TITLE: Uniqueness of the fusion ring for simple finite group
QUESTION [6 upvotes]: We know that the irreducible representations $R_i$ of a group $G$ can give rise to a fusion ring: $R_i\otimes R_j = \oplus_k N^{ij}_k R_k$.
I wonder if the following statement is true or not:
If $G$ is a simple finite group, then its fusion ring is different from the fusion rings of any other groups.

REPLY [8 votes]: The fusion ring, as a ring with basis, contains the same information as the character table. So your question, phrased in language more familiar to finite group theorists, is:

Is a finite simple group determined by its character table among all finite groups? 

Here is how to see the above claim. The easy direction is that if you know the character table then you know the fusion ring, as a ring with basis: this is clear since multiplication of characters corresponds to tensor product of representations. In the other direction, for varying $i$, the matrices $N^{ij}_k$ admit simultaneous eigenvectors (namely the vectors $g \mapsto \chi_V(g)$, where $V$ runs over all irreps), and the simultaneous eigenvalues for these eigenvectors are the character values of $R_i$. 
It's known that a finite simple group is determined by its character table among all finite simple groups. And the character table determines whether or not a finite group is simple: $G$ is simple iff the kernel of every nontrivial irreducible character (the elements $g$ such that $\chi(g) = \chi(1)$) is trivial, since kernels are always normal subgroups and every proper normal subgroup is contained in a kernel. 
So the final answer is yes.