Frobenius reciprocity

Remarks

The proof is a direct computation: insert the formula $\operatorname{Ind}_H^G \theta(g) = \frac{1}{\abs{H}}\sum_{x \in G,\ x^{-1}gx \in H} \theta(x^{-1}gx)$, substitute $k = x^{-1} g x$, and use that $\psi$ is constant on conjugacy classes. The module form holds over any field $F$: for an $F[H]$-module $W$ and an $F[G]$-module $V$, $\Hom_G(\operatorname{Ind}_H^G W, V) \cong \Hom_H(W, V|_H)$ as vector spaces. Example: for irreducible characters $\theta$ of $H$ and $\psi$ of $G$, the multiplicity of $\psi$ in $\operatorname{Ind}_H^G \theta$ equals the multiplicity of $\theta$ in $\psi|_H$. For $H = \set{1}$ and $\theta$ trivial this gives that each irreducible $\psi$ occurs $\psi(1)$ times in the regular representation.

Going deeper

For a student

In the language of categories, induction is left adjoint to restriction. The formula is used constantly to decompose induced characters, for instance in the proofs of Brauer's theorem on induced characters and Artin's induction theorem. It also gives Mackey-type computations and is the basis of Clifford theory.