I'm reading the chapter on group homology/cohomology in Rotman's "An Introduction to Homological Algebra". The book says the following (where $G$ is a group and $\mathbb{Z}G$ the group ring):
Exercise 9.3 on page 503 shows that every left $\mathbb{Z}G$-module $K$ is also a right $\mathbb{Z}G$-module if one defines $ax$ to be $x^{-1}a$, where $x \in G$ and $a \in K$. Thus, if $K$ and $L$ are $G$-modules, we can always adjust them so that $K \otimes_G L$ is defined.
Then, further along, the book defines the diagonal action on the tensor product:
If $M$ is a right $\mathbb{Z}G$-module and $B$ is a left $\mathbb{Z}G$-module, the diagonal action on $M \otimes_{G} B$ is given by $g(m \otimes b) = gm \otimes gb$, where $g\in G$, $m \in M$, and $b \in B$.
Since $M$ is a right $\mathbb{Z}G$-module, I take $gm$ to mean $mg^{-1}$. But then we have $$g(m \otimes b) = gm \otimes gb = m g^{-1} \otimes gb = m \otimes g^{-1} g b = m \otimes b.$$ This can't be right. What am I missing? Thanks!
Well, I guess that is an action of $G$: it's just the trivial action!
But I'm pretty sure that what Rotman meant to write is that if $M$ and $B$ are left $\mathbb{Z}G$-modules, then the diagonal action on $M\otimes_{\mathbb{Z}}B$ (the tensor product over $\mathbb{Z}$ rather than over $\mathbb{Z}G$) is given by $$g(m\otimes b)=gm\otimes gb.$$
The book has an analogous error (on page 519) for the diagonal action on what should be $\operatorname{Hom}_{\mathbb{Z}}(A,B)$, writing $\operatorname{Hom}_G(A,B)$ instead. This one is corrected in the errata at https://faculty.math.illinois.edu/~rotman/, but he doesn't seem to have caught the tensor product mistake.