Kourovka Notebook Problem 21.50

Definitions and mathematical statement

Let $T\ne1$ be a finite $3$-group, meaning that its order is a positive power of three. A subgroup $C\le T$ is characteristic if it is invariant under every automorphism of $T$. For any finite group $G$ containing $T$, say $T$ is a Sylow $3$-subgroup when $|T|$ is the largest power of three dividing $|G|$.

For a subgroup $C\le G$, its normalizer is $N_G(C)=\{g\in G:g^{-1}Cg=C\}$. The derived subgroup $G'$ is generated by all commutators $a^{-1}b^{-1}ab$ with $a,b\in G$. The question asks whether every such $T$ has one nontrivial characteristic subgroup $C$ with the following universal property: \[ \text{for every finite }G\text{ with }T\in\operatorname{Syl}_3(G), \qquad T\cap G'=T\cap N_G(C)'. \] Thus $C$ is chosen from $T$ before the finite overgroup $G$ is specified. Its choice cannot depend on the overgroup.

Because $C$ is characteristic in $T$, the group $T$ normalizes $C$ and lies in $N_G(C)$. The right-hand intersection is therefore meaningful inside the same copy of $T$, and it is always contained in the left one. The question asks for the opposite containment uniformly over all overgroups. The group $T=1$ is excluded explicitly because no nontrivial subgroup $C$ could then exist. Replacing one characteristic subgroup by a family is a different conclusion.

Short English statement

Does every nontrivial finite $3$-group have a characteristic subgroup whose normalizer controls its intersection with the derived subgroup in every Sylow overgroup?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2559 (KOU-21.50), 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 21.50. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.