2
$\begingroup$

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$_G(\theta)$ ?

$\endgroup$
3
  • $\begingroup$ But why it is also true when G/N is not solvable? Thanks for any help. $\endgroup$ – Vegetable Feb 1 '20 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 '20 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.. $\endgroup$ – Vegetable Mar 9 '20 at 13:24
2
$\begingroup$

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|.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.