rt.representation theory - Induction of tensor product vs. tensor product of inductions

Try using Frobenius reciprocity. Let $V$ and $W$ be two representations of $H$, and let $U$ be a representation of $G$. Consider first the space:
$$Hom_G left(U, Ind_H^G (V otimes W) right) cong Hom_H left( U, V otimes W right),$$
by Frobenius reciprocity.

On the other hand, one can consider the space:
$$Hom_G(U, Ind_H^G V otimes Ind_H^G W).$$
This is canonically isomorphic to
$$Hom_G(U otimes Ind_H^G V', Ind_H^G W),$$
where $V'$ denotes the dual representation of $V$. By Frobenius reciprocity again, this is isomorphic to:
$$Hom_H(U otimes Ind_H^G V', W).$$
This is canonically isomorphic to
$$Hom_H(U, (Ind_H^G V) otimes W).$$
Now, we are led to compare the two spaces:
$$Hom_H(U, V otimes W), quad Hom_H left( U, (Ind_H^G V) otimes W right).$$

There is a natural embedding of $V$ into $Res_H^G Ind_H^G V$. This gives a natural map:
$$iota: Hom_H(U, V otimes W) rightarrow Hom_H left( U, (Ind_H^G V) otimes W right).$$

Using complete reducibility, let us (noncanonically) decompose $H$-representations:
$$Res_H^G Ind_H^G V cong V oplus V^perp.$$
It follows that
$$Hom_H left( U, (Ind_H^G V) otimes W right) cong Hom_H left(U, V otimes W right) oplus Hom_H left( U, V^perp otimes W right).$$

It follows that $iota$ is injective. This explains (via Yoneda, if you like) why $Ind_H^G(V otimes W)$ is canonically a subrepresentation of $Ind_H^G V otimes Ind_H^G W$. It also explains that computation of "the rest" of $Ind_H^G V otimes Ind_H^G W$ -- the full decomposition into irreducibles -- requires Mackey theory: the decomposition of $Res_H^G Ind_H^G V$. There can be no neat answer, without performing this kind of Mackey theory.