Supervector spaces look a lot like the category of representations of $\mathbb{Z}/2\mathbb{Z}$ - the even part corresponds to the copies of the trivial representation and the odd part corresponds to the copies of the sign representation. This definition gives the usual morphisms, but it does not account for the braiding of the tensor product, which I've asked about [previously](https://mathoverflow.net/questions/4427/what-is-the-conceptual-significance-of-supercommutativity).


Since the monoidal structure on $\text{Rep}(G)$ comes from the comultiplication on the group Hopf algebra, which in this case is $H = \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$, it seems reasonable to guess that the unusual tensor product structure comes from replacing the usual comultiplication, which is $\Delta g \mapsto g \otimes g$, with something else, or alternately from placing some extra structure on $H$. What extra structure accounts for the braiding? 



Are supervector spaces the representations of a Hopf algebra?
The answer is yes, but comultiplication is not what you change. The symmetrizer (or braiding as you call it) is given by an $R$-element in $H \otimes H$ that makes $H$ into a [triangular Hopf algebra](http://en.wikipedia.org/wiki/Quasitriangular_Hopf_algebra). (The Wikipedia article says quasitriangular; triangular means that plus that the modified switching map $\widetilde{R}$ is an involution.)


The $R$ element is the correction to the usual switching map $x \otimes y \mapsto y \otimes x$. You can solve for it directly in terms of the usual description of the switching map on supervector spaces, and worry later about what axioms it satisfies. Let $1$ and $a$ be the group elements of $H$ and let $\epsilon$ and $\sigma$ be the dual vectors on $H$ which are the trivial and sign representations. Then
we want
$$(\epsilon \otimes \epsilon)(R) = 1 \quad (\epsilon \otimes \sigma)(R) = 1 \quad (\sigma \otimes \epsilon)(R) = 1 \quad (\sigma \otimes \sigma)(R) = -1,$$
because that is the correction in the modified switching map $x \otimes y \mapsto (-1)^{|x||y|} y \otimes x$. You can solve the linear system of equation to obtain
$$R = \frac{1 \otimes 1 + a \otimes 1 + 1 \otimes a - a \otimes a}{2}.$$


