Suppose $\chi$ is an irreducible character of a finite group $G$, and $H$ is a nontrivial abelian subgroup such that $\chi(1)=[G:H]$. Why does $H$ contain a nontrivial normal subgroup?
I understand that $$ \chi(1)=\chi|_H(1)\leq(\chi|_H,\chi|_G)\leq [G:H](\chi,\chi)=[G:H]=\chi(1) $$ So I have equality throughout. Since $(\chi|_H,\chi|_H)=[G:H]$, I know that $\chi$ vanishes on $G\setminus H$, so necessarily $\ker\chi$ and $Z(\chi)$ are in $H$ at least. These are normal in $G$, so if they're nontrivial, that would explain it. Since $\ker\chi\subseteq Z(\chi)$, I think it'd make sense to show $Z(\chi)$ is nontrivial. Does this happen to be true, or should I be looking for a different subgroup?
For simplicity, I'm working over $\mathbb{C}$.
This is exercise (2.17) of Isaacs' Character Theory of Finite Groups. With Lemma (2.29) of that book you can see that $\chi$ vanishes outside $H$. Now look at the subgroup $N = \langle g \in G : \chi(g) \neq 0 \rangle$. Obviously $N \subseteq H$ hence $N$ is abelian. This subgroup is normal (conjugation does not change the character value) and non-trivial (otherwise the irreducible $\chi$ would vanish outside the identity element which is nonsense).