Isaacs exercise 5.4 (Character Theory of Finite groups) Any advice/hints how to prove the following statement?
If $G$ is a finite group and $b(G)=\max\{\chi(1); \chi\in \mathrm{Irr}G\}$ is the maximal degree of irreducible characters and $H\leq G$, then $b(H)\leq b(G)\leq [G:H]b(H)$.
A character of $G$ restricted to $H$ may become reducible. A character of $H$ expanded to $G$ may have increased degree. If $V$ is a $\mathbb{C}H$-module, then $\dim_\mathbb{C}V^G=[G:H]\dim_\mathbb{C}V$. 
 A: If $\psi\in\mathrm{Irr} H$, then $\deg\psi^G=[G\!:\!H]\deg\psi$, so $\psi^G\neq\!0$ and thus there exists $\chi\in\mathrm{Irr} G$ such that by Frobenius reciprocity $[\psi^G,\chi]= [\psi,\chi_H]\neq0$ (because irreducible characters form an orthonormal basis). Therefore $\psi^G=m\chi+\ldots$ and $\chi_H=n\psi+\ldots$ for some $m,n\in\mathbb{N}$. But then $\chi(1)\leq m\chi(1)+\ldots= \psi^G(1)= [G\!:\!H]\psi(1)$ and $\psi(1)\leq n\psi(1)+\ldots= \chi_H(1)=\chi(1)$, so $\psi(1)\leq\chi(1)\leq[G\!:\!H]\psi(1)$.
Thanks to Jack Schmidt.
A: Actually, the idea above is correct, but it is not a proof of the given statement. You find a bound for some $\psi\in$Irr$(H)$ and the respective $\chi\in$ Irr$(G)$.
The upper bound. Let $\chi\in$ Irr$(G)$ with a maximal degree. We have $\chi_H$ is a character of $H$, so there is $\psi\in$ Irr$(H)$ such that $[\psi,\chi_H]\ne0$. Therefore, $[\psi^G,\chi]\ne0$ by the Frobenius reciprocity. We have $\psi^G=n\chi+\ldots$ for some $n$. Thus, $b(G)=\chi(1)\le\psi^G(1)=[G:H]\psi(1)\le[G:H]b(H)$.
The lower bound. Let $\psi\in$ Irr$(H)$ with a maximal degree. Since $\psi^G$ is a character of $G$, we have $[\psi^G,\chi]\ne0$ for some $\chi\in$ Irr$(G)$. Therefore, $[\psi,\chi_H]\ne0$ by the Frobenius reciprocity. We have, $\chi_H=n\psi+\ldots$ for some $n$ and $b(H)=\psi(1)\le\chi_H(1)=\chi(1)\le b(G)$.
