AI 中文总结
针对Ginosar提出的每个G-调和元组是否为Z-调和的问题，本文证明长度为4的界是紧的，给出了A5-调和但非Z-调和的5元组例子，还证明了与可能反例相关的一类5元组对任意群都不是G-调和的。
AI 中文摘要
正整数的$n$元组若存在$G$的子群，其指数为该元组元素且陪集可两两不交，则称为$G$-调和；若存在模数为该元组元素且两两不交的同余类，则称为$\boldsymbol{\text{Z}}$-调和。Ginosar提出问题：每个$G$-调和元组是否都是$\boldsymbol{\text{Z}}$-调和？Margolis与Schnabel证明长度不超过4的元组满足该结论，并分析了一类长度为5的元组，若其中任意一个是$G$-调和的，则会构成反例。本文证明该界4是紧的：$(6,6,6,10,15)$是$A_5$-调和但非$\boldsymbol{\text{Z}}$-调和；且实现该元组的5个两两不交陪集可仅用指数为6、10、15的陪集扩展为$A_5$的陪集划分，该划分的指数元组非$\boldsymbol{\text{Z}}$-调和，因指数重复，不与Herzog–Schönheim猜想矛盾。本文还证明，Margolis与Schnabel分析的、与可能反例相关的长度5元组中，没有任何一个对任意群$G$是$G$-调和的。
英文摘要
An $n$-tuple of positive integers is $G$-harmonic if there are subgroups of $G$ having those indices whose cosets can be chosen pairwise disjoint, and $\mathbb{Z}$-harmonic if there are pairwise disjoint residue classes with those moduli. Ginosar asked whether every $G$-harmonic tuple is $\mathbb{Z}$-harmonic. Margolis and Schnabel proved this for tuples of length at most $4$, and analysed a particular family of length-$5$ tuples that would yield a counterexample if any member were $G$-harmonic. We show that the bound $4$ is sharp: $(6,6,6,10,15)$ is $A_5$-harmonic but not $\mathbb{Z}$-harmonic. Moreover, the five pairwise disjoint cosets realising this tuple can be extended to a coset partition of $A_5$ using only cosets of indices $6$, $10$, and $15$. The index tuple of this partition is not $\mathbb{Z}$-harmonic; because its indices repeat, this does not contradict the Herzog--Schönheim conjecture. We also prove that no member of the length-$5$ family analysed by Margolis and Schnabel in connection with possible counterexamples is $G$-harmonic for any group $G$.
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