Let $G$ be a group. If $B$ and $A$ are $G$-modules, make $A \otimes_{\mathbb{Z}} B$ into a $G$-module with diagonal action:
$$
g(b \otimes a)=(g b) \otimes(g a) .
$$
If $A$ is a $G$-module, let $A_{0}$ be its underlying abelian group. Prove that $\mathbb{Z} G \otimes_{\mathbb{Z}} A_{0} \cong \mathbb{Z} G \otimes_{\mathbb{Z}} A$ as $G$-modules.
Hint. Define $f: \mathbb{Z} G \otimes_{\mathbb{Z}} A_{0} \rightarrow \mathbb{Z} G \otimes_{\mathbb{Z}} A$ by $g \otimes a \mapsto g \otimes g a$.