Kourovka Notebook Problem 21.112

Definitions and mathematical statement

A complete class $\mathcal X$ is a nonempty class of finite groups closed under subgroups, quotient groups, and extensions: if $N\trianglelefteq G$ and $N,G/N\in\mathcal X$, then $G\in\mathcal X$. Assume $\mathcal X$ is proper, so it does not contain every finite group. Its symmetric boundary is $n=\max\{r:S_r\in\mathcal X\}$, where $S_r$ is the permutation group on $r$ points. This maximum is finite because every finite group embeds in some symmetric group.

The Baer--Suzuki width $\operatorname{BS}(\mathcal X)$ is the least nonnegative integer $m$ with the following property: for every finite group $G$ and each conjugacy class $D\subset G$, if $\langle d_1,\ldots,d_m\rangle\in\mathcal X$ for every tuple $(d_1,\ldots,d_m)\in D^m$, then $\langle D\rangle\in\mathcal X$. Repetitions in the tuple are allowed. The source establishes finiteness of this width and bounds it in terms of the symmetric boundary.

For each positive integer $n\ne3$, set \[ f_+(n)=\max_{\partial_s\mathcal X=n}\operatorname{BS}(\mathcal X),\qquad f_-(n)=\min_{\partial_s\mathcal X=n}\operatorname{BS}(\mathcal X), \] where $\partial_s$ denotes symmetric boundary and the extrema range over all proper complete classes. Such classes exist for precisely the indicated $n$. Determine $f_+(4),f_+(5),f_+(6)$, and decide whether $f_-(n)=n$ for every positive $n\ne3$.

Short English statement

Find the largest Baer--Suzuki widths at symmetric boundaries four, five and six, and decide whether the smallest width always equals the boundary.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2621 (KOU-21.112), 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 21.112. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.