Does Tambara-Yamagami category admit a braiding when G is a non-abelian group? Tambara-Yamagami category is a fusion category which its simple objects are elements of a group and one element out of group. i.e : 
$$simple\;objects = G \cup \{m\}$$ 
The fusion rule of this category is :
$$m\times m=\sum_{i \in G} g_i, \hspace{0.5cm} g_i\times g_j=g_i*g_j(*:group\;action),\hspace{0.5cm}
m\times g_i=g_i\times m =m $$
As far as I know, according to [1],[2] there is a complete classification for braided Tambara-Yamagami category if we put G an abelian (2-)group.  
My question is about this category if G is a non-abelian group.
1-Tambara, Daisuke, and Shigeru Yamagami. "Tensor categories with fusion rules of self-duality for finite abelian groups." Journal of Algebra 209.2 (1998): 692-707.
2-Siehler, Jacob A. "Braided near-group categories." arXiv preprint math/0011037 (2000).
 A: The paper of Tambara-Yamagami classifies all semi-simple split tensor categories whose fusion rules are given by the rules you stated above. 
In theorem 3.2, they state that such categories are determined by only two invariants: The bicharacter $\chi:A\times A\rightarrow k^*$ and a special element $\tau\in k$ such that $|A|\tau^2=1$. Moreover, the any such category can be explicitly realized as $\mathcal{C}(\chi,\tau)$ (see definition 3.1).
Now above definition 3.1 you can read that the non-degeneracy of $\alpha_2$ implies that $A$ must be abelian. $\alpha_2$ is one of these functionals appearing from the tetrahedral transformations and must satisfy 
$$\alpha_2(ab,c)=\alpha_2(a,c)\alpha_2(b,c)$$
for all $a,b,c\in A$.
But the right-hand side of the above equation lives in $k$, thus $$\alpha_2(ab,c)=\alpha_2(b,c)\alpha_2(a,c)=\alpha_2(ba,c).$$
Together with non-degeneracy, you conclude that $ab=ba$, hence $A$ is abelian.
A very interesting application to this paper is the following: Consider the dihedral group $D_8$ of order 8 and the quaternion group $Q_8$. These groups are not isomorphic but have the same character table. It follows that the representation categories of $D_8$ and $Q_8$ are equivalent, but they are not monoidally equivalent. These two representation categories fit into the above framework. Both groups have 4 one-dimensional representations that can be identified with the group $\mathbb{Z}_2\oplus \mathbb{Z}_2$. Both also have an irreducible 2-dimensional representation which can be identified with $m$ in the fusion rules. Thus the group $A=\mathbb{Z}_2\oplus \mathbb{Z}_2$ corresponding to these representation categories is indeed an abelian group as  required. The difference of these two representation categories can be found deep in the monoidal structure (you can't see the difference from the fusion rules). It turns out that the element $\tau$ will see the difference.  You could wonder what that actually means.
