If $H$ is a normal subgroup of $G$ with character $\chi$, what is $\chi(ghg^{-1})$?

Question

Let $H\subset G$ be a normal subgroup of $G$ and suppose that $\chi$ and $\tilde{\chi}$ are irreducible characters of $H$ satisfying $$ \sum_{g\in G}\chi(ghg^{-1})=\sum_{g\in G}\tilde{\chi}(ghg^{-1}) $$ for all $h\in H$. Does it follows that $\chi=\tilde{\chi}$? I don't think it needs to hold, but I can't construct a proof nor with a counterexample

Answer 1

I think you want to assume $\chi$ to be irreducible; and we require more conditions on $\chi,\tilde{\chi}$ so that your equation will not hold for them.

Let $G=Q_8=\langle x,y\rangle$ and $H=\langle x\rangle\cong\mathbb{Z}/4$.

Let $\chi(x)=i$ and $\tilde{\chi}(x)=-i$, so $\chi\neq \tilde{\chi}$. But, $$\sum_{g\in G} \chi(ghg^{-1})=|H|(\chi(h)+\chi(h^{-1}))$$ and similarly for $\tilde{\chi}$. Moreover, although $\chi\neq \tilde{\chi}$, we have $\chi(h)+\chi(h^{-1})=\tilde{\chi}(h)+\tilde{\chi}(h^{-1})$ for all $h\in H$. So your equation is satisfied with this $\chi,\tilde{\chi}$.

Answer 2

Let $\chi$ and $\tilde{\chi}$ be irreducibles of different dimensions. (Which will be available in 'most examples'). Let $h=e$, the identity, then we get $|G|dim\chi=|G|dim\bar{\chi}$, a contradiction.