Kourovka Notebook Problem 21.57

Definitions and mathematical statement

Let $\mathcal X$ be a nonempty class of finite groups, each of odd order. Assume it is closed under isomorphism, subgroups, quotients and extensions. The last condition means that if $N\trianglelefteq K$ and both $N$ and $K/N$ belong to $\mathcal X$, then $K$ belongs to $\mathcal X$. Because the class is nonempty and subgroup-closed, it contains the trivial group.

For a finite group $G$, an $\mathcal X$-maximal subgroup is a subgroup $H\le G$ such that $H\in\mathcal X$ and there is no strictly larger subgroup $J\le G$ belonging to $\mathcal X$. This is maximality within the specified class, not necessarily maximality among all proper subgroups. If $G\in\mathcal X$, the subgroup $G$ itself is $\mathcal X$-maximal.

Given an $\mathcal X$-maximal subgroup $H$ of any finite group $G$ and a normal subgroup $N\trianglelefteq G$, must $H\cap N$ be $\mathcal X$-maximal in $N$? In explicit terms, does \[ H\cap N\le J\le N,\quad J\in\mathcal X \quad\Longrightarrow\quad J=H\cap N \] always hold?

Membership $H\cap N\in\mathcal X$ already follows from subgroup closure. The issue is its maximality. The ambient finite group may have even order and need not belong to $\mathcal X$. The odd-order restriction is imposed on the class members, and all allowed classes, normal subgroups and class-maximal choices of $H$ are quantified over.

Short English statement

Does intersecting a class-maximal subgroup with a normal subgroup preserve maximality for every extension-closed class of finite odd-order groups?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2566 (KOU-21.57), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.57. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.