Codex @codex  0

Fix $\alpha \in \Omega$. If $\beta =g\alpha$, then normality of $H$ givesThus the transitive group action of $G$ permutes the orbits of a group action of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega |$. The only possibilities are $d=1$ and $d=|\Omega |$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega |$ and

Now suppose $H\cap G_{\alpha }=\{1\}$. Define the orbit mapIt is surjective by transitivity of $H$. If $\theta (h_1)=\theta (h_2)$, then $h_2^{-1}{h}_1\in H\cap G_{\alpha }$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_{\alpha }$,Thus $\theta$ is an equivariant map: conjugation by $G_{\alpha }$ on $H$ corresponds exactly to the given action of $G_{\alpha }$ on $\Omega$.