Answer by Nicky Hekster for Normal subgroups and permutation characters
Use the following hint: $ker((1_H)^G)=core_G(H)$. Since $H$ is normal this kernel equals $H$. And $ker(\chi)$ contains $ker((1_H)^G)$.
View ArticleNormal subgroups and permutation characters
Consider $H\unlhd G$ and $\chi \in \text{Irr}(G) $ is an irreducible constituent of the the permutation character $(1_H)^G$, then $\chi_H = \chi(e)1_H$, where $e$ is the identity and $1_H$ is the...
View Article
