TITLE: Induction from normal subgroups
QUESTION [5 upvotes]: Let $G$ be a (finite) group and $N$ a normal subgroup. Given an irreducible representation $\pi$, how can I decompose $Ind_N^G \pi$?
I'd be happy also about a good reference for this.

REPLY [2 votes]: The easiest way I know is by modified group projectors technique
[Damnjanovic, Milosevic, J.Phys. A 28 (1995) 669-79]:
To shorten notation, $D$ is your induced rep of $\pi$ of $N$,  $d$ is arbitrary irreducible representation of $G$. You want to find frequency numbers $f_{d}$ in the decomposition (direct sum over irreducible reps of $G$):
$$D=\sum_{d} f_{d} d.$$  
So, find modified projector ($\otimes$ is tenzor product of matrices, $d^*$ conjugated representation $d$):
$$P_d=1/|N| \sum_n \pi(n) \otimes d^*(n).$$
Then: $f_{d}=\mathrm{Tr}\, P_{d}$    (Tr is trace). 
You should do this for all irreducible representations of $G$ (of course, you can stop when the dimension of $D$, $|D|=|d| |G|/|N|$, is exhausted).
Note that here only subduced (restricted) to $N$ irreducible reps of $G$ comes. Further, if you have infinite discrete group $N$, the problem can be reduced to generators only, and if it is Lie group, again only Lie generators are used.