Kourovka Notebook Problem 21.31

Definitions and mathematical statement

Let $N$ be a finite group and let $\operatorname{Aut}(N)$ be its group of automorphisms. Its holomorph is the semidirect product \[ \operatorname{Hol}(N)=N\rtimes\operatorname{Aut}(N),\qquad (n,\alpha)(m,\beta)=(n\alpha(m),\alpha\beta). \] It acts on the underlying set of $N$ by $(n,\alpha)\cdot x=n\alpha(x)$. This action is faithful. A subgroup $R\le\operatorname{Hol}(N)$ is regular if for every $x,y\in N$ exactly one element of $R$ sends $x$ to $y$. Equivalently, the action is transitive and every point stabilizer is trivial; hence $|R|=|N|$.

A group $K$ is soluble if its derived series $K^{(0)}=K$, $K^{(i+1)}=[K^{(i)},K^{(i)}]$ reaches $1$, where brackets mean the subgroup generated by commutators. The conjecture states \[ N\text{ finite soluble},\quad R\le\operatorname{Hol}(N)\text{ regular} \quad\Longrightarrow\quad R\text{ soluble}. \] The subgroup $R$ is not assumed normal in the holomorph, and need not be its evident translation subgroup $N\times\{\mathrm{id}\}$. Its multiplication is the multiplication inherited from the holomorph, not a multiplication transported from $N$ by an arbitrary set bijection.

Regularity specifies a particularly restrictive embedding of a group of order $|N|$ into the permutation group of $N$. The question does not claim that every group of that order is soluble. Both the algebraic solubility assumption and the regularity of this specific action must hold in each instance.

Short English statement

Must every regular subgroup of the holomorph of a finite soluble group itself be soluble?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2540 (KOU-21.31), 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 21.31. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.