Kourovka Notebook Problem 21.64

Definitions and mathematical statement

Let $p$ be a prime. A finite group $G$ is $p$-soluble if it has a normal series whose factors are either $p$-groups or $p'$-groups, where a $p'$-group has order prime to $p$. Its $p$-length $\ell_p(G)$ is the smallest number of nontrivial $p$-group factors in any such series; intervening $p'$-factors are not counted. All terms of a normal series are normal in its ambient group.

Choose a Sylow $p$-subgroup $P\le G$, so $|P|$ is the largest power of $p$ dividing $|G|$, and let $A\trianglelefteq P$ have exponent $p^e$. The exponent is the least positive integer annihilating every element by powering. The normal closure of $A$ in $G$ is \[ N=\langle A^g:g\in G\rangle, \qquad A^g=g^{-1}Ag. \] It is a normal $p$-soluble subgroup of $G$, although it need not be a $p$-group.

Is there a function $f$ depending only on $p$ and $e$ such that $\ell_p(N)\le f(p,e)$ for every choice of $G,P,A$ above? More strongly, can one choose a function $b(e)$ independent of the prime with $\ell_p(N)\le b(e)$ in all cases? The bounds must be independent of $|G|$, $|A|$, the index of $P$, and the number of conjugates needed to generate $N$. Normality of $A$ is assumed inside $P$ only, not inside $G$. The case $A=1$, with exponent $p^0=1$, has $\ell_p(N)=0$.

Short English statement

Is the p-length of the normal closure of a normal Sylow-subgroup part bounded by its exponent and the prime, or even by the exponent parameter alone?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2573 (KOU-21.64), 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 21.64. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.