ON ZERO-SUM SEQUENCES IN ℤ/nℤ ⊕ ℤ/nℤ

Weidong Gao ^{1}

Department of Computer Science and Technology, University of Petroleum, Beijing, Shuiku Road,

Changping, Beijing 102200, P.R. China

                        wdgao@public.fhnet.cn.net

Alfred Geroldinger ^{2}

Institut für Mathematik, Karl-FranzensUniversität, Heinrichstrasse 36, 8010 Graz, Austria

                     alfred.geroldinger@uni-graz.at

Received: 11/9/02, Accepted: 6/30/03, Published: 7/8/03

Abstract

It is well known that the maximal possible length of a minimal zero-sum sequence S in the group ℤ/nℤ⊕ℤ/nℤ equals 2n−1, and we investigate the structure of such sequences. We say that some integer n ≥ 2 has Property B, if every minimal zero-sum sequence S in ℤ/nℤ ⊕ ℤ/nℤ with length 2n − 1 contains some element with multiplicity n − 1. If some n ≥ 2 has Property B, then the structure of such sequences is completely determined. We conjecture that every n ≥ 2 has Property B, and we compare Property B with several other, already well-studied properties of zero-sum sequences in ℤ/nℤ ⊕ ℤ/nℤ. Among others, we show that if some integer n ≥ 6 has Property B, then 2n has Property B.

1. Introduction

In 1961, P. Erdös, A. Ginzburg and A. Ziv proved that every sequence S in ℤ/nℤ with length |S| ≥ 2n − 1 contains a zero-sum subsequence with length n [EGZ61]. Some years later, P. Erdös (for the special group ℤ/pℤ ⊕ ℤ/pℤ), H. Davenport (for general finite abelian groups) and P.C. Baayen formulated the following problem (see [MO67], [vEBK67]).

Problem 1: For a finite abelian group G, determine the smallest integer l ∈ ℕ such that every sequence S in G with length |S| ≥ l contains a zero-sum subsequence.

In subsequent literature, the integer l in Problem 1 has come to be known as the Davenport constant of G, and we will denote it by D(G). J.E. Olson and D. Kruyswijk determined independently its precise value for p-groups and for groups with rank at most two ([Ols69a], [Ols69b], [vEB69b]). In particular, we have D(ℤ/nℤ ⊕ ℤ/nℤ) = 2n − 1, which implies the Theorem of Erdös-Ginzburg-Ziv. However, for general finite abelian groups, even for groups with rank three or for groups of the form (ℤ/nℤ)^{r}, D(G) is still unknown (cf. [Gao00a], [GG03] [CFGS02] for recent developments).

1This work has been supported by NSFC with grant number 19971058 and by MOE of China with grant number 02047.

2Part of this manuscript was completed while the second author visited the University of Petroleum in Changping. He wishes to thank the Department of Computer Science and Technology for their support and hospitality.

The result of P. Erdös, A. Ginzburg and A. Ziv was also the starting point for much recent research devoted to the more general problem of studying subsequences of given sequences that have sum zero and satisfy some given additional property (see [Ham96], [Car96b], [HOO98], [GGH^{+}02], [Tha02a], [Tha02b], [Sch01] and the literature cited there). We give a precise formulation of some key questions of this type.

Problem 2: For a finite abelian group G, determine the smallest integer l ∈ ℕ such that every sequence S in G with length |S| ≥ l contains a zero-sum subsequence T such that

1. |T | ≤ exp(G),

2. |T | = exp(G),

3. |T | = |G|.

For general finite abelian groups only Problem 2.3 is solved (|G|+D(G)−1 is the required integer (see [Car96a] and [Gao96a] ). For finite cyclic groups 2.1 is obvious and 2.2 (resp. 2.3) is answered by the Erdös-Ginzburg-Ziv-Theorem. Now, suppose G = ℤ/nℤ ⊕ ℤ/nℤ with n ≥ 2. Then 3n−2 is the required integer in Problem 2.1 ([Ols69b], [GG99], Lemma 4.4). In 1983, A. Kemnitz conjectured that 4n − 3 is the required integer in Problem 2.2. Recent progress on this topic was made by L. Ronyai and W. Gao, but the conjecture is still open (see [Har73], [Kem83], [AD93], [Ron00], [Gao01a], [Els] and the literature cited there).

Let us consider the inverse questions associated with Problem 1 and Problem 2. Let G be a finite abelian group.

Problem 1*: Determine the structure of a sequence S with maximal length (i.e., |S| = D(G) − 1) which has no zero-sum subsequence.

Problem 2*: Determine the structure of a sequence S with maximal length which has no zero-sum subsequence T such that

1. |T | ≤ exp(G),

2. |T | = exp(G),

3. |T | = |G|.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A8 3

Let G = ℤ/nℤ with n ≥ 2. Then, obviously, a sequence S in G with maximal length which contains no zero-sum subsequence has the form S = (a + nℤ)^{n−1} for some a ∈ ℤ with gcd{a, n} = 1. This answers Problem 1* and Problem 2*.1. The structure of a sequence S in G with length |S| = 2n − k for ”small” k ≥ 2 which does not contain a zero-sum subsequence with length n was studied successfully by several authors (cf. [BD92], [Car92], [FO96], [Car96b], [Gao97]).

Let G = ℤ/nℤ ⊕ ℤ/nℤ with n ≥ 2. Problem 2*.1 was first tackled by P. van Emde Boas who asked for the structure of sequences S with length |S| = 3n − 3 which have no zero-sum subsequences with length at most n. This was motivated by investigations of Davenport’s constant for groups having rank three (see [vEB69b] and [Gao00a], Lemma 4.7). Problem 1* appears naturally in the theory of non-unique factorizations and it was first addressed in [GG99]. Problem 2*.2 was first considered by W. Gao in [Gao00b]. All three problems (1*, 2*.1 2*.2) are open; there are conjectures which would provide complete answers to these problems and some partial results supporting these conjectures (cf. the discussion after Definition 3.2).

This paper concentrates on Problem 1* (for sequences in ℤ/nℤ ⊕ ℤ/nℤ). We say that an integer n ≥ 2 has Property B, if every minimal zero-sum sequence S in ℤ/nℤ ⊕ ℤ/nℤ with length |S| = D(G) contains some element with multiplicity n − 1 (cf. Theorem 4.3 for various characterizations of this Property). We conjecture that every integer n ≥ 2 satisfies Property B. If this holds true, then, by Theorem 4.3, Problem 1* is completely answered. We show that Property B is closely related to (usually stronger than) several other already well-studied properties of sequences in ℤ/nℤ ⊕ ℤ/nℤ (cf. Theorems 5.3 and 6.2); after having introduced some additional terminology, we give a more detailed preview of our results after Definition 3.2. Among these results, we show that if some integer n ≥ 6 has Property B, then 2n has Property B ( Theorem 8.1).

2. Preliminaries

Let ℕ denote the set of positive integers, ℙ ⊂ ℕ the set of prime numbers and let ℕ_{0} = ℕ ∪ {0}. For a prime p ∈ ℙ let v_{p} : ℕ → ℕ_{0} denote the p-adic exponent whence n = ∏_{p∈ℙ} p^{v_{p}(n)} for every n ∈ ℕ. For a, b ∈ ℤ we set

[a, b] = {x ∈ ℤ | a ≤ x ≤ b}.

Throughout, all abelian groups will be written additively, and for n ∈ ℕ let C_{n} denote the cyclic group with n elements. Let G be a finite abelian group. There are n_{1}, …, n_{r} ∈ ℕ such that G ≅ C_{n_{1}} ⊕ … ⊕ C_{n_{r}} where either r = n_{1} = 1 or 1 < n_{1} | … | n_{r}. Then r = r(G) is the rank of the group and n_{r} = exp(G) its exponent.

Elements e_{1}, …, e_{r} ∈ G are called independent, if every equation of the form ∑_{i=1}^{r} m_{i}e_{i} = 0 with m_{1}, …, m_{r} ∈ ℤ, implies that m_{1}e_{1} = ⋯ = m_{r}e_{r} = 0. We say that (e_{1}, …, e_{r}) is a basis of G, if e_{1}, … e_{r} are independent and generate the group (equivalently, G = ⊕_{i=1}^{r}⟨e_{i}⟩).

Let G = C_{n} ⊕ C_{n} with n ≥ 2 and e_{1}, e_{2} ∈ G. Then (e_{1}, e_{2}) is a basis if and only if (e_{1}, e_{2} are independent with ord(e_{1}) = ord(e_{2}) = n) if and only if e_{1}, e_{2} generate G.

Let (e_{1}, e_{2}) be a basis of G. An endomorphism φ : G → G with

(φ(e_{1}), φ(e_{2})) = (e_{1}, e_{2}) · (\begin{matrix}a & b \\ c & d \end{matrix}) where a, b, c, d ∈ ℤ

is an automorphism if and only if (φ(e_{1}), φ(e_{2})) is a basis which is equivalent to gcd{ad− bc, n} = 1. Let f_{1} ∈ G with ord(f_{1}) = n. Then there are a, c ∈ ℤ with gcd{a, c, n} = 1 such that f_{1} = ae_{1} + ce_{2} and there are b, d ∈ ℤ with ad − bc ≡ 1 mod n whence (f_{1}, f_{2} = be_{1} + de_{2}) is a basis of G.

Let F(G) denote the free abelian monoid with basis G. An element S ∈ F(G) is called a sequence in G and will be written in the form

S = ∏_{i=1}^{l} g_{i} = ∏_{g∈G} g^{v_{g}(S)} ∈ F(G) where all v_{g}(S) ∈ ℕ_{0}.

For every g ∈ G we call v_{g}(S) the multiplicity of g in S, and a sequence T ∈ F(G) is a subsequence of S, if v_{g}(T) ≤ v_{g}(S) for every g ∈ G. The unit element 1 ∈ F(G) is called the empty sequence. We denote by

• |S| = l = ∑_{g∈G} v_{g}(S) ∈ ℕ_{0} the length of S,

• σ(S) = ∑_{i=1}^{l} g_{i} = ∑_{g∈G} v_{g}(S)g ∈ G the sum of S,

• supp(S) = {g_{i} | i ∈ [1, l]} = {g ∈ G | v_{g}(S) > 0} ⊂ G the support of S, and by

• Σ(S) = {∑_{i∈I} g_{i} | ∅ ≠ I ⊂ [1, l]} ⊂ G the set of sums of non-empty subsequences of S.

The sequence S is called

• zero-sumfree, if 0 ∉ Σ(S),

• a zero-sum sequence, if σ(S) = 0,

• a minimal zero-sum sequence, if it is a zero-sum sequence and every proper zero-sum subsequence is zero-sumfree,

• a short zero-sum sequence, if it is a zero-sum sequence with length |S| ∈ [1, exp(G)].

Every group homomorphism φ : G → H extends in a canonical way to a homomorphism φ : F(G) → F(H) where φ(S) = ∏_{i=1}^{l} φ(g_{i}) ∈ F(H). Obviously, φ(S) is a zero-sum sequence if and only if σ(S) ∈ ker(φ). If φ : G → G is an automorphism, then S is a (minimal) zero-sum sequence if and only if φ(S) is a (minimal) zero-sum sequence. Suppose G = C_{mn}^{r} with r, m, n ∈ ℕ_{≥2}. If φ : G → G denotes the multiplication by n, then clearly we have ker(φ) = {g ∈ G | ng = 0} ≅ C_{n}^{r} and φ(G) = nG ≅ C_{m}^{r}.

Davenport’s constant D(G) of G is defined as the maximal length of a minimal zero-sum sequence in G, equivalently this is the smallest integer l ∈ ℕ such that every sequence S ∈ F(G) with |S| ≥ l contains a zero-sum subsequence. It is easy to see that 1 + ∑_{i=1}^{r}(n_{i} − 1) ≤ D(G). J.E. Olson and D. Kruyiswijk proved independently that equality holds if r(G) ≤ 2 or G a p-group (see [Ols69a] and [vEB69b]). If S ∈ F(G) is zero-sumfree with length |S| = D(G) − 1, then Σ(S) = G \ {0} whence G = ⟨supp(S)⟩.

We shall frequently use the fact that in a cyclic group G with n ≥ 2 elements every minimal zero-sum sequence S ∈ F(G) with length |S| = n has the form S = g^{n} for some g ∈ G with ord(g) = n, and that a sequence S ∈ F(G) with length |S| = n − 1 is zero-sumfree if and only if S = g^{n−1} for some g ∈ G with ord(g) = n.

3. Sequences in C_{n} ⊕ C_{n}

In this section we give a key definition of various well-studied properties of sequences in ℤ/nℤ ⊕ ℤ/nℤ (Definition 3.2) and outline the program of the subsequent sections. Then we prepare the main tools which will be used throughout the whole paper (Lemma 3.3 to Lemma 3.14). Among them Theorem 3.7 may be of its own interest.

Lemma 3.1. Let n ≥ 2.

1. (Erdös-Ginzburg-Ziv-Theorem) Every sequence S ∈ F(C_{n}) with |S| ≥ 2n − 1 contains a zero-sum subsequence with length n.

2. Every sequence S ∈ F(C_{n} ⊕ C_{n}) with |S| ≥ 3n − 2 contains a short zero-sum subsequence.

Proof. 1. see [EGZ61] and [AD93] for a variety of proofs.

2. See [GG99], Lemma 4.4.

Definition 3.2. Let G = C_{n} ⊕ C_{n} with n ≥ 2. We say that n has

• Property B, if every minimal zero-sum sequence S ∈ F(G) with length |S| = 2n − 1 contains some element with multiplicity n − 1.

• Property C, if every sequence S ∈ F(G) with length |S| = 3n − 3 which contains no short zero-sum subsequence has the form S = a^{n−1}b^{n−1}c^{n−1} with some pairwise distinct elements a, b, c ∈ G of order n.

• Property D, if every sequence S ∈ F(G) with length |S| = 4n − 4 which contains no zero-sum subsequence of length n has the form S = a^{n−1}b^{n−1}c^{n−1}d^{n−1} with some pairwise distinct elements a, b, c, d ∈ G of order n.

• Property E, if every sequence S ∈ F(G) with length |S| = 4n−3 contains a zero-sum subsequence of length n.

We say that Property B (resp. C, D, E) is multiplicative if the following holds: if two integers m, n ∈ ℕ both satisfy Property B (resp. C, D, E), then so does their product mn.

Let G = C_{n} ⊕ C_{n} with n ≥ 2. It has been conjectured, that every integer n ≥ 2 satisfies each of the above Properties. If n has Property B, then, as we shall see in Theorem 4.3, this answers Problem 1* of the Introduction. If n has Property C, then by Lemma 3.1.2 this answers Problem 2*.1. A. Kemnitz conjectured that n has Property E (which answers Problem 2.2 of the Introduction) and if this holds true, then Property D answers the associated inverse problem.

It is immediately clear that 2 satisfies each of these Properties whence whenever it is convenient we restrict to integers n ≥ 3. Lemma 3.3 states that Properties C, D and E are multiplicative and that D implies C and E. The Properties C, D and E have been verified for 2, 3, 5 and 7 ([vEB69b], [vEB69a], [Kem83], [ST02]). Furthermore, E holds true for various classes of composite numbers (cf. [Gao96b], [Gao03], [Gao01b], [Tha01]). We are going to prove that 2, 3, 5 and 6 have Property B (Proposition 4.2), that (under some weak additional assumption) Property B implies Property C and that if some n ≥ 6 has Property B, then 2n has Property B (Theorem 8.1).

Lemma 3.3. Let n ≥ 2.

1. The Properties C, D and E are multiplicative.

2. Property D implies Properties C and E.

Proof. We set G = C_{n} ⊕ C_{n}.

1. In [Gao00b] it is proved that Properties C and D are multiplicative. The fact that Property E is multiplicative follows from a more general result of H. Harborth (cf. [Har73], Hilfssatz 2). For convenience we provide a simple proof.

Let m, n ∈ ℕ be two integers satisfying Property E. We have to verify that every sequence S ∈ G ≅ C_{mn} ⊕ C_{mn} with |S| ≥ 4mn − 3 has a zero-sum subsequence with length mn Let φ : G → G denote the multiplication by n and let S be a sequence in G with length |S| = 4mn − 3. Since every sequence in φ(G) ≅ C_{m} ⊕ C_{m} with length 4m − 3 contains a zero-sum subsequence of length m and since

4mn − 3 = (4n − 4)m + (4m − 3)

there exist t = 4n − 3 disjoint subsequences S_{1}, …, S_{t} of S with length |S_{i}| = m such that φ(S_{i}) has sum zero in φ(G) for every i ∈ [1, t]. Thus

T = ∏_{i=1}^{t} σ(S_{i})

is a sequence in ker(φ) ≅ C_{n} ⊕ C_{n}. Since n has Property E there exists some I ⊂ [1, t] with |I| = n such that ∏_{i∈I} σ(S_{i}) is a zero-sum subsequence of T. This implies that

S′ = ∏_{i∈I} S_{i} ∈ F(G)

is a zero-sum subsequence of S with length |S| = ∑_{i∈I} |S_{i}| = mn.

2. Suppose that n satisfies Property D and that n ≥ 3.

We first verify that n has Property C. Let S ∈ F(G) be a sequence with length |S| = 3n − 3 and suppose that S contains no short zero-sum subsequence. We consider the sequence T = 0^{n−1} · S. If T has a zero-sum subsequence T′ with |T′| = n, then T′ = 0^{k} · S′ with k ∈ [0, n − 1] and S′ | S whence S′ is a short zero-sum subsequence of S. Thus T has no zero-sum subsequence of length n, and the assertion follows.

Next we show that n satisfies Property E. Let S ∈ F(G) with length |S| = 4n − 3 and assume to the contrary that S contains no zero-sum subsequence of length n. Let g ∈ supp(S). Then |g^{−1} · S| = 4n − 4, and g^{−1} · S contains no zero-sum subsequence of length n whence g^{−1} · S = a^{n−1} · b^{n−1} · c^{n−1} · d^{n−1} for some a, b, c, d ∈ G. Thus there is some h ∈ supp(S) with v_{h}(S) ≥ 2. After changing notation if necessary, we suppose that v_{g}(S) ≥ 2 and that g = a. Thus a^{n} is a zero-sum subsequence of S, a contradiction.

Definition 3.4. Let G be a finite abelian group with exp(G) = n ≥ 2. Let s(G) ( resp. s_{0}(G)) denote the smallest integer l ∈ ℕ such that every sequence S ∈ F(G) with |S| ≥ l contains a zero-sum subsequence T with length |T | = n ( resp. with length |T | ≡ 0 mod n).

Hence, by definition, an integer n ≥ 2 has Property E if and only if s(C_{n}⊕C_{n}) = 4n−3.

Lemma 3.5. Let G be a finite abelian group with exp(G) = n ≥ 2. Then

D(G) + n − 1 ≤ s_{0}(G) ≤ min{s(G), D(G ⊕ C_{n})}.

Proof. If S ∈ F(G) is a zero-sumfree sequence with length |S| = D(G) − 1, then the sequence 0^{n−1} · S has no zero-sum subsequence with length divisible by n. Thus D(G) + n − 2 = |0^{n−1} · S| < s_{0}(G). By definition we have s_{0}(G) ≤ s(G).

Suppose that G ⊕ C_{n} = G ⊕ ⟨e⟩ . In order to verify that s_{0}(G) ≤ D(G ⊕ C_{n}), let S = ∏_{i=1}^{l} g_{i} ∈ F(G) with l = D(G ⊕ C_{n}). Then the sequence ∏_{i=1}^{l}(g_{i} + e) ∈ F(G ⊕ C_{n}) contains a zero-sum subsequence T with length |T | ≡ 0 mod n, and whence the same is true for S.

Lemma 3.6. Let m, n ∈ ℕ_{≥2} and suppose that s_{0}(C_{m} ⊕ C_{m}) = 3m − 2, s(C_{m} ⊕ C_{m}) ≤ 4m − 2 and that D(C_{n}^{3}) = 3n − 2. Then s_{0}(C_{mn} ⊕ C_{mn}) = 3mn − 2.

Proof. By Lemma 3.5 it remains to show that s_{0}(C_{mn} ⊕ C_{mn}) ≤ 3mn − 2. Let S = ∏_{i=1}^{l} g_{i} be a sequence in G = C_{mn} ⊕ C_{mn} with length l = 3mn − 2. We set H = G ⊕ C_{mn} = G ⊕ ⟨e⟩ and S^{H} = ∏_{i=1}^{l}(g_{i} + e). It is sufficient to prove that S^{H} has a non-empty zero-sum subsequence. Let φ : H → H denote the multiplication by n. Since 3mn − 2 = (3n − 4)m + (4m − 2) and s(C_{m} ⊕ C_{m}) ≤ 4m − 2, there exist 3n − 3 disjoint subsequences S_{1}, …, S_{3n−3} of S with length m such that all φ(S_{i}) have sum zero. Since | ∏_{i=1}^{3n−3} S_{i}^{−1}·S| = 3m−2 = s_{0}(C_{m}⊕C_{m}), there exists a subsequence S_{3n−2} of ∏_{i=1}^{3n−3} S_{i}^{−1}·S such that φ(S_{3n−2}) has sum zero and with |S_{3n−2}| ∈ {m, 2m}. For i ∈ [1, 3n−2] we denote by S_{i}^{H} the subsequence of S^{H} corresponding to S_{i} and obtain that σ(S_{i}^{H}) ∈ ker(φ). Thus ∏_{i=1}^{3n−2} σ(S_{i}^{H}) is a sequence in ker(φ) with length 3n − 2 = D(C_{n}^{3}). Therefore there exists some ∅ ≠ I ⊂ [1, 3n − 2] such that ∑_{i∈I} σ(S_{i}^{H}) = 0 whence ∏_{i∈I} S_{i}^{H} is a non-empty zero-sum subsequence of S^{H}.

Theorem 3.7. Let n ∈ ℕ with n ≥ 2.

1. If n is divisible by at most two distinct primes, then s_{0}(C_{n} ⊕ C_{n}) = 3n − 2.

2. If Property E holds for all prime divisors of n, then s_{0}(C_{n} ⊕ C_{n}) = 3n − 2.

Proof. 1. If n is a prime power, then the assertion follows from Lemma 3.5. If n is a product of two prime powers, then by the previous case and by [Gao01a] the assumptions of Lemma 3.6 are satisfied whence the assertion follows.

2. Suppose that n = ∏_{i=1}^{r} p_{i}^{k_{i}} with r, k_{1}, …, k_{r} ∈ ℕ and primes p_{1}, …, p_{r}. If Property E holds for p_{1}, …, p_{r}, then for every divisor 1 < d of n we have s(C_{d} ⊕ C_{d}) = 4d − 3 by Lemma 3.3.1. Using Lemma 3.6 we obtain the assertion by induction on r.

Lemma 3.8. Let G = C_{n} ⊕ C_{n} with n ≥ 2 and S ∈ F(G) a minimal zero-sum sequence with |S| = 2n − 1.

1. Then ord(g) = n for every g ∈ supp(S).

2. Let n be prime. Then each two distinct elements in supp(S) are independent and 3 ≤ |supp(S)| ≤ n + 1.

Proof. See [GG99], Proposition 6.3, Theorem 10.3 and Corollary 10.5.

Lemma 3.9. Let G = C_{n} ⊕ C_{n} with n ≥ 3 and e_{1}, e_{2} ∈ G distinct such that the sequence e_{1}^{n−2}e_{2}^{n−2} does not contain a short zero-sum subsequence. Then (e_{1}, e_{2}) is a basis of G.

Proof. Obviously, the assertion is true for n = 3. Suppose that n ≥ 4. Since ord(e_{i}) | n and ord(e_{i}) > n − 2 for i ∈ [1, 2], it follows that ord(e_{1}) = ord(e_{2}) = n. Thus it remains to show that e_{1}, e_{2} are independent. Let λ_{1}, λ_{2} ∈ [0, n − 1] such that λ_{1}e_{1} + λ_{2}e_{2} = 0. We have to verify that λ_{1} = λ_{2} = 0. Assume to the contrary, that λ_{1} + λ_{2} > n. Then λ_{1}, λ_{2} ∈ [2, n − 1] and T = e_{1}^{n−λ_{1}} · e_{2}^{n−λ_{2}} is a zero-sum subsequence of S with length |T | = 2n − (λ_{1} + λ_{2}) ∈ [1, n − 1], a contradiction. Thus λ_{1} + λ_{2} ≤ n. If λ_{1} = n − 1, then λ_{2} = 1, e_{1} = e_{2} and e_{1}^{n} is a short zero-sum subsequence of S, a contradiction. Thus λ_{1} ≤ n − 2, and similarly we obtain that λ_{2} ≤ n − 2. Therefore T = e_{1}^{λ_{1}} · e_{2}^{λ_{2}} is a zero-sum subsequence of S, which implies that λ_{1} + λ_{2} = |T | = 0.

Lemma 3.10 (Moser-Scherk). Let G be a finite abelian group and S ∈ F(G) a zerosumfree sequence. If S = ∏_{i=1}^{l} S_{i}, then |Σ(S)| ≥ ∑_{i=1}^{l} |Σ(S_{i})|.

Proof. See [MS55].

Lemma 3.11. Let G = C_{m} ⊕ C_{m} with m ≥ 2 and S ∈ F(G) a zero-sum sequence with |S| ≥ tm for some t ≥ 2. Then S may be written as a product of t non-empty zero-sum subsequences and at least t − 2 of these sequences are short.

Proof. We proceed by induction on t. If t = 2, then |S| ≥ 2m > D(G) whence S contains a zero-sum subsequence S_{1} with |S_{1}| ≤ 2m − 1 and the assertion follows. If t ≥ 3, then Lemma 3.1.2 implies that S contains a short zero-sum subsequence S_{1}. Since S_{1}^{−1}S is a zero-sum sequence with |S_{1}^{−1}S| ≥ (t − 1)m, the assertion follows by induction hypothesis.

Lemma 3.12. Let G = C_{m} ⊕ C_{m} with m ≥ 2 and S ∈ F(G) a zero-sum sequence with |S| = tm−1 for some t ≥ 3 which cannot be written as a product of t non-empty zero-sum subsequences.

1. Every short zero-sum subsequence of S has length m. In particular, we have 0 ∉ supp(S).

2. S has a product decomposition of the form S = ∏_{ν=0}^{t−2} S_{ν} where S_{0} is a minimal zerosum sequence with length 2m − 1 and S_{1}, …, S_{t−2} are short zero sum sequences.

3. If S = ∏_{ν=1}^{t−1} S_{ν} with zero-sum subsequences S_{1}, …, S_{t−1}, then at most m − 1 of these sequences are not short.

Proof. 1. Assume to the contrary that S contains a short zero-sum subsequence T with |T | ∈ [1, m − 1]. Then |T^{−1}S| ≥ (t − 1)m whence Lemma 3.11 implies that T^{−1}S may be written as a product of t − 1 non-empty zero-sum subsequences. Thus S may be written as a product of t non-empty zero-sum subsequences, a contradiction.

2. Applying Lemma 3.1.2 (t − 2)-times we see that S may be written in the form

S = S_{0} · ∏_{ν=1}^{t−2} S_{ν}

where S_{1}, …, S_{t−2} are zero-sum subsequences with length m. Thus S_{0} is a zero-sum subsequence with |S_{0}| = 2m − 1. Since S is not a product of t zero-sum subsequences, it follows that S_{0} is minimal.

3. Assume to the contrary, that S = ∏_{ν=1}^{t−1} S_{ν} where all S_{ν} are zero-sum subsequences and S_{1}, …, S_{m} are not short. Then T = ∏_{i=1}^{m} S_{i} is a zero-sum subsequence with length |T | ≥ m(m + 1). Thus by Lemma 3.11 T may be written as a product of m + 1 zero-sum subsequences whence S is a product of t zero-sum subsequences, a contradiction.

Lemma 3.13. Let G = C_{mn} ⊕ C_{mn} with m, n ∈ ℕ_{≥2} and φ : G → G the multiplication by n. If (e′_{1}, e′_{2}) is a basis of ker(φ), then there is a basis (e_{1}, e_{2}) of G such that me_{i} = e′_{i} for i ∈ [1, 2].

Proof. Suppose that G = ℤ/mnℤ × ℤ/mnℤ and e′_{i} = (a′_{i} + mnℤ, b′_{i} + mnℤ) with a′_{i}, b′_{i} ∈ [0, mn−1]∩mℤ for i ∈ [1, 2] such that (e′_{1}, e′_{2}) is a basis of ker(φ) = mℤ/mnℤ×mℤ/mnℤ. For i ∈ [1, 2] we set a_{i} = m^{−1}a′_{i}, b_{i} = m^{−1}b′_{i} and e_{i} = (a_{i} + mnℤ, b_{i} + mnℤ). Then ord(e_{1}) = ord(e_{2}) = mn and e_{1}, e_{2} are independent whence (e_{1}, e_{2}) is a basis of G.

Lemma 3.14. Let G = C_{mn} ⊕ C_{mn} with m, n ∈ ℕ_{≥2}, φ : G → G the multiplication by n and S ∈ F(G) a minimal zero-sum sequence with length |S| = 2mn − 1.

1. φ(S) is not a product of 2n zero-sum subsequences. Every short zero-sum subsequence of φ(S) has length m and 0 ∉ supp(φ(S)).

2. S has a product decomposition S = ∏_{ν=0}^{2n−2} S_{ν} where |S_{0}| = 2m − 1, |S_{1}| = … = |S_{2n−2}| = m and σ(S_{0}), …, σ(S_{2n−2}) ∈ ker(φ).

Proof. 1. Obviously, φ(S) is a zero-sum sequence in nG with length tm−1 where t = 2n. Assume to the contrary, that φ(S) can be written as a product of t non-empty zero-sum subsequences, say φ(S) = ∏_{ν=1}^{t} φ(S_{ν}). Then T = ∏_{ν=1}^{t} σ(S_{ν}) is a sequence in ker(φ). Since t = 2n > D(ker(φ)), T contains a proper zero-sum subsequence whence S contains a proper zero-sum subsequence, a contradiction. The remaining assertions follow from Lemma 3.12.1.

2. By 1. we may apply Lemma 3.12.2 to the sequence φ(S) (with t = 2n) whence the assertion follows.

4. Some characterizations of Property B

After some technical preparation we show that 2, 3, 4, 5 and 6 satisfy Property B and then we give some characterizations of Property B in Theorem 4.3.

Proposition 4.1. Let G = C_{n} ⊕ C_{n} with n ≥ 2.

1. If (e_{1}, e_{2}) is a basis of G and a_{1}, …, a_{n} ∈ ℤ with ∑_{ν=1}^{n} a_{ν} ≡ 1 mod n, then

S = e_{1}^{n−1} · ∏_{ν=1}^{n}(a_{ν}e_{1} + e_{2}) (∗)

is a minimal zero-sum sequence with |S| = D(G).

2. Let S ∈ F(G) be a minimal zero-sum sequence with |S| = D(G) and e_{1} ∈ G with v_{e_{1}}(S) = n − 1.

(a) If (e_{1}, e′_{2}) is a basis of G, then there exist some b ∈ [0, n − 1] with gcd{b, n} = 1

and a′_{1}, …, a′_{n} ∈ [0, n − 1] with ∑_{ν=1}^{n} a′_{ν} ≡ 1 mod n such that

S = e_{1}^{n−1} · ∏_{ν=1}^{n}(a′_{ν}e_{1} + be′_{2}).

(b) There exists a basis (e_{1}, e_{2}) of G such that S has the form (∗).

(c) If g, g′ ∈ supp(S) \ {e_{1}}, then g − g′ ∈ ⟨e_{1}⟩.

Proof. 1. Let S be a sequence of the form (∗). Then S has sum zero and length 2n − 1 = D(G). Let T be a non-empty zero-sum subsequence of S with e_{1} | T. Since e_{1}^{n−1} is zero-sumfree, there exists some i ∈ [1, n] such that (a_{i}e_{1} + e_{2}) | T. This implies that ∏_{i=1}^{n}(a_{i}e_{1} + e_{2}) divides T. Thus S = T whence S is a minimal zero-sum sequence.

2. Suppose S = e_{1}^{n−1} ∏_{i=1}^{n} g_{i}.

a) Let (e_{1}, e′_{2}) be a basis of G. Then for every i ∈ [1, n] we have

g_{i} = a′_{i}e_{1} + b_{i}e′_{2}

with a′_{i}, b_{i} ∈ [0, n − 1]. Since the sequence e_{1}^{n−1} · (a′_{i}e_{1}) ∈ F(⟨e_{1}⟩) is not zero-sumfree, it follows that b_{i} ≠ 0 for every i ∈ [1, n]. Assume to the contrary, that ∏_{i=1}^{n} b_{i}e′_{2} ∈ F(⟨e′_{2}⟩) is not a minimal zero-sum sequence. Then there exists some ∅ ≠ I ⊊ [1, n] such that ∏_{i∈I} b_{i}e′_{2} is a zero-sum sequence and hence

e_{1}^{n−1} ∏_{i∈I}(a′_{i}e_{1} + b_{i}e′_{2})

contains a zero-sum subsequence, a contradiction. Therefore, ∏_{i=1}^{n} b_{i}e′_{2} is a minimal zerosum sequence and thus it follows that b_{1}e′_{2} = ⋯ = b_{n}e′_{2} whence b_{1} = ⋯ = b_{n} = b ∈ [1, n − 1]. Since G = ⟨supp(S)⟩ , it follows that gcd{b, n} = 1.

b) Clearly, there exists some e′_{2} ∈ G such that (e_{1}, e′_{2}) is a basis of G whence S has the form described in 2. a). Then

(e_{1}, e_{2}) = (e_{1}, e′_{2}) · (\begin{matrix}1 & a′_{1} \\ 0 & b \end{matrix})

is a basis of G and for every ν ∈ [1, n] we obtain that

a′_{ν}e_{1} + be′_{2} = (a′_{ν} − a′_{1})e_{1} + e_{2}.

Since S is a zero-sum sequence, the required congruence is satisfied.

c) This follows immediately from b).

Proposition 4.2. The integers 2, 3, 4, 5 and 6 have Property B.

Remark: We have also verified that 7 has Property B, but we do not give this proof here.

Proof. Let G = C_{n} ⊕ C_{n} with n ≥ 2 and S = ∏_{i=1}^{l} g_{i}^{k_{i}} a minimal zero-sum sequence with length |S| = D(G) = 2n − 1, k_{1} ≥ ⋯ ≥ k_{l} ≥ 1, g_{1}, …, g_{l} pairwise distinct and |supp(S)| = l. We have to show that k_{1} = n − 1.

This is obvious for n = 2. If n = 3, then Lemma 3.8 implies that l ∈ [3, 4] whence k_{1} = 2.

Let n = 4. By Lemma 3.8 all elements in supp(S) have order 4, and clearly G has exactly 12 elements of order 4. Assume to the contrary that k_{1} ≤ 2. If k_{1} = 1, then l = |S| = 7, ∏_{i=1}^{6} g_{i} is zero-sumfree and {−g_{i}, g_{i} | i ∈ [1, 6]} are the twelve elements of order 4 whence g_{7} ∈ {−g_{i} | i ∈ [1, 6]} a contradiction. Thus k_{1} = 2 and S = h_{1}^{2} ∏_{i=2}^{6} h_{i} with h_{2}, …, h_{6} ∈ G not necessarily pairwise distinct. Since by Lemma 3.1.2 every sequence in C_{2} ⊕ C_{2} \ {0} with length 4 contains a short zero-sum subsequence with length two, every sequence S ∈ F(G \ {0}) with |S| ≥ 4 contains a subsequence S′ with |S′| = 2 and ord(σ(S′)) = 2. Thus after renumeration we may suppose that ord(h_{2} + h_{3}) = 2. Then h_{4} + h_{5} + h_{6} has order two and no proper subsum has order two. Considering the sequence h_{1} ·h_{4} ·h_{5} ·h_{6} we may suppose (after renumerating again) that ord(h_{1} +h_{4}) = 2. Therefore we obtain that h_{1} +h_{4} ∈ {h_{1} +h_{1}, h_{2} +h_{3}, h_{4} +h_{5} +h_{6}} whence either h_{1}h_{4}h_{2}h_{3} or h_{1}h_{5}h_{6} is a zero-sum subsequence of S, a contradiction.

Let n = 5. Lemma 3.8 implies that l ∈ [3, 6] whence k_{1} ≥ 2. Assume to the contrary that k_{1} ∈ [2, 3]. By Lemma 3.8 g_{1} and g_{2} are independent whence (g_{1}, g_{2}) is a basis of G.

Case 1: k_{1} = 3. Since |S| = 9 and l ≤ 6, it follows that k_{2} ≥ 2. If l = 3, then k_{2} = k_{3} = 3 and 0 = σ(S) = 3(g_{1} + g_{2} + g_{3}) implies that 0 = g_{1} + g_{2} + g_{3}, a contradiction to the fact that S is a minimal zero-sum sequence. Thus we have l ∈ [4, 6]. Since for every i ∈ [3, l] the sequence g_{1}^{3}g_{2}^{2}g_{i} is zero-sumfree, it follows that

g_{i} ∈ G \ ({0, g_{1}, g_{2}} ∪ Σ(−(g_{1}^{3}g_{2}^{2})) )
= {2g_{2}, g_{1} + g_{2}, g_{1} + 2g_{2}, g_{1} + 3g_{2}, g_{1} + 4g_{2},
2g_{1} + g_{2}, 3g_{1} + g_{2}, 4g_{1} + g_{2}, 2g_{1} + 2g_{2}, 3g_{1} + 2g_{2}, 4g_{1} + 2g_{2}}.

We argue step by step that none of the following elements lies in supp(S): g_{1} + 2g_{2}, g_{1} + 3g_{2}, g_{1} + 4g_{2}, 2g_{1} + 2g_{2}, 3g_{1} + 2g_{2} and 4g_{1} + 2g_{2}. To exclude the remaining cases we decide between k_{2} = 2 and k_{2} = 3 and obtain a contradiction to k_{1} = 3.

Case 2: k_{1} = 2. Since |S| = 9 and l ≤ 6, it follows that either (k_{1}, …, k_{l}) = (2, 2, 2, 1, 1, 1) or (k_{1}, …, k_{l}) = (2, 2, 2, 2, 1) whence

∏_{i=1}^{3} g_{i}^{2} · g_{4} · g_{5}

zero-sumfree. Since (g_{1}, g_{2}) is a basis of G, there are a, b, c, d, e, f ∈ [0, 4] such that g_{3} = ag_{1} +bg_{2}, g_{4} = cg_{1} +dg_{2} and g_{5} = eg_{1} +fg_{2}. Then Lemma 3.8 implies that a, b, c, d, e, f ∈ [1, 4]. Obviously, none of the pairs (a, b), (c, d), (e, f) lies in {(4, 4), (3, 3), (3, 4), (4, 3)}.

Since ∏_{i=1}^{3} g_{i}^{2} is zero-sumfree, it follows that (a, b) ∉ {(2, 4), (4, 2), (2, 2)}. Thus by symmetry it remains to consider the cases (a, b) ∈ {(1, 1), (1, 2), (1, 3), (1, 4)(2, 3)}. Discussing these five possibilities we obtain a contradiction.

Let n = 6. We proceed in two steps. First we show that k_{1} ≥ 4 and then we verify that k_{1} ≠ 4 whence k_{1} = 5 follows.

Assertion 1: k_{1} ≥ 4. Let φ : G → G denote the multiplication by 2 whence φ(G) = 2G ≅ C_{3} ⊕ C_{3} and ker(φ) ≅ C_{2} ⊕ C_{2}. We set S = ∏_{h∈φ(G)} S_{h} where φ(S_{h}) = h^{|S_{h}|}. Since by Lemma 3.14.1 every short zero-sum subsequence of φ(S) ∈ F(φ(G)) has length 3, we have |S_{0}| = 0 and if |S_{h}| > 0 for some h ∈ φ(G), then |S_{−h}| = 0. Thus S = ∏_{ν=1}^{t} S_{h_{ν}} with t ≤ \frac{1}{2}|φ(G) \ \{0\}| = 4 and |S_{h_{1}}| ≥ … ≥ |S_{h_{t}}| ≥ 1.

Frequently we shall use the fact that S does not have a proper subsequence of the form T = T_{1}T_{2}T_{3} with |T_{i}| ≥ 1 and σ(T_{i}) ∈ ker(φ) for i ∈ [1, 3]: because D(ker(φ)) = 3 the sequence ∏_{i=1}^{3} σ(T_{i}) ∈ F(ker(φ)) is not zero-sumfree whence T_{1}T_{2}T_{3} would not be zero-sumfree.

In particular, S does not have disjoint subsequences T_{1}, T_{2}, T_{3} where each T_{i} has length 3 and divides some S_{h_{j}} for some j ∈ [1, t]. This implies that |S_{h_{3}}| ≤ 2. From this we obtain, because t ≤ 4 and |S| = 11, that |S_{h_{1}}| ≥ 4.

Next we assert that

|supp(S_{h_{1}})| ≤ 2.

Assume to the contrary, that there are pairwise distinct elements x, y, z such that xyz | S_{h_{1}}. Since |S_{h_{1}}| ≥ 4, there is some w ∈ G such that wxyz | S_{h_{1}}. Since D(φ(G)) = 5, there exists a subsequence 1 ≠ T of (wxyz)^{−1}S such that φ(T) has sum zero whence σ(T) ∈ ker(φ) = {0, w + x + y, w + x + z, w + y + z}. Thus we obtain a proper zero-sum subsequence of S, a contradiction.

If |S_{h_{1}}| ≥ 7, then |supp(S_{h_{1}})| ≤ 2 implies that k_{1} ≥ 4. Hence it remains to consider the cases where |S_{h_{1}}| ∈ [4, 6]. We assume to the contrary that k_{1} < 4. Then |supp(S_{h_{1}})| = 2, say supp(S_{h_{1}}) = {α, β}. We distinguish two cases.

Case 1: |S_{h_{1}}| ∈ {5, 6}, say S_{h_{1}} = α^{3}β^{3} or S_{h_{1}} = α^{3}β^{2}. By Lemma 3.1.2 (applied to the group φ(G)) the sequence α^{2} ∏_{i=2}^{t} S_{h_{i}} contains a subsequence T_{3} with |T_{3}| ≤ 3 and σ(T_{3}) ∈ ker(φ), and Lemma 3.14.1 implies that |T_{3}| = 3.

We assert that there exists such a sequence T_{3} with v_{α}(T_{3}) > 0. Assume to the contrary that for all such sequences T_{3} we have v_{α}(T_{3}) = 0. Then there is no T′_{3} with |T′_{3}| = 3, σ(T_{3}) ∈ ker(φ), T′_{3} | β^{2} ∏_{i=2}^{t} S_{h_{i}} and v_{β}(T_{3}) > 0. We show that there exist sequences T_{1}, T_{2} such that T_{1}T_{2}T_{3} is a proper subsequence of S and σ(T_{1}), σ(T_{2}) ∈ ker(φ), which leads to a contradiction. If S_{h_{1}} = α^{3}β^{3}, then we set T_{1} = α^{3} and T_{2} = β^{3}. Suppose S_{h_{1}} = α^{3}β^{2}. Then T_{3}^{−1}S = α^{3}β^{2}γ_{1}γ_{2}γ_{3} contains a subsequence T_{2} with |T_{2}| ≤ 3 and σ(T_{2}) ∈ ker(φ). Since v_{α}(T_{2}) = v_{β}(T_{2}) = 0 we obtain T_{2} = γ_{1}γ_{2}γ_{3} and we set T_{1} = α^{3}.

Thus we have some sequence T_{3} = αγδ ∈ F(ker(φ)). Clearly, 2α + β and 2β + α are distinct non-zero elements of ker(φ). If α + γ + δ = 2β + α, then α^{2}β^{2}γδ is a proper zerosum subsequence of S. Hence α+γ +δ ≠ 2β +α and similarly α+γ +δ ≠ 2α+β whence ker(φ) \ {0} = {2α + β, 2β + α, α + γ + δ}. Since α ≠ β but φ(α + γ + δ) = φ(β + γ + δ), we have β + γ + δ ∈ {2β + α, 2α + β}. If β + γ + δ = 2α + β, then α^{2}β^{2}γδ is a proper zero-sum subsequence of S whence we infer that β + γ + δ = 2β + α = 3α ∈ ker(φ) (note that 2α = φ(α) = φ(β) = 2β = h_{1}) whence α^{3}βγδ is a proper zero-sum subsequence of S, a contradiction.

Case 2: |S_{h_{1}}| = 4. First we suppose that |S_{h_{2}}| = 4. If |supp(S_{h_{2}})| = 1, then we obtain k_{1} ≥ 4. Assume that |supp(S_{h_{2}})| > 1. Arguing as for supp(S_{h_{1}}) we obtain that |supp(S_{h_{2}})| = 2, say supp(S_{h_{2}}) = {γ, δ}. Then we may suppose without restriction that S_{h_{1}} ∈ {α^{3}β, α^{2}β^{2}} and S_{h_{2}} ∈ {γ^{3}δ, γ^{2}δ^{2}}. Thus we have either {3α, 2α+β} ⊂ ker(φ)\{0} or {2α + β, α + 2β} ⊂ ker(φ) \ {0}; and similarly, either {3γ, 2γ + δ} ⊂ ker(φ) \ {0} or {2γ + δ, γ + 2δ} ⊂ ker(φ) \ {0}. Therefore we obtain a proper zero-sum subsequence of S, a contradiction.

Suppose that |S_{h_{2}}| ≤ 3. Since t ≤ 4 and |S_{h_{3}}| ≤ 2, it follows that |S_{h_{2}}| = 3 and |S_{h_{3}}| = |S_{h_{4}}| = 2. Since |supp(S_{h_{1}})| = 2, S_{h_{1}} has two (not disjoint) subsequences V, V′ with |V | = |V′| = 3 and σ(V), σ(V′) ∈ ker(φ) \ {0} distinct. Thus for every subsequence T of S_{h_{2}}S_{h_{3}}S_{h_{4}} with σ(T) ∈ ker(φ), we obtain σ(T) = σ(V) + σ(V′). Since D(φ(G)) = 5, h_{2}^{2}h_{3}^{2}h_{4} contains a zero-sum subsequence whence S_{h_{2}}S_{h_{3}}S_{h_{4}} contains a subsequence T such that φ(T) | h_{2}^{2}h_{3}^{2}h_{4} and σ(T) ∈ ker(φ) \ {0}. Since h_{2}, h_{3}, h_{4} ∈ φ(G) ≅ C_{3} ⊕ C_{3} are pairwise distinct, we have h_{2}h_{3}h_{4} | φ(T). Since σ(T) has the same value for all such T, we infer that

S_{h_{2}} = γ^{3}, S_{h_{3}} = δ^{2} and S_{h_{4}} = ϵ^{2}.

By Lemma 3.1.2 the sequence h_{1}^{2}h_{2}^{2}h_{3}^{2}h_{4}^{2} ∈ F(φ(G)) contains a short zero-sum subsequence whence S has a subsequence T_{3} such that |T_{3}| = 3 and φ(T_{3}) | h_{1}^{2}h_{2}^{2}h_{3}^{2}h_{4}^{2}. If v_{h_{2}}(φ(T_{3})) = 0, then φ(T_{3}) = h_{1}h_{3}h_{4} and we set T_{1}, T_{2} such that φ(T_{2}) = φ(T_{3}) and φ(T_{1}) = h_{2}^{3}, which leads to a contradiction. If v_{h_{1}}(φ(T_{3})) = 0, then φ(T_{3}) = h_{2}h_{3}h_{4} and we set T_{1}, T_{2} such that φ(T_{1}) = h_{1}^{3} and φ(T_{2}) = φ(T_{3}), which leads to a contradiction.

Thus h_{1}h_{2} | φ(T_{3}) and after a suitable renumeration we may suppose that φ(T_{3}) = h_{1}h_{2}h_{3}. Since σ(T_{3}) ∈ {α + γ + δ, β + γ + δ} ⊂ ker(φ) \ {0} = {σ(V), σ(V′), 3γ}, we may suppose that

α + γ + δ ∈ {σ(V), σ(V′)}.

If S_{h_{1}} = αβ^{3}, then {σ(V), σ(V′)} = {α + 2β, 3β}, γ + δ = 2β or α + γ + δ = 3β whence either γ^{2}δ^{2}β^{2} or αγδβ^{3} is a proper zero-sum subsequence of S, a contradiction.

If S_{h_{1}} = α^{2}β^{2}, then {σ(V), σ(V′)} = {α + 2β, 2α + β}, γ + δ = 2β or γ + δ = α + β whence either γ^{2}δ^{2}β^{2} or γ^{2}δ^{2}αβ is a proper zero-sum subsequence of S, a contradiction.

If S_{h_{1}} = α^{3}β, then {σ(V), σ(V′)} = {3α, 2α + β}, γ + δ = 2α or γ + δ = α + β whence either γ^{2}δ^{2}α^{2} or γ^{2}δ^{2}αβ is a proper zero-sum subsequence of S, a contradiction.

Assertion 2: k_{1} ≠ 4. Assume to the contrary that k_{1} = 4. Let e_{2} ∈ G such that (g_{1} = e_{1}, e_{2}) is a basis of G. Then

S = e_{1}^{4} ∏_{i=1}^{7}(x_{i}e_{1} + y_{i}e_{2})

with x_{i}, y_{i} ∈ [0, 5] and (x_{i}, y_{i}) ≠ (1, 0) for all i ∈ [1, 7]. Since S is a minimal zero-sum sequence, it follows that for every ∅ ≠ I ⊊ [1, 7]

(∗)  ∑_{i∈I} y_{i} ≡ 0 mod 6 implies that ∑_{i∈I} x_{i} ≡ 1 mod 6.

In particular, this implies that y_{1}, …, y_{7} ∈ [1, 5]. Next we assert that for each two distinct i, j ∈ [1, 7] we have

y_{i} + y_{j} ≢ 0 mod 6. (∗∗)

Assume to the contrary, that this does not hold, say y_{6}+y_{7} ≡ 0 mod 6. Then x_{6}+x_{7} ≡ 1 mod 6. Clearly, ∏_{i=1}^{5} y_{i}e_{2} is a zero-sum sequence and (∗) implies that it is minimal. Thus it follows that ∏_{i=1}^{5} y_{i}e_{2} = (ye_{2})^{4} · (2ye_{2}) for some y ∈ [1, 5] with gcd{y, 6} = 1 (cf. for example [Ger90], Lemma 13). Thus y ∈ {1, 5} and without restriction we may suppose that y = y_{1} = … = y_{4} = 1 and y_{5} = 2. Therefore we have

∏_{i=1}^{7} y_{i}e_{2} = e_{2}^{4} · (2e_{2}) · (y_{6}e_{2}) · (y_{7}e_{2}).

Since y_{6} + y_{7} ≡ 0 mod 6, it follows that {y_{6}, y_{7}} ∩ [3, 5] ≠ ∅, say y_{6} ∈ [3, 5]. Since for every I ⊂ [1, 4] with |I| = 6 − y_{6} we have |I| · 1 + y_{6} ≡ 0 mod 6, (∗) implies that x_{6} + ∑_{i∈I} x_{i} ≡ 1 mod 6 whence x_{1} = x_{2} = x_{3} = x_{4} = x. If y_{6} = 5, then y_{7} = 1 whence x_{7} = x, a contradiction to 4 = k_{1} ≥ … ≥ k_{l} ≥ 1. Thus y_{6} ∈ [3, 4]. Then (∗) implies that x_{6} + (6 − y_{6})x ≡ 1 mod 6 and x_{7} + (6 − y_{7})x ≡ 1 mod 6 whence 2 ≡ x_{6} + x_{7} mod 6, a contradiction.

We consider the sequence

T = ∏_{i=1}^{7} y_{i}e_{2}.

By (∗∗) it follows that v_{3e_{2}}(T) ≤ 1, v_{e_{2}}(T)v_{5e_{2}}(T) = 0 and v_{2e_{2}}(T)v_{4e_{2}}(T) = 0. Without restriction we may suppose that v_{5e_{2}}(T) = 0. If v_{2e_{2}}(T) ≥ 4, say y_{1} = … = y_{4} = 2, then by (∗) we infer that x_{1} = … = x_{4} and 3x_{1} = x_{1} + x_{2} + x_{3} ≡ 1 mod 6, a contradiction. Thus v_{2e_{2}}(T) ≤ 3 and by a similar argument we obtain that v_{4e_{2}}(T) ≤ 3. Since v_{2e_{2}}(T)v_{4e_{2}}(T) = 0, it follows that v_{2e_{2}}(T) + v_{4e_{2}}(T) ≤ 3. This implies that

v_{e_{2}}(T) ≥ 3.

Suppose v_{2e_{2}}(T) = 3, say y_{1} = y_{2} = y_{3} = 1 and y_{4} = y_{5} = y_{6} = 2. Since 1 + 1 + 2 + 2 ≡ 0 mod 6, (∗) implies that x_{4} = x_{5} = x_{6}. Since y_{4} + y_{5} + y_{6} ≡ 0 mod 6, (∗) implies that 3x_{4} = x_{4} + x_{5} + x_{6} ≡ 1 mod 6, a contradiction. Thus we get v_{2e_{2}}(T) ≤ 2 and similarly v_{4e_{2}}(T) ≤ 2. Thus v_{2e_{2}}(T) + v_{4e_{2}}(T) ≤ 2, which implies that

v_{e_{2}}(T) ≥ 4.

Suppose v_{e_{2}}(T) = 4. If v_{2e_{2}}(T) = 2, then ∑_{i=1}^{7} y_{i} ≡ 0 mod 6 implies that v_{4e_{2}}(T) = 1, a contradiction. If v_{4e_{2}}(T) = 2, then ∑_{i=1}^{7} y_{i} ≡ 0 mod 6 implies that v_{0}(T) = 1, a contradiction. Thus v_{2e_{2}}(T) + v_{4e_{2}}(T) ≤ 1 whence 7 = ∑_{i=1}^{5} v_{ie_{2}}(T) ≤ 6, a contradiction. So we finally obtain that

v_{e_{2}}(T) ≥ 5, say T = e_{2}^{5} · (y_{6}e_{2}) · (y_{7}e_{2}).

If y_{6} ∈ [2, 5] and I ⊂ [1, 5] with |I| = 6 − y_{6}, then y_{6} + |I|.1 ≡ 0 mod 6 whence (∗) implies that x_{6} + ∑_{i∈I} x_{i} ≡ 1 mod 6 whence x_{1} = … = x_{5}, a contradiction to k_{1} = 4. Thus y_{6} = 1 and similarly y_{7} = 1. Thus v_{e_{2}}(T) = 7 and (∗) implies that x_{1} = … = x_{7}, a contradiction to 4 = k_{1} ≥ … ≥ k_{l} ≥ 1.

Theorem 4.3 (Characterization of Property B). Let G = C_{n} ⊕ C_{n} with n ≥ 2. Then the following statements are equivalent:

1. Every sequence S ∈ F(G) with length |S| = 3n − 3, which contains no zero-sum subsequence of length greater than or equal to n, has a subsequence of the form 0^{n−1}a^{n−2} for some a ∈ G.

2. Every zero-sumfree sequence S ∈ F(G) with length |S| = 2n − 2 contains some element with multiplicity at least n − 2.

3. Every minimal zero-sum sequence S ∈ F(G) with length |S| = 2n−1 contains some element with multiplicity n − 1.

4. For every minimal zero-sum sequence S ∈ F(G) with length |S| = 2n − 1 there exists some basis (e_{1}, e_{2}) of G and integers a_{1}, …, a_{n} ∈ [0, n − 1] with ∑_{ν=1}^{n} a_{ν} ≡ 1 mod n such that S = e_{1}^{n−1} ∏_{ν=1}^{n}(a_{ν}e_{1} + e_{2}).

Proof. 1. ⟹ 2. Let S ∈ F(G) be a zero-sumfree sequence with length |S| = 2n − 2. Then the sequence 0^{n−1}S contains no zero-sum subsequence of length greater than or equal to n. By assumption there exists some a ∈ G such that 0^{n−1}a^{n−2} divides 0^{n−1}S and the assertion follows.

2. ⟹ 3. Let S = ∏_{i=1}^{2n−1} g_{i} ∈ F(G) be a minimal zero-sum sequence. Then there are a, b ∈ G such that a^{n−2} | g_{2n−1}^{−1} · S and b^{n−2} | a^{−1} · S. Assume to the contrary that v_{g}(S) < n − 1 for all g ∈ G. Then a ≠ b and a^{n−2}b^{n−2} is a zero-sumfree subsequence of S. By Lemma 3.9 (a, b) is a basis of G whence S has the form

S = a^{n−2}b^{n−2} · (x_{1}a + y_{1}b) · (x_{2}a + y_{2}b) · (x_{3}a + y_{3}b)

with all x_{i}, y_{i} ∈ [0, n − 1]. Since S is a minimal zero-sum sequence, there exists some i ∈ [1, 3] such that x_{i}a + y_{i}b ∈ {a, b} and the assertion follows.

3. ⟹ 4. This follows from Proposition 4.1.

4. ⟹ 1. Let S ∈ F(G) be a sequence with length |S| = 3n−3 containing no zero-sum subsequence of length greater than or equal to n. Since |S| > D(G), S has a zero-sum subsequence. Let T denote a maximal zero-sum subsequence of S. Then U = T^{−1}S is zero-sumfree whence in particular we have |U| ≤ D(G)−1 = 2n−2. Therefore |T | ≥ n−1 whence |T | = n−1 and |U| = 2n−2. Therefore −σ(U)·U is a minimal zero-sum sequence with length 2n − 1 whence by assumption there exists a basis (e_{1}, e_{2}) of G such that

U = e_{1}^{r} · ∏_{i=1}^{2n−2−r}(x_{i}e_{1} + e_{2})

with r ∈ {n − 1, n − 2} and all x_{i} ∈ [0, n − 1]. Then T has the form

T = 0^{k} · ∏_{i=1}^{n−1−k}(u_{i}e_{1} + v_{i}e_{2})

with k ∈ [0, n − 1] and all u_{i}, v_{i} ∈ [0, n − 1] such that (u_{i}, v_{i}) ≠ (0, 0). If k = n − 1, then the assertion follows. Assume to the contrary, that k < n − 1.

We first show that

(∗) if ∅ ≠ I ⊂ [1, n − 1 − k] and ∑_{i∈I} v_{i}e_{2} = 0, then ∑_{i∈I} u_{i}e_{1} = 0.

Assume to the contrary, that ∅ ≠ I ⊂ [1, n − 1 − k], ∑_{i∈I} v_{i} ≡ 0 mod n and ∑_{i∈I} u_{i} ≢ 0 mod n. Let a ∈ [1, n − 1] such that a ≡ ∑_{i∈I} u_{i} mod n. We construct a zero-sum subsequence S′ of S with length |S′| ≥ n, which contradicts our assumption on S. Suppose r = n − 1. If a ≤ |I|, then set S′ = e_{1}^{n−a} ∏_{i∈I}(u_{i}e_{1} + v_{i}e_{2}), and if a > |I|, then set S′ = T ∏_{i∈I}(u_{i}e_{1} + v_{i}e_{2})^{−1} · e_{1}^{a}. Suppose r = n − 2. Since U is zero-sumfree, it follows that ∑_{i=1}^{n} x_{i} ≡ 1 mod n whence

S′ = e_{1}^{n−a−1} · ∏_{i∈I}(u_{i}e_{1} + v_{i}e_{2}) ∏_{i=1}^{n}(x_{i}e_{1} + e_{2})

is the required sequence.

Since T is a zero-sum sequence, there exists some J ⊂ [1, n − 1 − k] with |J| ≥ 2 such that ∏_{j∈J} v_{j}e_{2} is a minimal zero-sum sequence in ⟨e_{2}⟩. We assert that there exists some ∅ ≠ I ⊂ J such that

1 ≤ ∑_{i∈I} v_{i} ≤ n − |J|. (∗∗)

This obviously holds in case |J| = 2. Suppose that |J| ≥ 3. First we consider the case that at least two of the v_{i} are distinct, say J = [1, t] with 3 ≤ t ≤ n − 1 − k and v_{1} ≠ v_{2}. Then ∏_{i=1}^{t−1} v_{i}e_{2} is zero-sumfree and Lemma 3.10 implies that

|Σ(∏_{i=1}^{t−1} v_{i}e_{2})| ≥ |Σ(v_{1}e_{2} · v_{2}e_{2})| + ∑_{i=3}^{t−1} |Σ(v_{i}e_{2})| = 3 + (t − 3) = t = |J|,

whence (∗∗) holds. It remains to consider the case where there exists some v ∈ [0, n − 1] such that v_{j} = v for all j ∈ J. Then |J|ve_{2} = 0, ord(ve_{2}) < n, −e_{2} ∉ Σ((ve_{2})^{|J|−1}) and |Σ((ve_{2})^{|J|−1}))| = |J| − 1 whence (∗∗) holds.

Let ∅ ≠ I ⊂ J such that (∗∗) holds and let a = ∑_{i∈I} v_{i}. For every Z ⊂ [1, 2n − 2 − r] with |Z| = n − a let b = b_{Z} ∈ [1, n] such that b ≡ ∑_{i∈I} u_{i} + ∑_{i∈Z} x_{i} mod n. If r = n − 2 and for all such sets Z we have b_{Z} = 1, then x_{1} = … = x_{n}, a contradiction to U zerosumfree. Thus in case r = n − 2 we may choose Z such that b = b_{Z} ≠ 1. If r = n − 1, we choose any subset Z. Then, in both cases,

S′ = T · ∏_{j∈J}(u_{j}e_{1} + v_{j}e_{2})^{−1} ∏_{i∈I}(u_{i}e_{1} + v_{i}e_{2}) ∏_{i∈Z}(x_{i}e_{1} + e_{2}) · e_{1}^{n−b}

is a zero-sum subsequence of S with length

|S′| = n − 1 − |J| + |I| + n − a + n − b ≥ n − 1 − |J| + 1 + n − a ≥^{(∗∗)} n,

a contradiction.

5. Property B and ν(G)

The invariant ν(G) (see Definition 5.1) was introduced by van Emde Boas in 1969. It plays a key role in all investigations of Davenport’s constant of groups with rank three (see [vEB69b] and also [Gao00a], section 5 where for groups G of rank two ν(G) is studied in detail). The relationship between zero-sum problems in finite abelian groups and covering problems by proper cosets was recently investigated in [?]. The proof of the inequalities in Proposition 5.2.1 is straightforward. Up to now there is known no group G such that D(G) < ν(G) + 2.

Definition 5.1. Let G be a finite abelian group. Let ν(G) denote the smallest integer l ∈ ℕ_{0} such that for every zero-sumfree sequence S ∈ F(G) with |S| ≥ l there exists a subgroup H < G and some a ∈ G \ H such that G \ (Σ(S) ∪ {0}) ⊂ a + H.

Proposition 5.2. Let G be a finite abelian group.

1. ν(G) + 1 ≤ D(G) ≤ ν(G) + 2.

2. If G is cyclic or a p-group, then D(G) = ν(G) + 2.

Proof. 1. see [Gao00a], Lemma 3.3.

2. see [vEB69b], Proposition 1.19 and Theorem 2.8.

Theorem 5.3. Let G = C_{n} ⊕ C_{n} with n ≥ 2. If n satisfies Property B, then D(G) = ν(G) + 2.

Proof. If n ∈ [2, 3], then G is a p-group whence the assertion follows from Proposition 5.2.2. Suppose that n ≥ 4. By Proposition 5.2.1 we have ν(G) ≥ D(G) − 2. Hence it remains to show that for every zero-sum free sequence S ∈ F(G) with |S| ≥ D(G) − 2 there exists a subgroup H < G and some a ∈ G \ H such that G \ (Σ(S) ∪ {0}) ⊂ a + H.

Let S be such a sequence. If Σ(S) = G \ {0}, then the assertion is clear. Suppose that there exists some b ∈ G \ {0} such that −b ∉ Σ(S). Thus bS is a zero-sumfree sequence of length D(G) − 1 ≥ |bS| ≥ 1 + (D(G) − 2) = 2n − 2 and hence there is some a ∈ G, such that a · b · S is a minimal zero-sum sequence of length 2n − 1. By Proposition 4.1.2 there exists a basis (e_{1}, e_{2}) of G and a_{1}, …, a_{n} ∈ [0, n − 1] with ∑_{i=1}^{n} a_{i} ≡ 1 mod n such that

a · b · S = e_{1}^{n−1} ∏_{i=1}^{n}(a_{i}e_{1} + e_{2}).

Hence, up to enumeration, there are the following three possibilities for S.

Case 1: S = e_{1}^{n−1} ∏_{i=1}^{n−2}(a_{i}e_{1} + e_{2}). We assert that G \ (Σ(S) ∪ {0}) ⊂ −e_{2} + ⟨e_{1}⟩. Let g ∈ G \ (−e_{2} + ⟨e_{1}⟩). We have to verify that g ∈ Σ(S) ∪ {0}. There are λ_{1} ∈ [0, n − 1] and λ_{2} ∈ [0, n − 2] such that g = λ_{1}e_{1} + λ_{2}e_{2}, and obviously we have

g ∈ ∑_{i=1}^{λ_{2}}(a_{i}e_{1} + e_{2}) + ⟨e_{1}⟩ ⊂ Σ(S) ∪ {0}.

Case 2: S = e_{1}^{n−2} ∏_{i=1}^{n−1}(a_{i}e_{1} + e_{2}). We distinguish two subcases.

Case 2.1: a_{1} = … = a_{n−1} = a. Clearly, we obtain Σ(S)∪{0} = ⋃_{i=0}^{n−2}(ie_{1} +⟨ae_{1} +e_{2}⟩) whence G \ (Σ(S) ∪ {0}) ⊂ −e_{1} + ⟨ae_{1} + e_{2}⟩.

Case 2.2: |{a_{1}, …, a_{n−1}}| ≥ 2, say a_{1} ≠ a_{2}. We set a = ∑_{i=1}^{n−1} a_{i} and assert that

⋃_{i=0}^{n−2}(ie_{1} + ⟨ae_{1} − e_{2}⟩) = G \ (−e_{1} + ⟨ae_{1} − e_{2}⟩) ⊂ Σ(S) ∪ {0}.

Let i ∈ [0, n−2] and λ ∈ [0, n−1]. We have to verify that there exists some Λ ⊂ [1, n−1] with |Λ| = n − λ and some θ ∈ [0, n − 2] such that

ie_{1} + λ(ae_{1} − e_{2}) = θe_{1} + ∑_{j∈Λ}(a_{j}e_{1} + e_{2}).

If λ = 1, then Λ = [1, n − 1] and θ = i fulfill the requirements. Suppose λ > 1. We choose some Λ ⊂ [2, n − 1] with 2 ∈ Λ and |Λ| = n − λ. If i + ∑_{j∈Λ}(a − a_{j}) ≢ n − 1 mod n, then θ ∈ [0, n − 2] with θ ≡ i + ∑_{j∈Λ}(a − a_{j}) mod n fulfills the requirements. If i+∑_{j∈Λ}(a−a_{j}) ≡ n−1 mod n, we set Λ′ = (Λ\{2})∪{1} whence i+∑_{j∈Λ′}(a−a_{j}) ≢ n−1 mod n and there exists some θ having the required properties.

Case 3: S = e_{1}^{n−3} ∏_{i=1}^{n}(a_{i}e_{1} + e_{2}) with n ≥ 3. We distinguish two subcases.

Case 3.1: There exist i, j ∈ [1, n] such that a_{j} − a_{i} ≥ 2, say a_{2} − a_{1} ≥ 2. We assert that G\(Σ(S)∪{0}) ⊂ −e_{1}+{0}. Let g ∈ G\{0, −e_{1}}. We have to verify that g ∈ Σ(S). Let λ_{1}, λ_{2} ∈ [0, n−1] with g = λ_{1}e_{1} +λ_{2}e_{2} whence (λ_{1}, λ_{2}) ∉ {(0, 0), (n−1, 0)}. If λ_{2} = 0, then g ∈ Σ(e_{1}^{n−2}) ⊂ Σ(S) because ∑_{i=1}^{n}(a_{i}e_{1} + e_{2}) = e_{1}. Suppose that λ_{2} ∈ [1, n − 1]. We choose some Λ ⊂ [1, n] with |Λ| = λ_{2}, 1 ∈ Λ and 2 ∉ Λ. If Λ′ = (Λ \ {1}) ∪ {2}, then

g ∈ {(∑_{j∈Λ} a_{j}e_{1} + e_{2}) + ie_{1} | i ∈ [0, n − 3]} ∪ {(∑_{j∈Λ′} a_{j}e_{1} + e_{2}) + ie_{1} | i ∈ [0, n − 3]} ⊂ Σ(S).

Case 3.2: {a_{1}, …, a_{n}} = {a, a + 1} for some a ∈ [0, n − 2]. Then S = e_{1}^{n−3}(ae_{1} + e_{2})^{k}((a + 1)e_{1} + e_{2})^{n−k} for some k ∈ [1, n − 1], and since ka + (n − k)(a + 1) ≡ 1 mod n, it follows that k = n − 1. We assert that G \ (−e_{1} + ⟨ae_{1} + e_{2}⟩) ⊂ Σ(S) ∪ {0}. Since obviously, ⋃_{i=0}^{n−3}(ie_{1} + ⟨ae_{1} + e_{2}⟩) ⊂ Σ(S) ∪ {0}, it remains to check that (n − 2)e_{1} + ⟨ae_{1} + e_{2}⟩ ⊂ Σ(S) ∪ {0}. Let g = (n − 2)e_{1} + λ(ae_{1} + e_{2}) with λ ∈ [0, n − 1]. If λ = 0, then g = (n − 3)e_{1} + (n − 1)(ae_{1} + e_{2}) + ((a + 1)e_{1} + e_{2}) ∈ Σ(S). If λ > 0, then g − ((a + 1)e_{1} + e_{2}) = (n − 3)e_{1} + (λ − 1)(ae_{1} + e_{2}) ∈ Σ(e_{1}^{n−3}(ae_{1} + e_{2})^{n−1}) whence the assertion follows.

6. Property B implies Property C

In this section we show that, under some additional weak condition, Property B implies Property C. This was first done for prime numbers in [GG99]. For a ∈ ℤ we denote by |a|_{n} the positive integer in [1, n] such that a ≡ |a|_{n} mod n.

Proposition 6.1. Let G = C_{n} ⊕ C_{n} with n ≥ 2 and S = a^{n−1}b^{n−1} ∏_{i=1}^{n−1} c_{i} ∈ F(G) a sequence which does not contain a short zero-sum subsequence. If n satisfies Property B, then c_{1} = ⋯ = c_{n−1}.

Proof. For n = 2 there is nothing to do. Suppose that n ≥ 3 and let S be as above. Since a ≠ b, Lemma 3.9 implies that (e_{1} = a, e_{2} = b) is a basis of G whence S has the form

S = e_{1}^{n−1}e_{2}^{n−1} ∏_{i=1}^{n−1}(x_{i}e_{1} + y_{i}e_{2})

with x_{i}, y_{i} ∈ [1, n]. Since S has no short zero-sum subsequence, it follows that x_{i}, y_{i} ∈ [1, n − 1]. Furthermore, S has no zero-sum subsequence of length n or 2n > D(G) and the same is true for

S_{e_{2}} = (e_{1} − e_{2})^{n−1}0^{n−1} ∏_{i=1}^{n−1}(x_{i}e_{1} + (y_{i} − 1)e_{2}).

Therefore

(e_{1} − e_{2})^{n−1} ∏_{i=1}^{n−1}(x_{i}(e_{1} − e_{2}) + (x_{i} + y_{i} − 1)e_{2})

is zero-sumfree whence ∏_{i=1}^{n−1}(x_{i} + y_{i} − 1)e_{2} is zero-sumfree in ⟨e_{2}⟩ ≅ C_{n} which implies that

x_{1} + y_{1} ≡ ⋯ ≡ x_{n−1} + y_{n−1} mod n.

Since for every i ∈ [1, n − 1]

e_{1}^{n−x_{i}}e_{2}^{n−y_{i}}(x_{i}e_{1} + y_{i}e_{2})

is a zero-sum subsequence of S of length 2n + 1 − (x_{i} + y_{i}), it follows that x_{i} + y_{i} ≤ n. Thus

x_{1} + y_{1} = ⋯ = x_{n−1} + y_{n−1} = m

for some m ∈ [2, n]. Since ∏_{i=1}^{n−1}(x_{i} +y_{i} −1)e_{2} = ((m − 1)e_{2})^{n−1} is zero-sumfree, it follows that gcd{m − 1, n} = 1.

If m = 2, then x_{1} = y_{1} = ⋯ = x_{n−1} = y_{n−1} = 1 and the assertion is proved.

Suppose m = n. If ∏_{i∈I} x_{i}e_{1} is a zero-sum sequence for some ∅ ≠ I ⊂ [1, n − 1], then the same is true for ∏_{i∈I} y_{i}e_{2} and thus ∏_{i∈I}(x_{i}e_{1} + y_{i}e_{2}) would be a zero-sum sequence. Since S contains no short zero-sum subsequence, ∏_{i=1}^{n−1} x_{i}e_{1} is zero-sumfree whence x_{1} = ⋯ = x_{n−1}. Therefore y_{1} = ⋯ = y_{n−1} and the assertion is proved.

It remains to consider the case where m ∈ [3, n − 1]. Since gcd{m − 1, n} = 1, there is a unique t ∈ [1, n] such that t(m − 1) ≡ 1 mod n. Since m ∈ [3, n − 1], it follows that t ∈ [2, n − 2] whence |tm|_{n} = t + 1. Since t ≥ 2, it suffices to show that for every subset I ⊂ [1, n − 1] with |I| = t all x_{i} with i ∈ I are equal.

Let I ⊂ [1, n − 1] with |I| = t and consider the sequence

S_{I} = e_{1}^{n−|Σ_{i∈I}x_{i}|_{n}}e_{2}^{n−|Σ_{i∈I}y_{i}|_{n}} ∏_{i∈I}(x_{i}e_{1} + y_{i}e_{2}) .

Clearly, S_{I} is a zero-sum subsequence of S of length

|S_{I}| = 2n + t− | Σ_{i∈I}x_{i} |_{n} − | Σ_{i∈I}y_{i} |_{n}

= 2n + t− | Σ_{i∈I}x_{i} |_{n} − | tm − Σ_{i∈I}x_{i} |_{n}

= \begin{cases}2n + t− | tm |_{n}= 2n − 1 , | tm |_{n}>| Σ_{i∈I}x_{i} |_{n} \\ 2n + t − (n+ | tm |_{n}) = n − 1 , | tm |_{n}≤| Σ_{i∈I}x_{i} |_{n} \end{cases}

Since S has no short zero-sum subsequence, we infer that |S_{I}| = 2n − 1 and that S_{I} is a minimal zero-sum sequence.

Since t ≤ n − 2 and {x_{i}e_{1} + y_{i}e_{2} | i ∈ I} ∩ {e_{1}, e_{2}} = ∅, Property B implies that either

n− | Σ_{i∈I}x_{i} |_{n}= n − 1 or n− | Σ_{i∈I}y_{i} |= n − 1 .

Therefore by Proposition 4.1.2.a) either (y_{i} = 1 for all i ∈ I) or (x_{i} = 1 for all i ∈ I).

Theorem 6.2. Let G = C_{n} ⊕ C_{n} with n ≥ 2. Suppose that n satisfies Property B and that every sequence S ∈ F(G) with |S| ≥ 3n − 2 has a zero-sum subsequence of length n or 2n. Then n satisfies Property C.

Remark: If n has at most two distinct prime divisors or if Property E holds for all prime divisors of n, then every sequence S ∈ F(G) with |S| ≥ 3n−2 has a zero-sum subsequence of length n or 2n (see Theorem 3.7)

Proof. Since 2 satisfies Property C, we may suppose that n ≥ 3. Let S ∈ F(G) be a sequence with length |S| = 3n − 3 which does not contain a short zero-sum subsequence. By assumption the sequence 0.S contains a zero-sum subsequence of length n or 2n whence S contains a zero-sum subsequence T of length |T | ∈ {n − 1, n, 2n − 1, 2n}. Therefore |T | = 2n − 1 and T is a minimal zero-sum sequence. Hence by Property B there is some b ∈ G with b^{n−1} | T and thus

S = b^{n−1} ∏_{i=1}^{2n−2} c_{i} .

Since S has no zero-sum subsequence of length n or 2n, the same is true for

S_{b} = 0^{n−1} ∏_{i=1}^{2n−2}(c_{i} − b).

Therefore ∏_{i=1}^{2n−2}(c_{i} − b) is zero-sumfree and thus c ∏_{i=1}^{2n−2}(c_{i} − b) is a minimal zerosum sequence where c = − ∑_{i=1}^{2n−2}(c_{i} − b). Since n satisfies Property B, there are two possiblities. If there is some g ∈ G such that g^{n−1} | ∏_{i=1}^{2n−2}(c_{i} − b), then b^{n−1}(g + b)^{n−1} | S and the assertion follows from Proposition 6.1. Otherwise it follows that c^{n−2} divides ∏_{i=1}^{2n−2}(c_{i} − b), say c = c_{1} − b. Setting e_{1} = c_{1} = c + b and e_{2} = b we obtain that e_{1}^{n−2}e_{2}^{n−1} is a subsequence of S. By Lemma 3.9 (e_{1}, e_{2}) is a basis of G whence S has the form

S = e_{1}^{n−2}e_{2}^{n−1} ∏_{i=1}^{n}(x_{i}e_{1} + y_{i}e_{2})

with x_{i}, y_{i} ∈ [1, n]. Setting

S_{e_{2}} = 0^{n−1}(e_{1} − e_{2})^{n−2} ∏_{i=1}^{n}(x_{i}e_{1} + (y_{i} − 1)e_{2})

and arguing as above we infer that

(e_{1} − e_{2})^{n−2} ∏_{i=1}^{n}(x_{i}e_{1} + (y_{i} − 1)e_{2})

is zero-sumfree. Since

\begin{aligned}0 & = c_{1} − b + ∑_{i=1}^{2n−2}(c_{i} − b) = c_{1} + b + ∑_{i=1}^{2n−2} c_{i} \\ & = e_{1} + e_{2} + (n − 2)e_{1} + ∑_{i=1}^{n}(x_{i}e_{1} + y_{i}e_{2}) = (n − 1)(e_{1} − e_{2}) + ∑_{i=1}^{n}(x_{i}e_{1} + y_{i}e_{2}), \end{aligned}

we obtain that

(e_{1} − e_{2})^{n−1} ∏_{i=1}^{n}(x_{i}(e_{1} − e_{2}) + (x_{i} + y_{i} − 1)e_{2})

is a minimal zero-sum sequence.

Clearly, (e_{1} − e_{2}, e_{2}) is a basis of G whence ∏_{i=1}^{n}(x_{i} + y_{i} − 1)e_{2} is a minimal zero-sum sequence in ⟨e_{2}⟩ which implies that

x_{1} + y_{1} ≡ ⋯ ≡ x_{n} + y_{n} mod n.

If for some i ∈ [1, n] we have x_{i} = 1 and y_{i} = n, then e_{1}^{n−1}e_{2}^{n−1} | S and the assertion follows from Proposition 6.1. Suppose that all (x_{i}, y_{i}) ≠ (1, n). If x_{i} + y_{i} ≥ n + 1 for some i ∈ [1, n], then x_{i} ≥ 2 and

e_{1}^{n−x_{i}}e_{2}^{n−y_{i}}(x_{i}e_{1} + y_{i}e_{2})

is a zero-sum subsequence of S with length 2n + 1 − (x_{i} + y_{i}) ≤ n, a contradiction. Thus

x_{1} + y_{1} = ⋯ = x_{n} + y_{n} = m

for some m ∈ [2, n]. Since ∏_{i=1}^{n}(x_{i} + y_{i} − 1)e_{2} is a minimal zero-sum sequence, we infer that gcd{m − 1, n} = 1.

Suppose that m = n. There is some ∅ ≠ I ⊂ [1, n] such that Σ_{i∈I}x_{i}e_{1} = 0. This implies that Σ_{i∈I}y_{i}e_{2} = 0 whence ∏_{i∈I}(x_{i}e_{1} + y_{i}e_{2}) is a short zero-sum subsequence of S, a contradiction.

If m = 2, then x_{1} = y_{1} = … x_{n} = y_{n} = 1 whence ∏_{i=1}^{n}(x_{i}e_{1} + y_{i}e_{2}) is a short zero-sum subsequence of S, a contradiction.

Therefore we obtain that m ∈ [3, n − 1]. Let t ∈ [2, n] such that t(m − 1) ≡ 1 mod n and I ⊂ [1, n] be a subset with |I| = t and Σ_{i∈I}x_{i} ≢ 1 mod n. Then |Σ_{i∈I}x_{i}|_{n} ∈ [2, n], and arguing as in Proposition 6.1 we infer that

S_{I} = e_{1}^{n−|∑_{i∈I} x_{i}|_{n}}e_{2}^{n−|∑_{i∈I} y_{i}|_{n}} ∏_{i∈I}(x_{i}e_{1} + y_{i}e_{2})

is a minimal zero-sum subsequence of S with length |S_{I}| = 2n − 1. As in Propositon 6.1 we argue that either all x_{i} are equal to 1 or all y_{i} are equal to 1.

Therefore, for every subset I ⊂ [1, n] with |I| = t we have:

either (∑_{i∈I} x_{i} ≡ 1 mod n) or (all x_{i} are equal to 1) or (all x_{i} are equal to m − 1). (∗)

Assume to the contrary, that |{x_{1}, …, x_{n}}| ≥ 3, say |{x_{n−2}, x_{n−1}, x_{n}}| = 3. Since t − 1 ≤ n − 3, it follows that |{x_{j} + ∑_{i=1}^{t−1} x_{i} | n − 2 ≤ j ≤ n}| = 3, a contradiction to (∗).

Therefore, ∏_{i=1}^{n} x_{i}e_{1} = (xe_{1})^{u}(x′e_{1})^{v} with x, x′ ∈ [1, n] , x ≠ x′ , u + v = n and 0 ≤ v ≤ u. If v ≤ 1, then u ≥ n − 1 and Proposition 6.1 implies the assertion.

Assume to the contrary, that v ≥ 2. If t ≥ 3, one can choose u_{0} ∈ [2, u − 1] and v_{0} ∈ [1, v − 1] such that u_{0} + v_{0} = t because t ≤ n − 2 = u + v − 2. However,

u_{0}x + v_{0}x′ ≠ (u_{0} − 1)x + (v_{0} + 1)x′

which contradicts (∗). Hence we have t = 2, and (∗) implies that x + x′ ≡ 1 mod n. Thus x + x ≢ x + x′ ≡ 1 mod n whence (∗) implies that x ∈ {1, m − 1}. We argue in a similar way for x′ and obtain {x, x′} = {1, m − 1}. Therefore m = x + x′ ≡ 1 mod n, a contradiction to m ∈ [3, n − 1].

7. Zero-sum sequences S in C_{m} ⊕ C_{m} with length |S| = tm − 1

Let G = C_{mn}⊕C_{mn} with m, n ∈ ℕ_{≥2}, φ : G → G the multiplication by n and S ∈ F(G) a minimal zero-sum sequence with length |S| = D(G) = tm − 1 where t = 2n. Then by Lemma 3.14 φ(S) is a zero-sum sequence in nG ≅ C_{m} ⊕ C_{m} which is not a product of t = 2n zero-sum subsequences. It is the aim of this section to determine the structure of such sequences under the assumption that C_{m} ⊕ C_{m} has Property B.

Theorem 7.1. Let G = C_{m} ⊕ C_{m} with m ≥ 2. Suppose that m satisfies Property B and that every sequence T ∈ F(G) with |T | ≥ 3m − 2 has a zero-sum subsequence of length m or 2m. Let S ∈ F(G) be a zero-sum sequence with |S| = tm − 1 for some t ≥ 3 which cannot be written as a product of t non-empty zero-sum subsequences. Then there exists a basis (e_{1}, e_{2}) of G such that either

S = e_{1}^{sm−1} · ∏_{ν=1}^{(t−s)m}(a_{ν}e_{1} + e_{2})

where a_{1}, …, a_{(t−s)m} ∈ [0, m − 1] and s ∈ [1, t − 1] or

S = e_{1}^{s_{1}m} · (be_{1} + e_{2})^{s_{2}m−1} · e_{2}^{s_{3}m−1} · (be_{1} + 2e_{2})

where b ∈ [0, m − 1] with gcd{b, m} = 1 and s_{1}, s_{2}, s_{3} ∈ ℕ with s_{1} + s_{2} + s_{3} = t.

We are going to prove Theorem 7.1 by induction on t. Throughout this section, let all notations be as in Theorem 7.1.

Lemma 7.2. The assertion of Theorem 7.1 holds for t = 3.

Proof. Suppose that

S = ∏_{ν=1}^{l} g_{ν}^{k_{ν}}

where g_{1}, …, g_{l} ∈ G are pairwise distinct and k_{1} ≥ k_{2} ≥ … ≥ k_{l} ≥ 1. Since S does not contain three disjoint nonempty zero-sum subsequences, it follows that k_{2} ≤ m − 1. By Lemma 3.12.1 every short zero-sum subsequence of S has length m. Suppose there is some j ∈ [1, l − 1] such that k_{j} ≥ m − 1 and k_{j+1} ≥ m − 1.

We assert that either((g_{j}, g_{j+1}) is a basis of G) or (k_{1} = m − 1 and (g_{1}, g_{j})) is a basis of G). This is obviously true for m = 2. Suppose that m ≥ 3 and that (g_{j}, g_{j+1}) is not a basis of G. Then by Lemma 3.9 g_{j}^{m−1}g_{j+1}^{m−1} contains a short zero-sum subsequence T. Then T^{−1}S is a minimal zero-sum subsequence with length 2m − 1 containing some element g with multiplicity m − 1, say g = g_{i} with i ∈ [1, l] minimal. Note that g_{j}g_{j+1} | T^{−1}S. If g′ ∈ supp(T^{−1}S) \ {g}, then by Proposition 4.1(g, g′) is a basis of G whence the assertion follows.

We distinguish several cases.

Case 1: k_{2} < m − 1. By Lemma 3.12.2 S has a product decomposition of the form S = S_{0}S_{1} where S_{0} is a minimal zero-sum sequence with length 2m − 1 and S_{1} is a short zero-sum sequence. Since m has Property B, Theorem 4.3 implies that there exists a basis (e_{1}, e_{2}) of G such that

S = e_{1}^{m−1} · ∏_{ν=1}^{m}(a_{ν}e_{1} + e_{2}) · ∏_{ν=1}^{m}(x_{ν}e_{1} + y_{ν}e_{2})

with all x_{ν}, y_{ν}, a_{ν} ∈ [0, m − 1] and ∑_{ν=1}^{m} a_{ν} ≡ 1 mod m. Let ν ∈ [1, m]. It remains to verify that y_{ν} = 1. The sequence (x_{ν}e_{1}+y_{ν}e_{2})^{−1}·S contains a short zero-sum subsequence W (clearly, W ≠ ∏_{ν=1}^{m}(a_{ν}e_{1} + e_{2})) and W^{−1} · S is a minimal zero-sum sequence with length 2m − 1. Since max{v_{g}(W^{−1} · S) | g ∈ G} = m − 1 > k_{2}, it follows that

W^{−1} · S = e_{1}^{m−1} · (a_{μ}e_{1} + e_{2}) · (x_{ν}e_{1} + y_{ν}e_{2}) · ∏_{λ=1}^{m−2}(u_{λ}e_{1} + v_{λ}e_{2})

for some μ ∈ [1, m] and all u_{λ}, v_{λ} ∈ [0, m − 1]. Since W^{−1} · S is a minimal zero-sum sequence, Proposition 4.1.2.a) implies that y_{ν} = v_{1} = … = v_{m−2} = 1.

Case 2: k_{2} = m − 1 and k_{3} < m − 1. Then (g_{1} = e_{1}, g_{2} = e_{2}) is a basis of G, and we distinguish three subcases.

Case 2.1: k_{1} ≥ m + 1. The sequence

g_{1}^{−m} · S = e_{2}^{m−1} · e_{1}^{k_{1}−m} · ∏_{ν=1}^{2m−k_{1}}(x_{ν}e_{1} + y_{ν}e_{2})

is a minimal zero-sum sequence whence x_{1} = ⋯ = x_{2m−k_{1}} = 1 by Proposition 4.1.

Case 2.2: k_{1} = m. We have

S = e_{1}^{m} · e_{2}^{m−1} · ∏_{ν=1}^{m}(x_{ν}e_{1} + y_{ν}e_{2})

with all x_{ν}, y_{ν} ∈ [0, m − 1], ∑_{ν=1}^{m} x_{ν} ≡ 0 mod m and ∑_{ν=1}^{m} y_{ν} ≡ 1 mod m, say y_{1} ≠ 1. Since e_{1}^{−m} · S is a minimal zero-sum sequence, it follows that x_{1} = ⋯ = x_{m}. We verify that x_{1} = 1 which implies the assertion. Since k_{3} < m − 1, Theorem 6.2 implies that e_{1}^{−1} · (x_{1}e_{1} + y_{1}e_{2})^{−1} · S contains a short zero-sum subsequence W, and clearly we have |W | = m and e_{2}^{m−1} ∤ W. Then

W^{−1} · S = e_{1} · (x_{1}e_{1} + y_{1}e_{2}) · e_{2} · T for some T ∈ F(G)

is a minimal zero-sum sequence whence either v_{e_{1}}(W^{−1}·S) = m−1 or v_{e_{2}}(W^{−1}·S) = m−1. Since y_{1} ≠ 1, Proposition 4.1 implies that v_{e_{2}}(W^{−1} · S) = m − 1 and x_{1} = 1.

Case 2.3: k_{1} = m − 1. We have

S = e_{1}^{m−1} · e_{2}^{m−1} · ∏_{ν=1}^{m+1}(x_{ν}e_{1} + y_{ν}e_{2})

with all x_{ν}, y_{ν} ∈ [0, m−1], ∑_{ν=1}^{m+1} x_{ν} ≡ ∑_{ν=1}^{m+1} y_{ν} ≡ 1 mod m. If x_{1} = x_{2} = ⋯ = x_{m+1} = 1 or y_{1} = y_{2} = ⋯ = y_{m+1} = 1, then we are done. Assume to the contrary that this does not hold. Then there are i < j with x_{i} ≠ 1 and x_{j} ≠ 1 and there are i′ < j′ such that y_{i′} ≠ 1 and y_{j′} ≠ 1, say x_{1} ≠ 1 and y_{2} ≠ 1. Since k_{3} < m − 1, Theorem 6.2 implies that S · (x_{1}e_{1} + y_{1}e_{2})^{−1} · (x_{2}e_{1} + y_{2}e_{2})^{−1} contains a short zero-sum subsequence W. Then |W | = m and W^{−1} · S is a minimal zero-sum subsequence with length 2m − 1 which contains the sequence e_{1} · e_{2} · (x_{1}e_{1} + y_{1}e_{2}) · (x_{2}e_{1} + y_{2}e_{2}). Since k_{l} ≤ ⋯ ≤ k_{3} < m − 1, either v_{e_{1}}(W^{−1} · S) = m − 1 or v_{e_{2}}(W^{−1} · S) = m − 1. If v_{e_{1}}(W^{−1} · S) = m − 1, then Proposition 4.1 implies that y_{1} = y_{2} = 1, a contradiction. If v_{e_{2}}(W^{−1} · S) = m − 1, then Proposition 4.1 implies that x_{1} = x_{2} = 1, a contradiction.

Case 3: k_{2} = m − 1 and k_{3} = m − 1. Then k_{1} ∈ [m − 1, m + 1]. We distinguish two subcases.

Case 3.1: k_{1} ∈ {m, m + 1}. Then (g_{2} = e_{1}, g_{3} = e_{2}) is a basis of G. Thus

S = e_{1}^{m−1} · e_{2}^{m−1} · (ae_{1} + be_{2})^{k_{1}} · ∏_{ν=1}^{m+1−k_{1}}(x_{ν}e_{1} + y_{ν}e_{2})

where a, b and all x_{ν}, y_{ν} ∈ [0, m − 1]. If k_{1} = m + 1, then (m + 1)a + (m − 1) ≡ 0 mod m implies that a = 1, whence the assertion is proved. Suppose that k_{1} = m. Then ma + x_{1} + m − 1 ≡ 0 mod m and mb + y_{1} + m − 1 ≡ 0 mod m whence x_{1} = y_{1} = 1. If a = 1 or b = 1, then the assertion follows. Suppose that both a and b are distinct to 1. The sequence (ae_{1} + be_{2})^{−1} · S contains a short zero-sum subsequence W and clearly e_{1}^{m−1} ∤ W and e_{2}^{m−1} ∤ W. Thus W^{−1} · S is a minimal zero-sum sequence containing e_{1}, e_{2} and ae_{1} + be_{2}. Since a ≠ 1 and b ≠ 1, Proposition 4.1 implies that v_{e_{i}}(W^{−1} · S) < m − 1 for i ∈ [1, 2] whence v_{ae_{1}+be_{2}}(W^{−1} · S) = m − 1 and e_{1} − e_{2} ∈ ⟨ae_{1} + be_{2}⟩. This implies that b = m − a and gcd{b, m} = 1. If c ∈ [0, m − 1] with −ac ≡ 1 mod m and

(f_{1}, f_{2}) = (e_{1}, e_{2}) · (\begin{matrix}1 & a \\ 0 & b \end{matrix}) ,

then

S = f_{1}^{m−1} · (f_{1} + cf_{2})^{m−1} · f_{2}^{m} · (2f_{1} + cf_{2})

has form 2 (with basis (f_{2}, f_{1}) and s_{1} = 1).

Case 3.2: k_{1} = m − 1. Then {g_{1}, g_{2}, g_{3}} contains a basis {e_{1}, e_{2}} ⊂ G. Therefore we have

S = e_{1}^{m−1} · e_{2}^{m−1} · (ae_{1} + be_{2})^{m−1} · (x_{1}e_{1} + y_{1}e_{2}) · (x_{2}e_{1} + y_{2}e_{2})

with a, b, x_{1}, x_{2}, y_{1}, y_{2} ∈ [0, m−1] such that −1−a+x_{1}+x_{2} ≡ 0 mod m and −1−b+y_{1}+ y_{2} ≡ 0 mod m. The sequence S · (x_{1}e_{1} + y_{1}e_{2})^{−1} contains a short zero-sum subsequence W. Then W^{−1} · S is a minimal zero-sum sequence with contains the sequence

e_{1} · e_{2} · (ae_{1} + be_{2}) · (x_{1}e_{1} + y_{1}e_{2}).

If v_{e_{1}}(W^{−1} · S) = m − 1, then Proposition 4.1 implies that 1 = b = y_{1} whence y_{2} = 1 and we are done. If v_{e_{2}}(W^{−1}·S) = m−1, then Proposition 4.1 implies that 1 = a = x_{1} whence x_{2} = 1 and we are done. Suppose that v_{ae_{1}+be_{2}}(W^{−1} · S) = m − 1. Then Proposition 4.1 implies that e_{1} − e_{2} ∈ ⟨ae_{1} + be_{2}⟩ whence b = m − a and gcd{b, m} = 1. Furthermore, we have (1 − x_{1})e_{1} − y_{1}e_{2} = e_{1} − (x_{1} + y_{1}e_{2}) ∈ ⟨a(e_{1} − e_{2})⟩ which implies that y_{1} ≡ 1 − x_{1} mod m. We deal with the sequence S · (x_{2}e_{1} + y_{2}e_{2})^{−1} in a similar way. In the only remaining case we have b = m − a, y_{1} ≡ 1 − x_{1} mod m and y_{2} ≡ 1 − x_{2} mod m whence

S = e_{1}^{m−1} · e_{2}^{m−1} · (ae_{1} − ae_{2})^{m−1} · (x_{1}e_{1} + (1 − x_{1})e_{2}) · (x_{2}e_{1} + (1 − x_{2})e_{2})

If c ∈ [0, m − 1] such that −ac ≡ 1 mod m and

(f_{1}, f_{2}) = (e_{1}, e_{2}) · (\begin{matrix}1 & a \\ 0 & b \end{matrix}) ,

then

S = f_{1}^{m−1} · (f_{1} + cf_{2})^{m−1} · f_{2}^{m−1} · (f_{1} + (1 − x_{1})cf_{2}) · (f_{1} + (1 − x_{2})cf_{2})

has form 1 (with basis (f_{2}, f_{1}) and s = 1).

Lemma 7.3. Suppose t ≥ 4 and let T be a subsequence of S with length |T | = m + 1. Then there exists a zero-sum sequence W with T | W | S and a basis (e_{1}, e_{2}) of G such that

W = e_{1}^{sm−1} · ∏_{ν=1}^{(3−s)m}(a_{ν}e_{1} + e_{2})

where s ∈ [1, 2] and a_{1}, …, a_{(3−s)m} ∈ [0, m − 1] or

W = e_{1}^{m} · (be_{1} + e_{2})^{m−1} · e_{2}^{m−1} · (be_{1} + 2e_{2})

where b ∈ [0, m − 1] with gcd{b, m} = 1.

Proof. Since the sequence T^{−1}S has length |T^{−1}S| = (t − 1)m − 2, Lemma 3.1.2 implies that T^{−1}S has t − 3 disjoint short zero-sum subsequences S_{1}, ⋯, S_{t−3}, and by Lemma 3.12 all of them have length m. Clearly, W = (S_{1} · . . . · S_{t−3})^{−1} · S does not contain three disjoint non-empty zero-sum subsequences and has length |W | = 3m − 1. Now the assertion follows from Lemma 7.2.

Proof of Theorem 7.1. The case t = 3 was handled in Lemma 7.2 whence we may suppose that t ≥ 4. We distinguish three cases.

Case 1: |supp(S)| ≥ 5. This implies that m ≥ 3. If m = 3, then there is some g ∈ G such that {−g, g} ⊂ supp(S) whence S contains a nonempty zero-sum subsequence of length 2, a contradiction to Lemma 3.12.1. So it follows that m ≥ 4. Let T be a subsequence of S with |T | = m + 1 and |supp(T)| ≥ 5. By Lemma 7.3 there exist a zero-sum sequence W with T | W | S and a basis (e_{1}, e_{2}) of G such that

W = e_{1}^{s′m−1} · ∏_{ν=1}^{(3−s′)m}(a_{ν}e_{1} + e_{2})

with s′ ∈ [1, 2], a_{1}, …, a_{(3−s′)m} ∈ [0, m−1], a_{1}, a_{2}, a_{3} pairwise distinct and ∑_{ν=1}^{(3−s′)m} a_{ν} ≡ 1 mod m. This implies that

S = e_{1}^{s′m−1} · ∏_{ν=1}^{(3−s′)m}(a_{ν}e_{1} + e_{2}) · ∏_{ν=1}^{l}(x_{ν}e_{1} + y_{ν}e_{2})

with all x_{ν}, y_{ν} ∈ [0, m − 1].

Assume to the contrary that there exists some y_{ν} ∉ {0, 1}, say y_{1} ∉ {0, 1}. Then the sequence

U = e_{1} · (a_{1}e_{1} + e_{2}) · (a_{2}e_{1} + e_{2}) · (a_{3}e_{1} + e_{2}) · (x_{1}e_{1} + y_{1}e_{2})

has length |U| = 5 ≤ m + 1 and |supp(U)| ≥ 5. The sequence U^{−1} · S contains (t − 3) disjoint short zero-sum subsequences and let T denote their product. Then V = T^{−1} · S has length 3m − 1, contains the sequence U and cannot be written as a product of three proper zero-sum subsequences. Since |supp(V)| ≥ |supp(U)| ≥ 5, Lemma 7.2 implies that there exists a basis (f_{1}, f_{2}) of G such that

V = f_{1}^{m−1} · ∏_{ν=1}^{2m}(b_{ν}f_{1} + f_{2}).

If we can verify that e_{1} = f_{1}, then (x_{1}e_{1} +y_{1}e_{2})−(a_{1}e_{1} +e_{2}) ∈ ⟨f_{1}⟩ whence (y_{1} −1)e_{2} = 0 and thus y_{1} = 1 gives the required contradiction. Assume to the contrary that e_{1} ≠ f_{1}. Since a_{1}, a_{2}, a_{3} are pairwise distinct, we may suppose that f_{1} ∉ {a_{1}e_{1} + e_{2}, a_{2}e_{1} + e_{2}} whence (a_{1} − a_{2})e_{1} ∈ ⟨f_{1}⟩ and e_{1} − (a_{1}e_{1} + e_{2}) = (1 − a_{1})e_{1} + e_{2} ∈ ⟨f_{1}⟩. This implies that f_{1} = z_{1}e_{1} + z_{2}e_{2} with gcd{z_{2}, m} = 1 whence a_{1} = a_{2}, a contradiction.

Therefore y_{ν} ∈ {0, 1} for all ν ∈ [1, l]. If y_{ν} = 0, then (x_{ν}e_{1}) · e_{1}^{m−x_{ν}} is a short zero-sum subsequence of S with length m − x_{ν} + 1 whence x_{ν} = 1. Therefore (x_{ν}, y_{ν}) = (1, 0) or (x_{ν}, y_{ν}) = (x_{ν}, 1) which implies that

S = e_{1}^{s′′} · ∏_{ν=1}^{|S|−s′′}(a_{ν}e_{1} + e_{2})

for some s′′ ≥ m − 1. Since S is a zero-sum sequence, it follows that |S| − s′′ ≡ 0 mod m whence s′′ = sm − 1 for some s ∈ [1, t − 1].

Case 2: |supp(S)| = 3. By Lemma 7.3 there exists a zero-sum subsequence W of S and a basis (e_{1}, e_{2}) of G such that

W = e_{1}^{s′m−1} · ∏_{ν=1}^{(3−s′)m}(a_{ν}e_{1} + e_{2})

with s′ ∈ [1, 2] and a_{1}, …, a_{(3−s′)m} ∈ [0, m − 1]. Since 3 ≤ |supp(W)| ≤ |supp(S)| = 3, it follows that supp(S) = supp(W) whence

S = e_{1}^{s′′} · ∏_{ν=1}^{|S|−s′′}(a_{ν}e_{1} + e_{2})

where all a_{ν} ∈ [0, m − 1] and s′′ ≥ m − 1. Since S is a zero-sum sequence, it follows that |S| − s′′ ≡ 0 mod m whence s′′ = sm − 1 for some s ∈ [1, t − 1].

Case 3: |supp(S)| = 4. This implies that m ≥ 3. Let T be a subsequence of S with |T | = 4 and |supp(T)| = 4. By Lemma 7.3 there exists a zero-sum sequence W with T | W | S and a basis (e_{1}, e_{2}) of G such that either

W = e_{1}^{s′m−1} · ∏_{ν=1}^{(3−s′)m}(a_{ν}e_{1} + e_{2})

where s′ ∈ [1, 2] and a_{1}, …, a_{(3−s′)m} ∈ [0, m − 1] or

W = e_{1}^{m} · (be_{1} + e_{2})^{m−1} · e_{2}^{m−1} · (be_{1} + 2e_{2})

where b ∈ [0, m − 1] with gcd{b, m} = 1. Clearly we have supp(S) = supp(W). Hence in the first case the assertion follows as in Case 2, and it remains to consider the case where

S = e_{1}^{u} · (be_{1} + e_{2})^{v} · e_{2}^{w} · (be_{1} + 2e_{2})^{q}

with u ≥ m, v ≥ m − 1, w ≥ m − 1, q ≥ 1 and u + v + w + q = tm − 1.

Assume to the contrary that q ≥ 2. If b = m − 1 and

(f_{1}, f_{2}) = (e_{1}, e_{2}) · (\begin{matrix}1 & −1 \\ 0 & 1 \end{matrix}) ,

then

S = f_{1}^{u} · f_{2}^{v} · (f_{1} + f_{2})^{w} · (f_{1} + 2f_{2})^{q}

whence we may suppose that b ∈ [1, m − 2]. If b = 1 and

(f_{1}, f_{2}) = (e_{1}, e_{2}) · (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) ,

then

S = f_{2}^{u} · (f_{1} + f_{2})^{v} · f_{1}^{w} · (2f_{1} + f_{2})^{q}

whence we may suppose that b ∈ [2, m − 2]. Thus there is some b′ ∈ [2, m − 2] such that b′ · b ≡ −1 mod m whence

e_{1} · (be_{1} + e_{2})^{b′−2} · e_{2}^{m−b′−2} · (be_{1} + 2e_{2})^{2}

is a short zero-sum subsequence of S with length m − 1, a contradiction.

Therefore we infer that q = 1. We write u = u_{1}m+u_{0}, v = v_{1}m+v_{0} and w = w_{1}m+w_{0} with u_{0}, v_{0}, w_{0} ∈ [0, m − 1] and set

M = e_{1}^{u_{0}} · (be_{1} + e_{2})^{v_{0}} · e_{2}^{w_{0}} · (be_{1} + 2e_{2}).

Clearly, we have |M| = u_{0} + v_{0} + w_{0} + 1 ≤ 3m − 2 and |M| ≡ |S| ≡ −1 mod m which implies that |M| = 2m − 1 and M is a minimal zero-sum sequence. If u_{0} = 0 then v_{0} = w_{0} = m − 1 and we are done. Assume to the contrary that u_{0} ∈ [1, m − 1]. The sequence

N = e_{1}^{u_{0}} · (be_{1} + e_{2})^{v_{0}} · e_{2}^{w_{0}+1}

contains a zero-sum subsequence

1 ≠ N′ = e_{1}^{u′} · (be_{1} + e_{2})^{v′} · e_{2}^{w′}.

Since M is a minimal zero-sum subsequence, it follows that w′ = w_{0} + 1. If v′ ≥ 1, then

e_{1}^{u′} · (be_{1} + e_{2})^{v′−1} · e_{2}^{w′−1} · (be_{1} + 2e_{2})

is a proper zero-sum subsequence of M, a contradiction. Thus v′ = 0 whence σ(N′) = 0 = u′e_{1} + (w_{0} + 1)e_{2}. This implies that u′ = 0 and w_{0} = m − 1. Since M is a zero-sum sequence, it follows that v_{0} = m − 1 and |M| = u_{0} + v_{0} + w_{0} + 1 = 2m − 1 implies that u_{0} = 0, a contradiction.

8. If n has Property B, then 2n has Property B

It is the aim of this section to prove the following theorem.

Theorem 8.1. Let n ∈ ℕ with n ≥ 6. If n satisfies Property B, then 2n satisfies Property B.

We start with two lemmata, which rest on Lemmata 3.11 to 3.14. Let S ∈ F(C_{mn} ⊕ C_{mn}), where m, n ∈ ℕ_{≥2}, be a minimal zero-sum sequence with length |S| = 2mn − 1. A product decomposition S = ∏_{ν=0}^{2n−2} S_{ν} having the properties described in Lemma 3.14.2 will be called a canonical product decomposition of S. If not stated otherwise, we always numerate the sequences in such a way that |S_{0}| = 2m − 1 and |S_{1}| = … = |S_{2n−2}| = m.

Lemma 8.2. Let G = C_{mn} ⊕ C_{mn} with m, n ∈ ℕ_{≥2}, φ : G → G the multiplication by n and S ∈ F(G) a minimal zero-sum sequence with length |S| = 2mn − 1. Suppose that n has Property B and let S = ∏_{ν=0}^{2n−2} S_{ν} be a canonical product decomposition of S. Then ∏_{ν=0}^{2n−2} σ(S_{ν}) is a minimal zero-sum sequence in ker(φ) and there exists a basis (e_{1}, e_{2}) of G such that

∏_{ν=0}^{2n−2} σ(S_{ν}) = (me_{1})^{n−1} · ∏_{i=1}^{r}(a_{i}me_{1} + me_{2})^{t_{i}}

where r ∈ [1, n], t_{1} ≥ … ≥ t_{r} ≥ 1, ∑_{i=1}^{r} t_{i} = n, a_{1}, …, a_{r} ∈ [0, n − 1] and ∑_{i=1}^{r} t_{i}a_{i} ≡ 1 mod n.

Proof. Clearly, ∏_{ν=0}^{2n−2} σ(S_{ν}) is a minimal zero-sum sequence in ker(φ) ≅ C_{n} ⊕ C_{n}. By Theorem 4.3 there exists a basis (e′_{1}, e′_{2}) of ker(φ) such that

∏_{ν=0}^{2n−2} σ(S_{ν}) = e′_{1}^{n−1} · ∏_{i=1}^{r}(a_{i}e′_{1} + e′_{2})^{t_{i}}

where r ∈ [1, n], t_{1} ≥ … ≥ t_{r} ≥ 1, ∑_{i=1}^{r} t_{i} = n, a_{1}, …, a_{r} ∈ [0, n − 1] and ∑_{i=1}^{r} t_{i}a_{i} ≡ 1 mod n. Thus the assertion follows from Lemma 3.13.

Lemma 8.3. Let G = C_{mn} ⊕ C_{mn} with m, n ∈ ℕ_{≥2}, φ : G → G the multiplication by n and S ∈ F(G) a minimal zero-sum sequence with length |S| = 2mn − 1. Suppose that n has Property B and let S = ∏_{ν=0}^{2n−2} S_{ν} be a canonical product decomposition such that in all decompositions

∏_{ν=0}^{2n−2} σ(S_{ν}) = (me_{1})^{n−1} · ∏_{i=1}^{r}(a_{i}me_{1} + me_{2})^{t_{i}},

derived in Lemma 8.2, t_{1} is minimal possible. Then we have

1. If (t_{1} = n−1 and n ≥ m+3) or (t_{1} ≤ n−m−1), then for every subsequence T of S with σ(T) ∈ ker(φ) and |T | = m, we have either σ(T) = me_{1} or σ(T) = ame_{1} +me_{2} for some a ∈ [0, n − 1].

2. If (t_{1} ∈ [n − m, n − 2] and n > 3m), then there exists some λ ∈ [0, 2n − 2] such that for every subsequence T of S_{λ}^{−1}S with σ(T) ∈ ker(φ) and |T | = m, we have either σ(T) = me_{1} or σ(T) = ame_{1} + me_{2} for some a ∈ [0, n − 1].

3. If n ≥ 6 and m = 2, then for every subsequence T of S with σ(T) ∈ ker(φ) and |T | = 2, we have either σ(T) = 2e_{1} or σ(T) = 2ae_{1} + 2e_{2} for some a ∈ [0, n − 1].

Proof. Let T be a subsequence of S ( resp. of S_{λ}^{−1}S for some λ ∈ [0, 2n − 2]) with σ(T) ∈ ker(φ) and |T | = m. Without restriction we may suppose that T ∉ {S_{1}, …, S_{2n−2}}. Let Γ_{1} ⊂ [0, 2n − 2] (resp. Γ_{1} ⊂ [0, 2n − 2] \ {λ}) be a minimal subset such that T divides ∏_{i∈Γ_{1}} S_{i}. We set Γ_{2} = [0, 2n − 2] \ Γ_{1}, W = T^{−1} ∏_{i∈Γ_{1}} S_{i} and l = |Γ_{1}|. By the minimality of Γ_{1} we obtain that l = |Γ_{1}| ≤ |T | = m. Furthermore, φ(W) is a zero-sum sequence with length

|W | = ∑_{i∈Γ_{1}} |S_{i}| − |T | ≥ |Γ_{1}| · m − m = (l − 1)m.

By Lemma 3.11 W = W_{1} · . . . · W_{l−3} · W′ where φ(W_{1}), …, φ(W_{l−3}) are short zero-sum sequences (in case l ≤ 3 we have W′ = W). Since S = ∏_{i∈Γ_{2}} S_{i} · T · W and since by Lemma 3.12.1 all short zero-sum sequences of φ(S) have length m, φ(W_{1}), …, φ(W_{l−3}) have length m.

Now we distinguish two cases. Firstly, we suppose that 0 ∉ Γ_{1}. Then 0 ∈ Γ_{2}, |W | = (l − 1)m and φ(W′) is a zero-sum sequence of length 2m. Hence W′ = W_{l−2}W_{l−1} where φ(W_{l−2}) and φ(W_{l−1}) are zero-sum sequences with length m. Secondly, we suppose that 0 ∈ Γ_{1}. Then |W | = lm − 1 whence |W′| = |W | − (l − 3)m = 3m − 1. Thus by Lemma 3.1.2 W′ = W_{l−2}W_{l−1} where φ(W_{l−2}) is a short zero-sum sequence of length m. Since φ(S) is not a product of 2n zero-sum subsequences, it follows that φ(W_{l}) is a minimal zero-sum sequence of length 2m − 1.

Therefore in both cases

S = ∏_{i∈Γ_{2}} S_{i} · T · ∏_{i=1}^{l−1} W_{i}

is a canonical product decomposition and

\overline{S} = (∏_{i∈Γ_{2}} σ(S_{i})) σ(T)σ(W_{1}) · . . . · σ(W_{l−1})

is a minimal zero-sum sequence in ker(φ).

1. (i) Suppose that t_{1} = n − 1 and n ≥ m + 3. By the minimality of t_{1} there are two distinct elements α, β ∈ ker(φ) each occuring exactly (n − 1)-times in the sequence \overline{S}. Assume to the contrary that {me_{1}, a_{1}me_{1} + me_{2}} ≠ {α, β}. If γ ∈ {me_{1}, a_{1}me_{1} + me_{2}} \ {α, β} then we infer that

2n − 1 = |\overline{S}| ≥ v_{γ}(\overline{S}) + v_{α}(\overline{S}) + v_{β}(\overline{S})
≥ (n − 1 − |Γ_{1}|) + (n − 1) + (n − 1) ≥ (n − 1 − m) + (2n − 2)
> 2n − 1,

a contradiction. Therefore, {me_{1}, a_{1}me_{1} + me_{2}} = {α, β} and the assertion follows.

1. (ii) Suppose that t_{1} ≤ n − m − 1. Then every element distinct to me_{1} occurs at most

t_{1} + (l − 1) ≤ n − m − 1 + (l − 1) ≤ n − 2

times in \overline{S}. Since n satisfies Property B, there is some element α occuring (n−1)-times in \overline{S} whence α = me_{1}. Thus either σ(T) = me_{1} or, by Proposition 4.1, σ(T) = ame_{1} + me_{2} for some a ∈ [0, n − 1].

2. Suppose that t_{1} ∈ [n − m, n − 2] and n > 3m. First we discuss how to choose a suitable λ ∈ [0, 2n − 2]. Since ∑_{j=1}^{r} t_{j}a_{j} ≡ 1 mod n, ∑_{j=1}^{r} t_{j} = n and t_{1} ≤ n − 2, it follows that there exists some j ∈ [2, r] such that a_{j} ≢ a_{1} + 1 mod n, say j = r. Choose λ ∈ [0, 2n − 2] such that σ(S_{λ}) = a_{r}me_{1} + me_{2}.

Let T be a subsequence of S_{λ}^{−1} · S with σ(T) ∈ ker(φ) and |T | = m. Since n satisfies Property B and by the minimality of t_{1}, there exist two elements α, β such that α occurs (n − 1)-times and β occurs at least t_{1} ≥ n − m times in the sequence \overline{S}. Assume to the contrary, that {α, β} ≠ {me_{1}, a_{1}me_{1} + me_{2}}. Then we infer that

2n − 1 = |\overline{S}| ≥ v_{α}(\overline{S}) + v_{β}(\overline{S}) + min\{v_{me_{1}}(\overline{S}), v_{a_{1}me_{1}+me_{2}}(\overline{S})\}
≥ (n − 1) + (n − m) + min{n − 1 − |Γ_{1}|, t_{1} − |Γ_{1}|}
≥ (n − 1) + (n − m) + (n − m − |Γ_{1}|)
≥ (n − 1) + (n − m) + (n − 2m)
> 2n − 1

a contradiction, since n > 3m. Thus we obtain that {α, β} = {me_{1}, a_{1}me_{1} + me_{2}}. Assume to the contrary that α = a_{1}me_{1} + me_{2} and β = me_{1}. Since

(α, β) = (me_{1}, me_{2}) · (\begin{matrix}a_{1} & 1 \\ 1 & 0 \end{matrix})

and gcd{a_{1}·0−1, n} = 1, it follows that {α, β} is a basis of ker(φ) ≅ C_{n}⊕C_{n}. Since σ(S_{λ}) occurs in \overline{S}, Proposition 4.1.2.a implies that there exist a, b ∈ [0, n−1] with gcd\{b, n\} = 1 and σ(S_{λ}) = aα + bβ. Since β occurs in \overline{S}, it follows that b = 1 and we obtain that

a_{r}me_{1} + me_{2} = σ(S_{λ}) = aα + bβ = a(a_{1}me_{1} + me_{2}) + me_{1} = (aa_{1} + 1)me_{1} + ame_{2}

whence a ≡ 1 mod n and a_{r} ≡ a_{1} + 1 mod n, a contradiction. Thus α = me_{1} whence Proposition 4.1 implies that σ(T) has the required form.

3. Suppose n ≥ 6 and m = 2. If t_{1} = n − 1 or t_{1} ≤ n − 3, the assertion follows from 1. Suppose that t_{1} = n − 2. Then

∏_{ν=0}^{2n−2} σ(S_{ν}) = (2e_{1})^{n−1} · (2a_{1}e_{1} + 2e_{2})^{n−2} · (2a_{2}e_{1} + 2e_{2}) · (2a_{3}e_{1} + 2e_{2})

where a_{1}, a_{2}, a_{3} ∈ [0, n − 1], a_{2} ≠ a_{1} ≠ a_{3}, and

\overline{S} = (∏_{i∈Γ_{2}} σ(S_{i})) σ(T)σ(W_{1}) = b_{0}^{n−1}b_{1}^{t′_{1}} · B

where b_{0}, b_{1} ∈ ker(φ), B ∈ F(ker(φ)) with |B| ≤ 2 and t′_{1} ≥ n − 2. We set Γ_{1} = {λ, μ} whence T W_{1} = S_{λ}S_{μ}.

If 2e_{1} ∉ \{σ(S_{λ}), σ(S_{μ})\}, then (2e_{1})^{n−1} | \overline{S} and the assertion follows by Proposition 4.1. If 2e_{1} = σ(S_{λ}) = σ(S_{ν}) and σ(T) ≠ 2e_{1}, then σ(W_{1}) ≠ 2e_{1}, v_{2e_{1}}(\overline{S}) = n − 3 ≥ 3 whence b_{1} = 2e_{1}, a contradiction to t′_{1} ≥ n − 2. If, say, σ(S_{λ}) = 2e_{1}, σ(S_{μ}) = 2a_{i}e_{1} + 2e_{2} for some i ∈ [1, 3] and σ(T) ∉ {2e_{1}, 2ae_{1} + 2e_{2}} for some a ∈ [0, n − 1], then σ(W_{1}) ∉ \{2e_{1}, 2ae_{1}+2e_{2}\} for any a ∈ [0, n−1] whence v_{2e_{1}}(\overline{S}) = n−2, n−3 ≤ v_{2a_{1}e_{1}+2e_{2}}(\overline{S}) ≤ n−2 , a contradiction to max\{v_{g}(\overline{S}) | g ∈ ker(φ)\} = n − 1.

Proof of Theorem 8.1. Let G = C_{2n} ⊕ C_{2n} with n ≥ 6 and suppose that n satisfies Property B. Let S ∈ F(G) be a minimal zero-sum sequence with length |S| = 4n − 1. We have to show that S contains some element with multiplicity 2n − 1.

Let φ : G → G denote the multiplication by n. By Lemmata 3.14 and 8.2 (with m = 2) S has a canonical product decomposition S = ∏_{ν=0}^{2n−2} S_{ν} where |S_{0}| = 3, |S_{1}| = … = |S_{2n−2}| = 2, and there exists a basis (f_{1}, f_{2}) of G such that

∏_{ν=0}^{2n−2} σ(S_{ν}) = (2f_{1})^{n−1} · ∏_{i=1}^{r}(a_{i}2f_{1} + 2f_{2})^{t_{i}} ∈ F( ker(φ) )

where r ∈ [1, n], t_{1} ≥ … ≥ t_{r} ≥ 1, ∑_{i=1}^{r} t_{i} = n, a_{1}, …, a_{r} ∈ [0, n − 1] and ∑_{i=1}^{r} t_{i}a_{i} ≡ 1 mod n. Suppose that t_{1} is minimal possible under all decompositions of this type.

Let (e_{1}, e_{2}) be any basis of G such that 2e_{1} = 2f_{1} and 2e_{2} ∈ 2f_{2} +⟨2f_{1}⟩. A basis having these properties will be called suitable. For i ∈ [1, 2] we denote by p_{i} : G = ⟨e_{1}⟩ ⊕ ⟨e_{2}⟩ → ⟨e_{i}⟩ the canonical projection, and we set nG = {0, α, β, γ} ≅ C_{2} ⊕ C_{2}.

By Lemma 3.14.1 we have 0 ∉ supp(φ(S)) whence S has the form S = S_{α}S_{β}S_{γ} where φ(S_{δ}) = δ^{|S_{δ}|} for every δ ∈ {α, β, γ}. Clearly, if ν ∈ [1, 2n−2], then S_{ν} divides S_{δ} for some δ ∈ {α, β, γ}, φ(S_{0}) = αβγ and |S_{δ}| ≡ 1 mod 2 for every δ ∈ {α, β, γ}. Let k, l, m ∈ ℕ_{0} such that |S_{α}| = 2k + 1, |S_{β}| = 2l + 1 and |S_{γ}| = 2m + 1.

Lemma 8.3.3 implies that for every subsequence T of S with σ(T) ∈ ker(φ) and |T | = 2 we have

(1) either σ(T) = 2e_{1} or σ(T) = 2ae_{1} + 2e_{2} for some a ∈ [0, n − 1].

Let δ ∈ {α, β, γ} and S_{δ} = ∏_{i=1}^{|S_{δ}|}(x_{i}e_{1} + u_{i}e_{2}) with all x_{i}, u_{i} ∈ [0, 2n − 1]. We assert that

|{u_{i} | i ∈ [1, |S_{δ}|]}| = 2. (2)

Assume to the contrary, that (2) does not hold. Then we may suppose without restriction that |{u_{1}, u_{2}, u_{3}}| = 3. Then u_{1}+u_{2}, u_{1}+u_{3} and u_{2}+u_{3} are pairwise distinct. However, (1) implies that u_{1} +u_{2} +2nℤ, u_{1} +u_{3} +2nℤ, u_{2} +u_{3} +2nℤ ∈ {2nℤ, 2+2nℤ}, a contradiction.

Therefore we obtain that

S_{α} = ∏_{i=1}^{k_{1}}(x_{i}e_{1} + ue_{2}) ∏_{i=1}^{k_{2}}(x_{k_{1}+i}e_{1} + u′e_{2}) where k_{1} ≥ k_{2} ≥ 0, k_{1} + k_{2} = 2k + 1,

S_{β} = ∏_{i=1}^{l_{1}}(y_{i}e_{1} + ve_{2}) ∏_{i=1}^{l_{2}}(y_{l_{1}+i}e_{1} + v′e_{2}) where l_{1} ≥ l_{2} ≥ 0, l_{1} + l_{2} = 2l + 1,
S_{γ} = ∏_{i=1}^{m_{1}}(z_{i}e_{1} + we_{2}) ∏_{i=1}^{m_{2}}(z_{m_{1}+i}e_{1} + w′e_{2}) where m_{1} ≥ m_{2} ≥ 0, m_{1} + m_{2} = 2m + 1,

and all x_{i}, y_{i}, z_{i}, u, u′, v, v′, w, w′ ∈ [0, 2n − 1]. Obviously, k_{1}, l_{1} and m_{1} are non-zero.

We assert that

k_{2}, l_{2}, m_{2} ∈ {0, 1}. (3)

Assume to the contrary that k_{2} ≥ 2. Then k_{1} ≥ k_{2} ≥ 2 and (1) implies that 2u + 2nℤ, 2u′ + 2nℤ ∈ {2nℤ, 2 + 2nℤ} whence u, u′ ∈ {0, 1, n, n + 1}. Since u ≠ u′, it follows that u + u′ ∈ {1, n, n + 1, n + 2, 2n + 1} whence u + u′ + 2nℤ ∉ {2nℤ, 2 + 2nℤ}, a contradiction to (1). Similarly, we argue for l_{2} and m_{2}.

Assume to the contrary, that at least two elements of {k_{1}, l_{1}, m_{1}} are equal to 1, say l_{1} = m_{1} = 1. This implies that l_{2} = m_{2} = 0, k_{1} = 4n − 1 − (k_{2} + l_{1} + l_{2} + m_{1} + m_{2}) ≥ 4n−4 ≥ 2, and by (1) we have 2u+2nℤ ∈ {2nℤ, 2+2nℤ}. The number of ν ∈ [0, 2n−2] for which p_{2}(S_{ν}) ≠ (ue_{2})^{2} is at most two. If 2u ≡ 0 mod 2n, then the multiplicity of 2e_{1} in the sequence ∏_{ν=0}^{2n−2} σ(S_{ν}) is at least (2n − 1) − 2 > n − 1, a contradiction. If 2u ≡ 2 mod 2n, then the multiplicity of 2e_{1} in the sequence ∏_{ν=0}^{2n−2} σ(S_{ν}) is at most two, a contradiction.

Next we assert that

2nℤ ∈ {2u + 2nℤ, 2v + 2nℤ, 2w + 2nℤ} ≠ {2nℤ}. (4)

Since for every ν ∈ [1, 2n − 2] S_{ν} divides S_{δ} for some δ ∈ {α, β, γ} and because of (3) , the number of ν ∈ [0, 2n − 2] for which p_{2}(S_{ν}) ∉ {(ue_{2})^{2}, (ve_{2})^{2}, (we_{2})^{2}} is at most four. Since the number of ν ∈ [0, 2n − 2] for which σ(S_{ν}) = 2e_{1} equals to n − 1 ≥ 5, it follows that 2nℤ ∈ {2u + 2nℤ, 2v + 2nℤ, 2w + 2nℤ}.

If 2u ≡ 2v ≡ 2w ≡ 0 mod 2n, then the number of ν ∈ [0, 2n−2] for which σ(p_{2}(S_{ν})) = 2e_{2}, is at most four, whence n ≤ 4 a contradiction.

Thus (4) holds and (1) implies the following facts: (k_{1} ≥ 2 ⇒ 2u + 2nℤ ∈ {2nℤ, 2 + 2nℤ}), (l_{1} ≥ 2 ⇒ 2v + 2nℤ ∈ {2nℤ, 2 + 2nℤ}) and (m_{1} ≥ 2 ⇒ 2w + 2nℤ ∈ {2nℤ, 2 + 2nℤ}). Assume to the contrary, that k_{1} ≥ 2, l_{1} ≥ 2, m_{1} ≥ 2 and that exactly two of the values 2u, 2v, 2w are congruent to zero modulo 2n, say 2u ≡ 2v ≡ 0 mod 2n and 2w ≡ 2 mod 2n. Since, by (1),

k_{2} = 0 or (k_{2} = 1 and u + u′ ≡ 2 mod 2n)
l_{2} = 0 or (l_{2} = 1 and v + v′ ≡ 2 mod 2n)
m_{2} = 0 or (m_{2} = 1 and w + w′ ≡ 0 mod 2n)

it follows that

k_{2} = 0 or (2u′ ≡ 4 mod 2n)
l_{2} = 0 or (2v′ ≡ 4 mod 2n)
m_{2} = 0 or (2w′ ≡ −2 mod 2n)

whence 2\overline{u}+2\overline{v} +2\overline{w} +2nℤ ∈ \{0, 4\}+\{0, 4\}+\{2, −2\}+2nℤ = \{2, 6, 10, −2\}+2nℤ where \overline{u} ∈ \{u, u′\}, \overline{v} ∈ \{v, v′\} and \overline{w} ∈ \{w, w′\}. Therefore 2σ(p_{2}(S_{0})) ∈ \{2, 6, 10, −2\}e_{2}. On the other hand we have σ(S_{0}) ∈ {2f_{1}, a_{i}2f_{1} + 2f_{2} | i ∈ [1, r]} whence σ(p_{2}(S_{0})) ∈ {0, 2e_{2}} and thus {0, 4} + 2nℤ ∩ {2, 6, 10, −2} + 2nℤ ≠ ∅, a contradiction to 2n ≥ 12.

Assume to the contrary that 1 ∈ {k_{1}, l_{1}, m_{1}}, say m_{1} = 1, and 2u ≡ 2v mod 2n. If 2u ≡ 2v ≡ 2 mod 2n, then the number of ν ∈ [0, 2n − 2] with σ(S_{ν}) = 2e_{1} is at most three, a contradiction. If 2u ≡ 2v ≡ 0 mod 2n, then the number of ν ∈ [0, 2n − 2] for which σ(p_{2}(S_{ν})) = 2e_{2} is at most four, a contradiction.

All these considerations show that we may suppose without restriction that k_{1} ≥ 2, l_{1} ≥ 2, 2u ≡ 0 mod 2n, 2v ≡ 2 mod 2n and (either 2w ≡ 2 mod 2n or m_{1} = 1).

Our next aim is to choose a special suitable basis (ẽ_{1}, ẽ_{2}). The number of ν ∈ [0, 2n−2] with p_{2}(S_{ν}) ≠ (ue_{2})^{2} but σ(p_{2}(S_{ν})) = 0 is at most three whence k_{1} ≥ 2(n − 1 − 3) = 2n − 8 ≥ 4. Since 2u ≡ 0 mod 2n, (1) implies that x_{i} + x_{j} ≡ 2 mod 2n for each two distinct i, j ∈ [1, k_{1}]. This implies that x_{1} = … = x_{k_{1}} = x ∈ [0, 2n − 1]. Lemma 3.8.1 implies that 2n = ord(xe_{1} + ue_{2}) whence gcd{x, u, 2n} = 1. Since (2xe_{1}) occurs in the sequence ∏_{i=0}^{2n−2} σ(S_{ν}), it follows that 2xe_{1} = 2e_{1} whence x ∈ {1, n + 1}.

If u = 0, then

(ẽ_{1}, ẽ_{2}) = (xe_{1}, e_{2}) = (e_{1}, e_{2}) · (\begin{matrix}x & 0 \\ 0 & 1 \end{matrix})

is a basis of G with 2ẽ_{i} = 2e_{i} for i ∈ [1, 2]. Suppose u = n. Then gcd{x, n} = 1 whence there are x′, n′ ∈ ℤ such that xx′ − nn′ = 1 and

(ẽ_{1}, ẽ_{2}) = (xe_{1} + ne_{2}, n′e_{1} + x′e_{2}) = (e_{1}, e_{2}) · (\begin{matrix}x & n′ \\ n & x′ \end{matrix})

is a basis of G with 2ẽ_{1} = 2e_{1} = 2f_{1} and 2ẽ_{2} ∈ 2e_{2} + ⟨2e_{1}⟩ ∈ 2f_{2} + ⟨2f_{1}⟩.

Thus (ẽ_{1}, ẽ_{2}) is a suitable basis and we may write all elements of S = S_{α}S_{β}S_{γ} with this new basis. We get new coordinates x̃_{i}, ỹ_{i}, z̃_{i}, ũ = 0, ũ′, ṽ, ṽ′, w̃ and w̃′. For simplicity of notation we omit all ˜∗, write (e_{1}, e_{2}) instead of (ẽ_{1}, ẽ_{2}) and so on. In new notation we obtain that

S = e_{1}^{k_{1}} · ∏_{i=1}^{k_{2}}(x_{k_{1}+i}e_{1} + u′e_{2}) · S_{β} · S_{γ}. (5)

We distinguish the cases m_{1} ≥ 2 and m_{1} = 1.

Case 1: k_{1} ≥ 2, l_{1} ≥ 2, m_{1} ≥ 2. Without restriction we suppose that l_{2} ≥ m_{2}. Recall that u = 0 and 2v ≡ 2w ≡ 2 mod 2n whence v, w ∈ {1, n + 1}.

We assert that

v = w. (6)

Assume to the contrary that v ≠ w. Since 2v ≡ 2w ≡ 2 mod 2n, it follows that {v, w} = {1, n + 1}. Since u′ ≠ u = 0, v ≠ v′, w ≠ w′, (1) implies that (k_{2} = 0 or u′ = 2), (l_{2} = 0 or v + v′ ≡ 0 mod 2n) and (m_{2} = 0 or w + w′ ≡ 0 mod 2n). Thus if \overline{u} ∈ \{u, u′\}, \overline{v} ∈ \{v, v′\} and \overline{w} ∈ \{w, w′\}, then \overline{u} + \overline{v} + \overline{w} ∈ \{0, 2\} + \overline{v} + \overline{w} ∈ {0, 2}+{v +w, v +w′, v′ +w, v′ +w′} = {0, 2}+{n+2, n, n−2} = {n−2, n, n+2, n+4}. Thus σ(p_{2}(S_{0})) ∈ {n−2, n, n+2, n+4}e_{2}. On the other hand we have σ(p_{2}(S_{0})) ∈ {0, 2e_{2}} whence {0, 2} + 2nℤ ∩ {n − 2, n, n + 2, n + 4} + 2nℤ ≠ ∅, a contradiction to n ≥ 6.

We distinguish six cases.

Case 1.1: k_{2} = l_{2} = m_{2} = 0. Then l_{1} + m_{1} is even.

We have

S = e_{1}^{k_{1}} · ∏_{i=1}^{l_{1}}(y_{i}e_{1} + ve_{2}) · ∏_{i=1}^{m_{1}}(z_{i}e_{1} + ve_{2}).

Since S is a zero-sum sequence, it follows that (l_{1}+m_{1})v ≡ 0 mod 2n whence l_{1}+m_{1} ≡ 0 mod 2n and l_{1} + m_{1} = 2n. Thus k_{1} = 2n − 1 and the assertion is proved.

Case 1.2: k_{2} = 0, l_{2} = m_{2} = 1. Then l_{1} + m_{1} is even.

Since v ≠ v′, w ≠ w′ and 2v ≡ 2w ≡ 2 mod 2n, (1) implies that v + v′ ≡ w + w′ ≡ 0 mod 2n. Since v = w, we infer that either

(v = w = 1 and v′ = w′ = 2n − 1) or (v = w = n + 1 and v′ = w′ = n − 1).

Since S is a zero-sum sequence, we have l_{1}v + m_{1}w + v′ + w′ ≡ 0 mod 2n. Therefore we obtain that (l_{1} + m_{1})v − 2 ≡ 0 mod 2n, l_{1} + m_{1} ≡ 2 mod 2n and l_{1} + m_{1} ∈ {2, 2n + 2}.

Since k_{1} = 4n − 1 − (l_{1} + m_{1} + l_{2} + m_{2}), we have l_{1} + m_{1} = 2n + 2 and k_{1} = 2n − 5. Therefore we obtain that either

S = e_{1}^{2n−5} ∏_{i=1}^{2n+2}(y_{i}e_{1} + e_{2}) · (d_{1}e_{1} − e_{2}) · (d_{2}e_{1} − e_{2})

or

S = e_{1}^{2n−5} ∏_{i=1}^{2n+2}(y_{i}e_{1} + (n + 1)e_{2}) · (d_{1}e_{1} + (n − 1)e_{2}) · (d_{2}e_{1} + (n − 1)e_{2})

where in both cases d_{1} = y_{l_{1}+1} and d_{2} = z_{m_{1}+1} ∈ [0, 2n − 1] whence d_{1} ≠ d_{2}.

We consider the first case. If T is a non-empty proper subsequence of ∏_{i=1}^{2n+2}(y_{i}e_{1} + e_{2}) · (d_{1}e_{1} − e_{2}) · (d_{2}e_{1} − e_{2}) such that σ(p_{2}(T)) = 0, then σ(p_{1}(T)) ∈ {1, 2, 3, 4}e_{1}. This implies that for every i ∈ {1, 2} and every j ∈ [1, 2n + 2] we have d_{i} + y_{j} + 2nℤ ∈ {1, 2, 3, 4} + 2nℤ. Since d_{1} + d_{2} + y_{j′} + y_{j} + 2nℤ ∈ {1, 2, 3, 4} + 2nℤ and n ≥ 6, it follows that d_{i} + y_{j} + 2nℤ ∈ {1, 2, 3} + 2nℤ. If d_{1} + y_{j} ≡ 3 mod 2n, then d_{2} + y_{i} ≡ 1 mod 2n for all i ∈ [1, 2n + 2] \ {j} whence y_{1} = … = y_{j−1} = y_{j+1} = … = y_{2n+2} and (y_{1}e_{1} + e_{2})^{2n} is a zero-sum subsequence of S, a contradiction. The same

argument works for d_{2} + y_{j}.

Thus d_{i} + y_{j} + 2nℤ ∈ {1, 2} + 2nℤ for all i ∈ [1, 2] and all j ∈ [1, 2n + 2]. This implies that |{y_{1}, …, y_{2n+2}}| ≤ 2, say ∏_{i=1}^{2n+2} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1} ≥ h_{2} ≥ 0. Since S is a minimal zero-sum sequence, it follows that h_{2} ≥ 3. After a suitable renumeration we may suppose that d_{1} +y_{1} ≡ 1 mod 2n. Then it follows that d_{2} +y_{1} ≡ 2 mod 2n, d_{1} +y_{2} ≡ 2 mod 2n and d_{2} + y_{2} ≡ 1 mod 2n whence 2y_{1} ≡ 3 − d_{1} − d_{2} ≡ 2y_{2} mod 2n. For i ∈ [1, 2] we choose even h′_{i} ∈ [0, h_{i}] with h′_{1} + h′_{2} = 2n. Then

h′_{1}y_{1} + h′_{2}y_{2} ≡ h′_{1}y_{1} + \frac{h′_{2}}{2} (2y_{1}) ≡ y_{1}(h′_{1} + h′_{2}) ≡ 0 mod 2n

whence (y_{1}e_{1} + e_{2})^{h′_{1}} · (y_{2}e_{1} + e_{2})^{h′_{2}} is a zero-sum subsequence of S, a contradiction.

Arguing in a similar way in the second case we obtain again a contradiction.

Case 1.3: k_{2} = 0, l_{2} = 1, m_{2} = 0. Then l_{1} + m_{1} is odd.

As in Case 1.2 we have v ≠ v′, 2v ≡ 2 mod 2n and v + v′ ≡ 0 mod 2n. Since S is a zero-sum sequence, we have 0 ≡ l_{1}v + m_{1}w + v′ ≡ (l_{1} + m_{1})v − v mod 2n whence l_{1} + m_{1} ≡ 1 mod 2n and thus l_{1} + m_{1} = 2n + 1. Therefore we obtain that either

S = e_{1}^{2n−3} ∏_{i=1}^{2n+1}(y_{i}e_{1} + e_{2}) · (de_{1} − e_{2})

or

S = e_{1}^{2n−3} ∏_{i=1}^{2n+1}(y_{i}e_{1} + (n + 1)e_{2}) · (de_{1} + (n − 1)e_{2})

where d = y_{l_{1}+1} ∈ [0, 2n − 1].

We consider the first case. If T is a non-empty proper subsequence of ∏_{i=1}^{2n+1}(y_{i}e_{1} + e_{2}) · (de_{1} − e_{2}) such that σ(p_{2}(T)) = 0, then σ(p_{1}(T)) ∈ {1, 2}e_{1}. Thus d + y_{i} + 2nℤ ∈ {1, 2} + 2nℤ for every i ∈ [1, 2n + 1]. This implies that |{y_{1}, …, y_{2n+1}}| = 2, and we set ∏_{i=1}^{2n+1} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1} + h_{2} = 2n + 1. Since S is a minimal zero-sum sequence, it follows that h_{1}, h_{2} ∈ [2, 2n − 1]. After a suitable renumeration we may suppose that d + y_{1} ≡ 1 mod 2n, and clearly we have h_{1}y_{1} + h_{2}y_{2} + d ≡ 3 mod 2n. Therefore, d + y_{2} ≡ 2 mod 2n, h_{1}y_{1} + h_{2}y_{2} − y_{1} ≡ 2 mod 2n, h_{1}y_{1} + h_{2}y_{2} − y_{2} ≡ 1 mod 2n, y_{2} − y_{1} ≡ 1 mod 2n, h_{1}y_{1} + (2n − h_{1})y_{2} ≡ 1 mod 2n, h_{1}(y_{1} − y_{2}) ≡ 1 mod 2n whence h_{1} ≡ −1 mod 2n. This implies that h_{1} = 2n − 1 and the assertion is proved.

Arguing in a similar way in the second case we obtain again the assertion.

Case 1.4: k_{2} = l_{2} = m_{2} = 1. Then l_{1} + m_{1} is even.

Since 0 = u ≠ u′ and (u + u′ ≡ 0 or u + u′ ≡ 2 mod 2n), it follows that u′ = 2. As in Case 1.2 we infer that either

(v = w = 1 and v′ = w′ = 2n − 1) or (v = w = n + 1 and v′ = w′ = n − 1).

Since S is a zero-sum sequence, we have l_{1}v+m_{1}w+u′+v′+w′ ≡ 0 mod 2n, (l_{1}+m_{1})v ≡ 0 mod 2n and l_{1} + m_{1} = 2n. Thus k_{1} = 4n − 1 − (l_{1} + m_{1} + k_{2} + l_{2} + m_{2}) = 2n − 4 and

S = e_{1}^{2n−4} ∏_{i=1}^{2n}(y_{i}e_{1} + ve_{2}) · (d_{1}e_{1} − ve_{2}) · (d_{2}e_{1} − ve_{2}) · (d_{3}e_{1} + 2e_{2})

where d_{1}, d_{2}, d_{3} ∈ [0, 2n−1] and d_{1} ≠ d_{2}. Arguing as in Case 1.2 we obtain a contradiction.

Case 1.5: k_{2} = l_{2} = 1, m_{2} = 0. Then l_{1} + m_{1} is odd.

As in Case 1.4 we conclude that u′ = 2, v + v′ ≡ 0 mod 2n and either

v = w = 1 or v = w = n + 1.

Since S is a zero-sum sequence, we have u′ + l_{1}v + m_{1}w + v′ = 2 + (l_{1} + m_{1} − 1)v ≡ 0 mod 2n whence l_{1} + m_{1} = 2n − 1 and

S = e_{1}^{2n−2} ∏_{i=1}^{2n−1}(y_{i}e_{1} + ve_{2}) · (d_{1}e_{1} − ve_{2}) · (d_{2}e_{1} + 2e_{2})

where d_{1} = x_{k_{1}+1} and d_{2} = y_{l_{1}+1} ∈ [0, 2n−1]. For every i ∈ [1, 2n−1] we have d_{1} +y_{i} ≡ 1 mod 2n. This implies that y_{1} = … = y_{2n−1}, and the assertion follows.

Case 1.6: k_{2} = 1, l_{2} = m_{2} = 0. Then l_{1} + m_{1} is even.

As in Case 1.4. we conclude that u′ = 2. Since S is a zero-sum sequence, we infer that (l_{1} + m_{1})v + 2 ≡ 0 mod 2n whence l_{1} + m_{1} = 2n − 2. This implies that k_{1} = 2n, a contradiction.

Case 2: k_{1} ≥ 2, l_{1} ≥ 2, m_{1} = 1.

Since m_{1} ≥ m_{2} and m_{1} + m_{2} is odd, it follows that m_{2} = 0. Recall that l_{1} + l_{2} is odd and that 2v ≡ 2 mod 2n whence v ∈ {1, n + 1}. We distinguish four cases.

Case 2.1: k_{2} = l_{2} = 0. Then l_{1} is odd.

We have

S = e_{1}^{k_{1}} · ∏_{i=1}^{l_{1}}(y_{i}e_{1} + ve_{2}) · (z_{1}e_{1} + we_{2}),

p_{2}(S_{0}) = 0 · (ve_{2}) · (we_{2}), σ(p_{2}(S_{0})) ∈ {0, 2e_{2}}, and since S is a zero-sum sequence, we infer that l_{1}v + w ≡ 0 mod 2n.

Firstly, we suppose that σ(p_{2}(S_{0})) = 2e_{2}. Then v+w ≡ 2 mod 2n and (l_{1}−1)v+2 ≡ 0 mod 2n. Thus it follows that l_{1} + 1 ≡ 0 mod 2n whence l_{1} = 2n − 1. This implies that k_{1} = 2n − 1 and the assertion is proved.

Secondly, we suppose that σ(p_{2}(S_{0})) = 0. Then v + w ≡ 0 mod 2n, (l_{1} − 1)v ≡ 0 mod 2n whence l_{1} = 2n + 1 and k_{1} = 2n − 3. Then z_{1}e_{1} + y_{i}e_{1} ∈ {e_{1}, 2e_{1}} for all i ∈ [1, 2n + 1] and, after renumeration, ∏_{i=1}^{2n+1} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1}, h_{2} ∈ [2, 2n − 1] and h_{1} + h_{2} = 2n + 1. Without restriction we suppose that z_{1} + y_{1} ≡ 1 mod 2n. Then z_{1} + y_{2} ≡ 2 mod 2n, h_{1}y_{1} + h_{2}y_{2} − y_{1} ≡ 2 mod 2n, h_{1}y_{1} + h_{2}y_{2} − y_{2} ≡ 1 mod 2n, y_{2} −y_{1} ≡ 1 mod 2n, h_{1}y_{1} +(2n−h_{1})y_{2} ≡ 1 mod 2n and h_{1}(y_{1} −y_{2}) ≡ 1 mod 2n. Thus h_{1} ≡ −1 mod 2n, h_{1} = 2n − 1 and the assertion is proved.

Case 2.2: k_{2} = 0 and l_{2} = 1. Then l_{1} is even.

Then v + v′ ≡ 0 mod 2n, and we have either p_{2}(S_{0}) = 0 · (ve_{2}) · (we_{2}) or p_{2}(S_{0}) = 0·(v′e_{2})·(we_{2}). Since S is a zero-sum sequence, we have 0 ≡ k_{1}u+l_{1}v+v′+w ≡ (l_{1}−1)v+w mod 2n.

Case 2.2.1: p_{2}(S_{0}) = 0 · (ve_{2}) · (we_{2}). Then v + w + 2nℤ ∈ {2nℤ, 2 + 2nℤ}.

Firstly, we suppose that v + w ≡ 0 mod 2n. Then 0 ≡ (l_{1} − 2)v mod 2n, l_{1} − 2 ≡ 0 mod 2n whence l_{1} = 2n + 2. Therefore

S = e_{1}^{2n−5} · ∏_{i=1}^{2n+2}(y_{i}e_{1} + ve_{2}) · (y_{2n+3}e_{1} − ve_{2}) · (z_{1}e_{1} − ve_{2}).

Since y_{2n+3} ≠ z_{1}, we may argue as in Case 1.2 and obtain a contradiction.

Secondly, we suppose that v + w ≡ 2 mod 2n. Thus v + w ≡ 2 ≡ 2v mod 2n whence v = w. Thus we obtain that 0 ≡ l_{1}v mod 2n and l_{1} = 2n. Therefore

S = e_{1}^{2n−3} · ∏_{i=1}^{2n}(y_{i}e_{1} + ve_{2}) · (y_{2n+1}e_{1} − ve_{2}) · (z_{1}e_{1} + ve_{2}).

Thus S has the same form as in the second part of Case 2.1 and the assertion follows.

Case 2.2.2: p_{2}(S_{0}) = 0 · (v′e_{2}) · (we_{2}). Then v′ + w + 2nℤ ∈ {2nℤ, 2 + 2nℤ}.

Firstly, we suppose that v′ + w ≡ 0 mod 2n. Since v + v′ ≡ 0 mod 2n, we obtain v = w, l_{1}v ≡ 0 mod 2n and l_{1} = 2n. Thus we come to a situation which we have already discussed.

Secondly, we suppose that v′ +w ≡ 2 mod 2n. Thus 2v ≡ 2 ≡ w −v mod 2n, w ≡ 3v mod 2n, 0 ≡ (l_{1}+2)v mod 2n, l_{1} = 2n−2 which implies k_{1} = 4n−1−(l_{1}+l_{2}+m_{1}+m_{2}) = 4n − 1 − (2n − 2 + 1 + 1) = 2n − 1 and the assertion is proved.

Case 2.3: k_{2} = 1 and l_{2} = 0. Then l_{1} is odd.

Since 0 = u ≠ u′ and (u+u′ ≡ 0 or u+u′ ≡ 2 mod 2n), it follows that u′ = 2. Since S is a zero-sum sequence and 2v ≡ 2 mod 2n, we infer that 0 ≡ u′ +l_{1}v +w ≡ (l_{1} +2)v +w mod 2n.

Case 2.3.1: p_{2}(S_{0}) = 0 · (ve_{2}) · (we_{2}). Then v + w + 2nℤ ∈ {2nℤ, 2 + 2nℤ}.

Firstly, we suppose that v + w ≡ 0 mod 2n. Then 0 ≡ (l_{1} + 1)v mod 2n, 0 ≡ l_{1} + 1 mod 2n and l_{1} = 2n − 1. Then k_{1} = 2n − 2 and

S = e_{1}^{2n−2} · (x_{k_{1}+1}e_{1} + 2e_{2}) · ∏_{i=1}^{2n−1}(y_{i}e_{1} + ve_{2}) · (z_{1} − ve_{2}).

Since S is a minimal zero-sum sequence, it follows that z_{1} + y_{i} ≡ 1 mod 2n for every i ∈ [1, 2n − 1] whence y_{1} = … = y_{2n−1} and the assertion is proved.

Secondly, we suppose that v + w ≡ 2 mod 2n. Thus v + w ≡ 2 ≡ 2v mod 2n whence v = w, and we obtain that (l_{1} + 3)v ≡ 0 mod 2n, l_{1} + 3 ≡ 0 mod 2n and l_{1} = 2n − 3. Then k_{1} = 4n − 1 − (k_{2} + l_{1} + l_{2} + m_{1} + m_{2}) = 2n, a contradiction.

Case 2.3.2: p_{2}(S_{0}) = (2e_{2}) · (ve_{2}) · (we_{2}). Then 2 + v + w + 2nℤ ∈ {2nℤ, 2 + 2nℤ}.

Firstly, we suppose that 2+v +w ≡ 2 mod 2n. Then (l_{1} +1)v ≡ 0 mod 2n, l_{1} +1 ≡ 0 mod 2n and l_{1} = 2n − 1. Then k_{1} = 2n − 2 and

S = e_{1}^{2n−2} · (x_{k_{1}+1}e_{1} + 2e_{2}) · ∏_{i=1}^{2n−1}(y_{i}e_{1} + ve_{2}) · (z_{1} − ve_{2}).

Now the assertion follows as in Case 2.3.1.

Secondly, we suppose that 2+v+w ≡ 0 mod 2n. Then w ≡ −3v mod 2n, (l_{1}−1)v ≡ 0 mod 2n and l_{1} = 2n + 1. Then k_{1} = 2n − 4 and

S = e_{1}^{2n−4} · (x_{k_{1}+1}e_{1} + 2e_{2}) · ∏_{i=1}^{2n+1}(y_{i}e_{1} + ve_{2}) · (z_{1}e_{1} − 3ve_{2}).

Since S is a minimal zero-sum sequence, it follows that y_{i} + x_{k_{1}+1} + z_{1} + 2nℤ ∈ {1, 2, 3} + 2nℤ for every i ∈ [1, 2n + 1] whence |{y_{1}, …, y_{2n+1}}| ≤ 3. Since S is a minimal zero-sum sequence, it follows that |{y_{1}, …, y_{2n+1}}| > 1.

Suppose that |{y_{1}, …, y_{2n+1}}| = 2, say ∏_{i=1}^{2n+1} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1}, h_{2} ∈ [1, 2n] and h_{1} + h_{2} = 2n + 1. Since S is a minimal zero-sum sequence, we have h_{1}, h_{2} ∈ [2, 2n − 1]. If {h_{1}, h_{2}} = {2, 2n−1}, then we are done. Assume to the contrary that h_{1}, h_{2} ∈ [3, 2n−2]. Since z_{1} + 3y_{1}, z_{1} + 2y_{1} + y_{2}, z_{1} + y_{1} + 2y_{2}, z_{1} + 3y_{2} are congruent to 1, 2 or 3 modulo 2n, it follows that either 2y_{1} ≡ 2y_{2} mod 2n or 3y_{1} ≡ 3y_{2} mod 2n. On the other hand, we have distinct i, j ∈ {1, 2} such that y_{i} −y_{j} +2nℤ = (y_{i} +x_{k_{1}+1} +z_{1})−(y_{j} +x_{k_{1}+1} +z_{1})+2nℤ ∈ {1, 2} + 2nℤ, a contradiction.

Suppose that |{y_{1}, …, y_{2n+1}}| = 3, say ∏_{i=1}^{2n+1} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}}y_{3}^{h_{3}} with h_{1}, h_{2}, h_{3} ∈ [1, 2n−1] and h_{1}+h_{2}+h_{3} = 2n+1. After renumerating if necessary we obtain that x_{k_{1}+1}+z_{1}+y_{i} ≡ i mod 2n for every i ∈ {1, 2, 3} whence y_{2} ≡ y_{1}+1 mod 2n and y_{3} ≡ y_{1}+2 mod 2n. Since S is a minimal zero-sum sequence, we obtain that z_{1}+y_{ν_{1}}+y_{ν_{2}}+y_{ν_{3}}+2nℤ ∈ {1, 2, 3}+2nℤ for every subsequence y_{ν_{1}}y_{ν_{2}}y_{ν_{3}} of ∏_{ı=1}^{2n+1} y_{i}. If h_{1} ≥ 3, then y_{1}^{3}, y_{1}^{2}y_{2}, y_{1}^{2}y_{3} and y_{1}y_{2}y_{3} are subsequences of ∏_{ı=1}^{2n+1} y_{i}, but their sums 3y_{1}, 2y_{1}+y_{2}, 2y_{1}+y_{3} and y_{1}+y_{2}+y_{3} are pairwise incongruent modulo 2n, a contradiction. Thus h_{1} ≤ 2. Similarly, if h_{3} ≥ 3, then then we get sums 3y_{3}, 2y_{3} + y_{2}, 2y_{3} + y_{1} and y_{1} + y_{2} + y_{3} which are pairwise incongruent modulo 2n, a contradiction. Thus h_{1}, h_{3} ∈ [1, 2] and h_{2} ≥ 2n − 3 > 3. If h_{1} ≥ 2, then we get sums 2y_{1} + y_{2}, 2y_{1} + y_{3}, y_{1} + y_{2} + y_{3}, 3y_{2} which are pairwise incongruent modulo 2n, a contradiction whence h_{1} = 1. Similarly, we obtain that h_{3} = 1. Thus h_{2} = 2n − 1 and the assertion is proved.

Case 2.4: k_{2} = l_{2} = 1. Then l_{1} is even.

As in Case 2.3 we have u′ = 2 ≡ 2v mod 2n. Since v + v′ ≡ 0 mod 2n and S is a zero-sum sequence, we infer that 0 ≡ u′ + l_{1}v + v′ + w ≡ (l_{1} + 1)v + w mod 2n. Then

S = e_{1}^{k_{1}} · (x_{k_{1}+1}e_{1} + 2ve_{2}) · ∏_{i=1}^{l_{1}}(y_{i}e_{1} + ve_{2}) · (y_{l_{1}+1}e_{1} − ve_{2}) · (z_{1}e_{1} + we_{2})

whence p_{2}(S_{0}) has one of the following four forms: 0·(ve_{2})·(we_{2}), 0·(−ve_{2})·(we_{2}), (2e_{2})· (ve_{2}) · (we_{2}), (2e_{2}) · (−ve_{2}) · (we_{2}). We distinguish four cases.

Case 2.4.1: p_{2}(S_{0}) = 0 · (ve_{2}) · (we_{2}). Then v + w + 2nℤ ∈ {0, 2} + 2nℤ.

Suppose v + w ≡ 0 mod 2n. Then l_{1}v ≡ 0 mod 2n whence l_{1} = 2n and

S = e_{1}^{2n−4} · (x_{2n−3}e_{1} + 2ve_{2}) · ∏_{i=1}^{2n}(y_{i}e_{1} + ve_{2}) · (y_{2n+1}e_{1} − ve_{2}) · (z_{1}e_{1} − ve_{2}).

Since y_{2n+1} ≠ z_{1}, this situation has already been discussed in Case 1.4.

Suppose v + w ≡ 2 mod 2n. Then v = w, 0 ≡ (l_{1} + 2)v mod 2n and l_{1} = 2n − 2. Thus

S = e_{1}^{2n−2} · (x_{2n−1}e_{1} + 2ve_{2}) · ∏_{i=1}^{2n−2}(y_{i}e_{1} + ve_{2}) · (y_{2n−1}e_{1} − ve_{2}) · (z_{1}e_{1} + ve_{2})

with z_{1} ∉ {y_{1}, …, y_{2n−2}}. Thus either y_{2n−1}+y_{1} ≢ 1 mod 2n or y_{2n−1}+z_{1} ≢ 1 mod 2n, and S has a proper zero-sum subsequence, a contradiction.

Case 2.4.2: p_{2}(S_{0}) = 0 · (−ve_{2}) · (we_{2}). Then −v + w + 2nℤ ∈ {0, 2} + 2nℤ.

If v = w, then we obtain a contradiction as in the second part of Case 2.4.1.

Suppose that −v + w ≡ 2 mod 2n. Then w ≡ 3v mod 2n, 0 ≡ (l_{1} + 4)v mod 2n, l_{1} = 2n − 4 and k_{1} = 4n − 1 − (k_{2} + l_{1} + l_{2} + m_{1} + m_{2}) = 2n, a contradiction.

Case 2.4.3: p_{2}(S_{0}) = (2e_{2}) · (ve_{2}) · (we_{2}). Then 2 + v + w + 2nℤ ∈ {0, 2} + 2nℤ.

If 2 + v + w ≡ 2 mod 2n, then we argue as in the first part of Case 2.4.1.

Suppose 2 + v + w ≡ 0 mod 2n. Then w ≡ −3v mod 2n, 0 ≡ (l_{1} − 2)v mod 2n, l_{1} = 2n + 2 and

S = e_{1}^{2n−6} · (x_{2n−5}e_{1} + 2ve_{2}) · ∏_{i=1}^{2n+2}(y_{i}e_{1} + ve_{2}) · (y_{2n+3}e_{1} − ve_{2}) · (z_{1}e_{1} − 3ve_{2}).

Since S is a minimal zero-sum sequence, it follows that |{y_{1}, …, y_{2n+2}}| > 1.

Suppose that |{y_{1}, …, y_{2n+2}}| ≥ 3. Then without restriction we may suppose that y_{2n+3} + y_{2n+2} + 2nℤ ∈ {3, 4, 5} + 2nℤ. If |{y_{1}, …, y_{2n+1}}| ≥ 3, say |{y_{1}, y_{2}, y_{3}}| = 3, then z_{1} + y_{1} + y_{4} + y_{5}, z_{1} + y_{2} + y_{4} + y_{5}, z_{1} + y_{3} + y_{4} + y_{5} are pairwise distinct whence z_{1}+y_{j}+y_{4}+y_{5}+2nℤ ∈ {3, 4, 5}+2nℤ for some j ∈ [1, 3] and y_{2n+3}+z_{1}+y_{2n+2}+y_{j}+y_{4}+y_{5} ∈ {6, 7, 8, 9, 10} + 2nℤ whence S contains a proper zero-sum subsequence, a contradiction. Thus we may suppose that ∏_{i=1}^{2n+1} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1}, h_{2} ∈ [2, 2n − 1]. Assume to the contrary that h_{1}, h_{2} ∈ [3, 2n − 2]. If 3y_{1}, 2y_{1} + y_{2}, y_{1} + 2y_{2} are pairwise distinct, we obtain a contradiction as before. Hence 2y_{1} ≡ 2y_{2} mod 2n. Since 2[\frac{h_{1}}{2} ] + 2[\frac{h_{2}}{2} ] = 2n, it follows that 2[\frac{h_{1}}{2} ]y_{1} + 2[\frac{h_{2}}{2} ]y_{2} ≡ 2ny_{1} ≡ 0 mod 2n whence (y_{1}e_{1} + ve_{2})^{2[ \frac{h_{1}}{2} ]}) · (y_{2}e_{1} + ve_{2})^{2[ \frac{h_{2}}{2} ]}) is a zero-sum subsequence of S, a contradiction.

Suppose that |{y_{1}, …, y_{2n+2}}| = 2, say ∏_{i=1}^{2n+2} y_{i} = y_{1}^{h_{1}}y_{2}^{h_{2}} with h_{1}, h_{2} ∈ [3, 2n − 1]. Assume to the contrary that h_{1}, h_{2} ∈ [4, 2n−2]. Since y_{1}+y_{2n+3}+2nℤ, y_{2}+y_{2n+3}+2nℤ ∈ [1, 5]+2nℤ are distinct, we may suppose that y_{2n+3}+y_{1}+2nℤ ∈ [2, 5]+2nℤ. Then the four numbers z_{1} + 3y_{1}, z_{1} + 2y_{1} + y_{2}, z_{1} + y_{1} + 2y_{2}, z_{1} + 3y_{2} are congruent to 1, 2 or 3 modulo 2n (otherwise, the sum of one of these elements and y_{2n+3} + y_{1} would not lie in [1, 5] modulo 2n). Thus 2y_{1} ≡ 2y_{2} mod 2n or 3y_{1} ≡ 3y_{2} mod 2n. If 2y_{1} ≡ 2y_{2} mod 2n, we obtain a contradiction as above. Suppose 3y_{1} ≡ 3y_{2} mod 2n. Then 3 | n, 3[\frac{h_{1}}{3} ] + 3[\frac{h_{2}}{3} ] = 3n, 3[\frac{h_{1}}{3} ]y_{1} + 3[\frac{h_{2}}{3} ]y_{2} ≡ 2ny_{1} ≡ 0 mod 2n whence (y_{1}e_{1} + ve_{2})^{3[ \frac{h_{1}}{3} ]}) · (y_{2}e_{1} + ve_{2})^{3[ \frac{h_{2}}{3} ]}) is a zero-sum subsequence of S, a contradiction.

Case 2.4.4: p_{2}(S_{0}) = (2e_{2}) · (−ve_{2}) · (we_{2}). Then 2 − v + w + 2nℤ ∈ {0, 2} + 2nℤ.

If 2 − v + w ≡ 2 mod 2n, then v = w and we obtain a contradiction as in the second part of Case 2.4.1.

Suppose that 2 − v + w ≡ 0 mod 2n. Then 0 ≡ 2v − v + w ≡ v + w mod 2n and we argue as in part one of Case 2.4.1.

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