Groups, partitions and representation functions

Transcription

1 Groups, partitions and representation functions Sándor Z. Kiss, Eszter Rozgonyi, Csaba Sándor bstract Let G be a finite (abelian) group. In this paper we determine all subsets G such that the number of solutions of g = x + y, x, y equals to the number of solutions of g = x + y, x, y G \. We discuss some related problems. 000 MS Mathematics subject classification number: 11B34. Key words and phrases: groups, representation functions. 1 Introduction Let X be a semigroup, written additively. Let 1,..., h be subsets of X and let x be an element of X. We define the ordered representation function R h (x) = #{(a 1,..., a h ) 1 h : a a h = x}. If i = for i = 1,..., h, then we write R (1),h (x) = #{(a 1,..., a h ) : a i : a a h = x}. Let X be an abelian semigroup, written additively. For X, let h denote the set of all h-tuples of. Two h-tuples (a 1,..., a h ) h and (a 1,..., a h ) h are equivalent if there Institute of Mathematics, Budapest University of Technology and Economics, MT-BME Lendület Future Internet Research Group, H-159 B.O. Box, Hungary; Computer and utomation Research Institute of the Hungarian cademy of Sciences, Budapest H-1111, Lágymányosi street 11; kisspest@cs.elte.hu; This author was supported by the OTK Grant No. K77476 and No. NK Institute of Mathematics, Budapest University of Technology and Economics, H-159 B.O. Box, Hungary, reszti@math.bme.hu, The work reported in the paper has been developed in the framework of the project "Talent care and cultivation in the scientific workshops of BME" project. This project is supported by the grant TÁMOP B-10/ This author was supported by the OTK Grant No. K Institute of Mathematics, Budapest University of Technology and Economics, H-159 B.O. Box, Hungary, csandor@math.bme.hu. This author was supported by the OTK Grant No. K

2 is a permutation α : {1,..., h} {1,..., h} such that a α(i) = a i for i = 1,..., h. Two other representation functions arise often and naturally in additive number theory. The unordered representation function,h (x) counts the number of equivalence classes of h-tuples (a 1,..., a h ) such that a a h = x. The unordered restricted representation function R (3),h (x) counts the number of equivalence classes of h-tuples (a 1,..., a h ) of pairwise distinct elements of such that a a h = x. lternative definitions for (x) and R(3) (x) are the following. Denote by D (x) = #{a : a, a + a = x} then and (x) = 1 R(1) (x) + 1 D (x) R (3) (x) = 1 R(1) (x) 1 D (x) Let N be the set of nonnegative integers. Let X = N. nswering a question of Sárközy, Lev [1] and independently Sándor [] characterized all subsets N such that (n) = N\ (n) or R(3) (n) = R(3) N\ (n) from a certain point on. The precise theorems are the following. Theorem (Lev, Sándor, 004). Let X = N. Let N be a positive integer. The equality (n) = R() N\ (n) holds for n N 1 if and only if [0, N 1] = N and m m, m + 1 m for m N. Theorem (Lev, Sándor, 004). Let X = N. Let N be a positive integer. The equality R (3) (n) = R(3) N\ (n) holds for n N 1 if and only if [0, N 1] = N and m m, m + 1 m for m N. Similar statement can not be formulated for the representation function R (1) (n) because R (1) (n) is odd if and only if n, therefore either R(1) (m) or R(1) N\ (m) is odd. nontrivial result is the following in this direction. Theorem 1. Let X = N. The equality R (1) +B (n) = R(1) N\+N\B (n) holds from a certain point on if and only if N \ ( B)) = B < The modular questions were solved by Chen and Yang [3]. Theorem (Chen, Yang, 01). Let X = Z m. The equality R (1) (n) = R(1) N\ (n) holds for all n Z m if and only if m is even and = m/.

3 Theorem (Chen, Yang, 01). Let X = Z m. For i {, 3}, the equality R (i) = R(i) Z (n) m\ holds for all n Z m if and only if m is even and t t + m/ for t = 0, 1,..., m/ 1. We extend the first theorem to arbitrary finite group G and the second theorem to finite abelian group. Theorem. Let X = G be a finite group. The equality R +B (g) = R G\+G\B (g) holds for all g G if and only if + B = G direct consequence is the following. Corollary 1. Let X = G be a finite group. The equality R (1) (g) = R(1) G\ (g) holds for all g G if and only if G is even and = G /. Theorem 3. Let X = G be a finite abelian group. For i {, 3}, the equality R (i) (g) = R (i) G\ (g) holds for all g G if and only if D (g) = D G\ (g) for every g G. Proof Proof of Theorem 1. Denote by S(x) the generating function of a subset S N, i.e. S(x) = x s. The generating function of R +B (n) is (x)b(x), while the generating s S function of R N\+N\B (n) is ( 1 (x))( 1 B(x)). Hence the condition R 1 x 1 x +B(n) = R N\+N\B (n) holds from a certain point on is equivalent to 1 (x)b(x) ( 1 x (x))( 1 B(x)) = p(x), 1 x where p(x) is a polynomial. This is equivalent to (x) + B(x) = 1 + p(x)(1 x). (1) 1 x Let (x) + B(x) = n=0 c nx n, where c n = 0, 1 or. The equation (1) holds if and only if c n = 1 except for finitely many integer n and the number of n for which c n = 0 is equal to the number of n for which c n =. This is equivalent to the condition N \ ( B)) = B <. Proof of Theorem. For a given S G denote by χ S its characteristic function, that is χ S (g) = 1 if g S and χ S (g) = 0 if g S for every g G. Then χ G\S = 1 χ S. 3

4 Obviously, for every g G we have R +B (g) R G\+G\B (g) = χ (c)χ B ( c + g) χ G\ (c)χ G\B ( c + g) = c G c G χ (c)χ B ( c+g) (c))(1 χ B ( c+g)) = c G c G(1 χ χ (c)+ χ B ( c+g) 1 = c G c G c G + B G. Hence if R +B (g) = R G\+G\B (g) for every g G then + B = G and if + B = G then R +B (g) = R G\+G\B (g) for every g G. Proof of Theorem 3. We only prove the case i =, because the proof of case i = 3 are very similar. Obviously, for every fixed g G we have = #{(a, y) : a, y G, a + y = g} = #{(a, y) : a, y, a + y = g} + #{(a, y) : a, y G \, a + y = g} = R + (g) + R +G\ (g) = (g) D (g) + R +G\ (g), thus Similarly, (g) = D (g) 1 R +G\(g). G \ = #{(x, b) : x G, b G \, x + b = g} = #{(x, b) : x G \, b G \, x + b = g} + #{(x, b) : x, b G \, x + b = g} = R G\+G\ (g) + R +G\ (g) = G\ (g) D G\(g) + R +G\ (g), thus G\ (g) = 1 G \ + 1 D G\(g) 1 R +G\(g). Hence for every g G we have (g) R() G\ (g) = D (g) ( 1 G \ + 1 D G\(g)) = 1 G + 1 (D (g) D G\ (g)). 4

5 Suppose that (g) = R() G\ (g) for every g G. Then we have ( ) + 1 = (g) = ( ) G \ + 1 G\ (g) =, therefore = G \, that is = G. Hence we get that D (g) = D G\ (g) for every g G. Finally, suppose that D (g) = D G\ (g) for every g G. Then we have = D (g) = D G\ (g) = G \, therefore = G, thus we get that (g) = R() G\ (g) for every g G, which completes the proof. References [1] Sándor, Csaba, Partitions of natural numbers and their representation functions. Integers 4 (004), 18, 5 pp. [] Lev, Vsevolod F., Reconstructing integer sets from their representation functions. Electron. J. Combin. 11 (004), no. 1, Research Paper 78, 6 pp. [3] Yang, Quan-Hui; Chen, Feng-Juan, Partitions of Z m with the same representation functions. ustralas. J. Combin. 53 (01),