Character theory - exercise 5.16 from Isaacs Hi I am trying to solve the following exercise. Let $H$ be maximal subgroup of a finite group $G$ and let $\chi=(1_H)^G$. Let $\psi$ be a non-principal irreducible constituent of $\chi$. Then $Ker \psi=Ker \chi$. ** $Ker \chi=Ker (1_H)^G=(Ker 1_H)_G=H_G$ where $H_G$ is the core of $H$. Since $\psi$ is a constituent of $\chi$, it follows that $Ker \chi \le Ker\psi$. What I need to show is $Ker\psi\le Ker\chi$.

Let $K = \ker \psi$. Then, since $H$ is maximal in $G$, we have $HK=H$ or $KH=G$.

If $HK=H$ then $K \le H$, so $K \unlhd G$ implies $K \le H_G$, which is what we want,

But if $HK=G$, then $\psi_H$ is irreducible. Then by Frobenius Reciprocity, $\langle \chi,\psi \rangle = \langle 1_H,\psi_H \rangle > 0$, so $\psi_H=1_H$ and hence $\psi=1$, contradiction.