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How Frobenius Reciprocity can help us to solve these two problems: Let $ H $ be a subgroup with index $ m $ in the finite group $ G $. Let $ F $ be an algebraic closed field of characteristic $ 0 $. If $ \chi $ is an irreducible $ F $-character of $ G $ with degree $ n $, then there is an irreducible $ F $-character $\varphi $ with degree at least $ n/m $. If $ H $ is abelian, then no irreducible $ F $-character of $ G $ has degree greater than $ m $.

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  • $\begingroup$ You asked the same question earlier. By Frobenius Reciprocity, any irreducible constituent $\varphi$ of $\chi_H$ must have degree at least $n/m$, and if $H$ is abelian then $\varphi$ has degree $1$, so $n \le m$. $\endgroup$ – Derek Holt May 23 '15 at 10:31
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For the second part (no Frobenius Reciprocity needed!): let $\chi \in Irr(G)$, $H \leq G$. Then $$[\chi_H, \chi_H]= \frac{1}{|H|}\sum_{h \in H}|\chi(h)|^2 \leq [G:H]\cdot \frac{1}{|G|} \sum_{g \in G}|\chi(g)|^2=[G:H]\cdot[\chi,\chi]=[G:H].$$ Now, if $H$ is abelian, then $\chi_H=\sum_{\lambda \in Irr(H)} a_\lambda \lambda$, with $a_\lambda$ non-negative integers. Note that all the $\lambda$'s are linear characters. Then $$[\chi_H,\chi_H]=\sum_\lambda a_\lambda^2 \geq \sum_\lambda a_\lambda=\chi_H(1)=\chi(1),$$ since the $a_\lambda$ are non-negative integers. It follows that $\chi(1) \leq [G:H].$

Now for the first part, you probably meant that $\varphi \in Irr(H)$. Take $\varphi$, to be an irreducible constituent of $\chi_H$. So $[\chi_H,\varphi] \geq 1$. By Frobenius Reciprocity, $[\chi_H,\varphi]=[\chi,\varphi^G] \geq 1$. So $\chi(1) \leq \varphi^G(1)=[G:H]\varphi(1)$. Hence, $\varphi(1) \geq \frac{\chi(1)}{[G:H]},$ and you are done.

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  • $\begingroup$ Why downvoting this? $\endgroup$ – Nicky Hekster May 23 '15 at 18:18

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