A Series of Regular Hadamard Matrices Dean Crnković Abstract Let p and p 1 be prime powers and p 3 (mod 4). Then there exists a symmetric design with parameters (4p,p p,p p). Thus there exists a regular Hadamard matrix of order 4p. AMS classification numbers: 05B0, 05B05 Keywords: regular Hadamard matrix, symmetric design, Menon design.
1 Introduction A -(v, k, λ) design is a finite incidence structure (P, B, I), where P and B are disjoint sets and I P B, with the following properties: 1. P = v;. every element of B is incident with exactly k elements of P; 3. every pair of distinct elements of P is incident with exactly λ elements of B. The elements of the set P are called points and the elements of the set B are called blocks. If P = B = v and k v, then a -(v, k, λ) design is called a symmetric design. A Hadamard matrix of order m is an (m m)-matrix H = (h i,j ), h i,j { 1, 1}, satisfying HH T = H T H = mi, where I is the unit matrix. A Hadamard matrix is regular if the row and column sums are constant. It is well known that the existence of a symmetric (4u, u u, u u) design is equivalent to the existence of a regular Hadamard matrix of order 4u (see [4, Theorem 1.4 p. 80]). Such symmetric designs are called Menon designs. If n + 1 and n 1 are prime powers there exists a symmetric Hadamard matrix with constant diagonal of order n (see [4, Corollary 5.1 p. 34]). T. Xia, M. Xia and J. Seberry proved (see [5]) the following statement: When k = q 1, q, q 1 q, q 1 q 4, q q 3 N, q 3 q 4 N, where q 1, q and q 3 are prime powers, q 1 1 (mod 4), q 3 (mod 8), q 3 5 (mod 8), q 4 = 7 or 3, N = a 3 b t, a, b = 0 or 1, t 0 is an arbitrary integer, there exist regular Hadamard matrices of order 4k. 1
The existence of some regular Hadamard matrice of order 4p, when p is a prime, p 7 (mod 16), is established in [3]. According to [5] and [3], there are just two values of k 100 for which the existence of a regular Hadamard matrix of order 4k is still in doubt, k = 47 and k = 79. Nonzero Squares in Finite Fields Let p be a prime power, p 3 (mod 4) and F p be a field with p elements. Then a (p p) matrix D = (d ij ), such that 1, if (i j) is a nonzero square in F p, d ij = 0, otherwise. is an incidence matrix of a symmetric (p, p 1, p 3 4 ) design. Such a symmetric design is called a Paley design (see []). Let D be an incidence matrix of a complementary symmetric design with parameters (p, p+1, p+1 4 ). Since 1 is not a square in F p, D is a skew-symmetric matrix. Further, D has zero diagonal, so D +I p and D I p, where I p is an (p p) identity matrix, are incidence matrices of symmetric designs with parameters (p, p+1, p+1 4 respectively. Matrices D and D have the following properties: D D T = (D I p )(D + I p ) T = p + 1 4 J p p + 1 4 I p, [ D D I p ] [ D I p D ] T = p 1 J p p 1 I p, [ D D ] [ D + I p D I p ] T = p 1 J p, [ D D ] [ D I p D I p ] T = p 1 J p, ) and (p, p 1, p 3 4 ),
where J p is the all-one matrix of dimension (p p). Let Σ(p) denote the group of all permutations of F p given by x aσ(x) + b, where a is a nonzero square in F(p), b is any element of F(p), and σ is an automorphism of the field F(p). Σ(p) is an automorphism group of symmetric designs with incidence matrices D, D + I p, D and D I p (see [, pp. 9]). If p is a prime, Σ(p) is isomorphic to a semidirect product Z p : Z p 1. Let q be a prime power, q 1 (mod 4), and C = (c ij ) be a (q q) matrix defined as follows: 1, if (i j) is a nonzero square in F q, c ij = 0, otherwise. C is a symmetric matrix, since 1 is a square in F q. There are as many nonzero squares as nonsquares in F(q), so each row of C has q 1 elements equal 1 and q+1 zeros. Let i j and C i = [c i1...c iq ], C j = [c j1...c jq ] be the i th and the j th row of the matrix C, respectively. Then C i Cj T = q 1 4, if c ij = c ji = 0, q 1 4 1, if c ij = c ji = 1. The matrix C I q has the same property. Let i j and C i = [c i1...c iq ], C j = [c j1...c jq ] be the i th and the j th row of the matrix C, respectively. Then C i C T j = q 1 4, if c ij = c ji = 0, q 1 4 + 1, if c ij = c ji = 1. The matrix C + I q has the same property. Further, C (C + I q ) T = C (C I q ) T = q 1 4 J q + q 1 4 I q, 3
C (C I q ) T = q 1 4 J q q 1 4 I q, (C + I q ) C T = q + 3 4 J q q 1 4 I q, [ C C + I q ] [ C C + I q ] T = q 1 J q + q + 1 I q, [ C C I q ] [ C C I q ] T = q 1 J q + q + 1 I q, [ C C + I q ] [ C C I q ] T = q + 1 J q q + 1 I q. Σ(q) acts as an automorphism group of incidence structures with incidence matrices C, C + I q, C and C I q. 3 Regular Hadamard Matrices Let H = (h ij ) and K be m n and m 1 n 1 matrices, respectively. Their Kronecker product is a mm 1 nn 1 matrix h 11 K h 1 K... h 1n K h 1 K h K... h n K H K =... h m1 K h m K... h mn K For v N we denote by j v the all-one vector of dimension v, by 0 v the zero-vector of dimension v, and by 0 v v the zero-matrix of dimension v v. Theorem 1 Let p and p 1 be prime powers and p 3 (mod 4). Then there exists a symmetric design with parameters (4p, p p, p p). Proof Put q = p 1. Then q 1 (mod 4). Let D, D, C, C be defined as above. Define a (4p 4p ) matrix M in the following way: 4
M = 0 0 T q jp q T 0 T p q 0 q 0 q q (C I q ) jp T C jp T (C + I q ) D C D j p q C j p + + C (D I p ) (C I q ) D C (D + I p ) (C + I q ) (D I p ) 0 p q (C + I q ) j p + + (C I q ) (D I p ) C D Let us show that M is an incidence matrix of a Menon design with parameters (4p, p p, p p). It is easy to see that M J 4p = (p p)j 4p. We have to prove that M M T = (p p)j 4p + p I 4p. Using properties of the matrices D, D, C and C which we have menitioned before, one computes that the product of block matrices M and M T is: pq (p p)jq T (p p)jpq T (p p)jpq T (p p)j q (p p)j q + (p p)j q pq (p p)j q pq p I q M M T (p p)j pq = (p p)j pq (p p)j pq q + (p p)j pq pq p I pq (p p)j pq (p p)j pq (p p)j pq q (p p)j pq pq + p I pq 5
where J m n is the all-one matrix of dimension m n. Thus, M M T = (p p)j 4p + p I 4p which means that M is an incidence matrix of a symmetric design with parameters (4p, p p, p p). Corollary 1 Let p and p 1 be prime powers and p 3 (mod 4). Then there exists a regular Hadamard matrix of order 4p. That proves, in particular, that there exists a regular Hadamard matrix of order 4 79 = 4964. Incidence matrices of the Menon designs from Theorem 1 lead us to conclusion that the groups Σ(p) Σ(p 1) act as automorphism groups of these designs, semistandardly with one fixed point (and block), one orbit of length p 1, and two orbits of length p p. If p and p 1 are primes, then Σ(p) Σ(p 1) = (Z p : Z p 1) (Z p 1 : Z p 1 ), and the derived designs of the Menon designs from Theorem 1 with respect to the first block, i.e., the fixed block for an automorphism group (Z p : Z p 1) (Z p 1 : Z p 1 ), are cyclic. That proves the following corollary: Corollary Let p and p 1 be primes and p 3 (mod 4). Then there exists a cyclic -(p p, p p, p p 1) design having an automorphism group isomorphic to (Z p : Z p 1) (Z p 1 : Z p 1 ). Parameters of Menon designs belonging to the series described in this paper, 6
for p 100, are given below. table 1. Table of parameters for p 100 p q = p 1 4p Parameters of Menon Designs 3 5 36 (36,15,6) 7 13 196 (196,91,4) 19 37 1444 (1444,703,34) 7 53 916 (916,1431,70) 31 61 3844 (3844,1891,930) 79 157 4964 (4964,1403,616) References [1] H. Evangelaras, C. Koukouvinos and J. Seberry, Applicatons of Hadamard matrices, Journal of Telecommunications and Information Technology, Vol. (003) pp. 10. [] E. Lander, Symmetric Designs: An Algebraic Approach, Cambridge University Press, Cambridge (1983). [3] K. H. Leung, S. L. Ma and B. Schmidt. New Hadamard matrices of order 4p obtained from Jacobi sums of order 16, preprint. 7
[4] W. D. Wallis, A. P. Street and J. S. Wallis, Combinatorics: Room Squares, Sum-Free Sets, Hadamard matrices, Springer-Verlag, Berlin-Heidelberg- New York (197). [5] T. Xia, M. Xia, and J. Seberry, Regular Hadamard matrices, maximum excess and SBIBD, Australasian Journal of Combinatorics, Vol. 7 (003) pp. 63 75. 8
Affiliation of author: Dean Crnković Department of Mathematics Faculty of Philosophy Omladinska 14 51000 Rijeka Croatia E-Mail: deanc@mapef.pefri.hr 9