ φ(x)) φ(xy) (11) where φ : G → A is some 1-cochain.

We now deﬁne a map α : M → M 0 with α(aσ(x)) = aφ(x)σ0(x) for all x ∈ G and a ∈ A. Then for all aσ(x) and bσ(y) ∈ M, we have the following:

−1 α(aσ(x)bσ(y)) = α(aσ(x)bσ(x) µσ(x, y)σ(xy))

= α(a(x ∗ b)µσ(x, y)σ(xy))

0 = a(x ∗ b)µσ(x, y)φ(xy)σ (xy)

0 = aφ(xy)(x ∗ b)µσ(x, y)σ (xy) since a, φ(xy) and x ∗ b ∈ A

−1 −1 0 = aφ(xy)(x ∗ b)µσ(x, y)φ(x)φ(x) (x ∗ φ(y))(x ∗ φ(y)) σ (xy)

−1 −1 0 = aφ(x)(x ∗ b)(x ∗ φ(y))µσ(x, y)((x ∗ φ(y))φ(xy) φ(x)) σ (xy)

0 0 −1 0 0 −1 0 = aφ(x)(σ (x)bσ (x) )(σ (x)φ(y)σ (x) )µσ0 (x, y)σ (xy) by (11)

= aφ(x)(σ0(x)bσ0(x)−1)(σ0(x)φ(y)σ0(x)−1)(σ0(x)σ0(y)σ0(xy)−1)σ0(xy)

= (aφ(x)(σ0(x))(bφ(y)σ0(y))

= α(aσ(x))α(bσ(y))

Thus, α is a homomorphism. Moreover, for all a ∈ A and x ∈ G,

• α(a) = α(aσ(1)) = aφ(1)σ0(1) = a

• π(aσ(x)) = x = π0(aφ(x)σ0(x)) = π0(α(aσ(x))

This shows that α is a homomorphism that makes the diagram shown above commute. Thus, [E] = [E0], implying that θ is injective.

We now show that θ is surjective. Let f : G × G → A be a normalised two cocycle. Deﬁne M to be the set A × G. We endow M with the following binary operation:

(a, x)·(b, y) := (a(x ∗ b)f(x, y), xy)

26 for all a, b ∈ A and x, y ∈ G. It can be shown that M forms a group under the above operation with ∼ A × {0} E E and E/A = G. Thus we can construct the following short exact sequence, making E an extension of A by G:

E : 1 A i M π G 1

In the above diagram, i(a) := (a, 1) for all a ∈ A and π((a, x)) := x for all a ∈ A and x ∈ G.

We deﬁne a lift σ : G → M with σ(x) = (1, x) for all x ∈ G. Clearly σ is a normalised lift. We now note the following. For all x ∈ G and a ∈ A,

i−1(σ(x)i(a)σ(x)−1) = i−1((1, x)(a, 1)(1, x)−1)

= i−1((1, x)(a, 1)(x−1 ∗ f(x, x−1)−1, x−1))

= i−1(((x ∗ a)f(x, 1), x)(x−1 ∗ f(x, x−1)−1, x−1))

= i−1(((x ∗ a), x)(x−1 ∗ f(x, x−1)−1, x−1))

= i−1((x ∗ a)(x ∗ (x−1 ∗ f(x, x−1)−1))f(x, x−1), 1)

= i−1(x ∗ a, 1)

= x ∗ a

Thus, the action of G on A induced by E matches the G module structure of A. This implies that [E] ∈ opext(G, A). For all x, y ∈ G, we also have the following:

−1 −1 µσ(x, y) = i (σ(x)σ(y)σ(xy) )

= i−1((1, x)(1, y)(1, xy)−1)

= i−1((1f(x, y), xy)(1, xy)−1)

= i−1((f(x, y), xy)(1, xy)−1)

= i−1((f(x, y), xy)(1((xy)−1 ∗ (f(xy, (xy)−1)−1)), (xy)−1))

= i−1((f(x, y)f(xy, (xy)−1)−1f(xy, (xy)−1), 1)

= i−1(f(x, y), 1)

= f(x, y)

Thus θ([E]) = [µσ] = [f] and we conclude that θ is surjective.

Corollary E.2. Let A be a trivial G-module. Then the elements of H2(G, A) are in one to one correspondence with {[E]|E is a central extension of A by G}.

27 Proof. Assume [E] ∈ opext(G, A) where E is the following extension:

E : 1 A M π G 1.

Let σ be some lift from G to M. Then, for all x ∈ G and a ∈ A, σ(x)aσ(x)−1 = a, or equivalently σ(x)a = aσ(x). Thus, for all bσ(y) ∈ M where b ∈ A and y ∈ G, and for all a ∈ A, we have that abσ(y) = baσ(y) = bσ(y)a. This shows that E is a central extension of A by G.

We now assume that [E0] is s.t E0 is a central extension of A by G:

0 E0 : 1 A M 0 π G 1.

Let σ0 : G → M 0 be a lift. Note that for all a ∈ A and x ∈ G, σ0(x)aσ0(x)−1 = σ0(x)σ0(x)−1a = a. Thus, [E0] ∈ opext(G, A). The result follows by theorem 3.5.

e 2 e We now return to showing that Ext(Zpi i ,U) = H (Zpi i ,U), or more generally that Ext(C,U) = H2(C,U) for any cyclic group C where the action of C on U is assumed to be trivial. We let 2 n 2 C = {c, c , ..., c } for some n ∈ N and [f] ∈ H (C,U), then [f] = θ([E]), where E is some central extension of U by C:

E : 1 U M π C 1.

x Let σ : C → M be some normalised lift. Then, [µσ] = [f]. Let m1, m2 ∈ M. Then m1 = u1σ(c ) and y m2 = u2σ(c ) for some u1, u2 ∈ U and x, y ∈ N. Thus,

x y m1m2 = u1σ(c )u2σ(c )

x y = u2u1σ(c )σ(c )

x y = u2u1σ(c) uxσ(c) uy for some ux, uy ∈ U

y x = u2u1σ(c) uyσ(c) ux

y x = u2u1σ(c )σ(c )

y x = u2σ(c )u1σ(c )

= m2m1.

x y Hence, M is abelian. This implies that for all c , c ∈ C where x, y ∈ N,

x y x y x y −1 y x y x −1 y x µσ(c , c ) = σ(c )σ(c )σ(c c ) = σ(c )σ(c )σ(c c ) = µσ(c , c ).

28 So, µσ is symmetric. Since [f] = [µσ], we have [f] ∈ Ext(C,U). This shows that Ext(C,U) =

2 e 2 e H (C,U). Applying this to our speciﬁc case, we ﬁnd that Ext(Zpi i ,U) = H (Zpi i ,U).

Appendix F

We prove Lemma 4.6.

e en Proof. (i) Let [f1, r1] 1 ...[fn, rn] ∈ [F,R] for some n ∈ N, where ei ∈ {±1} for all i. Note that

e en since R E F , we have [f1, r1] 1 ...[fn, rn] ∈ R for all i. This implies [F,R] ⊆ R. Since [F,R] E F by Lemma 4.2, and [F,R] ⊆ R ⊆ F , we have [F,R] E R.

0 0 (ii) Since [F,R] E F by Lemma 4.2, and [F,R] ⊆ (R ∩ F ) ⊆ F , it follows that [F,R] E (R ∩ F ).

0 0 (iii) Clearly (R ∩ F ) ≤ R. Now let [x1, y1]·...·[xn, yn] ∈ (R ∩ F ) where xi, yi ∈ F and n ∈ N. We can

show that for all r ∈ R and for each [xi, yi], we have the following:

−1 −1 r[xi, yi] = [rxir , ryir ]r

Thus, −1 −1 −1 −1 r[x1, y1]·...·[xn, yn] = [rx1r , ry1r ]·...·[rxnr , rynr ]r,

−1 −1 −1 −1 0 Clearly [rx1r , ry1r ]·...·[rxnr , rynr ] ∈ F . −1 −1 −1 −1 −1 Note that [rx1r , ry1r ]·...·[rxnr , rynr ] = r[x1, y1]·...·[xn, yn]r ∈ R. −1 −1 −1 −1 0 0 Thus, [rx1r , ry1r ]·...·[rxnr rynr ] ∈ (R ∩ F ). This implies that (R ∩ F ) E R.

0 0 0 0 0 (iv) Since F E F by Lemma 4.2, and F ⊆ RF ⊆ F , it follows that F E RF .

Appendix G

2 The matrices corresponding to coboundaries in Z (G, C2) are exactly those of the form:

M∂φ = [∂φ(gi, gj)]1≤i,j≤4,

29 where φ : G → C2 is a 1-cochain. Using the binary operation of C2, we can simplify the above matrix to the following one:

1 a b ab   1 1 1 1 1     a  1 1 φ(a)φ(b)φ(ab) φ(a)φ(ab)φ(b)  M∂φ =     b  1 φ(b)φ(a)φ(ab) 1 φ(b)φ(ab)φ(a)    ab 1 φ(ab)φ(a)φ(b) φ(ab)φ(b)φ(a) 1

Since C2 is abelian, we can let γ := φ(a)φ(b)φ(ab) = φ(a)φ(ab)φ(b) = φ(b)φ(a)φ(ab) = φ(b)φ(ab)φ(a) = φ(ab)φ(a)φ(b) = φ(ab)φ(b)φ(a).

2 Thus, we ﬁnd there are exactly two coboundaries in Z (G, C2). These have matrix representations of the form: 1 a b ab   1 1 1 1 1     a  1 1 γ γ  M∂φ =     b  1 γ 1 γ    ab 1 γ γ 1 where γ ∈ {±1}.

Appendix H

We evaluate the value of µσ (as deﬁned on page 14) at all possible arguments:

−1 2 • Let f = 1, a, b. Then, µσ(f, f) = σ(f)σ(f)σ(1) = f S = S.

• Let f = 1, a, b, ab. Then, µσ(1, f) = S = µσ(f, 1) since µσ is a normalised cocycle.

−1 −1 −1 • µσ(a, b) = σ(a)σ(b)σ(ab) = abb a S = S.

−1 2 −1 2 • µσ(a, ab) = σ(a)σ(ab)σ(b) = a bb S = a S = S.

−1 2 −1 −1 2 • Since R/S is central in F/S, µσ(ab, b) = σ(ab)σ(b)σ(a) = ab a S = aa b S = S.

30 −1 −1 • µσ(b, ab) = σ(b)σ(ab)σ(a) = baba S. Since R/S is central in F/S, we also have the following:

[b−1, a]−1baba−1S = a−1bab−1baba−1S

= a−1ba2ba−1S

= a−1b2a−1a2S

= a−2b2a2S

= b2S

= S,

−1 which implies that µσ(b, ab) = [b , a]S.

−1 0 • µσ(ab, ab) = σ(ab)σ(ab)σ(1) = ababS. Since R/S is central in F/S, and since F ≤ R, we have the following:

abab[b−1, a]−1S = ababa−1bab−1S

= ababa−1b−1bbab−1S

= ababa−1b−1b2ab−1S

= abab−1aba−1b−1b2S

= abb−1aba−1ab−1b2S

= a2b2S

= S,

−1 which implies that µσ(ab, ab) = [b , a]S.

−1 −1 −1 −1 • µσ(b, a) = σ(b)σ(a)σ(ba) = σ(b)σ(a)σ(ab) = bab a S. Since R/S is central in F/S, and since F 0 ≤ R, we have the following:

bab−1a−1[b−1, a]−1S = bab−1a−1a−1bab−1S

= a−1bab−1bab−1a−1S

= a−1baab−1a−1S

= a−1ba2b−1a−1S

= a−1bb−1a−1a2S

= a−1a−1a2S

= S,

31 −1 which implies that µσ(b, a) = [b , a]S.

−1 −1 0 • µσ(ab, a) = σ(ab)σ(a)σ(b) = abab S. Since R/S is central in F/S, and since F ≤ R, we have the following:

abab−1[b−1, a]−1S = abab−1a−1bab−1S

= aab−1a−1bbab−1S

= a2b−1a−1b2ab−1S

= a2b2b−1a−1ab−1S

= a2S

= S,

−1 which implies that µσ(ab, a) = [b , a]S.