NON-CANONICAL EXTENSIONS OF ERDŐS-GINZBURG-ZIV THEOREM^{1}

R. Thangadurai

Stat-Math Division, Indian Statistical Institute, 203, B. T. Road, Kolkata 700035, INDIA

                          thanga v@isical.ac.in

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_{i_{1}} + a_{i_{2}} + ⋯ + a_{i_{n}} 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}}_{i=1}^{ℓ} 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.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A07 2

EGZ Theorem. (cf. [12]) Given a sequence S in Z_{n} of length 2n − 1, one can extract a zero subsequence of length n in Z_{n}.

1Mathematics subject classification (1991): 05D05, 20D10

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 1947.

Corollary 1.1 Let A_{1}, A_{2}, ⋯, A_{h} be non-empty subsets of Z_{p}. Then,

|A_{1} + A_{2} + ⋯ + A_{h}| ≥ min{p, ∑_{i=1}^{h} |A_{i}| − h + 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_{i_{1}}, g_{i_{2}}, ⋯, g_{i_{n}} from the given sequence satisfies g_{i_{1}} + g_{i_{2}} + ⋯ + g_{i_{n}} = 0 in G.

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_{i_{1}} + ⋯ + a_{i_{n}} = 0 in G. 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_{i_{1}} + a_{i_{2}} + ⋯ + a_{i_{p}} ≡ 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.

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_{i_{1}} + a_{i_{2}} + ⋯ + a_{i_{p}} ≡ 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_{2} + ⋯ + A_{p}| ≥ min{p, ∑_{i} |A_{i}| − p + 1} = min{p, 2p − 1 − p + 1} = p

⟹ A_{1} + A_{2} + ⋯ + A_{p} = Z_{p}.

Thus the result follows. □

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 1997. 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 3.1.1 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}.

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 3.1.1.

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 > n which is a contradiction to the assumption. This proves the claim.

Since s + t > n and 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} = − ∑_{i=1}^{n−k} 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 + 1 in Z_{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.

(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, \underbrace{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 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 ≤ \frac{1}{2}(n − [n/2] + 1). Then, the statement (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 ≤ \frac{1}{2}(n − [n/2] + 1). Either T has a zero subsequence of length n 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 ≤ \frac{1}{2}(n − [n/2] + 1), we get n − 2k + 1 ≥ [n/2]. Therefore, T has one element of Z_{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 3.2.1 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 3.2.1 was first proved by Yuster and Peterson (cf. [32]) and also proved by Bialostocki and Dierker (cf. [5]).

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 : \underbrace{a, a, ⋯, a}_{r times} where r ≤ n − 1.

(ii) S : \underbrace{a, a, ⋯, a}_{n−2 times}, 2a.

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, then nothing to prove. If a is repeated n − 3 times exactly, then S can consist only of two distinct residue classes modulo n. Then by above mentioned result of Hamidoune, it follows that the second residue has to be 2a and the result follows. □

At this juncture, we should mention a beautiful result of Gao and Geroldinger (cf. [22]) as follows.

A result of Gao and Geroldinger. Let n ≥ 4 be a natural number. Let S be a zero-free sequence in Z_{n} of length at least (n + 3)/2. Then, there exists a ∈ Z_{n} of order n and a appears in S at least n/6 + 13/12 times.

Using this result and the above techniques, we can prove the following theorem (we skip the proof here).

Theorem 3.5 Let S be a sequence in Z_{n} of length at least n − 1 + (n + 3)/2. Suppose S does not have any zero subsequence of length n. Then there exist a ≠ b ∈ Z_{n} such that both appears in S at least n/6 + 13/12 times and either a or b is of order n.

Theorem 3.6 Let n, k be positive integers such that n − 2k ≥ 1. Let S be a sequence in Z_{n} of length 2n − k − 1 such that at most one element in S appears more than n − 2k times. Then there exists a zero subsequence of S of length n.

Proof. Suppose not, that is S does not have zero subsequence of length n. Then by the statement (III) in Theorem 3.2 we get a contradiction and hence the result. □

We give a short proof of the following known theorem.

Theorem 3.7 Let S be a sequence in Z_{n} of length 2n − 1. The sequence S has exactly one zero subsequence of length n. in Z_{n} if and only if there exists a and b in Z_{n} such that a appears n times and b appears n − 1 times in S.

Proof. The converse is easy. We shall prove the other implication. Let

S : a_{1}, a_{2}, ⋯, a_{2n−1}

be the given sequence. By EGZ theorem, there is a zero subsequence of length n. We let a_{1} + a_{2} + ⋯ + a_{n} ≡ 0 (mod n), if necessary by renaming the indices. Let S_{1} be a subsequence of S of length 2n − 2 such that a_{1} does not appear in S_{1}. Clearly by hypothesis, S_{1} does not have any zero subsequence of length n. Therefore by Corollary 3.2.1, S_{1} consists of two distinct elements a and b of Z_{n} each appearing n−1 times. Now, let us consider a subsequence S_{2} of S such that a_{n} = a does not appear in S_{2}. Clearly, S_{2} has length 2n − 2 and it cannot have a zero subsequence of length n, by the hypothesis. Therefore, it forces that a_{1} has to be a and hence the theorem. □

4. Concluding Remarks

In this section, we shall discuss about the analoguous situation of the non-canonical extensions of EGZ Theorem for the group Z_{p} ⊕ Z_{p}. For the detail history and status of the following conjectures, one may refer to [35].

There is no result so far known which is an analogue of N-C Extension 1 for the group Z_{p} ⊕ Z_{p}. The analogue of EGZ Theorem for the group Z_{p} ⊕ Z_{p} is the following conjecture of Kemnitz.

Conjecture 4.1 (Kemnitz, 1983, [25]) Let S be a sequence in Z_{p} ⊕ Z_{p} of length 4p − 3. Then there exists a zero subsequence T of S of length p.

Conjecture 4.1 is yet to be resolved. For the analogue of Corollary 3.2.1, we have the following conjecture of Gao.

Conjecture 4.2 (W. D. Gao, 2000, [21]) Let S be sequence in Z_{p} ⊕ Z_{p} of length 4p − 4. Suppose that S does not contain any zero subsequences of length p. Then S consists of four distinct elements a, b, c and d in Z_{p} ⊕ Z_{p} each of them appearing in S exactly p − 1 times.

Conjecture 4.2 is also open. The analogue of Remark 2 (i) is the following conjecture of van Emde Boas.

Conjecture 4.3 (van Emde Boas, 1969, [36]) Let S be a sequence in Z_{p} ⊕ Z_{p} of length 3p − 3. If S does not contain any zero subsequences of length at most p, then S consists of three distinct elements a, b and c of Z_{p} ⊕ Z_{p} each appearing exactly p − 1 times.

Of course, Conjecture 4.3 is also open till now. The analogue of the statement (I) in Theorem 3.2 when k = 1 is the following conjecture of Gao and Geroldinger.

Conjecture 4.4 (Gao and Geroldinger, 1998, [23]) If S is a minimal zero sequence in Z_{p} ⊕ Z_{p} of length 2p − 1, then there exists a ∈ Z_{p} ⊕ Z_{p} such that a is appearing in S at least p − 1 times.

As we have seen in the last section, it natural to ask for the inter relationships between these four conjectures.

Conjecture 4.2 implies Conjecture 4.1 was proved by Gao (cf. [21]). Conjecture 4.4 implies Conjecture 4.3 was proved by Gao and Geroldinger (cf. [23]). Conjecture 4.2 implies Conjecture 4.3 was proved by the author (cf. [35]). Also, in [35] it has been proved the following partial implication.

Theorem 4.4 (Thangadurai, 2001, [35]) Assume Conjecture 4.3. Let S be a sequence of the form

\underbrace{a, a, ⋯, a,}_{s times}a_{1}, a_{2}, ⋯, a_{4p−4−s}

where a, a_{i} ∈ Z_{p} ⊕ Z_{p} for all i = 1, 2, ⋯, 4p − 4 − s and s > [\frac{p−3}{2}] . Suppose S does not contain a zero subsequence of length p. Then S consists of four distinct elements a, b, c and d in Z_{p} ⊕ Z_{p} each of them appearing exactly p − 1 times. In other words, S satisfies Conjecture 4.2.

References

[1] N. Alon, Subset sums, J. Number Theory, 27 (1987), 196-205.

[2] N. Alon, A. Bialostocki and Y. Caro, The extremal cases in Erdős-Ginzburg-Ziv theorem, Unpublished.

[3] D. D. Anderson, Factorization in integral domains, Lecture Notes in Pure and Applied Mathematics, Marcel Dekker, 189, 1997.

[4] A. Bialostocki and M. Lotspeich, Some developments on the Erdős-Ginzburg-Ziv theorem I., Sets, graphs and numbers (Butapest, 1991), 97-117, Colloq. Math. Soc. János Bolyai, 60, North-Holland, Amsterdam, 1992.

[5] A. Bialostocki and P. Dierker, On Erdős-Ginzburg-Ziv theorem and the Ramsey numbers for stars and matchings, Discrete Mathematics, 110 (1992), 1-8.

[6] J. D. Bovey, P. Erdős and I. Niven, Conditions for zero-sum modulo n, Canad. Math. Bull., 18 (1975) 27-29.

[7] Y. Caro, Zero-sum problems – A survey, Discrete Mathematics 152(1996), 93-113.

[8] Y. Caro, On zero-sum Ramsey Numbers–Stars, Discrete Mathematics 104 (1992) 1-6.

[9] A. L. Cauchy, Recherches sur les nombers, J. Ecole Polytechniques, 9 (1813), 99-123.

[10] H. Davenport, On the addition of residue classes, J. London Math. Soc., 22 (1947), 100-101.

[11] G. T. Diderrich and H. B. Mann, Combinatorial problems in finite abelian groups, in: A survey of combinatorial theory, edited by J. N. Srivastava, North-Holland, [12] P. Erdös, A. Ginzburg and A. Ziv, Theorem in the additive number theory, Bull.

Res. Council Israel, 10 F(1961), 41-43.

[13] P. Erdös and R. L. Graham, Old and new problems and results in combinatorial number theory, Monographie N. 28 de L’Enseignement Mathématique, Univerité de Genéve, 1980.

[14] C. Flores and O. Ordaz, On sequences with zero sum in Abelian groups. Volume in homage to Dr. Rodolfo A. Ricabarra (Spanish), 99–106, Vol. Homenaje, 1, Univ. Nac. del Sur, Baha Blanca, 1995.

[15] W. D. Gao, Addition theorems for finite abelian group, J. Number Theory, 53(1995), 241-246.

[16] W. D. Gao, On zero-sum subsequences of restricted size, J. Number Theory, 61(1996), 97-102.

[17] W. D. Gao, A combinatorial problem on finite Abelian group, J. Number Theory 58 (1996), 100-103.

[18] W. D. Gao, An addition theorem for finite cyclic groups, Discrete Math., 163 (1997), no. 1-3, 257-265.

[19] W. D. Gao, On the number of zero sum subsequences, Discrete Math. 163 (1997), no. 1-3, 267–273.

[20] W. D. Gao, Addition Theorems and Group Rings, J. Comb. Theory, Series A, (1997), 98-109.

[21] W. D. Gao, Two zero-sum problems and multiple properties, J. Number Theory, 81(2000), 254-265.

[22] W. D. Gao and A. Geroldinger, On the structure of zero-free sequences, Combinatorica, 18 (1998), 519-527.

[23] W. D. Gao and A. Geroldinger, On long minimal zero sequences in finite abelian groups, Periodica Mathematica Hungarica, Vol. 38 (3)(1999), 179-211.

[24] W. D. Gao and Y. X. Yang, Note on a combinatorial constant, J. Math. Res. and Expo. 17 (1997), 139-140.

[25] A. Kemnitz, On a lattice point problem, Ars Combinatorica, 16b(1983), 151-160.

[26] H. B. Mann, Two addition theorems, J. Combinatorial Theory, 3 (1967), 233-235.

[27] H. B. Mann, Addition theorems: the addition theorems of group theory and number theory, Interscience Publishers, J. Wiley & Sons, 1965.

[28] H. B. Mann, Additive group theory- a progress report, Bull. Amer. Math. Soc., 79 (1973), 1069-1075.

[29] M. B. Nathanson, Additive Number Theory, GTM 165, Springer, 1996.

[30] J. E. Olson, On a combinatorial problem of finite Abelian groups I and II, J. Number Theory, 1(1969), 8-10, 195-199.

[31] J. E. Olson, On a combinatorial problem of Erdös, Ginzburg and Ziv, J. Number Theory, 8(1976), 52-57.

[32] B. Peterson and T. Yuster, A generalization of an addition theorem for solvable groups, Can. J. Math., vol XXXVI, No. 3 (1984) 529-536.

[33] Shan Zun, A conjecture in elementary number theory, Adv. Math. (Beijing), 12, No. 4, (1983), 299-301.

[34] B. Sury, The Chevalley-Warning theorem and a combinatorial question on finite groups, Proc. Amer. Math. Soc., 127 (1999), 4, 951-953.

[35] R. Thangaduari, Interplay between four conjectures on certain zero-sum problems, To appear in: Expositiones Math.

[36] P. van Emde Boas, A combinatorial problem on finite Abelian groups II, Z. W. (1969-007) Math. Centrum-Amsterdam.