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 that: if $G$ is 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 https://math.stackexchange.com/questions/1810173/frattini-property-of-a-subgroup

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


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.

  • $\begingroup$ Let $g\in G$. If $H$ satisfies the Frattini property in $G$ and $N \unlhd G$, then $HN/N$ satisfies the Frattini Property in $G/N$. Then by induction $HN/N$ prn $G/N$. Hence there exists $x\in \langle H, H^g\rangle$ such that $H^xN = H^gN$. Why can we assume that $HN = H^gN$? $\endgroup$ – R Maharaj Jun 4 '16 at 14:57
  • $\begingroup$ If I can find $y \in \langle H^x,H^g \rangle$ with $H^{xy}=H^g$ then $xy \in \langle H,H^g \rangle$ and we are done. So I can replace $H$ by $H^x$. $\endgroup$ – Derek Holt Jun 4 '16 at 15:54
  • $\begingroup$ $g\in N_G(HN)$ and since $H\leq HN \unlhd N_G(HN)$, by the Frattini property, $N_G(HN) \leq N_{N_G(HN)}(H)HN = N_{N_G(HN)}(H)N$ and thus $g = an$ where $a \in N_{N_G(HN)}(H)$ and $n\in N$. From this $H^g = H^{an}= H^n$, and since $n \in HN$, $H$ and $H^g$ are conjugate in $HN$. Why can we assume that $g\in HN$? $\endgroup$ – R Maharaj Jun 5 '16 at 12:12
  • $\begingroup$ Because $H$ and $H^g$ are conjugate in $HN$. $\endgroup$ – Derek Holt Jun 5 '16 at 19:13
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    $\begingroup$ It holds in $HN$ by induction if $HN < G$, which is why you can assume that $HN=G$. Sorry, but that's my final comment! $\endgroup$ – Derek Holt Jun 10 '16 at 15:25

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