Distributivity of tensor products over direct sums for group representations

Solution 1:

For the best argument, see the comment by Qiaochu Yuan.

Here is an alternative argument: Let $R$ be a comutative ring. I assume that you already know that the tensor product of $R$-modules commutes with direct sums (see below for the isomorphism). Now let $G$ be a group and let $A,B,C$ be representations of $G$ over $R$, i.e. $R$-modules with an action of $G$. Then the isomorphism of $R$-modules

$f : (A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C), ~ f(a \otimes b)=a \otimes (b,0),~ f(a \otimes c)=a \otimes (0,c)$

is also $G$-equivariant, because $f(g(a \otimes b))=f(ga \otimes gb)=ga \otimes (gb,0)=ga \otimes g(b,0) = g f(a \otimes b)$ and likewise $f(g(a \otimes c))=g f(a \otimes c)$.

Actually, $f$ commutes with the $G$-actions simply because $f$ is a natural isomorphism. Even more abstractly, such a morphism $f$ exists in any monoidal category with coproducts, which is here the category ${}^G \mathsf{Mod}(R)$ of representations of $G$ over $R$. Therefore, actually no computation is needed at all.

Here is a generalization: Let $\mathcal{C}$ be a monoidal category and let $G$ be an arbitrary small category. Then the category ${}^G \mathcal{C}$ of functors $G \to \mathcal{C}$ is again monoidal, the tensor product is defined objectwise. If $\mathcal{C}$ has direct sums which commute with $\otimes$, then the same is true for ${}^G \mathcal{C}$, for trivial reasons: For $A,B,C \in {}^G \mathcal{C}$ there is a canonical morphism $(A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C)$ in ${}^G \mathcal{C}$. That it is an isomorphism, may be checked objectwise, and thereby reduced to $\mathcal{C}$.