Let $G$ be finite group and $x\in G$. If $\chi$ is irreducible complex character and normal $N\subset G$ with $N\cap C_G (x) =\{1\}$, then
(1) $|C_G (x)| \leq |C_{G/N} (xN)|$
(2)$\chi(x)=0$ if kernel does not contain $N$
I proved (1) by mapping y to yN, but I don't have any idea regarding (2).. (in my textbook,author suggests considering orthogonal relation expressing centralizer of x but it didn't help to me)
Let me use the bar convention. Note the canonical bijection $$ \operatorname{Irr}(\overline{G}) \leftrightarrow \{\, \chi \in \operatorname{Irr}(G) \mid N \subseteq \ker \chi \,\}. $$ From (1) and the orthogonal relation, we have $$\begin{align*} \lvert C_{\overline{G}}(\overline{x})\rvert &\ge \lvert C_G(x) \rvert \\ &= \sum_{\chi \in \operatorname{Irr}(G)} \lvert \chi(x) \rvert^2 \\ &\ge \sum_{\substack{\chi \in \operatorname{Irr}(G)\\ N \subseteq \ker \chi}} \lvert \chi(x) \rvert^2 \\ &= \sum_{\psi \in \operatorname{Irr}(\overline{G})} \lvert \psi(\overline{x}) \rvert^2 \\ &= \lvert C_{\overline{G}}(\overline{x})\rvert. \end{align*}$$ Thus equality holds and we get the desired conclusion.