Let a finite group $G$ act doubly transitively on a finite set $X$, i.e., given $x,y,z,w\in X$ such that $x\neq y$ and $z\neq w$, there is a $g\in G$ such that $g.x=z$ and $g.y=w$. So, we can write $$\chi = \mathbf{1}_G + \theta$$ where $\chi$ is the trivial representation and $\theta$ is a character in which $\mathbf{1}_G$ does not appear. Show that $\theta$ is irreducible.
I have solved a previous problem in which I have proved $$\sum_{g\in G} \chi(g)=m|G|$$ where $m$ is the number of orbits of $G$ in $X$ and $G$ acts on $V$ by permuting its basis where $V$ is a vector space with elements of $X$ as basis.
I have also proved that the number of times $\mathbf{1}_G$ occurs in $V$ is the same as the number of orbits of $G$ in $X$. In particular, if $G$ acts transitively on $X$, then $\mathbf{1}_G$ occurs exactly once in $\chi$.
Now, for this problem, I have observed that a doubly transitive action is also a transitive action, and have proved that the action of $G$ on $X\times X$ has exactly two orbits. So, I need to prove $$\langle \chi, \chi \rangle = 2$$ which I feel shouldn't be hard, but I can't find a way to do it.
Let $\chi_{X\times X}$ be the permutation character of $G$ arising from its action on $X\times X$ and $\chi_X$ the permutation character for $X$. Then $\chi_{X\times X} = \chi_X^2$ since $g$ fixes $(x, y)$ iff it fixes $x$ and $y$, so the number of fixed points of $g$ on $X\times X$ is the square of the number of fixed points on $X$. You know there are two orbits in the $G$ action on $X\times X$, so your formula gives $\frac{1}{|G|}\sum_g \chi_X(g)^2 = 2$. Since $\chi$ is real-valued this is the same as $\frac{1}{|G|}\sum_g\chi_X(g) \overline{\chi_X(g)} = 2$, that is, $\langle \chi_X, \chi_X\rangle = 2$.