Kourovka Notebook Problem 10.61

Definitions and mathematical statement

Let $G$ be a group with identity $1$, let $H<G$ be a proper subgroup, and let $a\in H$ satisfy $a^2\ne 1$. Write $a^g=g^{-1}ag$, and let $\langle S\rangle$ denote the subgroup generated by a subset $S\subseteq G$.

The source refers to Problem 6.55 for the definitions of a Frobenius group, a Frobenius complement, and a Frobenius kernel. That definition is not included in the supplied record. These terms below therefore retain their source-intended meanings; in particular, no finiteness or splitting hypothesis is added. The exact conventions for possibly infinite Frobenius groups and their kernels require verification against Problem 6.55.

Assume that \[ \forall g\in G\setminus H,\quad F_g=\langle a,a^g\rangle \text{ is a Frobenius group admitting a complement }C_g \text{ with }a\in C_g. \] The complement $C_g$ is required to contain $a$; it is not required by this hypothesis to equal $\langle a\rangle$.

Define the family of source-intended Frobenius kernels \[ \mathcal K_a(G)=\left\{K:\text{ there exists a subgroup }F\le G \text{ which is a Frobenius group with complement }\langle a\rangle \text{ and corresponding kernel }K\right\}, \] and its set-theoretic union \[ U_a(G)=\bigcup_{K\in\mathcal K_a(G)}K. \] The question is whether, for every $(G,H,a)$ satisfying the stated hypotheses, \[ U_a(G)\le G. \] This union ranges over all such Frobenius subgroups of $G$, not merely the groups $F_g$. Its precise interpretation remains conditional on the missing definitions from Problem 6.55.

The supplied editors' comment, dated 2005, reports an affirmative answer under either of the following additional conditions:

1. The order of $a$ is even.
2. The order of $a$ is neither $3$ nor $5$, and $\langle a,a^g\rangle$ is finite for every $g\in G\setminus H$.

These are supplied historical assertions, not independently verified results in this draft. The registry labels the general problem as open.

Short English statement

Given a proper subgroup $H<G$ and $a\in H$ with $a^2\ne1$, assume that each $\langle a,a^g\rangle$, for $g\notin H$, is a Frobenius group with a complement containing $a$. Must the union of the kernels of all Frobenius subgroups of $G$ having complement exactly $\langle a\rangle$ be a subgroup? The source reports two affirmative special cases, but its referenced definitions for Frobenius groups and kernels must be checked.

Sources

• Supplied local UnsolvedMath registry record, permanent ID $700676$, exact source code catalogue entry, title Kourovka Notebook Problem 10.61; upstream revision d371ee917ac9647fbb06cd6fc529deec6f3dda82. The supplied status is open; the record was updated on 2026-10-16.
• The Kourovka Notebook (Unsolved Problems in Group Theory), Problem 10.61, proposed by A. I. Sozutov in the 10th issue (1986). The supplied background identifies the 21st issue edition (2026), arXiv:1401.0300v48. The collection description attributes its editing to E. I. Khukhro and V. D. Mazurov.
• Problem 6.55 of The Kourovka Notebook, cited by the supplied statement for definitions; its text is not supplied.
• A. I. Sozutov, Algebra and Logic, 34, no. 5 (1995), 295--305; supplied as a related reference. No article title is supplied.
• Editors' comment (2005): A. M. Popov and A. I. Sozutov, Algebra and Logic, 44, no. 1 (2005), 40--45, cited for the affirmative result when the order of $a$ is even. No article title is supplied.
• Editors' comment (2005): A. M. Popov, Algebra and Logic, 43, no. 2 (2004), 123--127, cited for the affirmative result when the order of $a$ is neither $3$ nor $5$ and all groups $\langle a,a^g\rangle$, with $g\notin H$, are finite. No article title is supplied.
• All references, attribution, and historical assessments above are transcribed from the supplied record and have not been independently verified.