15.39 (2002)
SolvedAxiomatizing the basic properties of subnormal subgroups, we say that a functor $\tau$ associating with every finite group $G$ some non-empty set $\tau(G)$ of its subgroups is an ETP-functor if
$\qquad$ 1) $\tau(A)^\varphi \subseteq \tau(B)$ and $\tau(B)^{\varphi^{-1}} \subseteq \tau(A)$ for any epimorphism $\varphi : A \rightarrow B$, as well as $\{ H \cap R \mid R \in \tau(G) \} \subseteq \tau(H)$ for any subgroup $H \leqslant G$;
$\qquad$ 2) $\tau(H) \subseteq \tau(G)$ for any subgroup $H \in \tau(G)$;
$\qquad$ 3) $\tau(G)$ is a sublattice of the lattice of all subgroups of $G$.
Let $\tau$ be an ETP-functor. Does there exist a hereditary formation $\mathfrak{F}$ such that $\tau(G)$ coincides with the set of all $\mathfrak{F}$-subnormal subgroups in any finite group $G$?
Progress
This is true for finite soluble groups (A. F. Vasil'ev, S. F. Kamornikov, Siberian Math. J., 42, no. 1 (2001), 25–33).
In general, not always (S. F. Kamornikov, Math. Notes, 89, no. 3–4 (2011), 340–348).
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