Show that $G$ is doubly transitive on $\Omega$ if $\dfrac{1}{|G|}\sum_{g\in G} \chi(g)^2=2$ where $\chi$ is the associated permutation character.

Question

Show that $G$ is doubly transitive on $\Omega$ if $\dfrac{1}{|G|}\sum_{g\in G} \chi(g)^2=2$ where $\chi$ is the associated permutation character. Do not assume $G$ is transitive.

An action of $G$ on $\Omega$ is doubly transitive if $G$ is transitive on the set of ordered pairs $(\alpha,\beta)$ with $\alpha\neq \beta$ with $$(\alpha,\beta)\cdot g=(\alpha\cdot g, \beta\cdot g).$$

I know by the Cauchy-Frobeius Theorem that the total number of orbits is given by $$n=\dfrac{1}{|G|}\sum_{g\in G} \chi(g).$$

But where do I go from here?

Answer 1

$\chi^2$ is the permutation character of $G$ on $\Omega^2$, so $G$ has $\frac{1}{|G|}\sum_g \chi (g)^2 =2$ orbits on $\Omega^2$. One of them is contained in $\{(x,x) : x \in \Omega \}$, and one of them is contained in $\{(x,y) \in \Omega^2 : x \neq y \}$. Since there are only two orbits, we have equality in both cases, and $G$ is doubly transitive.