Induction from normal subgroups 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.
 A: If you read about Clifford's theorem and its consequences in a text on representation theory (eg Curtis and Reiner: Representation Theory of finite groups and associative algebras Wiley,1962, or Isaacs Character Theory), that should help to answer your question. I presume you are thinking of complex representations, though Clifford theory can be done in a more general context.
The problem breaks into two (or possibly three, depending on your point of view), steps.
If the isomorphism type of $\pi$ is $G$-stable, then it is possible (in the complex case)
to replace $G$ by a central extension $H$ by a cyclic central subgroup so that $\pi$ extends to an irreducible representation of $H$. So if we consider the case that $\pi$ extends to a representation of $G$, the induced representation breaks up as the tensor product of (the extension of) $\pi$  and the regular representation of $G/N$. If $\pi$ is $G$-stable, but does not extend, we need to look at a twisted group algebra for $G/N$
instead, (this is explained in the texts referred to above).
In general, the elements of $G$ such that $\pi^{g} \cong \pi$ as a representation of $N$
form a subgroup of $G$ called the inertial subgroup of $\pi$, which I denote by $I$.
If $I = G$ we are in the case above. If not, then the irreducible components of $Ind_{N}^{G}(\pi)$ are in bijection with the irreducible components of ${\rm Ind}_{N}^{I}(\pi)$
and an irreducible component for $G$ is just induced from the corresponding component for $I$.
A: 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.
