Let $ N \unlhd G $ and suppose $G/N $ is abelian. Let $C$ be the group of linear characters of $G/N $ so that $C$ acts on $Irr(G)$ by multiplication. Let $\vartheta \in Irr(N)$. Show that
$\vartheta^G=f \sum \chi_i $ where $f$ is an integer and $\chi_i \in Irr(G)$ contistute an orbit under $C$.
Try: I tried using Gallagher's theorem but couldn't use it somehow that would give me that $\beta\chi$ is a constituent, where $\chi \in Irr(G) $ such that $\chi_N =\vartheta$ and $\beta \in Ιrr(G/N) $. But First I don't know if such $\chi$ exists. Second I also know that those $\beta\chi$ for $\beta$ running in $G/N$ would give me indeed irreducible characters so they would be ok for the $\chi_i$'s the problem asks for. Thats what I have thought so far.
COROLLARY (Gallagher) Let $N \unlhd G$ and let $\chi \in Irr(G)$ be such that
$\chi_N = θ \in Irr(N)$ . Then the characters $\beta\chi$ for $\beta \in Irr(G/N)$ are irreducible,
distinct for distinct $\beta$ and are all of the irreducible constituents of $\vartheta^G$.
Best Answer
If $\phi$ is a class function on $N$ and $\psi$ is a class function on $G$, there is the following well-known formula $(\phi \psi_N)^G=\phi^G\psi$. Then if $\chi$ is an irreducible character of $G$ over $\theta$, one has that $(\chi_N 1_{N})^G=\rho_{G/N} \chi$. Since $\theta$ is a constituent of $\chi_N$, $\theta^G$ is a constituent of $\rho_{G/N}\chi$. From this follows that the irreducible constituents of $\theta$ are of the form $\lambda \chi$, with $\lambda\in \hat{G/N}$. Then they have the same multiplicity in $\theta^G$ by Frobenius reciprocity (their restriction is always $\chi_N$). It is also clear from above that they are an orbit for the multiplicative action of $\hat{G/N}$.