Relation between the Frattini property and pronormal subgroups of solvable groups

Question

A subgroup $H$ of $G$ is said to satisfy the Frattini Property if for any subgroup $K$ and $L$ such that $H\leq K \unlhd L$ implies that $L \leq N_L(H)K$.

A subgroup is $H$ is pronormal in $G$ if for each $g \in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x = H^g$.

A theorem characterising pronormal subgroups of soluble groups was proved by T. Peng 'Pronormality in finite groups' which stated: if $G$ is finite soluble group, $H$ is pronormal in $G$ $\iff$ H satisfies the Frattini Property. I do not have access to this paper nor does my institution have access to this.

The $\Rightarrow$ direction is true in general since for any $g\in G$, $\langle H, H^g \rangle \leq H^{\langle g \rangle}$ and using my previous question.

For the $\Leftarrow$ direction, solvability of the group will be needed. I'm not sure how to proceed with proving this implication.

Answer 1

Personally I would try to persuade someone who does have access to the paper to e-mail it to me, but that might not be legal, so I shouldn't have said it.

I have not thought this out in detail, but I think the following approach will work.

Use induction on $|G|$. Let $N$ be a minimal normal subgroup of $G$, so $N$ is an elementary abelian $p$-group for some prime $p$. Apply inductive hypothesis to $G/N$ to get $HN$ and $H^gN$ conjugate by an element of $\langle H,H^g \rangle$. So now we can assume that $HN=H^gN$.

Then $g \in N_G(HN)$ and so the Frattini property implies that $H$ and $H^g$ are conjugate in $HN$, and hence we can assume that $g \in HN$ and $HN=G$.

So $H$ acts irreducibly on $N$, and hence either $G=H$ and we are done, or $H$ is a complement of $N$ in $G$ and a maximal subgroup of $G$. Then either $H=H^g$ or $\langle H,H^g \rangle=G$ and we are done.