Cocyclic Hadamard matrices over Latin rectangles

Cocyclic Hadamard matrices over Latin rectangles V. Álvarez, R.M. Falcón, M.D. Frau, F. Gudiel, B. Güemes Department of Applied Mathematics I., University of Seville, Spain ab s t r a c t In the literature, the theory of cocyclic Hadamard matrices has always been developed over finite groups. This paper introduces the natural generalization of this theory to be developed over Latin rectangles. In this regard, once we introduce the concept of binary cocycle over a given Latin rectangle, we expose examples of Hadamard matrices that are not cocyclic over finite groups but they are over Latin rectangles. Since it is also shown that not every Hadamard matrix is cocyclic over a Latin rectangle, we focus on answering both problems of existence of Hadamard matrices that are cocyclic over a given Latin rectangle and also its reciprocal, that is, the existence of Latin rectangles over which a given Hadamard matrix is cocyclic. We prove in particular that every Latin square over which a Hadamard matrix is cocyclic must be the multiplication table of a loop (not necessarily associative). Besides, we prove the existence of cocyclic Hadamard matrices over non-associative loops of order 2t+3, for all positive integer t > 0. 1. Introduction A(binary) Hadamard matrix H of order nis an n×narray with every entry either 1 or −1, which satisfies HHT=nIn, where Indenotes the identity matrix of order n. It is well-known that nhas to be necessarily 1,2 or a multiple of 4, but there is no certainty whether such a Hadamard matrix exists at every possible order. The Hadamard Conjecture asserts that there exists a Hadamard matrix of order 4tfor every natural number t. Two Hadamard matrices are said to be equivalent if they are equal up to permutations or negation of their row and columns. There exist so many different constructions for Hadamard matrices: Sylvester, Paley, Williamson, Ito, Goethals–Seidel, one and two circulant cores or cocyclic matrices, amongst others (see [15]). With respect to the latter, Horadam and de Launey introduced [16] the use of cocycles and cocyclic matrices within the theory of block designs and showed [17] that the cocyclic framework could provide a structural approach to resolve the Hadamard Conjecture. Since then, the theory of cocyclic Hadamard matrices has been widely developed, but always under the assumption of working on finite groups. Their distribution into equivalence classes is explicitly known for order n<40 [8]. This paper generalizes this theory by introducing the concept of cocyclic Hadamard matrices on r×nLatin rectangles. Latin squares were already used in the classical theory of Hadamard matrices to construct symmetric Hadamard matrices of order 36 [6,7,13]. The paper is organized as follows. In Section 2, we expose some preliminary concepts and results on cocyclic Hadamard matrices, Latin rectangles and quasigroups that are used throughout our study. We introduce in Section 3the concept of cocycle over a Latin rectangle and that of cocyclic Hadamard matrix over a Latin rectangle. Section 4deals with the set of Hadamard matrices that are cocyclic over a given Latin rectangle, whereas Section 5focuses on the reciprocal problem, that is, on the set of Latin rectangles over which a given Hadamard matrix is cocyclic. Finally, since this paper has a high dependence on notation, a glossary of symbols is shown in the Appendix. 2. Preliminaries Let us review some basic concepts and results on cocyclic Hadamard matrices, Latin rectangles and quasigroups that are used throughout the paper. We refer the reader to [9,11,15,23] for more details about these topics. 2.1. Cocyclic Hadamard matrices Let Gbe a finite group and let Z2be the finite set {±1}considered as the cyclic group of order 2. Any map ψ:G×G→Z2may be represented by the square matrix Mψ=(ψ(a,b))a,b∈G, whose rows and columns are indexed by the elements of the group Gunder some fixed ordering. If there exists a function φ:G→Z2such that ψ(a,b)=φ(ab), for all a,b∈G, so that the matrix Mψis Hadamard, then the latter is said to be a Hadamard matrix developed over the group G. Turyn [29] proved that, if G is cyclic, then Mψis Hadamard only if the order of Gis 4t2, for some odd positive integer t∈Nthat is not a prime power. The only known equivalence class of Hadamard matrices developed over a cyclic group Z4t2is represented by the following circulant matrix defined for t=1 ⎛ ⎜ ⎝ + + + − + + − + + − + + − + + + ⎞ ⎟ ⎠ where, from here on, the symbols +and −represent, respectively, the entries 1 and −1. The Circulant Hadamard Conjecture [27] asserts that there does not exist any Hadamard matrix over a cyclic group Z4t2, for t>1. It is known [5,22] that this is true for all positive integer t<11715 and also that there exist 948 open cases for t≤1013. In the literature, this fact has dissuaded the authors from searching for cocyclic Hadamard matrices over cyclic groups. Horadam and de Launey [16,17] introduced the concept of (2-dimensional binary) cocycle over a finite group Gas a map ψ:G×G→Z2such that ψ(ab,c)=ψ(a,b)ψ(b,c)ψ(a,bc),for all a,b,c∈G.(1) Particularly, ψ(1,1) =ψ(a,1) =ψ(1,a), for all a∈G. If ψ(1,1) =1, then the cocycle ψis said to be normalized. The corresponding matrix Mψis called (pure) cocyclic over the group G. If this is Hadamard, then it is said to be a (pure) cocyclic Hadamard matrix over the group G. Baliga and Horadam [2] dealt with the existence of cocyclic Hadamard matrices over the groups Zt×Z2 2,fortodd.Shortlyafter,Flannery[12]provedconstructivelytheexistenceofcocyclicHadamard matrices over the dihedral group D4t:= ⟨ a,b|a2t=b2=(ab)2=1⟩,(2) for all positive integer t≤11. He also established a series of conditions to ensure the existence of cocyclicHadamardmatricesoverabelian,dicyclic,anddihedralgroupsovercertain orders. Alternative constructions have recently been obtained by Álvarez et al. [1]. The Kronecker or direct product of two cocyclic Hadamard matrices is also cocyclic Hadamard over the direct product of the involved pair of groups [12]. More specifically, if ψand ψ′are respective cocycles over two finite groups Gand G′, then, it is known [10] that the map ψ⊗ψ′defined as ψ⊗ψ′((a,a′),(b,b′)) =ψ(a,b)ψ′(a′,b′),for all a,b∈G,a′,b′∈G′,(3) is a cocycle over the finite group G⊗G′. If both cocyclic matrices Mψand Mψ′are Hadamard, then the cocyclic matrix Mψ⊗ψ′, which coincides indeed with their direct product, Mψ⊗Mψ′, is also Hadamard. 2.2. Latin rectangles Let rand nbe two positive integers such that r≤n. An r×n Latin rectangle is an r×narray L=(li,j) in which each cell contains one symbol of the set [n] := {1, . . . , n}and such that each symbol occurs exactly once in each row and at most once in each column. This is a Latin square of order n if r=n, in which case, each symbol of the set [n]also occurs exactly once in each column. Hereafter, let Rr,n denote the set of r×nLatin rectangles. The direct product of two Latin rectangles L=(li,j)∈Rr,nand L′=(l′ i′,j′)∈Rr′,n′is a Latin rectangle L⊗L′∈Rrr′,nn′such that, for each (i,j,i′,j′)∈ [r] × [n] × [r′] × [n′], its cell ((i−1)r′+i′,(j−1)n′+j′)∈ [rr′]×[nn′]contains the symbol (li,j−1)n′+l′ i′,j′∈ [nn′]. Example 1. Let us consider the Latin rectangles L≡2 1 ∈R1,2and L′≡1 2 3 2 3 1 ∈R2,3. Then, L⊗L′≡4 5 6 1 2 3 5 6 4 2 3 1 ∈R2,6. Every Latin rectangle L=(li,j)∈Rr,nis uniquely determined by its set of entries E(L):= {(i,j,li,j)|i≤r,j≤n}.(4) The Latin rectangle Lis said to be reduced if its set of entries contains the triples (1,j,j) and (i,1,i), for all i∈ [r]and j∈ [n]; that is, if all the symbols of the set [n](respectively, [r]) are displayed in naturalorderwithinthefirst row (respectively, first column) of L. Thus, for instance, the Latin rectangleL′∈ R2,3in Example 1 is reduced. Its set of entries is E(L′)= {(1,1,1),(1,2,2),(1,3,3),(2,1,2),(2,2, 3),(2,3,1)}. Let sbe a positive integer such that r≤s≤n. It is said that a Latin rectangle L∈Rr,ncan be extended to a Latin rectangle L′∈Rs,nif E(L)⊆E(L′). Marshall Hall [14] proved that every Latin rectangle in Rr,ncan be extended to at least one Latin rectangle in Rs,n. Example 2. The following two Latin rectangles, L∈R1,8and L′∈R2,8, satisfy that E(L)⊆E(L′). L≡2 3 4 1 6 7 8 5 L′≡23416785 35687214 Let Sndenote the symmetric group on nelements. For each positive integer i≤r, the ith row of a Latin rectangle L=(li,j)∈Rr,ncan be interpreted as the permutation π∈Snsuch that π(j)=li,j, for all j≤n. The disjoint cycles in which such a permutation decomposes determine the cycles of the row. Thus, for instance, the first row of the Latin rectangle L′in Example 2 constitutes the permutation (1234)(5678) ∈S8, whereas its second row determines the permutation (136257)(48) ∈S8. Permutations of rows, columns and symbols of Latin rectangles preserve the set Rr,n. In this regard, two Latin rectangles L=(li,j) and L′=(l′ i,j) in Rr,nare said to be isotopic if there exists a triple (f,g,h)∈Sr×Sn×Snsuch that l′ f(i),g(j)=h(li,j), for all i≤rand j≤n. The set of entries of the isotope L′is E(L′)= {(f(i),g(j),h(li,j)) |(i,j,li,j)∈E(L)}. The triple (f,g,h) is an isotopism between Land L′. This constitutes an isomorphism if r=nand f=g=h. In this case, the Latin squares L and L′are said to be isomorphic. To be isotopic (respectively, isomorphic) are equivalence relations among Latin rectangles (respectively, Latin squares). Nowadays, the distribution of Latin squares into isotopism and isomorphism classes is known [18,19,25] for order up to 11, whereas that of r×nLatin rectangles into isotopism classes is known [24] for r≤n≤9. 2.3. Quasigroups Aquasigroup [26] is a pair Q=(S,·) formed by a set Sthat is endowed with a binary operation ·, such that the equations a·x=band y·a=bhave unique solutions for xand yin S, for all a,b∈S. Hence, Qhas left- and right-division, which are respectively denoted \and /(that is, x=a\bin the first equation, and y=b/ain the second one). Every associative quasigroup constitutes a group. The quasigroup Qis a loop if there exists a unit element e∈Ssuch that a·e=e·a=a, for all a∈S. The quasigroup Qis finite if the set Sis. In such a case, the order of the former is defined as the cardinality of the latter. The quasigroup Qis monogenic if it is generated by a single element a∈S. Every monogenic associative quasigroup is a cyclic group. From here on, we denote by Qnthe set of finite quasigroups having [n]as their finite set of symbols. Every Latin square L=(li,j)∈Rn,nconstitutes, therefore, the multiplication table of a finite quasigroup Q=([n],·)∈Qn, where i·j=li,j, for all i,j≤n. Besides, any reduced Latin square constitutes the multiplication table of a loop. Transpositions and isotopisms of Latin squares preserve the set Qn. In this regard, the transpose of a quasigroup Q=([n],·)∈Qnis the quasigroup Qt=([n],◦) that is defined so that i◦j=j·i, for all i,j≤n. Further, two quasigroups Q=([n],·) and Q′=([n],◦) are said to be isotopic if there exist three permutations f,gand hin Snsuch that f(i)◦g(j)=h(i·j), for all i,j∈ [n]. The triple (f,g,h) is an isotopism from Qto Q′. This constitutes an isomorphism if f=g=h, in which case the quasigroups are said to be isomorphic. 3. Cocycles over Latin rectangles From here on, let L=(li,j) be a Latin rectangle in Rr,n. We define the subset of symbols S(L):= [r]∪{li,j|i,j≤r} ⊆ [n].(5) As such, this set is composed partly of the row labels of the array Land partly of the symbols appearing in the left r×rsubarray within the latter (see Example 3). Remark in particular that S(L)= [n] whenever Lis a Latin square of order n. Example 3. Let L∈R1,8and L′∈R2,8be the Latin rectangles in Example 2. In order to determine the set S(L), we focus on the left 1 ×1 subarray of L, which we highlight below. 23416785 Accordingto (5), thesetS(L) is formedbytherow label1andthe symbol 2appearingin the highlighted subarray. That is, S(L)= {1}∪{2} = {1,2}. Similarly, in order to determine the set S(L′), we focus on the left 2 ×2 subarray of L′, which we highlight below. 234 1 6 7 8 5 356 8 7 2 1 4 Thus, the set S(L′) is formed by the row labels 1 and 2, and the symbols 2, 3 and 5 appearing in the highlighted subarray. Hence, S(L′)= {1,2,3,5}. Definition (5) enables us to introduce the concept of cocycle over a Latin rectangle. We say that a function ψ:S(L)× [n] → Z2is a cocycle over the Latin rectangle Lif ψ(li,j,k)=ψ(i,j)ψ(j,k)ψ(i,lj,k),for all i,j≤rand k≤n.(6) In particular, if Lis a Latin square of order n, which constitutes in turn the multiplication table of a quasigroup Q=([n],·)∈Qn, then Condition (6) is equivalent to ψ(i·j,k)=ψ(i,j)ψ(j,k)ψ(i,j·k),for all i,j,k≤n.(7) In such a case, we also say that the function ψis a cocycle over the quasigroup Q. This generalizes the classical notion of cocycle over a finite group, which arises when Qis associative. Example 4. Let us consider the Latin rectangle L=(li,j)≡2 1 ∈R1,2. In particular, S(L)= {1,2}. Then, the function ψ: [2] × [2] → Z2defined so that ψ(1,1) =1 and ψ(1,2) =ψ(2,1) =ψ(2,2) = −1 is a cocycle over L, because •ψ(l1,1,1) =ψ(2,1) = −1=ψ(1,1)ψ(1,1)ψ(1,2) =ψ(1,1)ψ(1,1)ψ(1,l1,1), and •ψ(l1,1,2) =ψ(2,2) = −1=ψ(1,1)ψ(1,2)ψ(1,1) =ψ(1,1)ψ(1,2)ψ(1,l1,2). From here on, we denote Z(L) the set of cocycles over the Latin rectangle L. The following result establishes lower and upper bounds for the size of such a set. Lemma 5. Let L ∈Rr,n. Then, |Z(L)|is an even positive integer such that 2≤ |Z(L)| ≤ 2|S(L)|·n. Proof. From (6), it is straightforwardly verified that a function ψis a cocycle over Lif and only if the function −ψis. Hence, |Z(L)|must be even. Further, its lower bound follows from the fact that both constant maps ψand −ψ, which are defined so that ψ(i,j)=1, for all (i,j)∈S(L)× [n], are always cocycles over L. Finally, the upper bound holds readily from the domain and image of every cocycle over an r×nLatin rectangle. □ The following example shows a family of reduced Latin rectangles whose set of cocycles fits the lower bound indicated in Lemma 5. Example 6. Let L=(l1,j) be the reduced Latin rectangle in the set R1,n. That is, l1,j=j, for all j≤n. Let ψ∈Z(L). According to (6), for each k≤n, we have that ψ(1,k)=ψ(l1,1,k)=ψ(1,1)ψ(1,k)ψ(1,l1,k)=ψ(1,1)ψ(1,k)ψ(1,k)=ψ(1,1). As a consequence, ψmust be a constant map and hence, |Z(L)| = 2. Unlike this lower bound, the upper bound in Lemma 5 is just fitted by the only Latin rectangle existing in the set R1,1. In order to prove it and also to determine the exact size of the set of cocycles of each Latin rectangle, we introduce here the concept of cocyclic degree-of-freedom of a Latin rectangle L∈Rr,nas the minimum number df(L) of pairs (i,j)∈S(L)× [n]that are required to determine uniquely such a cocycle by keeping in mind to this end the set of constraints derived from Condition (6). The following example illustrates this definition. Example 7. Let L∈R1,2be the Latin rectangle in Example 4. Condition (6) implies that every cocycle φ∈Z(L) satisfies that •φ(2,1) =φ(l1,1,1) =φ(1,1)φ(1,1)φ(1,l1,1)=φ(1,1)φ(1,1)φ(1,2) =φ(1,2), and •φ(2,2) =φ(l1,1,2) =φ(1,1)φ(1,2)φ(1,l1,2)=φ(1,1)φ(1,2)φ(1,1) =φ(1,2). Hence, φ(1,2) =φ(2,1) =φ(2,2). Observe that none constraint concerning φ(1,1) derives from Condition (6). As a consequence, df(L)=2 and hence, keeping in mind the image Z2of any cocycle over L, we have that |Z(L)| = 22=4. More specifically, we have that Z(L)= {φ1, φ2, φ3, φ4}, where •φ1=ψ, the cocycle defined in Example 4. •φ2(i,j)=1, for all (i,j)∈ [2]×[2]. •φ3= −φ1. •φ4= −φ2. This example enables us to ensure in particular that isotopisms of Latin rectangles do not preserve the size of their cocycles, because, from Example 6, the reduced Latin rectangle in the set R1,2(which is isotopic to L) has only two cocycles. As we have illustrated in Example 7, this notion of cocyclic degree-of-freedom is well-defined because it does not depend on any cocycle, but only on the entries of the Latin rectangle under consideration. Further, the following result holds straightforwardly from such a definition. Lemma 8. Let L ∈Rr,n. Then, |Z(L)| = 2df(L), where 1≤df(L)≤ |S(L)| · n. The following result indicates the only case for which the upper bounds of both Lemmas 5 and 8 are tight. Proposition 9. Let L ∈Rr,n. Then, df(L)= |S(L)| · n if and only if r =n=1. Proof. Under our assumptions, we already consider that 1 ≤r≤n. So, suppose L=(li,j)∈Rr,nto be such that n≥2. According to the definition of the cocyclic degree-of-freedom of L, we have that df(L)= |S(L)| · nif and only if no constraints derive from Condition (6) apart from the trivial ones 1=1 and −1= −1. Keeping in mind the image Z2of any cocycle over L, this is possible if and only if, for each triple (i,j,k)∈ [r]×[r]×[n], one of the following three possibilities holds. 1. (li,j,k)=(i,j) and (j,k)=(i,lj,k). 2. (li,j,k)=(i,lj,k) and (i,j)=(j,k). 3. (li,j,k)=(j,k) and (i,j)=(i,lj,k). Observe that each one of the first two possibilities implies that i=j=k. Suppose, however, that i=j<k. This is possible because we consider n≥2. Then, the third possibility that we have just enumerated implies that li,i=i=li,k, which is not possible because of the Latin rectangle condition. Hence, n=1 and the result holds. □ Let us introduce now the notion of cocyclic matrix over a Latin rectangle L∈Rr,nin a similar way that the classical theory over finite groups does. To this end, suppose the symbols of the set S(L) to be relabeled by following the natural order as e1, . . . , e|S(L)|. We define the cocyclic matrix Mψof a cocycle ψ∈Z(L) as the |S(L)| × narray whose (i,j)th entry is ψ(ei,j), for all i≤ |S(L)|and j≤n. Thus, for instance, the Latin rectangles L∈R1,8and L′∈R2,8in Example 2 are respectively related to the cocyclic matrices (+ − − + + + − − − + − + + − + − )and ⎛ ⎝ + − − + + + − − − + − + + − + − − − + + + − − + + + + − + − − − ⎞ ⎠. Observe that the respective multiplication of both matrices with their corresponding transposes are eight times the identity matrix of order 2 and 4, respectively. Keeping this fact in mind, for each Latin rectangle L∈Rr,n, we define the set H(L):= {Mψ|ψ∈Z(L) and MψMt ψ=nI|S(L)|}.(8) The following result follows straightforwardly from this definition. Lemma 10. Let L ∈Rr,nbe such that |S(L)| = n. Then, every matrix in H(L)is Hadamard. Under the assumptions of Lemma 10, we call any of the matrices in the set H(L) a cocyclic Hadamard matrix over the Latin rectangle L. Example 11. The matrix ⎛ ⎜ ⎝ ++−− +−+− −++− −−−− ⎞ ⎟ ⎠ is cocyclic Hadamard over the Latin rectangle L≡2 3 4 1 3 4 1 2 ∈R2,4. Observe, in particular, that S(L)= {1,2,3,4}and hence, |S(L)| = 4. The following questions constitute a pair of open problems about the set of cocyclic Hadamard matrices over Latin rectangles that we study separately throughout the next two sections. Problem 12. Let Lbe a Latin rectangle in Rr,n. Does there exist a cocycle ψ∈Z(L) such that the cocyclic matrix Mψis Hadamard? Equivalently, when can we ensure that H(L)= ∅? Problem 13. Let Mbe a Hadamard matrix of order n. Does there exist a Latin rectangle L∈Rr,n such that M∈H(L)? If so, which is the minimum positive integer r∈Nso that the mentioned Latin rectangle exists? Before focusing on these questions, we expose some preliminary results on the topic. Lemma 14. Let L ∈Rr,n. If M ∈H(L)= ∅, then −M∈H(L). Proof. It is enough to consider the correspondence α:Z(L)→Z(L) that is defined so that α(ψ)= −ψ, for all ψ∈Z(L). Particularly, Mψis Hadamard if and only if M−ψ= −Mψis. □ Let L∈Rr,n. The following lemma ensures, for each i≤r, the existence of rdifferent cells within a common column of every cocyclic matrix over L, having all of them the same sign (1 or −1). The location of such cells is uniquely determined by the entries of the ith column of L. Lemma 15. Let M =(mi,j)be a cocyclic matrix over a Latin rectangle L =(li,j)∈Rr,n. For each row i of L, let j be the column of L in which the symbol i occurs. That is, li,j=i. Then, ml1,i,j= · · · = mlr,i,j=mi,j. Proof. Let ψ∈Z(L) be such that M=Mψ. Let (i,j)∈ [r] × [n]be such that li,j=i. The definition of Latin rectangle involves the existence of the column j, for all row i≤r. The result holds because, from Condition (6), we have that ψ(lk,i,j)=ψ(k,i)ψ(i,j)ψ(k,li,j)=ψ(k,i)ψ(i,j)ψ(k,i), for all k≤r, and hence, ψ(lk,i,j)=ψ(i,j). □ Example 16. Let us consider the Latin rectangle L=(li,j)≡2 1 3 4 3 4 2 1 ∈R2,4. Since l1,2=1, Lemma 15 ensures the existence of two cells within the second column of every cocyclic matrix M=(mi,j) over L, having both of them the value m1,2as sign. The location of both cells (more specifically, the rows in which they appear) is uniquely determined by the entries of the first column of L, that is, by the symbols 2 and 3. In particular, it must be m1,2=m2,2=m3,2. Similarly, since l2,3=2 and the second column of Lcontains the symbols 1 and 4, Lemma 15 implies that m1,3=m2,3=m4,3. The following cocyclic matrix over Lillustrates these facts. ⎛ ⎜ ⎝ + − + − − − + + + − − + + + + + ⎞ ⎟ ⎠ Now, we prove that the existence of a symbol j≤rwithin the jth column of a Latin rectangle L∈Rr,ninvolves the existence of a normalized row within every cocyclic matrix Mover L. Moreover, we indicate a necessary condition on the diagonal of Land Mensuring the existence of a row within Lhaving all its cycles of even length. Lemma 17. Let L =(li,j)∈Rr,nbe such that there exists a pair (e,j)∈ [r] × [r]satisfying that le,j=j. Let M =(mi,j)be a cocyclic matrix over L. Then, (a) me,k=me,j, for all k ≤n. (b) If there exists a positive integer i ∈ [r] \ {e}such that li,i=e and mi,i= −me,1, then all the cycles in the ith row of L have even length. Proof. Let ψ∈Z(L) be such that M=Mψ. Let k≤n. From the definition of Latin rectangle, there exists m≤nsuch that lj,m=k. Then, from Condition (6), we have that ψ(j,m)=ψ(e,j)ψ(j,m)ψ(e,k) and hence, ψ(e,k)=ψ(e,j). Now, in order to prove the second assertion, observe that Condition (6), together with assertion (a), involves that ψ(i,li,k)=ψ(e,1) ψ(i,i)ψ(i,k), for all k≤n. Thus, if ψ(i,i)= −ψ(e,1), then ψ(i,li,k)= −ψ(i,k), for all k≤n. As a consequence, every cycle of odd length in the ith row of L would give rise to a cycle of odd length with alternating signs (1 and −1) in the ith row of the cocyclic matrix Mψ, which is not possible. □ Example 18. Let us consider the Latin rectangle L=(li,j)≡1 2 3 4 3 1 4 2 ∈R2,4. Since the first column of Lcontains the symbol 1, Lemma 17.a ensures that every cocyclic matrix Mover Lhas a first row of 1′sor -1′s. The following cocyclic matrix over Lillustrates this fact. M=(mi,j)≡⎛ ⎜ ⎝ + + + + + − − + + − − + − + + + ⎞ ⎟ ⎠ Since l2,2=1 and m2,2= −m1,1,Lemma 17.b implies that all the cycles in the second row of L should have even length. In fact, this is true, because such a row determines the cycle (1342). We focus now on the relationship among the entry set, the set of cocycles and the set of cocyclic Hadamard matrices that are associated to two given Latin rectangles. We deal in particular with the case in which both Latin rectangles are isotopic or transpose of each other. Lemma 19. Let L ∈Rr,nand L′∈Rs,nbe such that r ≤s≤n. (a) If E(L)⊆E(L′), then i. S(L)⊆S(L′). ii. If S(L)=S(L′), then Z(L′)⊆Z(L)and H(L′)⊆H(L). (b) If r =s and L is isotopic to L′by means of an isotopism (f,g,g)∈Sr×Sn×Snsuch that f (i)=g(i), for all i ≤r, then there exists a 1-1correspondence between the sets Z(L)and Z(L′). (c) If r =s=n and L and L′are two Latin squares that are transpose of each other, then there exists a 1-1correspondence between the sets Z(L)and Z(L′). Proof. The first assertion follows straightforwardly from (5)and (6). Now, in order to prove the second assertion, it is enough to define the correspondence α:Z(L)→Z(L′) such that α(ψ)(i,j)=ψ(g−1(i), g−1(j)), for all ψ∈Z(L) and i,j∈ [n]. The result follows straightforwardly from (6) and the fact that gis a relabeling of the symbols of the sets [r]and {r+1, . . . , n}, in both cases by means of the same set of symbols under consideration. Table 1 Cocyclic Hadamard matrices over r × n Latin rectangles, with r ≤ n ≤ 4. r n L df(L) Cocyclic Hadamard matrices in H(L) 1 1 1 1 + 2 21 2 + − −− 2 2 12 21 2 + + +− 4 1243 3421 5 ++++ +−−+ +−+− ++−− ++++ −+−+ +−−+ ++−− ++++ −++− +−+− −−++ ++++ +−+− +−−+ −−++ 2341 3412 4 ++−− +−+− −++− −−−− +−−+ −+−+ −−++ ++++ 2134 3421 2 +−+− −−++ +−−+ ++++ 2314 3421 3 +−−+ −+−+ −−−− ++−− 2314 4123 2 +−+− −−++ −++− ++++ 3 4 1234 2143 3412 5 ++++ ++−− +−−+ +−+− ++++ ++−− +−+− +−−+ ++++ +−+− +−−+ ++−− ++++ +−+− ++−− +−−+ ++++ +−−+ +−+− ++−− ++++ +−−+ ++−− +−+− 1234 2143 3421 4 ++++ ++−− +−−+ +−+− ++++ ++−− +−+− +−−+ 2341 3412 4123 4 ++−− +−+− −++− −−−− +−−+ −+−+ −−++ ++++ 4 4 1234 2143 3412 4321 5 ++++ ++−− +−−+ +−+− ++++ ++−− +−+− +−−+ ++++ +−+− +−−+ ++−− ++++ +−+− ++−− +−−+ ++++ +−−+ +−+− ++−− ++++ +−−+ ++−− +−+− 1234 2143 3421 4312 4 ++++ ++−− +−−+ +−+− ++++ ++−− +−+− +−−+ Finally, in order to prove the third assertion, it is enough to define the correspondence α:Z(L)→ Z(L′) such that α(ψ)(i,j)=ψ(j,i), for all ψ∈Z(L) and i,j≤n. To see it, suppose Land L′to be the respective multiplication tables of two quasigroups Q=([n],·) and Q′=([n],◦). Then, from (7), we have that α(ψ)(i◦j,k)α(ψ)(i,j)α(ψ)(j,k)α(ψ)(i,j◦k)=ψ(k,j·i)ψ(j,i)ψ(k,j)ψ(k·j,i)=1. □ In case of being r=s=n, the isotopism (f,g,g) that is described in Lemma 19.b constitutes an isomorphism of Latin squares. In this regard, from here on and by an abuse of notation, we call such an isotopism an isomorphism of r×nLatin rectangles and we say that any two Latin rectangles related by any such an isotopism are isomorphic. This constitutes an equivalence relation among r×nLatin rectangles. Thus, for instance, the Latin rectangle 3 4 1 2 4 1 2 3 is isomorphic to the Latin rectangle L∈R2,4in Example 11 by means of the isomorphism ((12),(12)(34),(12)(34)) ∈S2×S4×S4. The following result holds straightforwardly from Lemma 19. Proposition 20. Let L ∈Rr,nand L′∈Rs,nbe such that r ≤s≤n. If r =s and L is isomorphic to L′, then there exists a 1-1correspondence between the sets H(L)and H(L′). From Proposition 20, the distribution of r×nLatin rectangles into isomorphism classes constitutes a first approximation to determine the set of cocyclic Hadamard matrices over the set Rr,n. By means of an exhaustive search, Table 1 illustrates the case r≤n≤4, where both Latin rectangles and Hadamard matrices are written row after row in a single line. Besides, keeping in mind Lemma 14, we only enumerate those cocyclic Hadamard matrices having its first entry equal to 1. We also indicate the cocyclic degree-of-freedom of each Latin rectangle under consideration. over which the Hadamard matrix H≡⎛ ⎜ ⎝ + + + + + − − + + − + − + + − − ⎞ ⎟ ⎠ is cocyclic. In what follows, we study the negation of either a proper subset of rows or a proper subset of columns within Hin order to determine the existence of new Hadamard matrices that are also cocyclic over L. Firstly, we study the possible negation of rows. According to Lemma 33.a, the negation of the third or fourth row of Hrequires also the negation of its first or second row. Further, Lemma 33.b implies that the negation of the first (respectively, second) row of Hrequires the negation of its third (respectively, fourth) row. Both possibilities are valid in this case. That is, both Hadamard matrices ⎛ ⎝ − − − − + − − + − + − + + + − − ⎞ ⎠and ⎛ ⎝ + + + + − + + − + − + − − − + + ⎞ ⎠. are cocyclic over L. Now, we analyze the possible negation of columns. From Lemma 34.a, the negation of the first or second columns of Hinvolves the negation of all its columns. Now, from Lemma 34.b, the negation of both third and fourth columns of Himplies the negation of the first column. This is because l1,3=4, or also, because l1,4=3. Then, again from Lemma 34.a, all the columns should be negated. As a consequence, the only candidates for proper subsets of columns in Hto be negated are N= {3}and N= {4}. In both cases, however, it can be checked that the resulting Hadamard matrix is not cocyclic over L. 5. Latin rectangles over which a Hadamard matrix is cocyclic This section deals with both questions exposed in Problem 13. With respect to the second question, observe that Lemma 25 enables us to ensure that the minimum possible positive integer r∈Nto which that question referred must be such that r≤n≤r+r2. This minimum is reached, for instance, for the following Hadamard matrix of order 8 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ + + + + + + + + + + + + − − − − + + − − − − + + + + − − + + − − + − − + − + + − + − − + + − − + + − + − − + − + + − + − + − + − ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ which is cocyclic over the following two 3 ×8 Latin rectangles L1≡ 1 2 3 4 7 8 5 6 2 5 8 3 1 6 7 4 4 7 6 2 8 3 1 5 and L2≡ 1 2 3 4 7 8 5 6 2 5 8 3 1 6 7 4 4 7 6 1 8 3 2 5 These two Latin rectangles constitute indeed the only two distinct isomorphism classes of the set R3,8that give rise to such a cocyclic matrix. Observe that both of them contain the eight symbols in their first three columns. Let us expose some preliminary results dealing with the first question in Problem 13. Firstly, we establishhowanormalizedcolumn within a cocyclic Hadamard matrix over a Latin rectangle connects signs of pair of entries within certain rows. Lemma 36. Let M =(mi,j)be a cocyclic Hadamard matrix of order n over a Latin rectangle L =(li,j)∈ Rr,nsuch that M has a normalized column e. That is, mi,e=m1,e, for all i ≤n. Then, for each row i in L, we have that mi,j=mi,lj,e, for all j ≤r. Proof. Let ψ∈Z(L) be such that Mψ=M. The result holds because, from Condition (6), we have that ψ(li,j,e)=ψ(i,j)ψ(j,e)ψ(i,lj,e), for all i,j≤r, and hence, ψ(i,j)=ψ(i,lj,e). □ The following result characterizes the existence of a normalized row e≤rwithin a cocyclic Hadamard matrix over an r×nLatin rectangle. Proposition 37. A cocyclic Hadamard matrix over a Latin rectangle L =(li,j)∈Rr,nhas a normalized row e ≤r if and only if lej =j, for all j ≤r. Proof. The sufficiency holds from Lemma 17.a. In order to prove the necessary condition, let us suppose the existence of a positive integer j≤rsuch that le,j= j. Let ψ∈Z(L). Then, from Condition (6), we have that ψ(le,j,k)=ψ(e,j)ψ(j,k)ψ(e,lj,k)=ψ(j,k), for all k≤n. This would imply the existence of two distinct rows in Mwith all their respective entries being equal, which contradicts the fact that Mis a Hadamard matrix. □ Example 38. In order to illustrate the previous results, observe, for instance, that any Latin rectangle L=(li,j)∈Rr,n, with r≥3, that is related to the Hadamard matrix indicated at the beginning of this section, satisfies that l2,1=2. To see it, let M=(mi,j) denote that matrix. From Lemma 36, we have that m2,2=m2,l2,1and m3,2=m3,l2,1. The former is possible only if l2,1∈ {5,6,7,8}, whereas the latter is possible only if l2,1∈ {3,4,5,6}. Hence, l2,1∈ {1,2}. But, from Proposition 37, it must be l1,1=1 and thus, since Lis a Latin rectangle, it must be l2,1=2. A similar reasoning involves that l3,1∈ {3,4}. The following example illustrates how Lemma 36 and Proposition 37 also enable us to ensure that not every Hadamard matrix of order nis cocyclic over an r×nLatin rectangle. Example 39. Let us consider the Hadamard matrix M=(mi,j)=⎛ ⎝ + + + + − − + + − + + − + − + − ⎞ ⎠ Suppose L=(li,j)∈Rr,4to be such that M∈H(L). In particular, Lemma 25 involves that 2 ≤r≤4. Now, from Proposition 37, it must be l1,1=1 and l1,2=2. As a consequence, l1,3∈ {3,4}and hence, m2,l1,3=1. Nevertheless, from Lemma 36, it should be m2,l1,3=m2,1= −1. Therefore, there does not exist such a Latin rectangle L. The just exposed examples suggest the question of which entries never appear in the possible related Latin rectangles over which a Hadamard matrix is cocyclic. In this regard, the following result holds. Lemma 40. Let M =(mi,j)be a Hadamard matrix of order n over a Latin rectangle L =(li,j)∈Rr,n. Then, (a) For each i ≤r, the value of li,imust be in the set VM(i):= {t≤n|mi,t=mt,i}.(9) (b) Let i,k≤r and j ≤n be such that i = j. Then, the value of li,jmust be in the set V− M(i,j;k):= {t≤n|mlk,i,jmk,imi,jmk,t=1}.(10) As a consequence, it must be in the set V− M(i,j):= ⋂ k≤r V− M(i,j;k).(11) (c) Let i,j≤r be such that i = j, and let k ≤n. Then, the value of li,jmust be in the set V+ M(i,j;k):= {t≤n|mt,kmi,jmj,kmi,lj,k=1}.(12) As a consequence, it must be in the sets V+ M(i,j):= ⋂ k≤n V+ M(i,j;k).(13) VM(i,j):= V− M(i,j)∩V+ M(i,j).(14) Proof. The result follows straightforwardly from Condition (6). Thus, for instance, the first assertion follows from the fact that ψ(li,i,i)=ψ(i,i)ψ(i,i)ψ(i,li,i)=ψ(li,i,i)ψ(i,li,i), for all i≤n. The other two assertions hold similarly. □ Example 41. In order to illustrate Lemma 40, let us consider the Hadamard matrix M=(mi,j)=⎛ ⎝ + + + + − + + − − − + + + − + − ⎞ ⎠. Suppose the existence of a Latin rectangle L=(li,j)∈Rr,4, with 2 ≤r≤4, such that M∈H(L). Similarly to Example 39, we have that l1,1=1 and l1,2=2. These conditions imply, respectively, that V+ M(2,1;1) = {1,4}and V+ M(2,1;2) = {3,4}. Thus, V+ M(2,1) = {4}and hence, from Lemma 40.c, we have that l2,1=4. As a consequence, since Lis a Latin rectangle, it must be l2,2∈ {2,4}, which contradicts that l2,2∈VM(2) = {2,4}according to Lemma 40.a. Therefore, there does not exist such a Latin rectangle L. We have made use of all the previous results to show in Table 2 an r×nLatin rectangle that gives rise to each equivalent class of Hadamard matrices of order n≤8. The latter are written row after row, where each row is represented in hexadecimal form. To this end, after replacing each −1 by 0, the resulting row in binary form is translated to its equivalent hexadecimal form. 6. Conclusions and further work In this paper, we have introduced the concept of cocyclic Hadamard matrices over Latin rectangles and we have exposed a series of preliminary results that enable us to determine the existence of such matrices and that of Latin rectangles over which a given Hadamard matrix is cocyclic. More specifically, we have proved that every Latin square related to a Hadamard matrix is the multiplication table of a loop. Particularly, we have proved the existence of cocyclic Hadamard matrices over nonassociative loops of order 2t+3, for all positive integer t>0. Furthermore, we have exposed examples of Hadamard matrices that are not cocyclic over finite groups but they are over Latin rectangles. In Section 5, we have also shown a pair of examples that enable us to ensure that not every Hadamard matrix of order nis cocyclic over an r×nLatin rectangle. According to this fact, we propose as further work to deal with the following question, which constitutes a natural generalization of Problem 13. Problem 42. Let Mbe a Hadamard matrix of order nfor which no Latin rectangle L∈Rr,nsatisfies that M∈H(L). Does there exist, however, a Hadamard matrix M′in the same equivalence class of M for which one such a Latin rectangle can be found? The following question also arises in a natural way as an open problem. Problem 43. Let us consider an equivalence class of Hadamard matrices such that none of them are cocyclic over any finite group. Does there exist, however, a Hadamard matrix within such a class that is cocyclic over a Latin rectangle? Even if the theory that we have presented in this paper derives from the classical notion of pure cocyclic Hadamard matrix over a finite group, let us remark the existence of a more general definition of cocyclic development over finite groups [10,16,17], whose natural generalization to Latin squares can be similarly studied. In this regard, we say that a Hadamard matrix M=(mi,j) is cocyclically developed over a Latin square L =(li,j)∈Rn,nif there exists a cocycle ψ: [n] × [n] → Z2over Land a map φ: [n] → Z2so that mi,j=ψ(i,j)·φ(li,j),for all i,j≤n.(15) In such a case, we call φadevelopment function over L. Analogously to the cocyclic development over finite groups, if ψ(i,j)=1, for all i,j≤n, then we say that the Hadamard matrix Mis quasigroup developed over L. This is group developed over Lif the latter constitutes the multiplication table of a group. The following result ensures that the existence of quasigroup developed Hadamard matrices over a given Latin square only depends on the isotopism class of the latter, and not on its isomorphism class, as it happens with cocyclic Hadamard matrices over Latin squares (see Proposition 20). Lemma 44. Quasigroup development is preserved by row and column permutations of Hadamard matrices, and by isotopisms of Latin squares. Proof. Let M=(mi,j) be a Hadamard matrix of order nthat is quasigroup developed over a Latin square L=(li,j) of the same order by means of a development function φ. For each triple of permutations (f,g,h)∈Sn×Sn×Sn, let us consider (a) the Hadamard matrix M′=(mf(i),g(j)), (b) the Latin square L′=(l′ i,j) such that l′ f(i),g(j)=h(li,j), for all i,j≤n, and (c) the function φ′=φ◦h−1: [n] → Z2. Then, the Hadamard matrix M′is quasigroup developed over the Latin square L′by means of φ′, because m′ i,j=mf−1(i),g−1(j)=φ(lf−1(i),g−1(j))=φ(h−1(l′ i,j)) =φ′(li,j),for all i,j≤n.□ The interest of dealing with cocylic development over Latin squares as further work derives, therefore, from the existence of Latin squares over which no cocyclic Hadamard matrix exists, but over which cocyclically developed Hadamard matrices do. Thus, for instance, even if Theorem 28 implies that no cocyclic Hadamard matrix exists over the Latin square L≡ 1 3 2 4 2 4 3 1 3 1 4 2 4 2 1 3 ∈R4,4, it is straightforwardly verified that the Hadamard matrix ⎛ ⎝ + + + − + − + + + + − + − + + + ⎞ ⎠ is quasigroup developed over Lby means of the development function φover Lsuch that φ(1) = φ(2) =φ(3) =1= −φ(4). Observe in particular that Lconstitutes the multiplication table of a non-associative quasigroup that is not a loop. As such, this example illustrates the fact that the cocyclic development is even less restrictive with respect to the assumptions that Latin squares have to satisfy. Thus, for instance, keeping in mind the previous example, the following result holds. Proposition 45. There exists a quasigroup developed Hadamard matrix over each Latin square of order four. Proof. Let L=(li,j)∈R4,4. For any given symbol sin L, let M=(mi,j) be the matrix of order four such that mi,j= −1, if li,j=s, and mi,j=1, otherwise. The Latin square condition of Limplies that exactly one -1 exists within each row and each column of M. The dimension of the array makes this condition to be enough in order to ensure that Mis Hadamard. Moreover, the matrix Mis quasigroup developed over Lby means of the development function φover Lsuch that φ(i)=1, for all i= s, and φ(s)= −1. □ There exist, however, some remarkable notions and results on group development theory that cannot be generalized as such to quasigroup development theory, because of their dependence on the associative property of the underlying structure. Thus, for instance, the concept of coboundary over a group Gas a cocycle of the form ψ(a,b)=φ(a)φ(b)φ(ab), for all a,b∈G, and all normalized function φ:G→Z2, involves the function φto satisfy that φ((ab)c)=φ(a(bc)), for all a,b,c∈G. A similar condition over a non-associative loop would involve φto be the trivial normalized function and hence, the only well-defined coboundary over such a loop would be the trivial cocycle. As a consequence, every result in group development theory that uses coboundaries either in its statement or its proof cannot be generalized in a natural way to quasigroup development theory. Such a dependence on the associative property occurs, for instance, in the majority of results concerning the interactions of group development theory with Hadamard equivalence and also with automorphism groups of Hadamard matrices [8]. Further work is, therefore, required to establish possible generalizations of both interactions. A possible approach in this regard consists of generalizing the notion of cocycle itself in order to adjust it to a balanced equation that regulates the quasigroup under consideration. Recall here that an equation that is satisfied by all the elements of a quasigroup is said to be balanced if each one of its variables appears precisely once on each side of the equation. Thus, for instance, the equation (ab)c=a(bc) that regulates the associative property of every group is balanced. Let us illustrate some aspects of this proposal by means of a specific balanced equation. Let Qebe the set of quasigroups Q=([n],·) verifying the balanced equation e: (i·j)·k=(k·j)·i,for all i,j,k≤n.(16) This is indeed an abelian group whenever Qis a loop. We say that a function ψ: [n] × [n] → Z2is an e-cocycle over a quasigroup Q=([n],·)∈Qeif ψ(i·j,k)=ψ(i,j)ψ(k,j)ψ(k·j,i),for all i,j,k≤n.(17) We say that an n×nmatrix M=(mi,j) is e-cocyclic over Q if there exists an e-cocycle ψover Qsuch that mi,j=ψ(i,j), for all i,j≤n. Example 46. The Hadamard matrices M1≡⎛ ⎝ + + + + + + − − + − − + + − + − ⎞ ⎠and M2≡⎛ ⎝ + + + + + + − − − + − + − + + − ⎞ ⎠ are e-cocyclic over the quasigroup Qhaving as multiplication table the Latin square 1 2 3 4 2 1 4 3 4 3 1 2 3 4 2 1 ∈R4,4. This quasigroup is not a loop. Moreover, it is non-abelian and non-associative. We say that a function ψ: [n] × [n] → Z2is an e-coboundary over a quasigroup Q=([n],·)∈Qe if there exists a function φ: [n] → Z2such that ψ(i,j)=φ(i)φ(j)φ(i·j), for all i,j≤n. In particular, ψis an e-cocycle over Q, because Condition (17) holds if and only if φ((i·j)·k)=φ((k·j)·i), for all i,j,k≤n. At this point, we propose as further work a detailed analysis to adapt in the context of the set Qeall those results in group development that deal with coboundaries. Let us finish the exposition of this approach with an illustrative result that adjusts the necessary condition of Lemma 1 in [8] for quasigroups within the set Qe. Here, we say that a matrix M=(mi,j) is quasigroup developed over a quasigroup Q=([n],·)∈Qeif there exists a function φ: [n] → Z2 such that mi,j=φ(i·j), for all i,j≤n. Lemma 47. Let Q =([n],·)∈Qe. For each positive integer k ≤n, we define the permutation matrix Tk=(δi·k j), where δi·k jdenotes the Kronecker delta. If an e-cocyclic matrix M over Q is quasigroup developed over Q, then TkMTk=Mt, for all k ≤n, where Mtdenotes the transpose of M. Proof. Let ψbe a cocycle over Qsuch that M=Mψ. Let k≤n. Then, •TkMTk=Mψ1, where ψ1(i,j)=ψ(i·k,j/k), for all i,j≤n. •Mt=Mψ2, where ψ2(i,j)=ψ(j,i), for all i,j≤n. Now, let φ: [n] → Z2be such that ψ(i,j)=φ(i·j), for all i,j≤n. Then, ψ1(i,j)=ψ(i·k,j/k)=φ((i·k)·j/k)=φ((j/k·k)·i)=φ(j·i)=ψ(j,i)=ψ2(i,j), for all i,j≤n. Hence, TkMTk=Mt.□ Example 48. In Example 46, we have that •TkM1Tk=Mt 1if and only if k=1. •TkM2Tk=Mt 2if and only if k=2. Hence, from Lemma 47, we can ensure that none of the e-cocyclic Hadamard matrices M1and M2are quasigroup developed over Q. Of course, a much deeper analysis is required to deal with the different aspects derived from these new concepts. Moreover, similar studies related to other types of balanced equations different from e are also necessary as further work. Particularly, the characterization of those equations that give rise to group isotopes could have special interest concerning the interaction of quasigroup development with Hadamard equivalence. We refer the interested reader to [3,4,20,21] for more details about the study of balanced equations on quasigroups. From all the previous remarks and proposals, we realize that this paper only scratches the surface of a vast world where there is still much to explore. The results here presented constitute the starting point from which we can ensure that Latin rectangles and quasigroups arise as an interesting alternative to delve into the theory of cocyclic Hadamard matrices. Acknowledgments The authors want to express their gratitude to the anonymous referees for their pertinent comments and suggestions, which have permitted to improve the readability and understandability of the paper. Particularly, we are very grateful to the anonymous reviewer who has suggested us Problem 43 and the possibility of dealing with the cocyclic development over Latin squares, which we have introduced in Section 6as further work. All authors are partially supported by the research project FQM-016 from Junta de Andalucía (JJAA), Spain. Appendix. Glossary of symbols E(L) The set of entries of a Latin rectangle L. H(L) The set of cocyclic Hadamard matrices over a Latin rectangle L. LnThe set of loops of order nhaving 1 as unit element. [n]The set {1, . . . , n}. QnThe set of finite quasigroups having [n]as their finite set of symbols. Rr,nThe set of r×nLatin rectangles. S(L) The set [r]∪{li,j|i,j≤r}, where L=(li,j)∈Rr,n. Z(L) The set of cocycles over a Latin rectangle L. References [1] V. Álvarez, J.A. Armario, M.D. Frau, F. Gudiel, B. Güemes, A. Osuna, On D4t-cocyclic Hadamard matrices, J. Combin. Des. 24 (2016) 352–368. [2] A. Baliga, K.J. Horadam, Cocyclic Hadamard matrices over Zt×Z2 2, Australas. J. Combin. 11 (1995) 123–134. [3] V.D. Belousov, Balanced identities in quasigroups, Mat. Sb. (N.S.) 70 (112) (1966) 55–97 (in Russian). [4] V.D. Belousov, Quasigroups with completely reducible balanced equations, Mat. Issled. 83 Issled. Teor. Binarnykh i n-arnykh Kvazigrupp (1985) 11–25, 134 (in Russian). [5] P. Borwein, M.J. Mossinghoff, Wieferich pairs and Barker sequences, II, LMS J. Comput. Math. 17 (2014) 24–32. [6] K.A. Bush, Construction of symmetric Hadamard matrices. A survey of combinatorial theory, in: Proc. Internat. Sympos. Colorado State Univ. Fort Collins, Colo., North-Holland, Amsterdam, 1971, pp. 81–83, 1973. [7] F.C. Bussemaker, J.J. Seidel, Symmetric Hadamard matrices of order 36, Ann. New York Acad. Sci. 175 (1970) 66–79. [8] P.Ó. Catháin, M. Röder, The cocyclic Hadamard matrices of order less than 40, Des. Codes Cryptogr. 58 (2011) 73–88. [9] W. De Launey, D.L. Flannery, Algebraic Design Theory, in: Mathematical Surveys and Monographs, American Mathematical Society, 2011. [10] W. De Launey, K.J. Horadam, A weak difference set construction for higher dimensional designs, Des. Codes Cryptogr. 3 (1993) 75–87. [11] J. Dénes, A.D. Keedwell, Latin Squares and Their Applications, Academic Press, New York-London, 1974. [12] D.L. Flannery, Cocyclic Hadamard matrices and Hadamard groups are equivalent, J. Algebra 192 (1997) 749–779. [13] J.M. Goethals, J.J. Seidel, Strongly regular graphs derived from combinatorial designs, Canad. J. Math. XXII (1970) 597–614. [14] M. Hall, An existence theorem for Latin squares, Bull. Amer. Math. Soc. 51 (1945) 387–388. [15] K.J. Horadam, Hadamard Matrices and Their Applications, Princeton University Press, Princeton, NJ, 2007. [16] K.J. Horadam, W. De Launey, Cocyclic development of designs, J. Algebraic Combin. 2 (1993) 267–290;J. Algebraic Combin. 3 (1994) (erratum). [17] K.J. Horadam, W. De Launey, Generation of cocyclic Hadamard matrices, Math. Appl. 325 (1995) 279–290. [18] A. Hulpke, P. Kaski, P.R.J. Östergård, The number of Latin squares of order 11, Math. Comp. 80 (2011) 1197–1219. [19] G. Kolesova, C.W.H. Lam, L. Thiel, On the number of 8 ×8 Latin squares, J. Combin. Theory Ser. A 54 (1990) 143–148. [20] A. Krapež, M.A. Taylor, Belousov equations on quasigroups, Aequationes Math. 34 (1987) 174–185. [21] A. Krapež, M.A. Taylor, Quasigroups satisfying balanced but not Belousov equations are group isotopes, Aequationes Math. 42 (1991) 37–46. [22] K.H. Leung, B. Schmidt, The anti-field-descent method, J. Combin. Theory Ser. A 139 (2016) 87–131. [23] S. Mac Lane, Homology, Springer-Verlag, Berlin, 1995. [24] B.D. McKay, Isomorph-Free Exhaustive Generation, J. Algorithms 26 (1998) 306–324. [25] B.D. McKay, A. Meynert, W. Myrvold, Small Latin squares, quasigroups, and loops, J. Combin. Des. 15 (2007) 98–119. [26] R. Moufang, Zur Struktur von Alternativkörpern, Math. Ann. 110 (1935) 416–430. [27] H.J. Ryser, Combinatorial Mathematics, Wiley, New York, 1963. [28] J.J. Sylvester, Thoughts on inverse orthogonal matrices, simultaneous sign successions, and tessellated pavements in two or more colours, with applications to Newton’s rule, ornamental tile-work, and the theory of numbers, Phil. Mag. 34 (1867) 461–475. [29] R.J. Turyn, Character sums and difference sets, Pacific J. Math. 15 (1965) 319–346.