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.



Uniqueness of the fusion ring for simple finite group
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](https://mathoverflow.net/questions/57375/character-tables-and-simple-groups) 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. 


