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

TL;DR

This paper determines the index of minimal zero-sum sequences of length five over cyclic groups of prime power order, extending previous results from prime order groups and characterizing when the index equals two.

Contribution

It generalizes the index determination for minimal zero-sum sequences from prime order to prime power order cyclic groups, providing explicit conditions for the index to be two.

Findings

The index of minimal zero-sum sequences of length five over prime power order groups is characterized.

If certain gcd conditions are met, the index is exactly two, with explicit sequence forms given.

The result extends known prime order cases to prime power order cyclic groups.

Abstract

Let $G$ be a finite cyclic group. Every sequence $S$ of length $l$ over $G$ can be written in the form $S=(x_1g)\cdot\ldots\cdot(x_lg)$ where $g\in G$ and $x_1, \ldots, x_l\in[1, \ord(g)]$, and the index $\ind(S)$ of $S$ is defined to be the minimum of $(x_1+\cdots+x_l)/\ord(g)$ over all possible $g\in G$ such that $\langle g \rangle =G$. Recently the second and the third authors determined the index of any minimal zero-sum sequence $S$ of length 5 over a cyclic group of a prime order where $S=g^2(x_2g)(x_3g)(x_4g)$. In this paper, we determine the index of any minimal zero-sum sequence $S$ of length 5 over a cyclic group of a prime power order. It is shown that if $G=\langle g\rangle$ is a cyclic group of prime power order $n=p^\mu$ with $p \geq 7$ and $\mu\geq 2$, and $S=(x_1g)(x_2g)(x_2g)(x_3g)(x_4g)$ with $x_1=x_2$ is a minimal zero-sum sequence with $\gcd(n,x_1,x_2,x_3,x_4,x_5)=1$, then $\ind(S)=2$ if and only if $S=(mg)(mg)(m\frac{n-1}{2}g)(m\frac{n+3}{2}g)(m(n-3)g)$ where $m$ is a positive integer such that $\gcd(m,n)=1$.