Math-1231 · mathematics · Group theory

Math-1231

Kourovka Notebook Problem 21.65

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

For a finite soluble group $L$, its Fitting height $h(L)$ is the least length $h$ of a normal series $1=L_0\trianglelefteq\cdots\trianglelefteq L_h=L$ having nilpotent factors; set $h(1)=0$. A group is nilpotent when its lower central series terminates at $1$. Solubility means that the derived series, obtained by repeatedly taking commutator subgroups, terminates.

Let $\phi\in\operatorname{Aut}(G)$ for a finite soluble group $G$. Write $m=|\phi|$ for the order of the automorphism and $s=|C_G(\phi)|$, where $C_G(\phi)=\{g\in G:\phi(g)=g\}$. Part (a) asks whether functions $f(m,s)$ and $a(m)$ exist such that every such pair $(G,\phi)$ has a subgroup $H\le G$ satisfying \[ [G:H]\le f(m,s),\qquad h(H)\le a(m). \] Part (b) asks the stronger height bound $h(H)\le b(\Omega(m))$, while still allowing the index bound to depend on $m,s$. Here $\Omega(m)$ is the number of prime factors of $m$ counted with multiplicity, with $\Omega(1)=0$. This is the composition length of the cyclic group $\langle\phi\rangle$.

All bounding functions must work for every group and automorphism with the indicated parameters. The subgroup may depend on both. It is not required to be normal or $\phi$-invariant. Neither question assumes the automorphism order is coprime to $|G|$. In particular, part (b) allows the primes themselves to vary while keeping their total multiplicity fixed.

Short English statement

Does a finite soluble group admit a bounded-index subgroup of Fitting height controlled solely by the order, or the composition length, of a given automorphism?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.