Let $G$ be a doubly transitive permutation group on $\Omega$ and let $\alpha,\beta\in \Omega$ with $\alpha\neq \beta$. Let $\phi\in Irr(G_\alpha)$ and assume that $\phi_{G_{\alpha\beta}}\in Irr(G_{\alpha\beta}).$ Show that $[\phi^G,\phi^G]\leq 2$.(Isaacs exercise 5.24)
The hint says to use the Mackey theorem.
Here is what I have gotten so far:
Since $G$ is doubly transitively on $\Omega$, we can decompose $G$ as $G={G_\alpha}\dot\cup{G_\alpha}g{G_\alpha}$ for any $g\notin G_\alpha$.
By the Mackey theorem, we can write $$ (\phi^G)|_{G_\alpha}=\phi+((\phi_g)|_{G_\alpha^g\cap G_\alpha})^{G_\alpha} $$ where $G_\alpha^g=g^{-1}G_\alpha g$ and $ \phi_g: G_\alpha^g\ni x\mapsto\phi(gxg^{-1}) $.
Since $\phi\in Irr(G_\alpha)$, we have by the Frobenius reciprocity: \begin{align} [\phi^G,\phi^G]_G&=[\phi,(\phi^G)|_{G_\alpha}]_{G_\alpha}\\ &=1+[\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}. \end{align} It suffices to show that $[\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}\le 1$.
Note that $\phi_{G_{\alpha\beta}}\in Irr(G_{\alpha\beta})$, we have \begin{align} 1=[\phi_{G_{\alpha\beta}} ,\phi_{G_{\alpha\beta}}]_{G_{\alpha\beta}}&=[\phi,(\phi_{G_{\alpha\beta}})^{G_{\alpha}}]_{G_\alpha}. \end{align} We know that \begin{align} [\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}&\le [(\phi_{G_{\alpha\beta}})^{G_{\alpha}},((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}\\ &=[\phi_{G_{\alpha\beta}},(((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha})|_{G_{\alpha\beta}}]_{G_\alpha\beta}. \end{align} Now it is sufficient to show that $(((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha})|_{G_{\alpha\beta}}$ has at most one irreducible constituent $\phi_{G_{\alpha\beta}}$ in $G_{\alpha\beta}$. Then I got stuck... Any hint? Thank you!
It would simplify notation if you wrote $G_{\alpha\beta}$ rather than $G_\alpha^g \cap G_\alpha$, where $\beta = g^\alpha$.
Then we want to prove that $[\phi,(\phi_g)_{G_{\alpha\beta}}^{G_\alpha}] \le 1$.
By Frobenius Reciprocity again, we have $[\phi,(\phi_g)_{G_{\alpha\beta}}^{G_\alpha}] = [\phi_{G_{\alpha\beta}},(\phi_g)_{G_{\alpha\beta}}]$.
But $\phi_{G_{\alpha\beta}}$ and $(\phi_g)_{G_{\alpha\beta}}$ are irreducible by assumption, so the result follows.