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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 which stated that: if $G$ is soluble group, $H$ is pronormal in $G$ $\iff$ H satisfies the Frattini Property

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 Frattini Property of a subgroup

For the $\Leftarrow$ direction, solvability of the group will be needed. Would induction on $|G|$ be the way of solving this implication?

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  • $\begingroup$ I think you already asked this question before, or at least a very similar one. That Frattini property seems to be a rather involved, very specific property and it seems to be not many around here can handle all the info. Try to look for some help perhaps in MathOverflow, $\endgroup$ – DonAntonio Jun 3 '16 at 19:14
  • $\begingroup$ You can only use induction on $|G|$ for a finite group $G$, and you do not appear to be assuming that $G$ is finite. $\endgroup$ – Derek Holt Jun 3 '16 at 19:23
  • $\begingroup$ In any case, if it is a published result then why don't you try and read the proof there? $\endgroup$ – Derek Holt Jun 3 '16 at 19:29
  • $\begingroup$ I don't have access to the paper which is in an Oxford Journal. My university which is the University of Kwa-Zulu Natal in South Africa also does not have library access to this paper. And yes, I missed out the crucial info of $G$ being finite $\endgroup$ – R Maharaj Jun 3 '16 at 19:47
  • $\begingroup$ @Joanpemo Thank you. I did not know that another Math forum apart from this one existed. $\endgroup$ – R Maharaj Jun 3 '16 at 19:52

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