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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.

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  • $\begingroup$ If you just type out operator names like $\ker$ and $\operatorname{Irr}$, $\TeX$ interprets them as juxtaposed variable names and formats them accordingly. To get the right font and spacing, you need to either use predefined commands like \ker, or, if there isn't a predefined command for an operator, you can use e.g. \operatorname{Irr}. $\endgroup$ – joriki Mar 23 '13 at 13:33
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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?

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  • $\begingroup$ Thanks for the answer, but i still don't get it. I don't see how to link Shur's Lemma with this. Could you detail this more? $\endgroup$ – Mark_Hoffman Mar 23 '13 at 14:27
  • $\begingroup$ Well anything in $C_{G}(H)$ is represented by a matrix commuting with (the image of) every element of $H$ in the representation affording $\chi.$ Since $H$ is represented irreducibly, Schur's Lemma tells you the form of any matrix commuting with (the images of) all the elements of $H.$ $\endgroup$ – Geoff Robinson Mar 23 '13 at 15:01
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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.

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