Let $p > 155$ be a prime and let $G$ be a cyclic group of order $p$. Let $S$be a minimal zero-sum sequence with elements over $G$, i.e., the sum ofelements in $S$ is zero, but no proper nontrivial subsequence of $S$ has sumzero. We call $S$ is unsplittable, if there do not exist $g$ in $S$ and $x,y\in G$ such that $g=x+y$ and $Sg^{-1}xy$ is also a minimal zero-sum sequence.In this paper we show that if $S$ is an unsplittable minimal zero-sum sequenceof length $|S|= \frac{p-1}{2}$, then$S=g^{\frac{p-11}{2}}(\frac{p+3}{2}g)^4(\frac{p-1}{2}g)$ or$g^{\frac{p-7}{2}}(\frac{p+5}{2}g)^2(\frac{p-3}{2}g)$. Furthermore, if $S$ is aminimal zero-sum sequence with $|S| \ge \frac{p-1}{2}$, then $\ind(S) \leq 2$.
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机译：假设$ p> 155 $是质数，而$ G $是订单$ p $的循环组。假设$ S $是元素大于$ G $的最小零和序列，即$ S $中的元素之和为零，但是$ S $的适当非平凡子序列都不具有sumzero。我们称$ S $是不可分裂的，如果在$ S $和$ x，y \ G $中不存在$ g $，使得$ g = x + y $和$ Sg ^ {-1} xy $也是最小零和序列。本文证明，如果$ S $是长度为$ | S | = \ frac {p-1} {2} $的不可分裂的最小零和序列，则$ S = g ^ { \ frac {p-11} {2}}（\ frac {p + 3} {2} g）^ 4（\ frac {p-1} {2} g）$或$ g ^ {\ frac {p- 7} {2}}（\ frac {p + 5} {2} g）^ 2（\ frac {p-3} {2} g）$。此外，如果$ S $是带有$ | S |的最小零和序列。 \ ge \ frac {p-1} {2} $，然后是$ \ ind（S）\ leq 2 $。
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