I want to prove the following formula:
$${\rm Ind}(W)\otimes E \cong {\rm Ind}(W \otimes{\rm Res}(E)),$$
where $H$ is a subgroup of a finite group $G$, $W$ is an irreducible representation of $H$, and $E$ is an irreducible representation of $G$.
I think the proof shouldn't use much more than Frobenius reciprocity. I tried breaking the ${\rm Ind}_H^G$ into tensor products:
$$(W \otimes_{\mathbb{C}[H]} \mathbb{C}[G]) \otimes_{\mathbb{C}[G]} E$$ and
$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} (W \otimes_{\mathbb{C}[H]} E)$$
but I'm not quite sure what to do when we have tensor products that are linear over two different rings in the same expression.
Best Answer
If you want to prove the result with tensor products; use bimodules:
Here the left action of $H$ on $U$ is the one you are given, the right action of $H$ on $U$ is $v \cdot h = h^{-1}v$. Similarly for the $(\mathbb{C}[H], \mathbb{C}[G])$-bimodule structure on $V$.
Then by associativity of tensor products of bimodules, we have an isomorphism $$(\mathbb{C}[G] \otimes U) \otimes V \cong \mathbb{C}[G] \otimes (U \otimes V)$$
of $(\mathbb{C}[G], \mathbb{C}[G])$-bimodules.
As a left $\mathbb{C}[G]$-module, the module on the left hand is isomorphic to $\operatorname{Ind}(U) \otimes V$. As a left $\mathbb{C}[G]$-module, the module on the right hand side is isomorphic to $\operatorname{Ind}(U \otimes V)$. Hence the desired isomorphism $$\operatorname{Ind}(U) \otimes V \cong \operatorname{Ind}(U \otimes V)$$ of left $\mathbb{C}[G]$-modules.
Of course, all of this works over any field, not just $\mathbb{C}$.