Orbit Stabilizer problem (I think) Let $G$ (finite) act transitively on the nonempty set $\Omega$. Show that if $\alpha \neq \beta$ are elements of $\Omega$ then $G_{\alpha}G_{\beta}$ is a proper subset of $G$, where $G_{\alpha}$ and $G_{\beta}$ are the stabilizers of $\alpha$ and $\beta$ in G, respectively.
Here's what I've done:
Let $\mathcal{O}$ denote the orbit of the elements from $G$ acting on $\Omega$. Then using the orbit stabilizer theorem and the fact that $|G_{\alpha}G_{\beta}| = \frac{|G_{\alpha}||G_{\beta}|}{|G_{\alpha}\cap G_{\beta}|}$ we can deduce that
\begin{equation}
|G_{\alpha}G_{\beta}|=\frac{|G|^{2}}{|\mathcal{O}|^{2}\cdot |G_{\alpha} \cap G_{\beta}|}.
\end{equation}
This is where I get stuck. It is clear that $|G_{\alpha}|=|G_{\beta}|$ so I would like to argue that $G_{\alpha}=G_{\beta}$ so that $G_{\alpha} \cap G_{\beta} =G_{\alpha}=G_{\beta}$. This would then give that $G_{\alpha}G_{\beta}=\frac{|G|}{\mathcal{O}}$ since $|G_{\alpha}|=\frac{|G|}{\mathcal{O}}$ in which case I could probably make an argument for why $|\mathcal{O}| > 1$. It might not be true that the stabilizers for $\alpha$ and $\beta$ are equal since I can't argue it in which case this would not work, obviously.
 A: $\newcommand{\Size}[1]{\vert #1 \rvert}$I think you are nearly there with your approach.
Suppose $G = G_{\alpha} G_{\beta}$. Then
$$\Size{G} =
|G_{\alpha}G_{\beta}| 
= 
\frac{\Size{G_{\alpha}} \cdot \Size{G_{\beta}}}{\Size{G_{\alpha}\cap G_{\beta}}},$$
so that
$$
\Size{\Omega}
=
\frac{\Size{G}}{\Size{G_{\beta}}} = \frac{\Size{G_{\alpha}}}{\Size{G_{\alpha}\cap G_{\beta}}}.
$$
This is telling you that $G_{\alpha}$ is transitive on $\Omega$ (by orbit-stabilizer, as $G_{\alpha} \cap G_{\beta}$ is the stabilizer of $\beta$ in the action of $G_{\alpha}$ on $\Omega$), which is only possible of course when $\Omega = \{ \alpha \}$. But by assumption $\Omega$ contains at least two distinct elements $\alpha, \beta$.

PS I am assuming everything's finite here, since you were taking orders etc. But the result is true in general. It is equivalent to showing that a group cannot be the product of a proper subgroup $H$ and one of its conjugates $H^{g}$. ($G$ being transitive, all stabilizers are conjugate.) You can find a proof in this spoiler - $G$ is claimed to be finite there, but the proof works in general.
