The least non-partitionable zero-sum subset for zero-sum triples in finite abelian groups

Let $G$ be a finite abelian group, and let $\mu (G)$ denote the least size of a subset $S\subseteq G$ with $3\mid \mathrm{\mid }S\mathrm{\mid }$, total sum zero, and no partition into zero-sum triples; put $\mu (G)=\mathrm{\infty }$ if no such subset exists. We prove the exact classification $\mu (G)=\mathrm{\infty }$ precisely for groups of order at most $8$ and for $G\cong C_3^2$, while $\mu (G)=6$ for every other finite abelian group. The cyclic special case gives $\mu ({\mathbb{Z}}_n)=\mathrm{\infty }$ for $1\le n\le 8$ and $\mu ({\mathbb{Z}}_n)=6$ for $n\ge 9$, answering the corresponding non-partite cyclic question. We also record a higher-uniformity interval construction which explains the large cyclic witnesses as the case $k=3$ of a general $k$-tuple obstruction.