I'm working on the following questions and I'm unable to make any progress on it. Can anyone offer any insight?
Let $G$ act transitively on a set $\Omega$ and suppose $G$ is finite. Define an action of $G$ on $\Omega \times \Omega$ by putting $(\alpha, \beta) \cdot g =(\alpha\cdot g, \beta\cdot g)$. Let $\alpha \in \Omega$. Show that $G$ has the same number of orbits on $\Omega \times \Omega$ as $G_{\alpha}$ does on $\Omega$.
On
With character theory this can be proved rather easily. Put $\chi(g)=\#\{\alpha \in \Omega: \alpha^ g=\alpha\}$. Then the function $\chi$ is called the permutation character of the action of $G$ on $\Omega$. It is a standard fact (basically the Burnside-Cauchy-Frobenius Lemma) that the number of orbits of $\Omega$ under the action of $G$ equals $[\chi,1_G]$. Write $G_{\alpha}$ for the subgroup of $G$ fixing $\alpha$. You need the following.
Lemma Let $G$ act transitively on $\Omega$. Then the number of orbits of $G_{\alpha}$ acting on $\Omega$ equals $[\chi,\chi]$.
For a proof see I.M. Isaacs, Character Theory of Finite Groups, (5.16).
Now the key observation is that the permutation character of $G$ acting on $\Omega \times \Omega$ is exactly the character $\chi^2$, since if $g \in G$ has $\chi(g)$ fixed points on $\Omega$, it has $\chi(g)^2$ fixed points on $\Omega \times \Omega$ – namely one for every pair of fixed points in the action on $\Omega$. Therefore $[\chi^2,1_G]$ is the number of orbits in the action of $G$ on $\Omega \times \Omega$. But obviously $[\chi^2,1_G]=[\chi,\chi]$ and by the Lemma we are done.
There is a map from $G_{\alpha}$ orbits on $\Omega$ to $G$ orbits on $\Omega \times \Omega$. Take an orbit $G_{\alpha}\beta$ and send it to the $G$ orbit of $(\alpha, \beta)$. If $\beta'$ is another representative of this orbit, $g \beta = \beta'$ with $g \in G_{\alpha}$, then $(\alpha, \beta')$ and $(\alpha, \beta)$ are in the same orbit since $g(\alpha, \beta) = (\alpha, \beta')$. So this map is well defined.
Now, show it is injective and surjective.