Let $G$ be a finite group. It admits finitely many irreducible complex representations $H_1, \dots, H_r$ which generate, for $\oplus$ and $\otimes$, the Grothendieck ring $\mathcal{G}(G)$ of $G$ (also called its fusion ring).

The tensor product of irreducible representations decomposes into direct sum as follows: $$H_i \otimes H_j = \bigoplus_k M_{ij}^k \otimes H_k$$ with $M_{ij}^k$ the multiplicity space. The Grothendieck ring is of multiplicity $m$ if $max_{i, j, k}(\dim(M_{ij}^k)) = m$.

For $G = A_5$ we get the following dimensions:

and for $G = A_1(7)$, we get:

We observe that $\mathcal{G}(A_5)$ is of multiplicity two, and $\mathcal{G}(A_1(7))$ of multiplicity three.

I've checked (with GAP) that for $\vert G \vert < 10^4$ (nonabelian simple), then $\mathcal{G}(G)$ is not multiplicity one.

**Question**: Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?

If no by using CFSG, can we expect a direct proof?

*Remark*: Suppose $\dim(H_1) \le \dim(H_2) \le \dots \le \dim(H_r)$, then there is the necessary condition:

If the multiplicity is one then $\dim(H_r)^2 \le \sum_i \dim(H_i)$. Perhaps there is no nonabelian finite simple group verifying this weaker condition, which is easier to check (I've used it for $\vert G \vert < 10^4$).