problem 6.7 in Character theory of finite groups

Question

Let $N$ be a normal subgroup of $G$, and $G/N$ is solvable. Let $\chi\in$Irr(G), $\theta\in$Irr($N$), with $[\chi_N, \theta]\neq0$. Show that $\chi(1)/\theta(1)$ devides $|G:N|$.

Note: The conclusion is valid even if $G/N$ is not solvable.

I can't solve it. here is some of my thinking.

Say $\chi_N=e\sum_{i=1}^t\theta_i$, where $\theta=\theta_1,\ldots,\theta_t$ are all conjugate character of $\theta$. We know that $t=|G: I_G(\theta)|$, and $\chi(1)/\theta(1)=et$. So I need show $e$ devides $|I_G(\theta):N|$. $\exists \phi\in $Irr(I$_G(\theta)$) such that $\phi_N=e\theta$ and $\phi^G=\chi$. So the question is: Does $\phi(1)/\theta(1)$ devide $|$I$_G(\theta):N|$, If I$_G(\theta)<G$, it's ture by induction.

But what if $G=$I$_G(\theta)$ ?

Answer 1

By “my thinking” above, we only need to solve it when $G=$I$_G(\theta)$.

Say $M$ is a maximal normal subgroup which contains $N$. Since $G/N$ is solvable, $|G/M|=p$ where $p$ is a prime. By corollary(6.19) in Character theory of finite groups by I.Isaacs, and the fact that $G=$I$_G(\theta)$, we know that $\chi_M$ is irreducible.

$N\leq M<G$, by induction, $\chi_{M}(1)/\theta(1)$ devides $|M:N|$, and $\chi_M(1)=\chi(1)$, so $\chi(1)/\theta(1)$ devides |G:N|.