For what $G$ is $Rep(D(S_3))_{ad}$ Grothendieck equivalent to $Rep(G)$?

Question

Given a fusion category $\mathcal C$, the Grothendieck Ring $K_0(\mathcal C)$ is the $\mathbb Z$-based ring whose basis elements correspond to isomorphism classes of simple objects and whose multiplication is given by

$$ X\times Y = \sum_Z N_{XY}^Z Z, N_{XY}^Z=\vert \mathcal C(X\otimes Y,Z)\vert $$

Two fusion categories $\mathcal C$ and $\mathcal D$ are said to be Grothendieck equivalent if $K_0(\mathcal C)\cong K_0(\mathcal D)$.

Given $\mathcal C$, the adjoint subcategory $\mathcal C_{ad}$ is the full fusion subcategory of $\mathcal C$ generated by $X\otimes X^*$, where $X$ is simple.

$Rep(D(S_3))$ has eight simple objects $\{1,\epsilon, \phi_{i=1,\ldots,4},\psi_\pm\}$ and $Rep(D(S_3))_{ad}$ is the subcategory generated by $\{1,\epsilon,\phi_{i=1,\ldots,4}\}$. Its Grothendieck ring is commutative and determined by

$$ \begin{align*} \epsilon \otimes \epsilon &\cong 1 \\ \epsilon \otimes \phi_i &\cong \phi_i \\ \phi_i \otimes \phi_i &\cong 1 \oplus \epsilon \oplus \phi_i \\ \phi_i \otimes \phi_j &\cong \phi_k \oplus \phi_l & i\neq j \neq k \neq l \\ \end{align*} $$

$Rep(D(S_3))$ is modular so $Rep(D(S_3))_{ad}$ is braided(properly premodular). $Rep(D(S_3))_{ad}$ also admits a braiding with S-matrix

$$ \left( \begin{array}{cccccc} 1 & 1 & 2 & 2 & 2 & 2 \\ 1 & 1 & 2 & 2 & 2 & 2 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ \end{array}\right). $$

This braiding is symmetric $(s_{ab}=d_a d_b)$ and so $Rep(D(S_3))_{ad}$ is equivalent as a fusion category to $Rep(G)$ for some finite group $G$. What is this $G$?

Answer 1

Despite my love of the finite group game, let me give an argument that doesn't use the classification of groups of order 18. The 1-dimensional objects correspond to representations of the abelianization. So your group must have abelianization $C_2$, and so its commutator subgroup must be a group of size 9. Note that this splits as a semidirect product because there's a 2-sylow subgroup.

A full tensor subcategory of $\mathrm{Rep}(G)$ which is closed under summands must be of the form $\mathrm{Rep}(G/N)$. (The proof of this is roughly the same as the proof that faithful representations tensor generate.) Thus your group must have $S_3$ as a quotient in four different ways. Hence the commutator subgroup must be elementary abelian and the $C_2$ must act on each of the factors by inversion (and thus on the whole thing by inversion).