Exercise 2.15 M.Isaacs' Character theory of finite groups I'm beggining to study character theory, and i'm  doing some problems from Isaacs' Character theory book.
I would need some help with this one:
(2.15): Let $\chi\in \operatorname{Irr}(G)$ be faithful and suppose $H\subseteq G$ and $\chi_{H} \in \operatorname{Irr}(H)$. Show that $C_{G}(H)=Z(G)$
I have only this...
I know i have to use lemma 2.27.
As $\chi$ is faithful, $\ker(\chi)=\{1\}$. From lemma 2.27, we could get that $Z(\chi)=Z(\chi)/\ker(\chi)=Z(G/\ker(\chi))=Z(G)$ and that $Z(G)$ is cyclic. But i'm stuck here.
Thank you very much for any help.
 A: Hint: Schur's Lemma. What do you know about the which commute with (the images of) all elements of $H$ in the associated (irreducible) representation of H?
A: 
Proposition Let $H$ be a subgroup of the finite group $G$, $\chi \in Irr(G)$ an irreducible complex character and assume that the restriction $\chi_H \in Irr(H)$. Then $C_G(H) \subseteq Z(\chi)$.
Corollary If $H \leq G$, $\chi \in Irr(G)$ a faithful irreducible character, such that $\chi_H \in Irr(H)$ then $Z(G)=C_G(H)$.

So let us prove this. Let $\mathfrak{X}$ be an $\mathbb{C}$-representation that affords $\chi$. Let $x \in C_G(H)$. Since the restriction to $H$ is irreducible and $\mathfrak{X}(x)$ commutes with every $\mathfrak{X}(h), h \in H$, and it follows from Lemma (2.25) of the book (which is basically Schur's Lemma) that $\mathfrak{X}(x)=\varepsilon I$ for some root of unity $\varepsilon \in \mathbb{C}$. By taking traces it follows that $|\chi(x)|=\chi(1)$, that is $x \in Z(\chi)$.
Now assume that $\chi$ is faithful (meaning $ker(\chi)=1$). Since in general, $Z(G/ker(\chi))=Z(\chi)/ker(\chi)$, we obtain $Z(\chi)=Z(G)$. The proposition then yields $C_G(H) \subseteq Z(G)$. But of course the other containment, $Z(G) \subseteq C_G(H)$ is always true. Hence $Z(G)=C_G(H)$ as wanted. (Note that $Z(G)$ must be cyclic in this case since $Z(\chi)/ker(\chi)$ is always cyclic).
A: Let $\phi$ be the representation of G affording $\chi$ and so $\phi_{H}$ will be the representation of H affording $\chi_{H}.$$\\$
Clearly Z(G) $\subseteq$ $C_{G}(H)$. Now let us suppose that this containment is proper and we come up with a contradiction. Let g $\in$ $C_{G}(H)$ but g $\notin$ Z(G).Now consider the subgroup generated by g and H. Call it K . Then consider $\chi_{K}$. Since H is a subgroup of K and $\chi_{H}$ is irreducible so, $\chi_{K}$ is irreducible.Also observe that g commutes with all elements of K. Now as in the hint above using Schur's lemma $\phi_{K}(g)$ is a scalar multiple of identity and hence $\phi(g)$ is so. But since $\phi$ is an isomorphism of G onto $\phi(G)$, center of G must correspond to center of $\phi(G)$. But a scalar matrix commute will all matrices and since g$\notin$Z(G) $\phi(g)$ cannot be a scalar matrix, and hence  it yields a contradiction thereby proving our claim.
