Show that ${\theta}^G \in Irr(G)$ iff $I_G(\theta) = N$, where $N \unlhd G$ and $ \theta \in Irr(N)$. Let $N \unlhd G$ and $ \theta \in Irr(N)$. Show that ${\theta}^G \in Irr(G)$ iff $I_G(\theta) = N$.
Where $I_G(\theta)$ is the stabilizer of $\theta$ in the action of $G$ on $Irr(N)$ defined by  $\theta^g:N \rightarrow {\mathbb C}$, $\theta^g(x)=\theta(gxg^{-1})$
 A: Hints:
I.
$<\theta^G,\theta^G>=\Sigma_{t\in T}<\theta^t,\theta>$, where $T$ is the set of double coset representatives with respect to $N$ and $N$, i.e. $G=\bigcup_{t\in T}NtN$ and $1_G\in T$.
II.
$\theta^G\in \text{Irr}(G)\iff<\theta^G,\theta^G>=1.$
III.
Show that, if $\theta^t\not=\theta,$ then $<\theta^t,\theta>=0$.  
A: Forget about double coset representatives and do it the Isaacs way: use Theorem(6.11) in your book, which describes the relationship between irreducible characters of the whole group and the inertia group lying over a certain irreducible character in a normal subgroup! Write $T=I_G(\theta)$. You have to prove that $\theta^G \in Irr(G)$ iff  $T=N$. If $T=N$ then according to (6.11)(a) and taking $\psi=\theta$, the induced character $\theta^G$ is irreducible. Conversely, assume $\chi := \theta^G \in Irr(G)$, and put $\psi=\theta^T$. Then $\psi \in Irr(T)$ since $\psi^G = (\theta^T)^G = \chi \in Irr(G)$. Since $\theta$ is trivially invariant in $T$, $\psi_N = e\theta$. But $e= [\psi_N,\theta] = [\psi, \theta^T]= 1$, so $\psi(1)=\theta(1)$. By definition $\psi(1)= \theta^T(1)=\theta(1)[T:N]$, whence $[T:N]=1$, that is $T=N$ and we are done.
