Kourovka Notebook Problem 20.83

Definitions and mathematical statement

Let $\mathfrak{X}$ be a class of finite groups closed under taking subgroups, homomorphic images, and extensions. Closure under extensions means that a finite group $E$ belongs to $\mathfrak{X}$ whenever it has a normal subgroup $D$ with $D\in\mathfrak{X}$ and $E/D\in\mathfrak{X}$.

An $\mathfrak{X}$-group is a group belonging to $\mathfrak{X}$. A subgroup $M\leq L$ is $\mathfrak{X}$-maximal in $L$ if $M\in\mathfrak{X}$ and there is no subgroup $U\leq L$ such that $M<U$ and $U\in\mathfrak{X}$. In particular, this does not require $M$ to be a maximal proper subgroup of $L$.

A subgroup $K\leq L$ is subnormal if there is a finite chain \[ K=K_0\trianglelefteq K_1\trianglelefteq\cdots\trianglelefteq K_r=L. \] For a finite group $K$, a subgroup $A\leq K$ is $\mathfrak{X}$-submaximal if there exist an injective homomorphism $\iota:K\to K^*$ and an $\mathfrak{X}$-maximal subgroup $M\leq K^*$ such that $\iota(K)$ is subnormal in $K^*$ and \[ \iota(A)=\iota(K)\cap M. \] The supplied definition calls $K^*$ a group without explicitly requiring it to be finite. This draft retains that wording; whether the intended definition restricts $K^*$ to finite groups requires confirmation.

Question. For every such class $\mathfrak{X}$, every finite group $G$, and every characteristic subgroup $N$ of $G$, suppose that all $\mathfrak{X}$-submaximal subgroups of $N$ are conjugate in $N$: for any two such subgroups $A,B\leq N$, there exists $n\in N$ such that $B=n^{-1}An$. Here characteristic means that $\alpha(N)=N$ for every automorphism $\alpha$ of $G$. Must it follow that, for every $\mathfrak{X}$-submaximal subgroup $H\leq G$, the subgroup \[ HN/N=\{hN:h\in H\}\leq G/N \] is $\mathfrak{X}$-submaximal in $G/N$?

Supplied qualifications. The source states that replacing the characteristic hypothesis by normality does not suffice, even if $N\in\mathfrak{X}$, or if $N$ contains no nontrivial $\mathfrak{X}$-subgroup. The latter condition means that $U\leq N$ and $U\in\mathfrak{X}$ imply $U=1$. The source also reports a positive answer when $N$ is the $\mathfrak{F}$-radical of $G$ for a Fitting class $\mathfrak{F}$. Here a Fitting class is a nonempty class of finite groups closed under normal subgroups and products of normal subgroups belonging to the class; its radical in $G$ is the largest normal subgroup of $G$ belonging to that class. These counterexample and positive-result assertions are supplied background, not independently verified results in this draft.

Short English statement

Suppose all $\mathfrak{X}$-submaximal subgroups of a characteristic subgroup $N$ of a finite group $G$ are conjugate within $N$. Must the image of every $\mathfrak{X}$-submaximal subgroup of $G$ be $\mathfrak{X}$-submaximal in $G/N$? The supplied record labels this problem open. The intended finiteness requirement on the ambient group in the definition needs confirmation.

Sources

• Local UnsolvedMath registry, permanent ID $701741$, exact source problem code catalogue entry, title Kourovka Notebook Problem 20.83. The supplied record labels the problem open; that status is not independently verified here. The record gives creation and update timestamps 2026-10-16T14:00:00Z.
• The Kourovka Notebook (Unsolved Problems in Group Theory), Problem 20.83, 20th issue (2022), also identified in the supplied background with the 21st issue edition (2026), arXiv:1401.0300v48. The supplied collection description names E. I. Khukhro and V. D. Mazurov as editors.
• Proposed by D. O. Revin and A. V. Zavarnitsine; proposed year supplied as 2022.
• The normal-subgroup counterexamples and the positive result for a Fitting-class radical are reported in the supplied statement without separate supporting references. No external source was consulted for this draft.