Inertia group and characters lying above a normal type

Let $G$ be finite, $N⊴G$, and $\theta \in \mathrm{Irr}(N)$, where $\mathrm{Irr}(N)$ denotes irreducible complex characters. Using Conjugate representations and conjugate characters on conjugate subgroups, set ${}^g\theta (n)=\theta (g^{-1}ng),I_G(\theta )=\{g\in G:{}^g\theta =\theta \}\mathrm{.}$ The subgroup $I_G(\theta )$ is the inertia group. For $N\le H\le G$, define $\mathrm{Irr}(H\mid \theta )=\{\psi \in \mathrm{Irr}(H):\theta \text{ occurs in }{\mathrm{Res}}_N^H\psi \}\mathrm{.}$ Equivalently the restriction has positive inner product with $\theta$, by The multiplicity of an irreducible summand is a character inner product. Such a character lies over $\theta$.

Normality (Normal subgroup: invariance under conjugation) ensures that the conjugates are again characters of $N$. Twisting by an automorphism preserves irreducibility. Direct substitution gives ${}^g({}^h\theta )={}^{gh}\theta$ and ${}^1\theta =\theta$. For $n_0\in N$, the matrices of $n_0^{-1}n{n}_0$ and $n$ are similar, so their traces coincide: ${}^{n_0}\theta =\theta$. Thus the action factors through $G\mathrm{/}N$ and its stabilizer satisfies $N\le I_G(\theta )\le G$. The stabilizer is a subgroup because products and inverses preserve a fixed point. Orbit representatives are indexed by left cosets $g{I}_G(\theta )$.