Algebra transitive action (Isaacs 4.1) 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 $\Omega$. 

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. 
