Let G be a finite group. Let $H\subseteq G$ be a subgroup of G and suppose $\varphi$ is a class function of $H$ and $\psi$ is a class function of $G$. Show that $(\varphi\psi_H)^G=\varphi^G\psi$.
This problem will be used frequently throughout the book, so it would be very nice to figure it out, but I can't prove it.
I tried to prove in this way:
The character $(\varphi\psi_H)^G$ is afforded to $(V\otimes W_H)^G$ and the character $\varphi^G\psi$ to $(V^G\otimes W)$. It remains to prove that $(V^G\otimes W)$ and $(V\otimes W_H)^G$ are isomorphic as G-modules, but I have no idea how to do it. In particular, it doesn't come in my mind which homomorphism I should use.
Thanks in advance.
Recall that given a class function $f$ of $H$, the induced class function $f^G$ is defined as $g \mapsto \sum_{s\in S} f_0(s^{-1}gs)$, where $S$ is a set of representatives for the cosets in $G/H$, and $f_0(h)=f(h)$ if $h\in H$ and $0$ otherwise.
Now applying this to $\phi \psi_H$, we get $g \mapsto \sum_{s\in S} \varphi_0(s^{-1}gs)(\psi_H)_0(s^{-1}gs)$. Notice that $\varphi_0(s^{-1}gs)(\psi_H)_0(s^{-1}gs)=\varphi_0(s^{-1}gs)\psi_H(s^{-1}gs)$ (just check the cases $s^{-1}gs\in H$ or $\not \in H$). In turn, $\psi_H(s^{-1}gs)=\psi(s^{-1}gs)$, and since $\psi$ is a class function for $G$, this equals $\psi(g)$. It follows that you can factor out $\psi(g)$ on the right, so you get $(\varphi \psi)^G=\varphi^G\psi$.