Finite groups G with Rep(G) Grothendieck equivalent to a modular category

Question

We refer to Chapter 8 of the book Tensor Categories for notions related to modular tensor categories and J.P. Serre for the basic theory of linear representations of finite groups over $\mathbb C$.

Let $G$ be a finite group, $\mathrm{Vec}_G^\omega$ be a category of finite dimensional $G$-graded vector spaces (potentially twisted by some non-trivial 3-cocycle $\omega$) and $\mathrm{Rep}(G)$ be the category of finite dimensional complex (for ease) semi-simple representations of $G$. The fusion rules of $\mathrm{Vec}_G^\omega$ (resp. $\mathrm{Rep}(G)$) are given by the product of elements (resp. irreducible characters) of $G$.

The number of conjugacy classes of $G$ (the class number) is equal to the number of its irreducible characters, but there is no "natural" bijection between these two sets (see this post), in particular, the character ring is not equivalent to the conjugacy class ring in general, but note that the equivalence holds in a specific case mentionned here, properly containing the abelian groups.

Two fusion categories are said to be 'Grothendieck equivalent' if their Grothendieck rings (i.e. the de-categorification of their monoidal structure) are equivalent as fusion rings. Let $A$ be a finite abelian group, then $\mathrm{Vec}_A^\omega$ and $\mathrm{Rep}(A)$ are Grothendieck equivalent.

Example 8.13.5 of 1 mentions way to make a modular tensor category using a finite abelian group $A$ and a non-degenerate quadratic form $q: A \rightarrow \mathbb C^*$. It is denoted $\mathcal C(A, q)$ and (see on page 205) is Grothendieck equivalent to $\mathrm{Rep}(A)$.

Thus, for every finite abelian group $A$ on which there exists a non-degenerate quadratic form, $\mathrm{Rep}(A)$ is Grothendieck equivalent to a modular tensor category. But it exists for everyone according to the answers of this post.

Question: Is there a classification or a group-theoretical characterization of the finite groups $G$ such that the tensor category $\mathrm{Rep}(G)$ is Grothendieck equivalent to a modular category? Is there a non-abelian one?

The paper On the classification of weakly integral modular categories shows that all integral modular categories of rank at most $7$ are pointed. It follows that for all non-abelian finite group $G$ of class number at most $7$ (as $S_3$ or $A_5$), $\mathrm{Rep}(G)$ is not Grothendieck equivalent to a modular category.

Answer 1

Here is a necessary condition for a group $G$ such that Rep($G$) is Grothendieck equivalent to a modular category:

there is a bijection between irreducible complex characters of $G$ and conjugacy classes of $G$ such that the size of a conjugacy class equals the square of dimension of the corresponding representation. In particular, the sizes of conjugacy classes are all squares, and the squares of degrees of irreducible characters divide the order of $G$.

Example: the Monster simple group $M$ has a character of degree 196,883; square of this degree does not divide the order of $M$. Thus there is no modular tensor category which is Grothendieck equivalent to Rep($M$).

In fact, I don't know a single non-abelian group $G$ satisfying the condition above.

One obtains the condition above as follows: it is well known that the columns of $S-$matrix of a modular tensor category ${\mathcal C}$ are proportional to various homomorphisms $K({\mathcal C})\to {\mathbb C}$ evaluated at basis elements; also the columns of the character table of $G$ are precisely all the homomorphisms $K({\mbox Rep}(G))\to {\mathbb C}$. Thus the $S-$matrix of a modular category Grothendieck equivalent to Rep($G$) can be obtained from the character table by normalizing and permuting the columns. Using the orthogonality relations for the characters, it is easy to compute that the normalization factors above are precisely square roots of the sizes of the conjugacy classes; since the $S-$matrix must be symmetric we get the condition.