Minimal zero-sum sequence of length five over finite cyclic groups of prime power order

preprint2014arXivOpen access

Abstract

Let GG be a finite cyclic group. Every sequence SS of length ll over GG can be written in the form S=(x1g)⋅…⋅(xlg)S=(x_1g)\cdot\ldots\cdot(x_lg) where g∈Gg\in G and $x_1, \ldots, x_l\in[1, \ord(g)]$, and the index $\ind(S)$ of SS is defined to be the minimum of $(x_1+\cdots+x_l)/\ord(g)$ over all possible g∈Gg\in G such that ⟨g⟩=G\langle g \rangle =G. Recently the second and the third authors determined the index of any minimal zero-sum sequence SS of length 5 over a cyclic group of a prime order where S=g2(x2g)(x3g)(x4g)S=g^2(x_2g)(x_3g)(x_4g). In this paper, we determine the index of any minimal zero-sum sequence SS of length 5 over a cyclic group of a prime power order. It is shown that if G=⟨g⟩G=\langle g\rangle is a cyclic group of prime power order n=pμn=p^μ with p≥7p \geq 7 and μ≥2μ\geq 2, and S=(x1g)(x2g)(x2g)(x3g)(x4g)S=(x_1g)(x_2g)(x_2g)(x_3g)(x_4g) with x1=x2x_1=x_2 is a minimal zero-sum sequence with gcd⁡(n,x1,x2,x3,x4,x5)=1\gcd(n,x_1,x_2,x_3,x_4,x_5)=1, then $\ind(S)=2$ if and only if S=(mg)(mg)(mn−12g)(mn+32g)(m(n−3)g)S=(mg)(mg)(m\frac{n-1}{2}g)(m\frac{n+3}{2}g)(m(n-3)g) where mm is a positive integer such that gcd⁡(m,n)=1\gcd(m,n)=1.

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