Why a different tensor product for left $G$-modules (group representations)? The category of vector (over $k$) representations of a group $G$ is isomorphic to the category of left $k[G]$-modules, where $k[G]$ is the group ring. This isn't just an equivalence of categories, but rather an isomorphism, which is stronger. However we define a tensor product of representations as $g(v\otimes w)= gv\otimes gw$, which differs from the tensor product over $k[G]$ of $k[G]$-modules. For various reasons, we have to define it this way. For starters, one cannot tensor two left modules if the base ring is not commutative.
But there's something unsettling about using a different tensor, about forgetting the base ring. Is there some deeper explanation for this discrepancy other than just "it works"?
 A: Left modules over an arbitrary noncommutative ring don't have a natural notion of tensor product. What is true generally is the following: if $R, S$ are two $k$-algebras ($k$ an arbitrary commutative ring) and $V, W$ two modules over $R, S$, then the tensor product $V \otimes_k W$ is naturally a module over $R \otimes_k S$ (by the functoriality of the tensor product).
In particular, the tensor product $V \otimes_k W$ of two $k[G]$-modules is naturally a $k[G] \otimes_k k[G]$-module. The way we turn it into a $k[G]$-module is by defining a map
$$k[G] \to k[G] \otimes_k k[G]$$
called the comultiplication. It sends $g$ to $g \otimes g$ and naturally gives $k[G]$ the structure of a bialgebra. In fact $k[G]$ is naturally even a Hopf algebra with antipode given by $g \mapsto g^{-1}$, and this lets us define duals of representations also. 
The comultiplication is an extremely natural map. In fact, for any set $S$ we have a canonical diagonal map
$$\Delta : S \to S \times S$$
sending $s$ to $(s, s)$, and this induces a comultiplication on the vector space $k[S]$ making it a coalgebra. So in some sense it is even more basic a structure than the multiplication on $k[G]$ and so there is no reason not to use it to define the tensor product. 
