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I would like to ask that : assume that $N$ is a normal subgroup of a finite group $G$, if $p$ is a prime divisor of $\chi(1)$ for some $\chi\in Irr(N)$, does it imply that $p$ divides $\varphi(1)$ for some $\varphi\in Irr(G)$?

Thanks in advance.

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1 Answer 1

Yes, it does. This is implied by Clifford theory: if $\tau$ is an irreducible character of $G$, and $N$ is a normal subgroup, then ${\rm Res}_{G/N}\tau = e\cdot \sum \chi_i$, where the irreducible characters $\chi_i$ form one orbit under the action of $G$ on the characters of $N$, and where $e\in \mathbb{N}$. In particular, all the $\chi_i$ have the same degree, and in particular $\deg\chi_i | \deg \tau$. So take your $\chi \in {\rm Irr}(N)$ with $p|\deg \chi$, and take $\tau$ to be any irreducible summand of ${\rm Ind}_{G/N}\chi$. By Frobenius reciprocity, $\chi$ is a summand of ${\rm Res}_{G/N}\tau$, and by Clifford theory therefore $p|\deg \chi | \deg \tau$.

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