On Hadamard Groups IV

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Journal of Algebra 234, 651 663 (2000) doi:10.1006/jabr.2000.8545, available online at http://www.idealibrary.

Ito Hachimae 2-1612, San Royal 301, Nagoya, Meito 465-0018, Japan E-mail: fn48@tcp-ip.or.

1 Journal of Algebra 234, (2000) doi: /jabr , available online at on On Hadamard Groups IV Noboru Ito Hachimae , San Royal 301, Nagoya, Meito , Japan Communicatedby Gernot Stroth Received April 18, 2000 dedicated to professor helmut wielandt on the occasion of his 90th birthday Let 2n be the set of 1 1 sequences of length 2n Then the purpose of the present paper is to introduce the concept of L-structure to elements of 2n and to rewrite the necessary and sufficient conditions for a pair of elements of 2n to be associated in terms of L-structure so that the condition becomes more visible. We hope that this makes it a little easier to construct an associated pair Academic Press 1. PRELIMINARIES Let G n be a dicyclic group of order 8n presented by G n = a b a 4n = e a 2n = b 2 and b 1 ab = a 1 where e is the identity of G n and n is odd. G n contains only one involution a 2n = b 2, and itis denoted by e.lett be a transversal of G n with respect to e.ift satisfies the property that T Tx =2n 1 for every elementof G n outside e, T is called an Hadamard subsetof G n and G n is called a Hadamard group. If G n is an Hadamard group of order 8n then a Hadamard matrix of order 4n will be constructed. For this see [2]. Now any T can be rewritten as T = A + bb where A and B are n- subsets of a. Further an element x in 1 may be restricted to an element /00 $35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.

2 652 noboru ito of a. For a proof see [4]. So 1 can be rewritten as A Ax + B Bx =2n for every element x of a outside e Moreover, A and B may be considered as elements of the set 2n of 1 1 sequences of length 2n on which the negacyclic translation σ acts so that if A = a 1 a 2 a 2n, then Aσ = a 2n a 1 a 2n 1 where a i equals 1 or 1, according to the exponent of a i as a power of a less than or greater equal to 2 n.leta and B be two elements of 2n Then the intersection number A B of A and B is defined by i a i = b i 1 i 2n Now 2 can be rewritten as A Aσ j + B Bσ j =2n for 1 j 2n 1 First we notice that Aσ 2n = A and that Aσ 4n = A for every element A of 2n Thus the order of σ equals 4n We assemble basic properties of intersection numbers. Lemma 1. Let A and B be elements of 2n Then we have the following. (i) A B = B A (ii) A B = Aσ i Bσ i for 1 i 4n 1 (iii) A Aσ 2n =0 (iv) A Aσ i = A Aσ 4n i for 0 i 4n 1 (v) A Aσ i + A Aσ 2n i =2n for 0 i 2n 1 (vi) A Aσ n =n (vii) A B = 2 A B n where A B denotes the usual inner product of A and B Proof. It is straightforward. Definition 1. Let A and B be in 2n. Then A and B are called associated if they satisfy 3. We also say that B is an associate of A. Proposition 1. Let A and B be elements of 2n Then A and B are associated if holds for 1 i n 1 Proof. By Lemma 1 we have that A Aσ i + B Bσ i =2n A Aσ i + A Aσ 2n i + B Bσ i + B Bσ 2n i =4n for 1 i 2n 1. Thus if 3 holds, then since n + 1 2n i 2n 1if1 i n 1 4 holds. Definition 2. Let A be an elementof 2n. Then A is called shortif there exists a positive integer m<4n such that Aσ m = A A is called long if A is notshort

3 hadamard groups 653 Proposition 2. Let A be a short element of 2n. Then A has the following form. There exists a divisor s of n such that m = 4s is the least positive integer such that Aσ m = A. Let A 2s be the subarray of length 2s of A consisting of a 1 a 2s. Then we have that A = A 2s A 2s A 2s A 2s A 2s where the number of subarrays consisting A equals t = n/s and A 2s is a long member of 2s Proof. Proposition 3. The proof is obvious. A short element A of 2n has no associate. Proof. Let A be as in Proposition 2. Then if B is an asociate of A, we have that A Aσ 4s + B Bσ 4s =2n Hence we have that B Bσ 4s =0 Bσ 4s = B and Bσ 8s = B Since gcd 4n 8s =4s we have a contradiction that Bσ 4s = B Let L 2n denote the subset of 2n consisting of all long elements of 2n. From now on we consider only L 2n. Definition 3. Let A be an elementof L 2n. Then the sequence int A = A Aσ A Aσ n 1 5 is called the intersection number sequence of A.Ifint A =int C then A and C are called equivalent. Seeking an associated pair A and B in L 2n, A and B can be replaced by any equivalent ones. However, we have been unable to determine the equivalentclass of an arbitrarily given elementof L 2n againstour desire. Definition 4. Let A be an elementof L 2n. Then a 4n-subset A Aσ Aσ 4n 1 is called the translation class of A. Since σ is a negacyclic translation, there exists Aσ i such that Aσ i 1 = Aσ i 2n = 1. From now on we mostly consider A such that a 1 = a 2n = 1 Definition 5. Let A be an elementof L 2n If a i a i+1 = 1 then i is called an alternating spot. Let be the alternating spots of A Then i 1 <i 2 < <i k A i 1 = a 1 a 2 a i 1 A i 2 = a i 1 +1 a i 2 A i k = a i k 1 +1 a 2n

4 654 noboru ito are called blocks of A. blk A denotes the number of blocks of A We notice that blk A is always odd, because a 1 = a 2n = 1 Further l A denotes the multiset of lengths of blocks of A Furthermore A may be written as A = i 1 i 2 i k 6 Definition 6. Let l A i denotes the number of blocks of A of length i Moreover, let l A i 1 i r denote the number of r-consecutive blocks of A of lengths i 1 i r where 1 r k n 1 2. STEP BY STEP APPROACH The calculation of A Aσ k becomes exponentially intricate as k increases. So we make the following definition in order to proceed step by step. Definition 7. Let A and B be elements of L 2n. Then they are called m-associated if they satisfy 4 up to m, where 1 m n 1. So they are associated if and only if they are n 1 -associated. Proposition 4. Let A and B be elements of L 2n. Then we have that A Aσ =2n blk A Hence A and B are 1-associated if and only if blk A +blk B =2n 7 holds. Proof. It is straightforward by Definitions 5 and 7. Definition 8. We introduce the notations o and x Both o and x can be equal to either 1 or 1 such that ox = xo = 1 Proposition 5. Let A and B be elements of L 2n. Then we have that A Aσ 2 =2n 2blk A +2l A 1 Hence A and B are 2-associated if and only if 7 and l A 1 +l B 1 =n 8 hold.

5 hadamard groups 655 Proof. First of all, by Definition 6 we get the following equations. il A i =2n i=1 l A i =blk A i= Now A Aσ 2 equals the sum of the portions of the shapes ooo and oxo in A. The latter obviously equals l A 1. The former, by 9 and 10, equals i 2 l A i =2n l A 1 2l A 2 2blk A i=3 Thus we have that By Proposition 4 the rest is clear. l A 1 l A 2 =2n 2 blk A +l A 1 A Aσ 2 =2n 2 blk A +2l A 1 Equation 8 implies a remarkable restriction on blk A and blk B when A and B are 2-associated. Proposition 6. Assume that an element A of L 2n possesses a 2- associate B. The it holds that 3n 1 2 Moreover, if n 3 (mod 4), then blk A n Proof. 3n 3 2 By 8 and 9 we have that blk A n blk A = i 1 l B i l B i i=2 i=2 12 By 8 and 10 we have that n l B 1 =2n blk A l B i 2n 2blk A i=2 Since n and blk A are odd, we getthe righthand side of 11 and 12 The lefthand sides of 11 and 12 follow from 7

6 656 noboru ito Proposition 7. Let A and B be elements of L 2n. Then we have that A Aσ 3 =2n 3 blk A +4l A 1 +2l A 2 2l A 1 1 Hence A and B are 3-associated if and only if they are 2-associated and holds. l A 2 +l B 2 =l A 1 1 +l B Proof. A Aσ 3 equals the sum of the portions of the shapes oooo, ooxo, oxoo and oxxo in A The fourth equals l A 2 By 9 and 10 the first equals i 3 l A i =2n l A 1 2l A 2 3l A 3 i=4 3 blk A l A 1 l A 2 l A 3 = 2n 3 blk A +2l A 1 +l A 2 In order to calculate the sums corresponding to the second and third shapes, it is useful to make the convention that the sum of the portions corresponding to a certain shape is denoted by the shape itself. Then we getequations like and oxoo + oxox = oxo ooxo + xoxo = oxo From these equations we get that the second and third sums equal l A 1 l A 1 1 By Propositions 4 and 5 the rest is clear. Remark 1. The use of the kind of the shape calculation in the proof of Proposition 7 is valid in general. It also reveals that we have to calculate the shape like oxox We notice that xoxo = oxox We also introduce the notation a 2n+k = a k for 1 k n because we have to look for up to a 2n+k when we calculate A Aσ k Proposition 8. Let A and B be elements of L 2n Then we have that A Aσ 4 =2n 4 blk A +6l A 1 +4l A 2 +2l A 3 4l A 1 1 2l A 1 2 2l A l A (15) Hence A and B are 4-associated if and only if they are 3-associated and holds. l A 3 +l B 3 +l A l B = l A 1 2 +l B 1 2 +l A 2 1 +l B 2 1 (16)

7 hadamard groups 657 Proof. A Aσ 4 equals the sum of the portions of the shapes ooooo, oooxo, ooxoo, ooxxo, oxooo, oxoxo, oxxoo and oxxxo The eighth equals l A 3 The sixth equals l A By 9 and 10 the first equals i 4 l A i =2n 4blk A +3l A 1 +2l A 2 +l A 3 i=5 From oooxo + xooxo = ooxo and from the proof of Proposition 7 we have that the second equals l A 1 l A 1 1 l A 2 1 From ooxoo + ooxox = ooxo ooxox + xoxox = oxox and from the proof of Proposition 7 we have that the third equals l A 1 2l A 1 1 +l A From ooxxo + xoxxo = oxxo and from the proof of Proposition 7 we have that the fourth equals l A 2 l A 1 2 The fifth is similar to the second and equals l A 1 l A 1 1 l A 1 2 The seventh is similar to the fourth and equals l A 2 l A 2 1 The restis clear by Propositions 4, 5, and 7. Remark 2. Now we hope that the mechanism of the shape calculation which we are using is quite clear. Example 1. It may be interesting to find extreme (in the sense of Proposition 6) associated pairs. The following pairs are associated. We use the notation in (6). (i) n = 3 A= and B = (ii) n = 5 A= 6, 2, 2 and B = 3, 1, 1, 2, 1, 1, 1. (iii) n = 7 A= 5 5, 1, 2, 1 and B = 3, 2, 2, 1, 1, 1, 1, 1, FROM A Aσ i TO A Aσ i+1 Now we consider relations between A Aσ i and A Aσ i+1 A Aσ i equals the number of shapes o i 1 o where m runs over all shapes of length m In the i + 1-stage o i 1 o branches out into two shapes, o i 1 oo and o i 1 ox Similarly o i 1 x branches out into two shapes, o i 1 xo and o i 1 xx Let α i+1 A β i+1 A γ i+1 A and δ i+1 A be

8 658 noboru ito the numbers of shapes o i 1 oo o i 1 ox o i 1 xo and o i 1 xx in A, respectively. Then we have that and A Aσ i =α i+1 A +β i+1 A (17) 2n A Aσ i =γ i+1 A +δ i+1 A (18) A Aσ i+1 =α i+1 A +γ i+1 A (19) 2n A Aσ i+1 =β i+1 A +δ i+1 A 20 Looking at the last two spots of o i 1 ox and o i 1 xo we see that β i+1 A +γ i+1 A =blk A 21 From (17) (21) we have that A Aσ i+1 A Aσ i =2γ i+1 A blk A 22 Proposition 9. even. A Aσ i is odd or even, according to whether i is odd or Proof. Since blk A is odd, itis clear by (22). Definition 9. Let A be an elementof L 2n Then A is called semiregular if A Aσ 2i 1 =n for every i such that 1 i n We notice that an associated pair corresponding to an Hadamard subset of Williamson type is semiregular. For this see the last section of the present paper. 4. MAIN RESULTS When we consider a partition of a positive integer, the order of summands should be taken into our consideration. Therefore, for instance, we consider that 3 = and 3 = are distinct. Moreover, we regard 1 + 2as 1 2. LetP i j be the set of such partitions of i into j summands. Theorem 1. Let 3 k n 1 Then we have that A Aσ k =2n kblk A +2 1 j+1 1 j k 1 j i k 1 x P i j k i l A x Proof. We rewrite o k 1 o in terms of the L-structure of A We divide o k 1 o into four portions: N 1 k =ox k 3 xo N 2 k =ox k 3 oo N 3 k =oo k 3 xo and N 4 k =oo k 3 oo Firstof all we getthat N 1 k = l A x 1 i k 1 andi is odd x P k 1 i

9 hadamard groups 659 As for N 2 k we use the equation ox k 3 oo + ox k 3 ox = ox k 3 o = ox k 4 oo + ox k 4 xo recurrently and we get that N 2 k =l A 1 l A 1 1 l A x + 3 i k 2 1 2m 1 i 2 x P i 1 2m 1 4 i k 1 1 2m i 1 x P i 1 2m l A x Since N 2 3 =N 3 3 we getthatn 3 k =N 2 k The case of N 4 k is a little more intricate. First let k 4 and put X k = oo k 3 ox Then using the equation oo k 3 ox + xo k 3 ox = o k 3 ox = oo k 4 ox + ox k 4 ox recurrently and using the formulae for N 1 i we getthat X k =X 4 l A x + 5 j k 4 j k 1 1 2m 1 2 2m j 2 x P j 1 2m 1 x P j 2 2m l A x Next we notice that N 4 3 =oooo = 2n 3 blk A +2l A 1 +l A 2 N 3 3 =ooxo = l A 1 l A 1 1 and that X 4 =oo 1 ox = blk A l A 1 l A 2 l A 3 +l A 1 1 l A Then using the equation oo k 3 oo + oo k 3 ox = oo k 4 oo + oo k 4 xo recurrently and using the formulae for X i and N 3 i we getthat N 4 k =2n kblk A +2 k 2 l A 1 + k 2 l A 2 k 3 l A l A x 4 i k + 4 j i x P j 2 2m 4 i k 3 i k 1 2 j i 2 x P j 2m 1 4 j i x P j 1 2m 1 l A x l A x 3 i k 1 3 j i 1 x P j 2m l A x Now we have only to show that the coefficient of x in P i j where k 1 i j in 1 i 4 N i k, equals 1 j+1 2 k i Since the result will depend only on i so we have only to count the number of times of the appearances of P i Now in N 1 k only P k 1 2m 1 appears with multiplicity 1 In 2N 2 k P i 1 2m where 3 i 1 k 2 together with l A 1 1 appears with mulitiplicity 2 and P i 1 2m 1 where 2 i 1 k 3 together with l A 1 appears with multiplicity 2 The restcomes from N 4 k without difficulty.

10 660 noboru ito Theorem 2. Let n k 3 Let A and B be two elements of L 2n Then A and B are associated if and only if the following conditions hold: blk A +blk B =2n l A 1 +l B 1 =n and for every k 3 l A x +l B x i is odd and 1 i k 1 = x P k 1 i i is even and 2 i k 1 x P k 1 i l A x +l B x Proof. Since A and B are k-associated for every k in n k 1 Theorem 2 follows easily from Theorem 1. Since the procedure is reversible, we only show the necessity. Put 2n = A Aσ k + B Bσ k In the equation of Theorem 1 consider the portion of i = 1 Then we get 2 k 1 l A 1 +l B 1 and since 2n = 4n 2kn + 2 k 1 n we getthe equation 0 = the equation in Theorem 1 under the condition i 2 Next consider the portion of the equation of i = 2 Then we getthe equation 2 l A x +l B x 2 l A x +l B x =0 x P 2 1 x P 2 2 Clearly we can go on up to k 1 and get the equation in Theorem REMARKS ON CASES OF PALEY, WILLIAMSON, AND QUADRATIC RESIDUE TYPES There exist two well-known infinite families of Hadamard subsets in dicyclic groups, in other words, associated pairs, namely Hadamard subsets of Paley type, which exist for all n such that 2n 1 are prime powers, and Hadamard subsets of quadratic residue type, which exist for all n such that 4n 1 are prime powers. The family of Hadamard subsets of Paley type is included in the family of Williamson type. However, as far as we know, integers n such that 2n 1 are notprime powers for which Hadamard subsets of Williamson type exist are so far rather scattered. For this see [1, 2, 6 8]. We already know that associated pairs corresponding to Hadamard subsets of Williamson type are semiregular (For this see [5, Proposition 2]). Proposition 10. Let A be a member of an associated pair corresponding to a Hadamard subset of Williamson. Then we have that blk A =n

11 hadamard groups 661 Proof. Put A = f 0 f 1 f 2n 1 Then we have that f i = 1 i+1 f 2n i for i = 1 2n 1 (for this see [5]). We count the number of alternating spots of A. Now exactly one of f n i 1 f n i and f n+i f n+i+1 is an alternating spot. Thus we have that blk A = n Proposition 11. Let A and B be an associated pair corresponding to a Hadamard subset of Paley type. Then (i) l A 1 l B 1 = n+1 n 1 and 2 2 (ii) A Aσ 2i =±2 and B Bσ 2i = 2 Proof. We may put A = c 0 c 1 c 2n 1 and B = c 0 c 1 c 2n 1 Namely they differ only at the first entry. (i) So l A 1 and l B 1 can differ atmostone. However, by Proposition 5 their sum equal n which is odd. (ii) Firstof all we have that A Aσ 2i + B bσ 2i = 0 On the other hand, the difference of two summands is either 4 or 4. By Proposition 9 A Aσ 2i cannotbe zero. This shows (ii). Finally we listknown (one for each n up to n = 45) associated pairs corresponding to Hadamard subsets of Paley, Williamson, or quadratic residue types. We choose the Paley type if possible; otherwise, we choose one of the Williamson type. If the Williamson type is not available, we choose the quadratic residue type. The three types are abbreviated as P, W, and Q. We use the abbreviated notation i j instead of j of i s. Example 2. n = 5 P A = B = n = 7 P A = B = n = 9 P A = B = n = 11 W A = B = n = 13 P A = B = n = 15 P A = B = n = 17 W A = B = n = 19 P A = B =

12 662 noboru ito n = 21 P n = 23 W n = 25 P n = 27 P n = 29 W n = 31 P n = 33 W n = 35 Q n = 37 P n = 39 W n = 41 P n = 43 W n = 45 P A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = A = B = n = 47 is the first number for which the existence of an associated pair is so far unknown.

13 hadamard groups 663 REFERENCES 1. D. Z. Dokovic, Williamson matrices of order 4n for n = , Discrete Math. 115 (1993), N. Ito, On Hadamard groups, J. Algebra 168 (1994), N. Ito, On Hadamard groups II, J. Algebra 169, (1994), N. Ito, Some results on Hadamard groups, in Groups Korea 94, , de Gruyter, Berlin/ New York, N. Ito, On Hadamard groups III, Kyushu J. Math. 51 (1997), W. D. Wallis, A. P. Street, and J. S. Wallis, Combinatorics: Room Squares, Sum-free Sets, Hadamard Matrices, Lecture Notes in Mathematics, Vol. 292, Springer-Verlag, Berlin/Heidelberg/New York, M. Yamada, On the Williamson type j matrices of order 4.29, 4.41, and 4.37, J. Combin. Theory A 27 (1979), K. Yamamoto, Combinatorics, Asakura, Tokyo, [In Japanese].

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer? Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative