Frobenius reciprocity, proof if $V$ irrep of $G$ then multiplicity of $V$ in regular rep of $G$ is $\dim(V)$ I know the standard proof of the fact that if $V$ is an irreducible representation of $G$ then the multiplicity of $V$ is the regular representation of $G$ is $\dim(V)$. Does there exist a proof using Frobenius reciprocity?
 A: There is a proof and it's pretty quick. Below, $\mathbb{F}$ is used to denote the trivial representation of the trivial group.
Let $R$ be the regular representation of a group G. Then, $R = \mathrm{Ind}_{1}^{G}(\mathbb{F})$. Let $V$ be an irreducible representation of $G$. Then, by Frobenius reciprocity
$$\dim \mathrm{Hom}_{G}(R, V) = \dim \mathrm{Hom}_{1}(\mathbb{F}, V) = \dim(V).$$
But the dimension on the left is precisely the multiplicity of $V$ in $R$ because $V$ is irreducible.
A: Let $H = (e)$, the trivial group in $G$, and let $W$ denote the trivial representation of $H$. Then $V = \text{Ind}_H^G W$ is the regular representation of $G$. If $U$ is any irreducible representation of $G$, we know by Frobenius Reciprocity that$$\langle V, U\rangle_G = \langle \text{Ind}_H^G W, U\rangle_G = \langle W, \text{Res}_H^G U\rangle_H.$$Since $H$ acts on $U$ by fixing every vector, it must be that $\text{Res}_H^G U = (\dim U)W$. Therefore,$$\langle W,\text{Res}_H^G U\rangle_H \cong (\dim U)\langle W, W\rangle_H = \dim U.$$It follows the multiplicity of $U$ in the regular representation of $G$ is $\dim U$.
