Character extension (${\chi }_N=\theta$)

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.