最小 $\mathcal{F}$-覆盖：超可解和可解情况
Minimum $\mathcal{F}$-covers: the supersolvable and metabelian cases
AI总结：
本文研究了超可解和可解群类中最小 $\mathcal{F}$-覆盖的存在性问题，证明在超可解和可解情况下不存在这样的最小覆盖。
摘要：
给定一个有限群集合 $\mathcal{F}$，若群 $G$ 中每一个 $\mathcal{F}$ 中的群都同构于 $G$ 的子群，则称 $G$ 为 $\mathcal{F}$-覆盖。此外，若不存在阶数小于 $|G|$ 的 $\mathcal{F}$-覆盖，则称 $G$ 为最小 $\mathcal{F}$-覆盖。在[Cameron P. J., et al., Minimal cover groups, J. Algebra 660 (2024)]中，作者提出了以下问题：对于哪些闭合于子群和直积下的群类 $\mathcal{X}$，若 $\mathcal{F}$ 是 $\mathcal{X}$-群的集合，则存在一个最小 $\mathcal{F}$-覆盖且该覆盖也是 $\mathcal{X}$-群？在本文中，我们给出了两种情况下否定回答：$\mathcal{X}\in \{``supersolvable
英文摘要：
Given a set $\mathcal{F}$ of finite groups, it is said that a group $G$ is an $\mathcal{F}$-cover if every group in $\mathcal{F}$ is isomorphic to a subgroup of $G$. Moreover, $G$ is a minimum $\mathcal{F}$-cover if there is no $\mathcal{F}$-cover whose order is less than $|G|$. In [Cameron P. J., et al., Minimal cover groups, J. Algebra 660 (2024)], the authors pose the following question: For which classes $\mathcal{X}$ of groups, closed under taking subgroups and direct products, is it true that, if $\mathcal{F}$ is a set of $\mathcal{X}$-groups, then there is a minimum $\mathcal{F}$-cover which is an $\mathcal{X}$-group? In this paper, we give a negative answer in two cases: $\mathcal{X}\in \{``supersolvable", ``metabelian"\}.$
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
