Let $G$ be a finite group and $\mathbb{k}$ is algebraically closed with characteristic zero. Let $H$ be an Abelian subgroup of $G$. Show that the degree of any irreducible representation $V$ of $G$ is bounded by the index of $H$ in $G$.

Given any irreducible representation $V$ of $G$, I know I can get a representation of $H$ by restricting: $V|_H$. But $H$ is Abelian, so $V|_H$ will be a direct sum of $1$-dimensional, and hence irreducible, representations of $H$...

I just don't see how am I going to be able to say something meaningful about the degree of a representation of $G$. Clearly it will have something to do with counting cosets, so it'd be nice if $H$ were normal, but that need not be the case (I'm guessing, given the way the problem is written.) Any help is appreciated

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    $\begingroup$ You might as well work over the complex field. Let $\chi$ be the character afforded by $V.$ Then $\langle \chi|_{H}, \chi|_{H} \rangle \geq \chi(1)$ since if we write $\chi|_{H} = \sum_{i=1}^{h}m_{i}\lambda_{i}$ where $h = |H|$ and the $\lambda_{i}$ are the irreducible (linear!) characters of $H,$ then $\langle \chi|_{H}, \chi|_{H} \rangle \geq \chi(1)$ since $\langle \chi|_{H}, \chi|_{H} \rangle= \sum_{i=1}^{h}m_{i}^{2}.$ On the other hand, this inner product is at most $[G:H]$ since $\chi$ is irreducible for $G$. $\endgroup$ – Geoff Robinson Oct 15 '12 at 1:01
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    $\begingroup$ char 0 is superfluous. Res(V) has a 1 dim sub by abelianality of H. So by adjunction V receives a map from a G-rep of dim |G/H|, surjective by irreducibility of V. $\endgroup$ – Peter McNamara Oct 15 '12 at 5:09
  • $\begingroup$ @GeoffRobinson: How does irreducibility of $\chi$ imply $\langle \chi|_H, \chi|_H\rangle \le [G:H]$. $\endgroup$ – Bey Oct 16 '12 at 0:42
  • $\begingroup$ @bartogian: What is adjunction? $\endgroup$ – Bey Oct 16 '12 at 0:44
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    $\begingroup$ @Bey: Because $\sum_{g \in G} |\chi(g)|^{2} = |G|$, so we certainly have $\sum_{h \in H} |\chi(h)|^{2} \leq |G|$, with equality if and only if $\chi$ vanishes identically outside $H.$ $\endgroup$ – Geoff Robinson Oct 16 '12 at 7:07

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