I want to prove the following:
Assume $G$ acts transitively on $\Omega$, and $H \leq G$ is a transitive subgroup. Show that for $\alpha \in \Omega, G=HG_\alpha$.
This is what I have so far:
Let $\alpha \in \Omega$.
Let $G_\alpha$ denotes the stabilizer of $\alpha$ in $G$, and $O_G(\alpha)$ denotes the orbit of $\alpha$ in $G$.
Clearly, $HG_\alpha\subseteq G$. So, we only need to show that $G \subseteq HG_\alpha $.
Let $g \in G$. As $H$ acts transitively on $\Omega,$ we have $ O_H(\alpha)=\Omega.$ Hence, $\alpha g=\alpha h$ for some $h \in H.$
So, $\alpha g h^{-1}=\alpha$. Thus, $g h^{-1} \in G_\alpha.$ Let $k=g h^{-1}.$ Hence, $g=kh \in G_\alpha H.$
But I want to say $g \in HG_\alpha.$ Did I do something wrong?
Best Answer
No, you're fine. Your next step is $G=G^{-1}=(G_{\alpha}H)^{-1}=H^{-1}{G_{\alpha}}^{-1}=HG_{\alpha}$, and you have what you want.