# problem 6.7 in Character theory of finite groups

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), it's ture by induction.

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

• But why it is also true when G/N is not solvable? Thanks for any help. – Vegetable Feb 1 '20 at 8:37
• According to a detailed version of Clifford's Theorem, in the case $t=1$, $e$ is the degree of an irreducible projective representation of $G/N$ which was shown by Schur to divide $|G/N|$. – Derek Holt Feb 1 '20 at 11:53
• thanks! This is very correct and helpful. This is even effective when $G$ is not solvable. I had not known projective representation when I posted this question.. – Vegetable Mar 9 '20 at 13:24

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