16.82 (2006)

Let $\mathcal{X}$ be a non-empty class of finite groups closed under taking homomorphic images, subgroups, and direct products. With every group $G \in \mathcal{X}$ we associate some set $\tau(G)$ of subgroups of $G$. We say that $\tau$ is a subgroup functor on $\mathcal{X}$ if:
$\qquad$ 1) $G \in \tau(G)$ for all $G \in \mathcal{X}$, and
$\qquad$ 2) for each epimorphism $\varphi : A \to B$, where $A, B \in \mathcal{X}$, and for any $H \in \tau(A)$ and $T \in \tau(B)$ we have $H^\varphi \in \tau(B)$ and $T^{\varphi^{-1}} \in \tau(A)$.

A subgroup functor $\tau$ is closed if for each group $G \in \mathcal{X}$ and for every subgroup $H \in \mathcal{X} \cap \tau(G)$ we have $\tau(H) \subseteq \tau(G)$. The set $\mathcal{F}(\mathcal{X})$ consisting of all closed subgroup functors on $\mathcal{X}$ is a lattice (in which $\tau_1 \leqslant \tau_2$ if and only if $\tau_1(G) \subseteq \tau_2(G)$ for every group $G \in \mathcal{X}$). It is known that $\mathcal{F}(\mathcal{X})$ is a chain if and only if $\mathcal{X}$ is a class of $p$-groups for some prime $p$ (Theorem 1.5.17 in S. F. Kamornikov and M. V. Sel'kin, Subgroups functors and classes of finite groups, Belaruskaya Navuka, Minsk, 2001 (Russian)).

Is there a non-nilpotent class $\mathcal{X}$ such that the width of the lattice $\mathcal{F}(\mathcal{X})$ is at most $|\pi(\mathcal{X})|$ where $\pi(\mathcal{X})$ is the set of all prime divisors of the orders of the groups in $\mathcal{X}$?