Induction, products, and tensor products $\DeclareMathOperator{\Ind}{Ind}$Let $G, G'$ be groups, and  $H'$ be a subgroup of finite index of $G'$.  Let $(\pi,V)$ be a representation of $G$, and $(\sigma',W)$ a representation of $H'$.  Consider the two representations
$$\pi \otimes \Ind_{H'}^{G'} \sigma'$$
$$\Ind_{G \times H'}^{G \times G'} \pi \otimes \sigma'$$
of $G \times G'$.  There is a $G \times G'$-linear map from the first to the second: if $f: G' \rightarrow W$ is an element of the underlying space of $\Ind_{H'}^{G'} \sigma'$ (so $f(h'g') = \sigma'(h')f(g')$, and $G'$ acts on $f$ by right translation), and $v \in V$, define 
$$F: G \times G' \rightarrow V \otimes W$$
$$F(g,g') = \pi(g)v \otimes f(g')$$
We see that $F$ lies in the underlying space of $\Ind_{G \times H'}^{G \times G'} \pi \otimes \sigma'$, and that $(v,f) \mapsto F$ is $\mathbb C$-bilinear, giving us a linear map of the underlying spaces
$$\pi \otimes \Ind_{H'}^{G'} \sigma' \rightarrow \Ind_{G \times H'}^{G \times G'} \pi \otimes \sigma'$$
which is easily seen to be $G \times G'$-linear.  Is this map an isomorphism?  It seems like it should be, but I haven't been able to prove it.
 A: You're aiming to show that 
$$ \hom_{G \times H'}(kG\otimes kG', V\otimes W) \cong \hom_G(kG,V) \otimes \hom _{H'}(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $G\times H'$-hom from $kG \otimes kG'$ to $V \otimes W$ can be written as a sum of homs of the form $\alpha \otimes \beta$ where $\alpha \in \hom_G(kG,V)$ and $\beta \in \hom_{H'}(kG',W)$.
To do this we'll make use of the fact that  $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ \hom_{G\times H'}(kG \otimes kH',V\otimes W) \cong \hom_G(kG,V)\otimes \hom_{H'}(kG,W)$$
because both sides identify with $V\otimes W$ in a canonical way.
Let $g_1,\ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kG\otimes kG' \cong \bigoplus_i kG \otimes kHg_i$:
$$\hom_{G\times H'}(kG \otimes kG', V\otimes W) = \bigoplus_i \hom_{G\times H'} (kG \otimes kHg_i,V\otimes W) \\ \cong \bigoplus_i \hom_G(kG,V)\otimes \hom_{H'}(kHg_i,W) \cong \hom_G(kG,V) \otimes \hom_{H'}(kG',W)$$
These isomorphisms (going from right to left) tell you that $\hom_{G\times G'}(kG\otimes kG' ,V\otimes W)$ is spanned by tensor products $\alpha\otimes\beta$ as above.  You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $\bigoplus$ would have to be a $\prod$.
