9 Frobenius Reciprocity

Theorem

Let $H<G$, and let $\chi ,\psi$ be characters of $G,H$, respectively. Then $⟨\psi {\uparrow }_H^G,\chi ⟩=⟨\psi ,\chi {\downarrow }_H^G⟩$.

\begin{proof} Note that

$$\begin{array}{rl}⟨\psi {\uparrow }_H^G,\chi ⟩& =\frac{1}{|G|}\sum _{g\in G}\psi {\uparrow }_H^G(g)\chi (g^{-1})=\frac{1}{|G|}\frac{1}{|H|}\sum _{g\in G}\sum _{x\in G}\psi (x^{-1}gx)\chi (g^{-1})\\ & =\frac{1}{|G||H|}\sum _{y\in G}\sum _{x\in G}\psi (y)\chi (x{y}^{-1}{x}^{-1})=\frac{1}{|G||H|}\sum _{y\in G}\sum _{x\in G}\psi (y)\chi (y^{-1})=\frac{1}{|H|}\sum _{y\in G}\psi (y)\chi (y^{-1})\\ & =\frac{1}{|H|}\sum _{y\in H}\psi (y)\chi (y^{-1})=⟨\psi ,\chi {\downarrow }_H^G⟩.\end{array}$$

Now we finish the proof. \end{proof}