Kourovka Notebook Problem 21.26

Definitions and mathematical statement

Let $G\ne1$ be a finite group, and list the distinct prime divisors of its order as $p_1,\ldots,p_k$. For each $i$, fix a Sylow $p_i$-subgroup $H_i$, meaning a subgroup whose order is the largest power of $p_i$ dividing $|G|$. For $g\in G$, use the conjugation convention $H_i^g=g^{-1}H_i g$ and define the finite family \[ \mathcal I_i=\{H_i\cap H_i^g:g\in G\}. \] A member $J\in\mathcal I_i$ is inclusion-minimal if no other member of $\mathcal I_i$ is a proper subgroup of $J$.

The problem asks whether there always exists one element $x\in G$ such that $H_i\cap H_i^x$ is inclusion-minimal in $\mathcal I_i$ for every $i$ simultaneously. Equivalently, \[ \forall i\ \forall g\in G,\qquad H_i\cap H_i^g\subseteq H_i\cap H_i^x \ \Longrightarrow\ H_i\cap H_i^g=H_i\cap H_i^x. \] Each family individually has a minimal member because it is finite. The assertion requires a single conjugator realizing minimal members for all prime divisors, even though the choices giving minimality for individual primes may initially differ.

Inclusion-minimal does not mean smallest cardinality, nor a least member contained in every other member. Several incomparable minimal intersections are possible. The Sylow subgroups are part of the given input and are fixed before $x$ is chosen. No solubility or simplicity assumption is imposed on $G$.

Short English statement

Can one conjugating element simultaneously make every chosen Sylow subgroup’s intersection with its conjugate inclusion-minimal?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2535 (KOU-21.26), 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 21.26. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.