# Frobenius Reciprocity and a character theory problem

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 $m$.

• 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 $n \le m$. – Derek Holt May 23 '15 at 10:31

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.