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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$_G(\theta)$ ?

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  • $\begingroup$ But why it is also true when G/N is not solvable? Thanks for any help. $\endgroup$ Feb 1, 2020 at 8:37
  • $\begingroup$ 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 $|G/N|$. $\endgroup$
    – Derek Holt
    Feb 1, 2020 at 11:53
  • $\begingroup$ 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.. $\endgroup$ Mar 9, 2020 at 13:24

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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|.

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