AI 中文总结
该研究构造一类循环-阿贝尔有限群，得到无扎森豪斯分解的增广保持自同构，证明第二、第三扎森豪斯猜想对这类群不成立，解决了相关问题。
AI 中文摘要
设r>1且gcd(r,30)=1，我们构造有限群Gᵣ=(C₅×C₃×Cᵣ)⋊W，其中|W|=32，Gᵣ'≅C_{60r}，同时构造增广保持自同构αᵣ∈Aut(ℤGᵣ)，该自同构无扎森豪斯分解。像集Yᵣ=αᵣ(Gᵣ)是一个标准化群基，其整体与Gᵣ非有理共轭，但Yᵣ的每个元素分别与Gᵣ的元素有理共轭。因此，有限循环-阿贝尔群的(ZC2)和(ZC3)不成立，解决了Margolis与del Río提出的问题。该构造统一扩展了Hertweck的例子：类保持障碍与整体胶合均独立于辅助因子的阶。该系列中最小的容许成员阶为3360，导出子群为C_{420}。
英文摘要
Let $r>1$ with $\gcd(r,30)=1$. We construct a finite group $G_r=(C_5\times C_3\times C_r)\rtimes W$, where $|W|=32$ and $G_r'\cong C_{60r}$, together with an augmentation-preserving automorphism $α_r\in\operatorname{Aut}(\mathbb{Z}G_r)$ having no Zassenhaus factorization. The image $Y_r=α_r(G_r)$ is a normalized group basis which is not rationally conjugate to $G_r$, although every element of $Y_r$ is individually rationally conjugate to an element of $G_r$. Consequently, (ZC2) and (ZC3) fail for finite cyclic-by-abelian groups, resolving a problem of Margolis and del Río. The construction extends Hertweck's example uniformly: both the class-preserving obstruction and the integral gluing are independent of the order of the auxiliary factor. The smallest admissible member of this family has order $3360$ and derived subgroup $C_{420}$.
Comments17 pages; no figures
