Induced character (${\mathrm{Ind}}_H^G\psi$)

For a character $\psi$ of $H\le G$, the induced character isThis is the character of a representation of $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}{V}_{\psi }$. Substituting this expression into the character inner product and putting $g=xh{x}^{-1}$ proves Frobenius reciprocity.

For a finite group $G$, a subgroup $H$, and a character $\chi$ of $H$, each element $h$ of $g^G\cap H$ has exactly $|C_G(g)|$ preimages under $x\mapsto x^{-1}gx$. This follows because one such preimage $x_0$ has full fiber $C_G(g)x_0$. Grouping the character formula for an induced representation by these fibers proves the formula. Grouping again by conjugacy classes of $H$ replaces each class by its size times its character value.

A monomial character is a character induced from a linear character of a subgroup. When discussing monomial finite groups, this property is required for every irreducible character.

Extend a class function $\psi$ on $H$ by zero outside $H$. For coset representatives $x_1,\dots ,x_m$ of $G/H$, ${\mathrm{Ind}}_H^G\psi (g)={\sum }_j{\psi }_0(x_j^{-1}g{x}_j)$. Each summand has squared normalized $G$-norm $\Vert \psi {\Vert }_H^2/m$. Pointwise Cauchy-Schwarz inequality and summation give $\Vert \mathrm{Ind}\psi {\Vert }_G^2\le m\Vert \psi {\Vert }_H^2$. If $\psi$ is an irreducible character, its norm is one, and the sum of squared induction multiplicities is at most the subgroup index.

An index-two subgroup is normal. Conjugation by an outside element gives an involution on its irreducible characters. If ${\psi }^g\ne \psi$, the squared norm of induction is one, so induction is irreducible of twice the degree and restriction is $\psi +{\psi }^g$. If ${\psi }^g=\psi$, the squared norm is two, so induction has exactly two distinct constituents, each once. Each restricts to a representation containing $\psi$, and their total degree is twice that of $\psi$, forcing both degrees to equal that of $\psi$. This proves the fixed-orbit versus two-element-orbit classification without assuming extension of $\psi$ in advance.