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I'm a starter at character theory. I'm trying to do this exercise:

(2.8) Let $\chi$ be a faithful character of a group $G$. Show that $H\subseteq G $ is abelian if and only if every irreducible constituent of $\chi_{H} $ is linear.

I think that one implication is easy, because if H is abelian, then every irreducible character of H is linear. So every irreducible constituent of $\chi_{H} $ will be linear. Am i right with this?

To prove the other implication, i need to see that $H$ is abelian ( this equals to see that it has as many conjugacy classes as elements). As the number of conjugacy classes is equal to the number of irreducible characters, it would be enough to prove that $H$ has $|H|$ irreducible characters. But how i get that?

Thank you very much for any help.

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    $\begingroup$ If all irreducible constituents of $\chi_H$ are linear, then the image of the restriction to $H$ of the corresponding representation is abelian, and so $[H,H]$ is in the kernel of $\chi$. But $\chi$ is faithful. $\endgroup$ – Derek Holt Mar 23 '13 at 13:01
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Your proof of the easier implication is perfectly correct. Your idea for the other direction is not bad, but I don't think it works this time. Derek Holt's comment provides a technique that is used again in chapter 3: you can compute the kernel of faithful reducible characters two different ways.

If $\chi_H$ is a sum of linear characters, then its kernel is the intersection of the kernels of those linear characters, but that kernel at least contains $[H,H]$. Since $\chi$ is faithful, $\chi_H$ is faithful and $[H,H] \leq \ker(\chi_H) = H \cap \ker(\chi) = 1$, so that $H$ is abelian.

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  • $\begingroup$ Thanks a lot for the explanation, Jack. $\endgroup$ – Mark_Hoffman May 23 '13 at 12:00

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