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NON-CANONICAL EXTENSIONS OF ERDŐS-GINZBURG-ZIV THEOREM 1 R. Thangadurai Stat-Math Division, Indian Statistical Institute, 203, B. T. Road, Kolkata , INDIA thanga Received: 11/28/01,

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NON-CANONICAL EXTENSIONS OF ERDŐS-GINZBURG-ZIV THEOREM 1 R. Thangadurai Stat-Math Division, Indian Statistical Institute, 203, B. T. Road, Kolkata , INDIA thanga Received: 11/28/01, Revised: 4/2/02, Accepted: 4/8/02, Published: 4/15/02 Abstract In 1961, Erdős-Ginzburg-Ziv proved that for a given natural number n 1 and a sequence a 1,a 2,,a 2n 1 of integers (not necessarily distinct), there exist 1 i 1 i 2 i n 2n 1 such that a i1 + a i2 + + a in is divisible by n. Moreover, the constant 2n 1 is tight. By now, there are many canonical generalizations of this theorem. In this paper, we shall prove some non-canonical generalizations of this theorem. 1. Introduction and Preliminaries Additive number theory, graph theory and factorization theory provide inexhaustible sources for combinatorial problems in finite abelian groups (cf. [27], [28], [13], [11], [29] and [3]). Among them zero sum problems have been of growing interest. Starting points of recent research in this area were the Theorem of Erdős-Ginzburg-Ziv (EGZ Theorem, for short) and a question of H. Davenport on an invariant which today carries his name. We shall denote the cyclic group of order n by Z n. A sequence S = {a i } l i=1 of length l in Z n, we mean a i Z n and a i s are not necessarily distinct, unless otherwise specified. Also, throughout this paper, writting n 1, we mean n is an arbitrary natural number and writting p, we mean an arbitrary prime number. We shall define some terminalogies as follows. A sequence S is called zero sequence if its sum is zero. A sequence S is called zero-free sequence if it contains no zero subsequence. A sequence S is called minimal zero sequence if S is a zero sequence; but any proper subsequence is zero-free. Now, we shall restate the EGZ theorem, using the above terminalogies, as follows. EGZ Theorem. (cf. [12]) Given a sequence S in Z n of length 2n 1, one can extract 1 Mathematics subject classification (1991): 05D05, 20D10 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 2 a zero subsequence of length n in Z n. We shall state the following known result which will be useful for our further discussion. Cauchy-Davenport inequality. Let A and B be two non-empty subsets of Z p. Then A + B min{p, A + B 1} where A + B = {x = a + b Z p : a, b Z p } and K denotes the cardinality of the subset K of Z p. This was first proved by Cauchy (cf. [9]) in 1813 and was rediscovered by Davenport (cf. [10]) in Corollary 1.1 Let A 1,A 2,,A h be non-empty subsets of Z p. Then, h A 1 + A A h min{p, A i h +1}. i=1 By today there are several extensions of Erdős-Ginzburg-Ziv theorem are known. All the known extensions are natural and we call them canonical extensions. In this paper, we shall prove several non-canonical extensions of EGZ theorem. 2. Canonical extensions of EGZ Theorem In this section, we shall survey the results which are natural generalization or extensions of EGZ Theorem and we call them as C-Extensions. The first natural generalization of EGZ in Z is the following due to Olson. C-Extension 1. (Olson, 1969, [30]) Suppose m k 2 are integers such that k m. Let a 1,a 2,,a m+k 1 be a sequence of integers. Then there exists a non-empty subset I of {1, 2,,m+ k 1}, such that I = m and i I a i 0 (mod k). If one view EGZ theorem as a statement over the solvable group Z n, then one can ask for the same in any finite group. Indeed, Olson proved that C-Extension 2. (Olson, 1976, [31]) Let g 1,g 2,,g 2n 1 be a sequence of length 2n 1 in a finite group (but not necessarily Abelian) G of order n. Then there exists elements g i1,g i2,,g in from the given sequence satisfies g i1 + g i2 + + g in =0in G. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 3 Conjecture 2.1 (Olson, 1976, [31]) The same conclusion holds in C-Extension 2, together with i 1 i 2 i n. In this direction, W. D. Gao (cf. [16]) proved in 1996 that if G is a non-cyclic solvable group and s = [11n/6] 1, then for any sequence a 1,a 2,,a s in G, we have i 1,i 2,,i n distinct such that a i1 + + a in =0inG. Other than this result, we know nothing about Conjecture 2.1. In C-Extension 2, 2n 1 may not be tight except for the group G = Z n. Indeed, if G is an abliean group (additively written) of order n, then Gao (cf. [17]) proved that n + D(G) 1 is the right constant in place of 2n 1 where D(G) (is the Davenport Constant), which is the least positive integer such that given any sequence S in G of length l(s) with l(s) D(G), there exists a zero subsequence T of S in G. One can easily see that when G = Z n, we have D(Z n )=n. There is yet another generalization of EGZ theorem as follows. EGZ theorem proves the existence of one zero subsequence of length n, whenever we consider a sequence in Z n of length 2n 1. C-Extension 3 (W. D. Gao, 1997, [19]) If a 1,a 2,,a 2n 1 be a sequence in Z n, then there exists at least n number of subsequences of length n having its sum a for any given a Z n provided no element occurs more than n times in the sequence. More over, there exists at least n +1 number of zero subsequences of length n in Z n unless only two elements x and y occur n and n 1 times respectively in that sequence. Indeed, chronologically, H. B. Mann (cf. [26]) in 1967 proved the existence of one subsequence of length p whose sum is g for a given g Z p, whenever we consider a sequence in Z p of length 2p 1. In 1996, W. D. Gao (cf. [20]) proved that C-Extension 3 for n = p prime. (Indeed, Sury (cf. [34]) gave a different proof of this fact). One should mention the result of Bialostocki and Dierker which is a stronger version of EGZ theorem; C-Extension 4. (Bialostocki and Dierker, 1992, [5]) If S = {a i } is a sequence in Z p of length 2p 1, then there are p indices 1 i i i 2 i p 2p 1 such that a i1 + a i2 + + a ip 0 (mod p). Moreover, if for two indices j, k we have a j a k (mod p), then we can choose i 1,i 2,,i p such that not both j and k are among them in that zero of length p. Indeed, we shall prove better result than C-Extension 4 as follows. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 4 C-Extension 4. If S = {a i } is a sequence in Z p of length 2p 1, then there are p indices 1 i i i 2 i p 2p 1 such that a i1 + a i2 + + a ip 0 (mod p). (1) Moreover, if s 2 distinct elements of Z p, say, a 1,a 2,,a s are in S, then we can choose the i 1,i 2,,i p such that only one of the indices from 1, 2,,s appears among i 1,i 2,,i p and satisfying (1). Proof. If one of the element of S is repeated more than p times, then the result follows trivially. Assume that none of the elements of S appears more than p 1 times. Let A 1 = {a 1,a 2,,a s } Z p and the remaining a i s be distributed into non-empty p 1 incongruent classes modulo p, say, A 2,A 3,,A p 1. Then by Cauchy-Davenport inequality, we have A 1 + A A p min{p, i A i p +1} = min{p, 2p 1 p +1} = p Thus the result follows. = A 1 + A A p = Z p. 3. Non-Canonical extensions of EGZ Theorem In this section, we prove non-canonical generalizations of EGZ Theorem and we call them as N-C Extensions. N-C Extension 1. Let S be a sequence in Z n of length at least n. Let h = h(s) = max a S g(a) where g(a) denote the number of times a Z n appearing in S. Then there is a zero subsequence of length less than or equal to h. This theorem was proved by Gao and Yang (cf. [24]) in We shall prove that this theorem indeed implies EGZ Theorem. Theorem 3.1 N-C Extension 1 implies EGZ theorem. Before going to the proof of Theorem 3.1, we prove the following lemma. Lemma Let S be a sequence in Z n of length 2n 1. Suppose there is an element a Z n such that a is appearing in S at least [n/2] times, then there is a zero subsequence of S of length n in Z n. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 5 Proof. Let S be a sequence in Z n of length 2n 1. Suppose S consists of an element a Z n which is repeated s [n/2] number of times. If s n, then the result is obvious. Let us assume that s n 1. Consider the translated sequence S a in which 0 is repeated s number of times. The length of the subsequence T 1 of S a which consists of all the non-zero elements of S a is 2n 1 s n. Since D(Z n )=n, the sequence T 1 contains a zero subsequence say T 2. Let the length of T 2 be t 2. Clearly, 2 t 2 n. Choose T 2 such that it has the maximal length t 2. Also, note that if s + t 2 n, then we can extract a zero of length n in S a which in turn produces a zero subsequence of length n in S. Thus we can assume that s + t 2 n.note that [n/2]+1 t 2 n. If not, that is, t 2 [n/2]. Then after omitting T 2 from T 1, the length of the sequence T 1 \T 2 is at least n and hence it contains a zero subsequence say T 3 with length t 3. Clearly, t 3 t 2 [n/2], which implies, t 2 + t 3 n, which contradicts to the maximality of T 2. Hence [n/2]+1 t 2 n is true. Since we have at least [n/2] zeros out side T 2, by adding appropriate number of zeros to T 2, we get a zero sequence of length n in S a which in turn produces a zero sequence of length n in S. Proof of Theorem 3.1. Let S be a sequence in Z n of length 2n 1. Suppose S consists of an element a Z n which is repeated maximum number of s times. If s n, then nothing to prove. Case (i) ([n/2] s n 1) This case is covered by Lemma Case (ii) (2 s [n/2] 1) Consider the translated sequence S a inwhich 0 is repeated s times. The length of the subsequence T 1 of S a which consists of all the non-zero elements of S a is 2n 1 s n + n [n/2]. Since D(Z n )=n, there exists a zero subsequence of T 2 of length t. Choose T 2 having the maximum length. Then, it follows that (apply the same argument given in the begining of the proof of Lemma 3.1) [n/2] + 1 t n. Claim. s + t n. Suppose not, that is, s + t n.now delete the subsequence T 2 from T 1. Then the length of the deleted sequence, say T 3, is 2n 1 s t 2n 1 (n 1) = n +1. By N-C Extension 1, there exists a zero subsequence of length less than or equal to s in T 3. Therefore there exists a subsequence T 4 of T 3 such that the length of T 4, say t 1 is less than or equal to s. Since T 2 is maximal with respect to its length less than or equal to n, it is clear that t + t 1 n.if 2 t 1 s [n/2] 1, then n t 1 +1 t n 1. Since t 1 s, n t 1 +1 n s + 1 which implies that s + t nwhich is a contradiction to the assumption. This proves the claim. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 6 Since s + t nand t n, we can add appropriate number of zeros to T 2 so as to get a zero sequence of length n. Before going into the further discussions, we shall prove the following theorem. Theorem 3.2. Let n and k be positive integers such that 1 k (n +2)/3. Then, the following statments are equivalent. (I) Let S be a minimal zero sequence of Z n of length n k +1. Then there exists a Z n such that a appears in S at least n 2k +2 times. (II) Let S be a zero-free sequence in Z n of length n k. Then one element of S is repeated at least n 2k +1 times. (III) Let S be a sequence of Z n of length 2n k 1. Suppose S does not have a zero subsequence of length n. Then there exist a b Z n such that a and b appear in S at least n 2k +1 times. Remark. The statement (I) was proved by the author in 2001 (cf. [35]). The statement (II) was proved by Bovey, Erdős and Niven in 1975 (cf. [6]). Also, all the three statements are valid for n 2k 1. But the equivalence is valid only for the range 1 k (n+2)/3. Proof. (I) (II) Assume that (I) is true. Consider a zero-free sequence S = {a i } in Z n of length n k. Let a n k+1 = n k i=1 a i and S 1 be the sequence consisting of all the elements a i together with a n k+1. Then, S 1 is a minimal sequence of length n k +1 in Z n. For, if any proper subsequence, say, T of S together with a n k+1 is a zero subsequence of S 1, then the deleted sequence S\T is a zero subsequence of S which is a contradiction. Hence S 1 is a minimal zero sequence of length n k +1inZ n. Now, by the statement (I), there exists a Z n such that a is repeated in S 1 at least n 2k + 2 times. Thus, the element a is repeated in S at least n 2k + 1 times. Assume that (II) is true. Consider a minimal zero sequence S = {a i } in Z n of length n k+1. Let S 1 be the sequence obtained from S by deleting the element a n k+1. Clearly, S 1 is a zero-free sequence in Z n of length n k. Therefore, by the statement (II), there exists a Z n such that a is repeated in S 1 at least n 2k +1 times. Now, let S 2 be a zero sequence in Z n of length n k, obtained from S by deleting the element a a n k+1 (if a = a n k+1, then nothing to prove). Again by the statement (II), there exists an element b Z n such that b is repeated n 2k + 1 times in S 2. If a b, then the length of S would be at least 2n 4k +2 n 2k +1. This forces that n 2k 1. That is, k (n +1)/2, which is a contradition to the assumption that k (n +2)/3. Therefore, a = b and hence S has an element a Z n which is repeated in S at least n 2k + 2 times. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 7 (II) (III) Assume that (II) is true. Let S be a sequence in Z n of length 2n k 1 satisfying the hypothesis. Suppose a is an element of S which appears maximum number of, say h, times in S. Consider the translated sequence S a. Let S 1 be the subsequence of S a such that it consists of all non-zero elements of S a. Claim. There exists zero-free subsequence T of S 1 of length n k. Assume the contrary. Suppose every subsequence T of S 1 of length n k has a zero subsequence. Let M be one such zero subsequence of S 1. Choose M such that M has the maximal length. Since every subsequence of length n k of S 1 has a zero subsequence, it is clear that 2n k 1 h M n k 1= M n h. If M n, then by adding appropriate number of zeros to M to get a zero subsequence of length n in S a which in turn produces a zero subsequence of length n in S (this is because we have h zeros out side S 1 ). This is a contradiction to the assumption. Therefore, M n.in this case, by N-C Extension 1, we can get a zero subsequence of length h, in M. Therefore, inductively, deleting the zero subsequences M 1,M 2,,M r from M so that we can make n h M n where M is the sequence obtained from M after deleting those sequences M i s. This is a contradiction as before. This contradiction implies that there is a zero-free subsequence T of S 1 of length n k. Therefore, by the statement (II), T has an element which is repeated at least n 2k + 1 times. Since 0 appears in S a maximum number of times, h n 2k +1. Therefore two distinct elements of Z n in S a which appears at least n 2k +1 times. Hence S has two distinct elements of Z n such that both appears in S at least n 2k + 1 times. Assume that (III) is true. Let S be a zero-free sequence of Z n of length n k. Let S 1 : S, 0, 0,, 0 } {{ } n 1 times be sequence in Z n of length 2n k 1. Clearly S 1 does not contain a zero subsequence of length n. Therefore by the statement (III), we know, S 1 consists of two distinct elements of Z n such that both appears n 2k + 1 times. Since 0 appears n 1 times, it is clear that S consists of one element of Z n which is repeated at least n 2k + 1 times. Remark 1. The statements (II) and (III) are equivalent for all n and k such that n 2k 1. Also, in the statement (III), there is a moreover part. That is, we can prove that, in the conclusion of the statement (III), S can consists of at most k + 1 distinct residue classes modulo n. This is because of the following. In [4], it is proved that if any sequence R in Z n of length 2n m + 1 consists of m distinct residue classes modulo n, then R contain a INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 8 zero subsequence of length n. Since the length of the given sequence S is 2n k 1 and S doesn t have any zero subsequence of length n, it follows that S can contain at most k + 1 distinct residue classes modulo n. Theorem 3.3 Let k be an integer with 1 k 1 (n [n/2]+1). Then, the statement 2 (III) in Theorem 3.2 implies EGZ Theorem. Proof. Let S be a sequence in Z n of length 2n 1. Let T be a subsequence of length 2n 1 k where 1 k 1 (n [n/2] + 1). Either T has a zero subsequence of length n 2 or it doesn t have. If T has such a zero subsequence, then nothing to prove. If T doesn t have any zero subsequence of length n, then by the statement (III) in Theorem 3.2, T consists of two distinct elements of Z n each appearing n 2k + 1 times. Since k lies in 1 k 1(n [n/2]+1), we get n 2k +1 [n/2]. Therefore, T has one element of Z 2 n repeating at least [n/2] times. Then by Lemma 3.1, we get the required zero subsequence of length n. N-C Extension 2. Let n and k be positive integers such that 1 k (n +2)/3. Let S be a minimal zero sequence of Z n of length n k +1. Then there exists a Z n such that a appears in S at least n 2k +2 times. Proof. This is nothing but (I) of Theorem 3.2. Since in Theorem 3.2, it has been proved that (I) (II) (III) and by Theorem 3.3, we get the result. For the simillar reasons, the following two statements are also true. N-C Extension 3. Let n and k be positive integers such that n 2k 1. Let S be a zero-free sequence in Z n of length n k. Then one element of S is repeated at least n 2k +1 times. N-C Extension 4. Let n and k be positive integers such that n 2k 1. Let S be a sequence of Z n of length 2n k 1. Suppose S does not have a zero subsequence of length n. Then there exist a b Z n such that a and b appear in S at least n 2k +1 times. The statement (III) in Theorem 3.2 is the generalization of the following results. Corollary Any sequence S in Z n of length 2n 2 having no zero subsequence of length n consists of two distinct elements in Z n each appearing exactly n 1 times. Proof. Put k = 1 in the statement (III) in Theorem 3.2, we get the result. Corollary was first proved by Yuster and Peterson (cf. [32]) and also proved by Bialostocki and Dierker (cf. [5]). INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 9 When k = 2 in Theorem 3.2, we get any sequence in Z n of length 2n 3 which does not have any zero subsequence of length n will have two distinct elements of Z n each appearing at least n 3 times. Indeed, a better result was proved by C. Flores and O. Ordaz (cf. [14]) as follows. A Result of Flores and Ordaz. (cf. [14]) Suppose S is any sequence in Z n of length 2n 3 such that S has no zero subsequence of length n. Then there exists a, b Z n such that Z n is generated by b a and a appearing n 1 times in S and one of the following conditions hold; (i) b appearing exactly n 2 times. (ii) b appearing exactly n 3 times in S and also, 2b a appearing exactly once in S. Remark 2 (i) By putting k = 1 in the statement (II) of Theorem 3.2, we get, if S is a zero-free sequence in Z n of length n 1, then S consists of only one element a Z n which is appearing n 1 times. Since S is zero-free, it is clear that the order of a has to be n. (ii) Now, we put k = 2 in the statement (II) in Theorem 3.2. If a zero-free sequence S in Z n of length n 2, we get less information about the structure of S. Using the following result of Hamidoune, we see in the following Proposition 3.4, in this case, the structure of S. A Result of Hamidoune. (See for instance, Lemma 2.3 in [14]) Let S be a zero-free sequence in Z n of length at least n 2. Also assume that S consists of at most 2 distinct residue modulo n. Then the length of S is at most n 1 and will have one of the following form; (i) S : a, a,,a where r n 1. } {{ } r times (ii) S : a, a,,a, 2a. } {{ } n 2 times Using this, we shall prove the following Proposition. Proposition 3.4 Let S be a zero-free sequence in Z n of length n 2. Then S consists of an element a Z n which is repeated either n 2 times or n 3 times and 2a appearing exactly once. Proof. By putting k = 2 in the statement Theorem 3.2, we get the sequence S has one element a Z n which is repeated at least n 3 times. If a is repeated n 2 times, INTEGERS: ELECTRONIC JOURNAL OF

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