Kourovka Notebook Problem 21.54

Definitions and mathematical statement

Let $G$ be a finite soluble group, meaning its iterated commutator subgroups eventually become trivial. Suppose a subgroup $S\le\operatorname{Aut}(G)$ is isomorphic to $S_3$, with generators $\sigma,\rho$ satisfying $\sigma^2=\rho^3=1$ and $\sigma\rho\sigma=\rho^{-1}$. Write $g^s=s(g)$ and $[g,s]=g^{-1}g^s$ for $g\in G,s\in S$. Define \[ M(G)=\{[g,\sigma]:g\in G\}. \] The triality assumption is the identity $m\,m^\rho m^{\rho^2}=1$ for every $m\in M(G)$, in that order. The set $M(G)$ need not be a subgroup.

Assume also $G=[G,S]$, where $[G,S]$ is generated by all $[g,s]$. For positive integers $d,n$, suppose there exist $m_1,\ldots,m_d\in M(G)$ such that \[ G=\langle m_i^s:1\le i\le d,\ s\in S\rangle, \qquad m^n=1\quad\text{for every }m\in M(G). \] The exponent condition is imposed on all members of $M(G)$, not merely the selected generators.

For a finite soluble group, its Fitting height is the least length of a normal series with nilpotent factors; a group is nilpotent when its lower central series terminates at $1$. Is there a function $f(d,n)$ such that $h(G)\le f(d,n)$ for every triality group satisfying these conditions? The bound must be independent of $|G|$ and of the particular automorphisms realizing triality.

Short English statement

For finite soluble triality groups generated by $d$ triality elements and their images, does bounded exponent on those elements bound Fitting height in terms of $d$ and $n$?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2563 (KOU-21.54), 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 21.54. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.