$\renewcommand{\Res}{\mathrm{Res}} \renewcommand{\Ind}{\mathrm{Ind}}$ Let $H$ be a subgroup of a finite group $G$, $N$ be a normal subgroup of $G$ and $(\rho,W)$ be a representation of $H$. Then $$W^{H \cap N} := \{w \in W \mid \rho(x)w=w,\;\forall x \in H \cap N\}$$ provides a representation of $H/(H \cap N)$, and $(\Ind_H^G W)^N$ provides a representation of $G/N$.
I want to show that $$ (\Ind_H^G W)^N \;\cong\; \Ind_{H/(H \cap N)}^{G/N}(W^{H \cap N}) $$ as representations of $G/N$.
In other words, I want to show $$V:= (k[G] \otimes_{k[H]} W)^N \;\cong\; k[\overline G] \otimes_{k[\overline H]} W^{H \cap N} =: V'$$ as $k[\overline{G}]$-modules, where $\overline G=G/N,\overline H=H/(H \cap N)$.
What I tried:
I wanted to define $$\psi : V \to V', \qquad g \otimes w \mapsto [g]_N \otimes w$$ and $$\psi^{-1} : V' \to V, \qquad [g]_N \otimes w \mapsto g \otimes w$$ and then show that they are $k[\overline G]$-homomorphisms, inverse of each other.
It shouldn't be too hard to show that the definition of $\psi^{-1}([g]_N \otimes w)$ doesn't depend on the choice of $g$. The most difficult is to show that this behaves well with tensor product. I had trouble to define properly a $k[H]$-biadditive map $k[G] \times W \to V'$ ($(g, w) \mapsto [g]_N \otimes w$ doesn't work because $w$ may not lie in $W^{H \cap N}$. Moreover, $ k[\overline G] \otimes_{k[\overline H]} W$ has no meaning since $W$ is not a $k[\overline H]$-module.)
So the main question is: how to construct properly the isomorphisms $\psi,\psi^{-1}$ (for instance, how to show, or avoid showing, that $$a:= \sum_{i=1}^n \left(\sum_{g \in G} a_g^i g \;\otimes w_i \right) = \sum_{j=1}^m \left(\sum_{g \in G} b_g^i g \;\otimes u_j \right) =: b \implies \psi(a)=\psi(b)$$ ?)
Thanks!