Suppose that the finite group $G$ acts on the set $\Omega$ of size $n$. let $\alpha \not= \beta$ in $\varOmega$ have stabilisers $A$ and $B$.
Prove that $|\alpha^G| |\beta^A| |A\cap B|=|G|$.
So I think I have proved this but I would like to check my answer.
We have that $A=G_\alpha=\{g \in G : \alpha^g=\alpha\}$ and $B=G_\beta=\{g \in G : \beta^g=\beta\}$. Since $G$ is finite by the OST we know that $|\alpha^G| |A|=|G|$ and $|\beta^A||A_\beta |=|A|$.
$A_\beta=\{g \in A : \beta^g|=\beta\}=\{g \in G : \alpha^g=\alpha\text{ AND } \beta^g=\beta\}=A\cap B$. So $|A_\beta|=|A\cap B|$
Hence $|\beta_A||A_\beta|=|A|$ becomes $|\beta^A||A \cap B|=|A|$
$$|\beta_A||A \cap B|=|A|\text{ implies that }|\beta_A|=|A|/|A\cap B|\tag{*}$$
Recall that $|\alpha^G||A|=|G|$ and $|\alpha^G| |\beta^A| |A\cap B|=|G|$ then we sub in (*) and obtain $|\alpha^G||A|=|G|$, hence $|\alpha^G| |\beta^A | |A\cap B|=|G|$ as required.
Thanks
By the orbit-stabilizer theorem $$|\alpha^G| = \frac{|G|}{|G_{\alpha}|}$$ $$|\beta^A| = |\beta^{G_{\alpha}}| = \frac{|G_{\alpha}|}{|G_{\beta}\cap G_{\alpha}|}$$ So $$|\alpha^G| |\beta^A| |A\cap B|= \frac{|G|}{|G_{\alpha}|} \cdot \frac{|G_{\alpha}|}{|G_{\beta}\cap G_{\alpha}|} \cdot |G_{\alpha}\cap G_{\beta}| = |G|$$