0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.