TITLE: Tannakian Formalism for the Quaternions and Dihedral Group
QUESTION [8 upvotes]: It is a basic fact in representation theory of finite groups over complex numbers that the character tables of $Q_8$ and $D_8$ are identical.  I believe, this implies that the corresponding categories of representations are equivalent (as tensor categories). 
On the other hand, Tannakian Formalism tells us that we can reconstruct a finite group $G$ from its category of representations $\mathbf{Rep}_G$ together with the natural (foregetfull) fibre functor $F_G: \mathbf{Rep}_G\rightarrow \mathbf{Vect}$. Namely, $G$ is canonically isomorphic to the tensor automorphisms of the tensor functor $F_G$. 
This implies that the fibre functors $F_{D_8}$ and $F_{Q_8}$ are different. How can one see this explicitly in terms of representations of $D_8$ and $Q_8$?

REPLY [16 votes]: Let $V_D$ and $V_Q$ be the two dimensional simple representations of $D_4$ and $Q_8$ respectively. Let $1_D$ and $1_Q$ denote their trivial representations.
Suppose that there is a tensor equivalence between $\mathbf{Rep}(D_4)$ and $\mathbf{Rep}(Q_8)$ commuting with the fibre functor to $\mathbf{Vect}_\mathbb{C}$. This equivalence sends $1_D$ to $1_Q$ (as they're the unit object) and sends $V_D$ to $V_Q$ (as they're the unique simple of dimension 2).
In particular there is a $\mathbb{C}$-linear isomorphism $g$ from $V_D$ to $V_Q$. Consider $$g\otimes g:V_D\otimes V_D\to V_Q\otimes V_Q.$$
It must send the unique copy of $1_D$ in $V_D\otimes V_D$ to the unique copy of $1_Q$ in $V_Q\otimes V_Q$.
It is easy to see that there is no such $g$. The slickest way I can see to prove this is to note that the flip map $v\otimes w\mapsto w\otimes v$ acts by -1 on $1_Q$ and by 1 on $1_D$.

REPLY [12 votes]: The categories ${\rm Rep}(Q_8)$ and ${\rm Rep}(D_8)$ are not equivalent as tensor categories. They have the same Grothendieck ring, but they have non equivalent associators. As far as I am aware, it is an open problem to classify all tensor categories which have the same Grothendieck ring as ${\rm Rep}(Q_8)$ (there are at least two).