This asks: Let $\varphi^{G} = \chi \in {\mathrm{Irr}}(G)$ with $\varphi \in {\mathrm{Irr}}(H)$. Show that $Z(\chi) \subseteq H$.
Here is my solution:
Fix a right coset transversal $T$ of $H$ in $G$ with $1 \in T$. Let $x \in Z(\chi)$. Then $$|\varphi^{G}(x)| = \varphi^{G}(1) = |T| \varphi(1)$$ and $\varphi^{G}(x) = \sum_{t \in T} \varphi^{\circ}(txt^{-1})$.
If $T(x) = \{ t \in T ~:~ txt^{-1} \in H \}$, then by triangle inequality we have
$$|T| \varphi(1) = |\varphi^{G}(x)| = |\sum_{t \in T(x)} \varphi(txt^{-1})| \leq |T(x)| \varphi(1)$$
This proves $|T| \leq |T(x)|$ and hence $T(x) = T$. Therefore $txt^{-1} \in H$ for every $t \in T$. In particular, we have $x \in H$.
I fail to see where am I using any of the characters are irreducible. Any idea?
Best Answer
Your proof is OK and yes you do not need $\chi$ being irreducible. Analogous to $(5.11)$ Lemma in Isaacs' book one can prove that if $\varphi$ is a character of a subgroup $H$ of $G$, then $$core_G(Z(\varphi))=Z(\varphi^G)=\bigcap_{x \in G}Z(\varphi)^x.$$ It follows that $Z(\varphi^G) \subseteq H$.