If $G$ is a finite solvable group with chief series $1 = G_0 \leq \ldots \leq G_n = G$ (so each $G_i \unlhd G$ and if $G_{i-1} < H < G_i$ then $H$ is not normal in $G$) and $I \subseteq \{ 1,\ldots,n\}$. Does there exist a subgroup $H \leq G$ such that $G_{i-1} H \geq G_i$ when $i \in I$ and $G_i \cap H \leq G_{i-1}$ when $i \notin I$?
In other words, can I choose which chief factors a subgroup covers and avoids?
(If so, can we describe how to find such a subgroup? If not, are there easy properties $I$ needs to have, like containing all the Frattini chief factors?)
Here are some example sets $I$ that we can use:
- $I_p = \{ i : [G_i:G_{i-1}] \text{ is divisible by } p \}$, $H$ is a Sylow $p$-subgroup
- $I_{p'} = \{ i : [G_i:G_{i-i}] \text{ is not divisible by } p \}$, $H$ is a Sylow $p$-complement
- $I_{\pi} = \{ i : [G_i:G_{i-1}] \text{ is a $\pi$-number} \}$, $H$ is a Hall $\pi$-subgroup
- $I_c = \{ i : G_i/G_{i-1} \leq Z(G/G_{i-1}) \}$, $H$ is a Sylow system normalizer
- $I_\Phi = \{ i : G_i/G_{i-1} \leq \Phi(G/G_{i-1}) \}$, then $H$ is a pre-Frattini subgroup
Indeed these can almost by used as definitions for the $H$ (there is a technical condition about behaving well with a Sylow system for the last one or two, I think). All of these subgroups exist in every finite soluble group.
Do the allowable $I$ at least form a Boolean algebra? $I_c \cup I_\Phi$ is allowable (by Hawkes, though I don't know the details). What about the complement of $I_c$?