Normal subgroups and permutation characters

Question

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 trivial character.

So far, I have just managed to use Frobenius Reciprocity to get that $\langle (1_H)^G, \chi \rangle = \langle 1_H, \chi_H \rangle$

Answer 1

Use the following hint: $ker((1_H)^G)=core_G(H)$. Since $H$ is normal this kernel equals $H$. And $ker(\chi)$ contains $ker((1_H)^G)$.