Restriction of a character (${\mathrm{Res}}_H^G\chi$)

For a subgroup $H\le G$, the restriction of a $G$-character $\chi$ to $H$ is the $H$-character obtained by evaluating the same function only on elements of $H$.

An extension of $\theta \in \mathrm{Irr}(N)$ from $N◃G$ is a character $\chi$ of $G$ whose restriction is exactly $\theta$. It is necessarily irreducible, because a nontrivial invariant subspace for $G$ would also be one for $N$. An ordinary extension can fail to exist even when $\theta$ is invariant under $G$.

Let $N◃G$ and let $\theta \in \mathrm{Irr}(N)$ be invariant under $G$. If $\theta (1)$ is coprime to $[G:N]$, then $\theta$ extends to $G$ if and only if its determinant character extends to a linear character of $G$. In particular, if the determinant order is also coprime to the index, there is a unique extension whose determinant order is coprime to the index. When $G/N$ is a p-group and $N=O^p(N)$, every invariant irreducible character of degree prime to $p$ satisfies this latter determinant condition.

Under the invariant, coprime-degree and determinant hypotheses of the coprime-degree determinant extension theorem, without assuming $G/N$ solvable, each cyclic-quotient subgroup $N⟨g⟩$ has a unique determinant-normalized extension. Use its value at $g$ to define $f$. Uniqueness makes $f$ conjugacy-invariant. On an elementary subgroup $E$, it agrees with the restriction of the extension to $NE$, since $NE/N$ is a solvable quotient of $E$. Brauer's characterization of characters therefore makes $f$ a generalized character, with $f_N=\theta$.

Suppose $N◃G$, $G/N$ is solvable, $\theta \in \mathrm{Irr}(N)$ is invariant under $G$, and $\theta (1)$ is coprime to $[G:N]$. If a linear character $\mu$ of $G$ restricts to the determinant character $\det \theta$, then there is a unique character extension $\chi \in \mathrm{Irr}(G)$ with determinant $\mu$. This is the solvable-quotient extension result supplied for the exam's application.

For $N◃G$ and $\theta \in \mathrm{Irr}(N)$, set ${\theta }^g(n)=\theta (g^{-1}ng)$. Its inertia group is its stabilizer under this conjugation action and contains $N$. Write $\mathrm{Irr}(G\mid \theta )$ for the irreducible characters of $G$ whose restriction to $N$ contains $\theta$.

For $N◃G$, $\theta \in \mathrm{Irr}(N)$ and $T=I_G(\theta )$, induction gives the displayed bijection. The inverse takes the $\theta$-isotypic component of an irreducible $G$-module, viewed as a $T$-module. Induction produces distinct $N$-isotypic components indexed by $G/T$. Any nonzero $G$-submodule meets one of them and hence, by translation and irreducibility over $T$, contains them all. The same argument shows that the inverse component is irreducible and determines the original inducing module uniquely.

If $\chi$ is an irreducible character of a finite group $G$ and $H\le G$, thenEquality holds exactly when $\chi$ vanishes on $G\setminus H$.

Let $H◃G$ have index two, and let $\lambda$ be the nontrivial linear character of $G/H$ inflated to $G$. For an irreducible character $\chi$ of $G$, either ${\mathrm{Res}}_H^G\chi$ is irreducible and $\chi \ne \chi \lambda$, or the restriction is a sum of two distinct irreducible characters and $\chi =\chi \lambda$.