How is ${}_{kG}\mathcal{M}$ a tensor category? Let $G$ be a monoid and $k$ be a commutative unital ring. We consider the algebra $$kG = \left\{\sum_{g \in G}' \alpha_g g: \alpha_g \in k\right\}$$
with its usual operations. Some notes I'm reading says that the category ${}_{kG}\mathcal{M}$ of $kG$-modules is a tensor category for $\otimes$ without giving further information.
Question: What is the tensor product $\otimes$ here? Given left $kG$-modules $M,N$, I believe $M \otimes N$ denotes the tensor product over the commutative ring $k$ where $M$ and $N$ get a $k$-module from the canonical embedding $k \to kG$. But how do we define a sensible action on $M \otimes N?$
We can try $$g. (m\otimes n) = (gm) \otimes n$$ but this is not symmetric so probably not what I'm looking for!
 A: When you have a free $k$-module $A$, it has the structure of a cocommutative coalgebra, i.e. $k$-linear maps $\delta : A \to A \otimes_k A$ and $\epsilon : A \to k$ such that the following equations are satisfied:
$$\begin{aligned}
(\epsilon \otimes_k \textrm{id}_A) \circ \delta & = \textrm{id}_A \\
(\textrm{id}_A \otimes_k \epsilon) \circ \delta & = \textrm{id}_A \\
(\delta \otimes_k \textrm{id}_A) \circ \delta & = (\textrm{id}_A \otimes_k \delta) \circ \delta \\
\sigma \circ \delta & = \delta
\end{aligned}$$
(Here, $\sigma : A \otimes_k A \to A \otimes_k A$ is the transposition $a \otimes b \mapsto b \otimes a$.)
Explicitly, suppose $A$ has a $k$-basis $X$. We define $\delta$ and $\epsilon$ on $X$ as follows and extend $k$-linearly:
$$\begin{aligned}
\delta (x) & = x \otimes x \\
\epsilon (x) & = 1
\end{aligned}$$
It is easy to check that this does make $A$ into a cocommutative coalgebra. Thus, when $A$ also has the structure of a $k$-algebra that makes $\delta : A \to A \otimes_k A$ and $\epsilon : A \to k$ $k$-algebra homomorphisms – for instance, if the multiplication and unit are defined by $k$-linearly extending a monoid structure on the basis $X$ – we can restrict scalars to turn any $(A \otimes_k A)$-module into an $A$-module. In particular, given two $A$-modules $M$ and $N$, we have an obvious $(A \otimes_k A)$-action on $M \otimes_k N$, namely,
$$(a \otimes b)(m \otimes n) = (a m) \otimes (b n)$$
and this yields an $A$-action on $M \otimes_k N$. Coassociativity of the comultiplication $\delta$ makes $\otimes_k$ an associative monoidal product on the category of $A$-modules, and cocommutativity makes it symmetric.
