© Revista de Matemática: Teoría y Aplicaciones 2021 28(1) : 39–53 cimpa – ucr issn: 1409-2433 (Print), 2215-3373 (Online)

doi: https:// doi. org/ 10.15517/ rmta. v28i1. 33773

existence conditions for k-barycentric olson constant

condiciones de existencia para la constante de olson k-baricéntrica

Luz Marchan^{∗}  Oscar Ordaz^{†}  José Salazar^{‡}

Felicia Villarroel^{§}

Received: 16/Jul/2018; Revised: 25/Jun/2020; Accepted: 30/Oct/2020

∗Escuela Superior Politécnica del Litoral, Departamento de Matemáticas, Facultad de Ciencias Naturales, Guayaquil, Ecuador. E-Mail: luzelimarchan@gmail.com

†Universidad Central de Venezuela, Departamento de Matemáticas y Laboratorio LaTecS, Centro ISYS, Facultad de Ciencias, Caracas, Venezuela. E-Mail: oscarordaz55@gmail.com

‡Universidad de Oriente, Departamento de Matemáticas, Núcleo Sucre, Cumaná, Venezuela. E-Mail: jsalazar@udo.edu.ve

§Universidad de Oriente, Cumaná, Departamento de Matemáticas, Núcleo Sucre, Venezuela. E-Mail: feliciavillarroel@gmail.com

40 l. marchan — o. ordaz — j. salazar — f. villarroel

Abstract

Let (G, +) be a finite abelian group and 3 ≤ k ≤ |G| a positive integer. The k-barycentric Olson constant denoted by BO(k, G) is defined as the smallest integer ℓ such that each set A of G with |A| = ℓ contains a subset with k elements {a_{1}, …, a_{k}} satisfying a_{1} + ⋯ + a_{k} = ka_{j} for some 1 ≤ j ≤ k. We establish some general conditions on G assuring the existence of BO(k, G) for each 3 ≤ k ≤ |G|. In particular, from our results we can derive the existence conditions for cyclic groups and for elementary p-groups p ≥ 3. We give a special treatment over the existence condition for the elementary 2-groups.

Keywords: finite abelian group; zero-sum problem; baricentric-sum problem; Davenport constant; k-barycentric Olson constant.

Resumen

Sean (G, +) un grupo abeliano finito y 3 ≤ k ≤ |G| un entero positivo. La constante de Olson k-baricéntrica, denotada por BO(k, G), se define como el menor entero positivo ℓ tal que todo conjunto A de G con |A| = ℓ contiene un subconjunto con k elementos {a_{1}, …, a_{k}} que satisface a_{1}+⋯+a_{k} = ka_{j} para algún 1 ≤ j ≤ k. Establecemos algunas condiciones generales sobre G asegurando la existencia de BO(k, G) para cada 3 ≤ k ≤ |G|. En particular, a partir de nuestros resultados podemos determinar las condiciones de existencia para los grupos cíclicos y para los p-grupos elementales con p ≥ 3. Damos un tratamiento especial a la condición de existencia para los 2-grupos elementales.

Palabras clave: grupos abelianos finitos; problemas de suma-cero; problemas de suma baricéntricas; constante de Davenport; constante k-baricéntrica de Olson.

Mathematics Subject Classification: 11B30, 11B75, 94B05, 94B65, 51E22.

1 Introduction

We recall some standard terminology and notation. We denote by ℕ the positive integers and we set ℕ_{0} = ℕ ∪ {0}. For abelian groups, we use additive notation and we denote the neutral element by 0. For n ∈ ℕ, let C_{n} denotes a cyclic group of order n. For each finite abelian group there exists 1 < n_{1} | ⋯ | n_{r} such that G ≅ C_{n_{1}} ⊕ ⋯ ⊕ C_{n_{r}}. The integer n_{r} is called the exponent of G, denoted exp(G). The integer r is called the rank of G, denoted r(G). For a prime p, the p-rank of G, denoted r_{p}(G), is the smallest number i such that n_{i} is divisible by p. For a prime number p we denote by 𝔽_{p} the field with p elements.

existence conditions for k-barycentric olson constant 41

We say that G is a p-group if its exponent is a prime power and we say that G is an elementary p-group if the exponent is a prime (except for the trivial group). Let G be an abelian finite group. The sumset of two subsets A and B of G will be denoted by A + B = {a + b : a ∈ A ∧ b ∈ B}. We denote the sum of the elements of a subset S of G by σ(S). Furthermore, for an integer k, let ∑_{k}(A) = {σ(B) : B ⊆ A ∧ |B| = k}. Finally, for t an integer, we denote by t · A the set of multiples t · A = {ta : a ∈ A}.

For a finite abelian group (G, +) and 3 ≤ k ≤ |G| a positive integer, the k-barycentric Olson constant denoted by BO(k, G) is the smallest ℓ such that each set A with |A| = ℓ over G has a subset with k elements {a_{1}, …, a_{k}} satisfying a_{1} + ⋯ + a_{k} = ka_{j} for some 1 ≤ j ≤ k. This set with k elements is called a k-barycentric set and a_{j} is called its barycenter. Notice that a k-barycentric set can be written as a weighted zero-sum set that is:

a_{1} + ⋯ + (1 − k)a_{j} + ⋯ + a_{k} = 0.

So that the k-barycentric Olson constant can be seen as a classical example of a weighted zero-sum constant over a finite abelian group. This constant together with related invariants have been studied in the literature [5, 6]. The aim of the present work is to establish conditions on G for the existence of BO(k, G) ≤ |G| for each 3 ≤ k ≤ |G|. That is to say, for each 3 ≤ k ≤ |G| there exists a k-barycentric set.

Existence conditions of the k-barycentric Olson constant with 3 ≤ k ≤ |G| were initially considered in [14] with the study on cyclic groups using the Orbits Theory. In [13] Ordaz, Plagne and Schmid researched on the existence conditions of BO(k, G) with |G| − 2 ≤ k ≤ |G| over finite abelian groups G in general; their results were Lemma 1 and Proposition 1. In case there are no k-barycentric sets in G we write BO(k, G) = |G| + 1.

Lemma 1 ([13] , Lemma 3.1) Let G be a finite abelian group. Then

σ(G) = \{ \begin{matrix}b^{∗} & if r_{2}(G) = 1 and b^{∗} denote the only element with order 2, \\ 0 & in other case. \end{matrix}

Hence we have that:

BO(|G|, G) = \{ \begin{matrix}|G| + 1 & if r_{2}(G) = 1, \\ |G| & in other case. \end{matrix}

The following result gives the values of BO(|G|−1, G) and BO(|G|−2, G).

Proposition 1 ([13] , Proposition 3.2) Let G be a finite abelian group. Then for |G| ≥ 2, we have:

BO(|G| − 1, G) = \{ \begin{matrix}|G| − 1 & if r_{2}(G) = 1, \\ |G| + 1 & in other case. \end{matrix}

and for |G| ≥ 3, we have:

BO(|G| − 2, G) = \{ \begin{matrix}|G| − 2 & if |G| is odd, \\ |G| + 1 & if exp(G) = 2 or |G| = 4, \\ |G| − 1 & in other case. \end{matrix}

In the Lemma 1 is determine the conditions of existence of BO(|G|, G) and in the Proposition 1 is determine the conditions of existence of BO(k, G) with |G| − 2 ≤ k ≤ |G| − 1.

In the same order of ideas of the above results, the main goal of our paper is to show that the finite abelian groups G with r_{2}(G) = 0 and the finite abelian groups G with r_{2}(G) = 1 contain a k-barycentric set for each 3 ≤ k ≤ |G| − 3. Notice that the cyclic groups C_{n} are members of these groups since r_{2}(C_{n}) = 0 if and only if n is odd and r_{2}(C_{n}) = 1 if and only if n is even. Similarly, elementary p-groups with p ≠ 2, are members of the above groups since r_{2}(C_{p}^{m}) = 0. In consequence our results solve completely the existence conditions of the k-barycetric Olson constant, for cyclic groups and for elementary p-groups. It is clear that the elementary 2-groups are outside the above groups and then we have a special consideration for its existence conditions for BO(k, C_{2}^{m}). As a second goal in our investigation, for some G and k, we give an exact value for BO(k, G) when it exists. For example, we show that BO(|G| − 3, G) = |G| − 2 for the abelian groups G with r_{2}(G) = 1, |G| ≥ 8 and non multiple of 3. Moreover, we show that BO(3^{m} − 3, C_{3}^{m}) = 3^{m} − 2, in this case r_{2}(C_{3}^{m}) = 0.

The organization of the paper besides this introduction and the conclusion, is as follows: a first section on preliminaries, a second section on existence conditions for general finite abelian groups and finally, a third section on some existence conditions for elementary 2-groups.

2 Preliminaries

In this section we give some previous and useful results.

Remark 1 Let G be a finite abelian group. Then

i. r_{2}(G) = 0 if and only if |G| is odd.

ii. r_{2}(G) = 1 implies that |G| is even. Let b^{∗} ∈ G be the only element of order 2. It is clear that for cyclic groups we have the equivalence r_{2}(C_{n}) = 1 if and only if n is even. Also we have that r_{2}(C_{p}^{m}) = 0 for p ≠ 2. Moreover, if t = r_{2}(G) ≥ 1, then |G| is even and G has 2^{t} − 1 elements of order 2.

Proposition 2 Let G be a finite abelian group with |G| ≥ 8 such that r_{2}(G) = 1 and 3 ∤ |G|. Then.

i. −3 · G = G.

ii. Let a ∈ G and S_{a} = {x ∈ G : 2x = a}. Then |S_{a}| ≤ 2.

Proof. i. Let ϕ : G → −3 · G be given by ϕ(a) = −3a where −3 · G = {3(−a) : a ∈ G}. Let y = 3(−a) ∈ G, then exits a ∈ G such that ϕ(a) = −3a = 3(−a) = y, therefore ϕ is surjective. Assuming that ϕ(a_{1}) = ϕ(a_{2}), then −3a_{1} = −3a_{2}, so that, 3(a_{1} − a_{2}) = 0. Since 3 ∤ |G|, then a_{1} = a_{2}, i.e., ϕ is injective. Then |G| = | − 3 · G|. Since −3 · G ⊆ G and G is finite, then −3 · G = G. ii. Assuming we have three different elements a_{1}, a_{2}, a_{3} ∈ S_{a}, then 2a_{1} = 2a_{2} and 2a_{1} = 2a_{3}, in consequence 2(a_{1} − a_{2}) = 0 and 2(a_{1} − a_{3}) = 0.

Since a_{1}, a_{2}, a_{3} are different, then a_{1}−a_{2} = b^{∗} and a_{1}−a_{3} = b^{∗}, where b^{∗} is the only element of order 2 in G. Hence a_{2} = a_{3}, contradiction. So that |S_{g}| ≤ 2.

We have the following result:

Proposition 3 If m ≥ 2, then 3^{m} − 2 ≤ BO(3^{m} − 3, C_{3}^{m}).

Proof. Let A = C_{3}^{m} \{−a, −b, 0} be a (3^{m} −3)-subset over C_{3}^{m} with a+b ≠ 0. Since σ(C_{3}^{m}) = 0 and σ(C_{3}^{m}) = σ(A) + σ(A^{c}) where A^{c} = {−a, −b, 0}, then σ(A) = −σ(A^{c}) ⇒ σ(A) = −(−a − b + 0) ⇒ σ(A) = a + b ≠ 0. Moreover we have that (3^{m}−3)a = (3^{m}−3)(3a) = (3^{m−1}−1)0 = 0 for all a ∈ A ⊂ C_{3}^{m} since 3x = 0 for all x ∈ C_{3}^{m}. Therefore, there exists a (3^{m} − 3)-subset A over C_{3}^{m} such that σ(A) ≠ (3^{m} −3)a for all a ∈ A, i.e., 3^{m} −2 ≤ BO(3^{m} −3, C_{3}^{m}).

We need the following result:

Proposition 4 Let A be a k-subset of C_{2}^{m} such that 3 ≤ k ≤ 2^{m}.

i. If k is even, then A is a k-barycentric set if and only if σ(A) = 0.

ii. If k is odd, then A is a k-barycentric set if and only if σ(A) ∈ A.

iii. Let A^{c} = C_{2}^{m} \ A the complement of A, then |A| = 2^{m} − |A^{c}|.

iv. σ(A) = σ(A^{c}).

v. 0 ∉ ∑_{2} C_{2}^{m}.

Proof. It follows directly.

The following lemma guarantees the existence of k-sets of zero-sum with 4 ≤ k ≤ \frac{|G|}{2} − 1 in a finite abelian group.

Lemma 2 ([3] , Lemma 7.1) Let G be a finite abelian group de orden |G| ≥ 2.

1. There exists a squarefree zero sequence S ∈ F (G) with |S| = |G| − 1.

2. Let 0 ≠ g_{0} ∈ G and 1 ≤ k ≤ \frac{|G|}{2} − 1 with k ≠ 2, if G is an elementary 2group. Then there exist a squarefree zero sequence S ∈ F (G) with g_{0} ∤ S and |S| = k.

The following corollary is a consequence of the above lemma.

Corollary 1 Let G is an elementary 2-group de orden |G| ≥ 3 such that 0 ≠ x ∈ G and 4 ≤ k ≤ \frac{|G|}{2} − 1. Then there exist a k-set A of zero-sum in G such that x ∉ A.

3 Existence conditions of BO(k, G) for general abelian groups

Let G be a finite. In the following two theorems, the values r_{2}(G) = 0 or r_{2}(G) = 1 are considered to give an existence condition in the order G to have a k-barycentric set, for each 3 ≤ k ≤ |G| − 3. Notice that from Remark 1 the parity of |G| is used and depends on r_{2}(G) = 0 or r_{2}(G) = 1. Observe that the fact r_{2}(G) = 0 means that for each g ∈ G we have −g ≠ g. The results provided in this section allow us to establish the existence of BO(k, G) with 3 ≤ k ≤ |G| − 3 for cyclic groups and elementary p-groups. A relationship between the Harborth g(G) and the k-barycentric Olson BO(k, G) constants is presented. From these relations, we give exact values of BO(k, G) for some groups where g(G) exists. Finally we identify some conditions on certain groups G in order to provide the exact values of BO(|G| − 3, G).

Theorem 1 Let G be a finite abelian group such that r_{2}(G) = 0 and 3 ≤ k ≤ |G| − 3. Then BO(k, G) ≤ |G|.

Proof. Assuming |G| ≥ 9. Let A be a zero-sum set of G such that |A| = 3 with 0 ∉ A and we consider B = {−a : a ∈ A}. Notice that the sets A ∪ {0}, A \ {a} ∪ B \ {−a}} ∪ {0} for some a ∈ A and A ∪ B ∪ {0} over G are k-barycentric, then BO(k, G) ≤ |G| for k = 4, 5 y 7.

Let C = G \ (A ∪ B ∪ {0}). Notice that since |G| ≥ 9 and also odd then |C| ≥ 2 is even. Moreover for all c ∈ C we can see that −c ∈ C, assuming the contrary, we have a contradiction. Hence there exists E ⊆ C with 2 ≤ |E| ≤ |C| conformed by elements a and its opposite. Since |E| is even then E ∪ A ∪ {0} or E ∪ A ∪ B ∪ {0} constitute the k-barycentric sets even or odd with barycenter 0, over G. Notice that 6 ≤ k ≤ |G| − 3 with k ≠ 7.

Moreover, since for all 0 ≠ g ∈ G the set {g, −g, 0} over G is a zero-sum then BO(3, G) ≤ |G|.

Now, we consider the finite abelian groups G of order 3, 5 and 7. Observe that these groups are cyclic. In what follows we consider the existence of BO(k, G). By Lemma 1 we have that BO(3, C_{3}) = 3, BO(5, C_{5}) = 5 and BO(7, C_{7}) = 7. Moreover by Proposition 1 we have that BO(4, C_{5}) and BO(6, C_{7}) does not exist and BO(3, C_{5}) = 3 and BO(5, C_{7}) = 5. Moreover, the 4-subset A = {0, 1, 2, 4} over C_{7} a zero-sum and 0 ∈ A, in consequence BO(4, C_{7}) ≤ 7 and for all 0 ≠ a ∈ C_{7} the 3-subset A = {0, a, −a} a zero-sum and 0 ∈ A, hence BO(3, C_{7}) ≤ 7.

The following two corollaries are a direct consequence of the above theorem.

Corollary 2 Let C_{n} be a cyclic group such that r_{2}(C_{n}) = 0 and 3 ≤ k ≤ n − 3. Then BO(k, C_{n}) ≤ n.

Corollary 3 Let C_{p}^{m} be a elementary p-group such that r_{2}(C_{p}^{m}) = 0 and 3 ≤ k ≤ p^{m} − 3. Then BO(k, C_{p}^{m}) ≤ p^{m}.

Theorem 2 Let G be a finite abelian group such that r_{2}(G) = 1 and 3 ≤ k ≤ |G| − 3 a positive integer. Then BO(k, G) ≤ |G|.

Proof. Assuming |G| ≥ 8. Let b^{∗} ∈ G the only element of order 2. Let A be a 3subset with zero-sum over G such that b^{∗} ∈ A, 0 ∉ A and B = {−a : a ∈ A} \ {b^{∗}}. It is clear that the sets A ∪ {0} and A \ {b^{∗}} ∪ B ∪ {0} over G are barycentric with barycenter 0. Hence BO(k, G) ≤ |G| for k = 4 and 5.

Consider now, the set C = G \ (A ∪ B ∪ {0}). By Remark 1 G is even and then since |A ∪ B ∪ {0}| = 6 we have that |C| ≥ 2 is even. Moreover for each c ∈ C we have −c ∈ C, assuming the contrary we have a contradiction. Therefore there exists E ⊆ C with zero-sum and 2 ≤ |E| ≤ |C| conformed by elements in C and its opposite. Hence the sets E ∪ A ∪ {0} and E ∪ A \ {b^{∗}} ∪ B ∪ {0} give the k-barycentric sets over G, k even and odd with barycenter 0 such that 6 ≤ k ≤ |G| − 3.

Moreover, since for all b^{∗} ≠ g ∈ G the set {g, −g, 0} of G has zero-sum then BO(3, G) ≤ |G|.

Now, we consider the finite abelian groups G of order 4 and 6. Observe that these groups are cyclic. In what follows we consider the existence of BO(k, G). By Lemma 1 we have that BO(4, C_{4}) and BO(6, C_{6}) does not exist. Moreover by Proposition 1 we have that BO(3, C_{4}) = 3 and BO(5, C_{6}) = 5. Moreover, for all 3 ≠ a ∈ C_{6} the 3-subset A = {0, a, −a} a zero-sum and 0 ∈ A, hence BO(3, C_{6}) ≤ 6.

The following corollary is a consequence of the above theorem.

Corollary 4 Let C_{n} be a cyclic group such that r_{2}(C_{n}) = 1 and 3 ≤ k ≤ n−3. Then BO(k, C_{n}) ≤ n.

Theorem 3 Let G be a finite abelian group with |G| ≥ 8, r_{2}(G) = 1 and 3 ∤ |G|. Then BO (|G| − 3, G) = |G| − 2.

Proof. Let b^{∗} ∈ G be the only element with order 2. Let A ⊆ G be such that |A| = |G| − 2. Assuming that A = G \ {a_{1}, a_{2}} and consider B = A \ [{b^{∗} + 2a_{1} − a_{2}, b^{∗} + 2a_{2} − a_{1}} ∪ S_{a_{1}+a_{2}−b^{∗}}]. Since |G| ≥ 8 then |B| = |A| − 2 − |S_{a_{1}+a_{2}−b^{∗}}| ≥ (|G| − 2) − 2 − 2 = |G| − 6 > 0. Hence B ≠ ∅.

Let b ∈ B ⊆ A be and consider the (|G| − 3)-subset A \ {b} of A and we will see that A \ {b} is a (|G| − 3)-barycentric set of A. We have that σ(A\{b}) = σ(A)−σ(b) = σ(G)−a_{1}−a_{2}−b = b^{∗}−a_{1}−a_{2}−b. Moreover, by Proposition 2 we have −3·G = G, then σ(A\{b}) = b^{∗}−a_{1}−a_{2}−b = −3c for some c ∈ G. If c = a_{1}, then b = b^{∗}+2a_{1}−a_{2} ∉ B, contradiction. If c = a_{2}, then b = b^{∗} +2a_{2} −a_{1} ∉ B, contradiction. if c = b, then 2b = a_{1} +a_{2} −b^{∗}, in consequence b ∈ S_{a_{1}+a_{2}−b^{∗}} ⊈ B, contradiction. Hence, c ∈ G\{a_{1}, a_{2}, b} = A\{b} and therefore σ(A \ {b}) = b^{∗} − a_{1} − a_{2} − b = −3c = (|G| − 3) c, for some c ∈ A \ {b}. Hence A \ {b} is a (|G| − 3)-barycentric set of A, i.e., BO (|G| − 3, G) ≤ |G| − 2.

Now we see, |G|−2 ≤ BO (|G| − 3, G). Consider the set B = G\[{0, b^{∗}}∪ S_{b^{∗}}]. Since |G| ≥ 8 then, |B| = |G| − 2 − |S_{b^{∗}}| ≥ |G| − 2 − 2 = |G| − 4 > 0. So that B ≠ ∅.

Let b ∈ B be, then 2b ≠ b and 2b ≠ b^{∗} since if 2b = b^{∗},b ∈ S_{b^{∗}}. Consider A = G \ {b^{∗}, b, 2b}, then |A| = |G| − 3 and σ(A) = σ(G \ {b^{∗}, b, 2b}) = σ(G) − b^{∗} − b − 2b = b^{∗} − b^{∗} − b − 2b = −3b. If σ(A) = −3c for some c ∈ A, then −3b = −3c, in consequence b = c, this is a contradiction with the fact that b ∉ A, that is to say, A it is not a (|G| − 3)-barycentric set of G. So that |G| − 2 ≤ BO (|G| − 3, G). Therefore, BO (|G| − 3, G) = |G| − 2.

The following corollary is a consequence of the above theorem.

Corollary 5 Let C_{n} be a cyclic group with n ≥ 8, r_{2}(C_{n}) = 1 and 3 ∤ n. Then BO (n − 3, G) = n − 2.

Theorem 4 Let m ≥ 2 be then we have that BO(3^{m} − 3, C_{3}^{m}) = 3^{m} − 2.

Proof. By Proposition 3 we have that 3^{m} − 2 ≤ BO(3^{m} − 3, C_{3}^{m}). Let A be a (k−2)-subset over C_{3}^{m}. If σ(A) ∈ A, then the (3^{m} −3)-subset B = A\{σ(A)} of A is a zero-sum. So that σ(B) = 0 = (3^{m} − 3)b for each b ∈ B. Hence B = A \ {σ(A)} is a (3^{m} − 3)-barycentric set.

Assuming that σ(A) ∉ A, then σ(A) ∈ A^{c} where A^{c} is a 2-subset over C_{3}^{m}. In consequence A^{c} = {σ(A), a} with σ(A) ≠ a. Since σ(C_{3}^{m}) = 0 and σ(C_{3}^{m}) = σ(A) + σ(A^{c}), then σ(A^{c}) = −σ(A) ⇒ σ(A) + a = −σ(A) ⇒ a + 2σ(A) = 0 = 3a ⇒ a = σ(A), a contradiction with the fact that σ(A) ≠ a. Therefore, BO(3^{m} − 3, C_{3}^{m}) = 3^{m} − 2.

In what follows we consider the Harborth constant and we give its relationship with the k-barycentric Olson constant.

Definition 1 Let G be a finite abelian group. The Harborth constant, denoted g(G), is defined as the smallest positive integer ℓ such that each set A ⊆ G with |A| = ℓ contains a subset B with |B| = exp(G) with zero-sum.

The following remark and theorem establishes a relationship between the Harborth constant and the zero-sum problem.

Remark 2 Kemnitz showed g(C_{p}^{2}) = 2p − 1 for p ∈ {3, 5, 7} in [9]. In particular, g(C_{3}^{2}) = 5. More recently Gao and Thangadurai [4] showed g(C_{p}^{2}) = 2p−1 for prime p ≥ 67 and g(C_{4}^{2}) = 9. In [2] we can find other values for elementary 3-group; for example g(C_{3}^{3}) = 10, g(C_{3}^{3}) = 21, g(C_{3}^{5}) = 46 and 112 ≤ g(C_{3}^{6}) ≤ 114 [1, 7, 8, 12].

Theorem 5 ( [11], Theorem 1.1)

g(C_{2} ⊕ C_{2n}) = \{ \begin{matrix}2n + 2 & if n is even, \\ 2n + 3 & if n is odd. \end{matrix}

The following result determines the exact values of BO(exp(G), G) in finite abelian groups G where g(G) there exists.

Theorem 6 Let G be a finite abelian group where g(G) exists. Then BO(exp(G), G) = g(G).

Proof. Let A ⊆ G be such that |A| = g(G), then there exits B ⊆ A with |B| = exp(G) such that σ(B) = 0. Therefore σ(B) = 0 = exp(G)b for all b ∈ B. Hence B is a (exp(G))-barycentric subset of A, so that BO(exp(G), G) ≤ g(G). Assuming that A ⊆ G with |A| = BO(exp(G), G), then A contains a (exp(G))-subset such that σ(B) = (exp(G))b = 0 for all b ∈ B. So that B is a (exp(G))-subset with zero-sum of A, that is to say g(G) ≤ BO(exp(G), G). Therefore, BO(exp(G), G) = g(G).

The following result gives the exact values of BO(exp(G) + 1, G) for finite abelian groups where g(G) exists and g(G) ≥ exp(G) + 1.

Theorem 7 Let G be a finite abelian group such that g(G) exists and g(G) ≥ exp(G) + 1. Then BO(exp(G) + 1, G) = g(G).

Proof. Let A ⊆ G be such that |A| = g(G) ≥ exp(G) + 1, then there exists B ⊆ A with |B| = exp(G) such that σ(B) = 0 . Now, since |A| ≥ |B|+1, then there exist some a ∈ A \ B. Let C = B ∪ {a} be then |C| = exp(G) + 1 and we have that σ(C) = σ(B) + σ({a}) = 0 + a = a = 0 + a = exp(G)a + a = (exp(G) + 1)a. Therefore C is a (exp(G) + 1)-barycentric subset of A, hence, BO(exp(G) + 1, G) ≤ g(G).

Assuming that A ⊆ G such that |A| = BO(exp(G) + 1, G), hence there exists B ⊆ A such that|B| = exp(G) + 1, hence σ(B) = (exp(G) + 1)b with b ∈ B. Let C = B \ b be a (exp(G))-subset of A such that σ(C) = σ(B) − σ{b} = (exp(G) + 1)b − b = exp(G)b + b − b = 0. Therefore C ⊆ A is a (exp(G))-subset with a zero-sum, in consequence g(G) ≤ BO(exp(G) + 1, G). Therefore, BO(exp(G) + 1, G) = g(G).

The following corollary is a consequence of Theorem 6 and Remark 2.

Corollary 6

BO(3, C_{3}^{2}) = 5, BO(3, C_{3}^{3}) = 10, BO(3, C_{3}^{4}) = 21,BO(3, C_{3}^{5}) = 46 and 112 ≤ BO(3, C_{3}^{6}) ≤ 114.

The following corollary is a consequence of Theorem 7 and Remark 2.

Corollary 7

BO(4, C_{3}^{2}) = 5, BO(4, C_{3}^{3}) = 10, BO(4, C_{3}^{4}) = 21, BO(4, C_{3}^{5}) = 46 and 112 ≤ BO(4, C_{3}^{6}) ≤ 114.

The following corollary is a consequence of Theorem 7 and Remark 2.

Corollary 8 BO(2n, C_{2} ⊕ C_{2n}) = \{ \begin{matrix}2n + 2 & if n is even, \\ 2n + 3 & if n is odd. \end{matrix}

The following corollary is a consequence of Theorem 7 and Theorem 5.

Corollary 9 BO(2n + 1, C_{2} ⊕ C_{2n}) = \{ \begin{matrix}2n + 2 & if n is even, \\ 2n + 3 & if n is odd. \end{matrix}

Theorem 8 ([13], Theorem 4.2) Let p ≥ 7 be an prime number and \frac{p+1}{2} ≤ k ≤ p − 3. Then BO(k, C_{p}) = k + 1.

Theorem 9 ([13], Theorem 4.3) Let p ≥ 7 be an integer prime number and k = \frac{p−1}{2}. Then

BO(k, C_{p}) = \{ \begin{matrix}k + 1 & if the multiplicity order of 2 module p is odd \\ k + 2 & if it is even. \end{matrix}

4 Existence conditions of BO(k, G) for elementary 2-groups

Let C_{2}^{m} be an elementary 2-group of order 2^{m}. From the results cited in [13] we have that: BO(2^{m}, C_{2}^{m}) = 2^{m}, BO(2^{m} − 1, C_{2}^{m}) = 2^{m} + 1 and BO(2^{m}−2, C_{2}^{m}) = 2^{m}+1. In this section we study the existence of BO(k, C_{2}^{m}) for 3 ≤ k ≤ 2^{m} − 3. In some cases when BO(k, C_{2}^{m}) exists, we give its exact value.

The following result is a consequence of Proposition 4 and Corollary 1.

Corollary 10 Let k be an even integer such that 4 ≤ k ≤ 2^{m−1} − 1. Then BO(k, C_{2}^{m}) < 2^{m}.

The following result provides the existence of BO(2^{m−1}, C_{2}^{m}).

Theorem 10 BO(2^{m−1}, C_{2}^{m}) ≤ 2^{m}.

Proof. Assuming that BO(2^{m−1}, C_{2}^{m}) = 2^{m} + 1, i.e., each (2^{m−1})-subset A of C_{2}^{m} verifies σ(A) ≠ 0 and σ(A^{c}) ≠ 0. If 0 ∈ A, then the (2^{m−1} − 1)-subset B = A \ {0} verifies that σ(B) ≠ 0, that is to say, there is no (2^{m−1} − 1)- subset B with zero-sum over C_{2}^{m}. Else 0 ∈ A^{c}, then the (2^{m−1} − 1)-subset C = A^{c} \ {0} verifies that σ(C) ≠ 0; hence, there is no (2^{m−1} − 1)-subset C with zero-sum over C_{2}^{m}; then a contradiction with Lemma 2. Therefore, BO(2^{m−1}, C_{2}^{m}) ≤ 2^{m}.

The following result gives the existence of BO(k, C_{2}^{m}) for even k and 2^{m−1} + 2 ≤ k ≤ 2^{m} − 4.

Theorem 11 Let k be an even number such that 2^{m−1} + 2 ≤ k ≤ 2^{m} − 4. Then BO(k, C_{2}^{m}) ≤ 2^{m}.

Proof. Assuming that BO(k, C_{2}^{m}) = 2^{m} + 1, i.e., each k-subset A over C_{2}^{m} is not barycentric, in consequence σ(A) ≠ 0 and σ(A^{c}) ≠ 0. Notice that |A^{c}| = 2^{m} − k is an even integer and we have that 4 ≤ 2^{m} − k ≤ 2^{m−1} − 1. So that C_{2}^{m} does not contain a (2^{m} − k)-subset A^{c} with zero-sum, therefore a contradiction with Corollary 10. So that, BO(k, C_{2}^{m}) ≤ 2^{m}.

The following results follow from the last three results .

Corollary 11 Let k be an even integer such that 4 ≤ k ≤ 2^{m} − 4. Then BO(k, C_{2}^{m}) ≤ 2^{m}.

In order to complete the existence of BO(k, C_{2}^{m}), we need to show that BO(k, C_{2}^{m}) ≤ 2^{m} for all even integers 3 < k ≤ 2^{m} − 3.

The following result shows the inexistence of BO(3, C_{2}^{m}).

Theorem 12 BO(3, C_{2}^{m}) = 2^{m} + 1.

Proof. Assuming that BO(3, C_{2}^{m}) ≤ 2^{m}, that is to say, there exists a 3-subset A in C_{2}^{m} such that σ(A) ∈ A. Let B = A \ {σ(A)} be a 2-subset in C_{2}^{m} such that σ(B) = 0 ∈ ∑_{2} C_{2}^{m}; therefore a contradiction with proposition 4.v. Hence, BO(3, C_{2}^{m}) = 2^{m} + 1 .

The following result gives the exact values of BO(2^{m} − 3, C_{2}^{m}).

Theorem 13 BO(2^{m} − 3, C_{2}^{m}) = 2^{m} − 3.

Proof. Assuming that BO(2^{m} − 3, C_{2}^{m}) = 2^{m} + 1, i.e., each (2^{m} − 3)-subset A over C_{2}^{m} is not barycentric, in consequence σ(A) ∉ A; hence σ(A) ∈ A^{c}. Since σ(A) = σ(A^{c}), then σ(A^{c}) ∈ A^{c}. So that, there exists a barycentric 3-subset A^{c} in C_{2}^{m}, that is to say, BO(3, C_{2}^{m}) ≤ 2^{m}; hence a contradiction with the fact that BO(3, C_{2}^{m}) = 2^{m} + 1. Therefore, BO(2^{m} − 3, C_{2}^{m}) = 2^{m} − 3.

To finalize the discussion on the existence conditions of BO(k, C_{2}^{m}) for the odd integers k in 5 ≤ k ≤ C_{2}^{m} − 5 we will use the following results:

Proposition 5 Let k be an even number and BO(k, C_{2}^{m}) = q. Then q > k.

Proof. Assuming that BO(k, C_{2}^{m}) = k, i.e., for each k-subset A over C_{2}^{m} we have that σ(A) = 0. Let A be a k-barycentric set over C_{2}^{m} such that 0 ∈ A and consider the (k − 1)-subset B = A \ {0} over C_{2}^{m}, hence σ(B) = σ(A) = 0. Let 0 ≠ c ∈ B^{c} be and consider the k-subset D = B ∪ {c} over C_{2}^{m}, hence σ(D) = σ(B) + σ({c}) = 0 + c = c ≠ 0. Therefore there exists a non barycentric k-subset D over C_{2}^{m}. Hence a contradiction with the fact that BO(k, C_{2}^{m}) = k. In consequence, q > k.

Theorem 14 Let k be an even integer such that 4 ≤ k ≤ 2^{m} − 4. If BO(k, C_{2}^{m}) = q, then BO(k + 1, C_{2}^{m}) = q.

Proof. Let A be a q-set over C_{m}^{2} and B a k-subset over A such that σ(B) = 0. Since |A| = q > k = |B|, then there exists a ∈ A \ B. Let us consider the set C = B ∪ {a}, notice that it is a (k + 1)-subset over A such that σ(C) = σ(B) + σ({a}) = 0 + a = a with a ∈ C, that is to say, C is a barycentric (k + 1)-subset in A. Hence BO(k + 1, C_{2}^{m}) ≤ q.

Assuming that A is a subset over C_{m}^{2} such that |A| = BO(k + 1, C_{2}^{m}), then A contains a (k + 1)-subset B such that σ(B) = (k + 1)b = kb + b = 0 + b = b for some b ∈ B. Let C = B \ {b} be the k-subset of A such that σ(C) = σ(B) − σ({b}) = b − b = 0. Hence C is a barycentric k-subset in A, i.e., q ≤ BO(k + 1, C_{2}^{m}). Hence, BO(k + 1, C_{2}^{m}) = q.

The following result is a direct consequence of the above theorem.

Corollary 12 Let k be an odd integer such that 5 ≤ k ≤ 2^{m} − 5. Then BO(k, C_{2}^{m}) ≤ 2^{m}.

The following result proves the increase of the values of BO(k, C_{2}^{m}) when k is odd and 4 ≤ k ≤ 2^{m} − 3.

Proposition 6 Let k be an odd integer in 4 ≤ k ≤ 2^{m} − 5. If BO(k, C_{2}^{m}) = q_{1} and BO(k + 1, C_{2}^{m}) = q_{2}, then q_{1} ≤ q_{2}.

Proof. Assuming that q_{2} < q_{1}. Let A be a q_{2}-set over C_{2}^{m} and B a barycentric (k + 1)-subset of A, that is to say, σ(B) = 0. Let b ∈ B be and consider the k-subset C = B \ {b} so that σ(C) = σ(B) − σ({b}) = 0 − b = −b = b, i.e., C is a barycentric k-subset of A. Hence BO(k, C_{2}^{m}) ≤ q_{2} < q_{1}, then a contradiction with the fact that BO(k, C_{2}^{m}) = q_{1}. Therefore, q_{1} ≤ q_{2}.

The following result is a direct consequence of the above result and Theorem 14.

Corollary 13 Let k be an integer such that 4 ≤ k ≤ 2^{m} − 4. Then BO(k + 1, C_{2}^{m}) ≥ BO(k, C_{2}^{m}).

5 Conclusions

The goal of the present paper was to continue with the work in [13] for 3 ≤ k ≤ |G| − 3. Our present main results are Theorem 1 and Theorem 2. The consequence of these two theorems were the complete existence conditions of cyclic groups and elementary p-groups. Moreover, in Section 4 the existence conditions for elementary 2-groups of our constant BO(k, G) was completely determined. The problem of the existence of BO(k, G) for all abelian groups G remains open, and also the problem of assigning exact values of the k-barycentric Olson constant when BO(k, G) exists; some examples are the interesting results given in Theorem 3 and Theorem 4. The relation between the Harborth and the k-barycentric Olson constants established in this paper could be a good option to provide their exact values.

Acknowledgements

The research of O. Ordaz is supported by the Postgrado de la Facultad de Ciencias de la U.C.V., the CDCH project, and the Banco Central de Venezuela.

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