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

A subgroup $H$ of $G$ satisfies the Frattini Property $\iff$ for any $x\in G$, there exists $y\in H^{\langle x\rangle}$ such that $H^y = H^x$

For the $\Leftarrow$ direction: Let $x\in G$ such that $y\in H^{\langle x\rangle}$ implies $H^y = H^x$. Then $y^{-1}x \in N_G(H)$ so $x\in H^{\langle x\rangle}N_G(H)$. Also $x \in \langle H, x\rangle$ then $x \in \langle H, x\rangle \cap H^{\langle x\rangle}N_G(H)$

Since $H^{\langle x\rangle} \unlhd \langle H, x\rangle$, by Dedekind's lemma $\langle H, x\rangle$ then $x \in \langle H, x\rangle \cap H^{\langle x\rangle}N_G(H) = H^{\langle x\rangle}(\langle H, x\rangle \,\cap \, N_G(H)) = H^{\langle x\rangle}N_{\langle H, x\rangle}(H) $

Also $H \leq H^{\langle x\rangle}N_{\langle H, x\rangle}(H) $. Hence $\langle H, x\rangle \leq H^{\langle x\rangle}N_{\langle H, x\rangle}(H) $, so $H$ satisfies the Frattini Property since $H\leq H^{\langle x\rangle} \unlhd \langle H, x\rangle$ . The following is my attempt on proving this

For the $\Rightarrow$ direction: Let $H$ satisfy the Frattini Property then $\langle H, x\rangle \leq H^{\langle x\rangle}N_{\langle H, x\rangle}(H) $

From this we have that $x\in H^{\langle x\rangle}N_{\langle H, x\rangle}(H)$ so $x = ya$ where $y\in H^{\langle x\rangle}$ and $a\in N_{\langle H, x\rangle}(H)$. I can't seem to show $H^y = H^x$

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    $\begingroup$ If you write it as $x \in N_{\langle H,x \rangle}(H)H^{\langle x \rangle}$ then you get $x=ay$, and it is clear. $\endgroup$ – Derek Holt Jun 3 '16 at 7:53

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