Let $N$ be a normal subgroup of a finite group $G$, $\chi\in\mathrm{Irr}(G)$, and $\psi$ be an irred constituent of $\chi_{N}$. Let $H$ be a subgroup containing $N$ such that $H/N$ is abelian and $\psi$ is extendible to $H$. Then there exists a $\mu\in\mathrm{Irr}(H)$ such that $\mu_{N}=\psi$ and $[\chi_{H},\mu]\neq0$.
Edit: My previous statement was wrong as it left out the necessary hypothesis that $H/N$ be abelian. See Marty Isaacs's answer.
Below is my old argument. When $H/N$ is not abelian it does not go through because it is not necessarily true that $\mu_{N}=\psi$; in general $\mu_{N}$ will be a number of copies of $\psi$. When $H/N$ is abelian there is a much simpler argument, as given in Marty Isaacs's answer.
The natural context for this statement is Clifford theory for finite groups (see Section 6 in Isaacs's book). To see why this is true, first note that $H$ must be a subgroup of the stabiliser $S$ of $\psi$ in $G$. Moreover, there exists a $\rho\in\text{Irr}(S)$ such that $\rho_{N}$ contains $\psi$ and $\rho^{G}=\chi$. Since $\rho_{N}$ contains $\psi$, there exists an irred constituent $\mu$ of $\rho_{H}$ such that $\mu_{N}$ contains $\psi$. Since $\psi$ is extendible to $H$, a theorem of Gallagher's (Isaacs's book (6.17)) implies that $\mu_{N}=\psi$. Since $\mu^{S}$ contains $\rho$, $\mu^{G}$ contains $\chi$, and so $\chi_{H}$ contains $\mu$.
