2010
DOI: 10.1016/j.jnt.2009.12.005
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Minimal zero sum sequences of length four over finite cyclic groups

Abstract: Text. Let G be a finite cyclic group. Every sequence S over G can be written in the form S = (n 1 g) · . . . · (n l g) where g ∈ G and n 1 , . . . ,n l ∈ [1, ord(g)], and the index ind(S) of S is defined to be the minimum of (n 1 + · · · + n l )/ord(g) over all possible g ∈ G such that g = supp(S) . The problem regarding the index of sequences has been studied in a series of papers, and a main focus is to determine sequences of index 1. In the present paper, we show that if G is a cyclic of prime power order s… Show more

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Cited by 25 publications
(28 citation statements)
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“…It was first addressed by Kleitman-Lemke (see the conjecture in [23, p. 344]), used as a key tool by Geroldinger [20, p. 736], and then investigated by Gao [13] in a systematical way. Since then it has received a great deal of attention (see, for example, [6,15,17,24,25,27,28,[32][33][34][35][36]39]). …”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…It was first addressed by Kleitman-Lemke (see the conjecture in [23, p. 344]), used as a key tool by Geroldinger [20, p. 736], and then investigated by Gao [13] in a systematical way. Since then it has received a great deal of attention (see, for example, [6,15,17,24,25,27,28,[32][33][34][35][36]39]). …”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…Thus, we may assume n > 1000. Moreover, it is known that S has index 1 if n has at most two distinct prime factors (see [8], [7] and [14]). Hence we may also assume that n has at least three distinct prime factors.…”
Section: Proof Of Theoremmentioning
confidence: 99%
“…In [8], Y. Li et al proved that if n is a prime power then Conjecture 1 is true. Later the case when n has two distinct prime factors was also proved (see [7] and [14]).…”
Section: Introductionmentioning
confidence: 99%
“…It has no zero entries, no sum of two entries is zero, and a + b − 1 is nonzero. It was shown in [5] that there is some invertible k ∈ Z/pZ such that (ak) p + (bk) p + (−k) p < p, equivalently (−ak) p + (−bk) p + (k) p < p, hence by Proposition 3.3, U is not Kummer. 2…”
Section: -Dimensional Kummer Spacesmentioning
confidence: 99%
“…The proof of this algebraic fact requires a nontrivial result from elementary number theory ( [5], see also [3]). Finally, we prove in Section 4 that p + 1 is the upper bound for the dimension of any Kummer subspace in the generic symbol algebra.…”
Section: Introductionmentioning
confidence: 99%