$\newcommand{\Z}{\mathbb{Z}}$ This identity occurs in Cassels and Frohlich, page 98. Let me recall the context: If $A, B$ are left $G$--module i.e. $\Z[G]$-module then we turn $A$ into right $G$-module by defining $a \cdot g = g^{-1} a$ so that tensor product $A \otimes_{\Z[G]} B$ makes sense. For any $G$-module $A$, we define $$A_G = \Z \otimes_{\Z[G]} A = A/I_G A$$ where $I_G$ is the augmentation ideal, generated by $g - 1$ for all $g \not= 1$. This is largest quotient of $A$ where $G$ acts identically. (Here, $\Z$ is a $(\Z[G], \Z[G])$-bimodule with trivial action on both sides.) The equation in question also reads $$A \otimes_{\mathbb{Z}[G]} B = \Z \otimes_{\Z[G]} (A \otimes B)$$ Note that the defined right action does NOT make $A$ into a $(\Z[G], \Z[G])$-bimodule since the associativity $g (a h) = (g a) h$ does not hold in general. So I do not expect the LHS to be a well-behaved $\Z[G]$-module with obvious action $g(a \otimes b) = ga \otimes b$. The obvious correspondence $a \otimes b \mapsto 1 \otimes (a \otimes b)$ does not seem to work: $g (a \otimes b) = ga \otimes b \mapsto 1 \otimes (ga \otimes b)$ and $g (1 \otimes (a \otimes b)) = 1 \otimes (a \otimes b)$ and one do not expect $ga \otimes b = a \otimes b$ in $A \otimes B$.
EDIT: I think I get this now. We always define $G$-action on $A \otimes_{\mathbb{Z}[G]} B$ and $A \otimes B$ to be $$g(a \otimes b) = ga \otimes gb,$$ not the one I expected from general tensor product $M \otimes_R N$ where $M$ is $(S,R)$-bimodule.
