Kourovka Notebook Problem 21.67

Definitions and mathematical statement

For a finite group $K$, let $d(K)$ be the least number of generators needed to generate $K$, with $d(1)=0$. Its Prüfer rank is \[ \operatorname{rk}(K)=\max\{d(L):L\le K\}. \] Thus every subgroup, not only $K$ itself, must be generated by at most this many elements. A bound on rank need not bound the order of $K$.

Let $G$ be a finite soluble group and $\phi\in\operatorname{Aut}(G)$. Set $m=|\phi|$ and $r=\operatorname{rk}(C_G(\phi))$, where $C_G(\phi)=\{g\in G:\phi(g)=g\}$. Soluble means that repeated commutator subgroups eventually become trivial. Its Fitting height $h(G)$ is the least length of a normal series with nilpotent factors. Nilpotent means that iterating $\gamma_{i+1}(L)=[\gamma_i(L),L]$, starting with $\gamma_1(L)=L$, eventually gives $1$.

Does there exist a function $f(m,r)$ such that $h(G)\le f(m,r)$ for all such $G$ and $\phi$? The bound must be independent of the order, exponent, and particular structure of the fixed-point subgroup, apart from its rank. It must also be independent of $|G|$.

No coprimality of $m$ and $|G|$ is assumed, and $m$ may have arbitrarily many prime divisors. The assertion concerns the whole group $G$; it does not permit passing to a subgroup or quotient to obtain the height bound. Rank zero is allowed when $C_G(\phi)=1$.

Short English statement

Can the Fitting height of a finite soluble group be bounded using only an automorphism order and the Prüfer rank of its fixed-point subgroup?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2576 (KOU-21.67), 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 21.67. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.