A fusion ring $\mathcal{F}$ is given by a finite set $B = \{b\_1,b\_2, \dots, b\_r \}$ such that $b\_i b\_j = \sum\_k n\_{i,j}^k b\_k$ with $n\_{i,j}^k \in \mathbb{Z}\_{\ge 0}$, satisfying axioms slightly augmenting the group axioms (see the details [here](https://mathoverflow.net/q/344079/34538)). 


The fusion ring $\mathcal{F}$ is called *noncommutative* if $\exists i,j$ with $b\_ib\_j\neq b\_jb\_i$. A *fusion subring* is given by a subset $B\_S=\{b\_s \ | \ s \in S\} \subseteq B$ with $1 \in S = S^\*$ and $\forall i,j \in S$ then $n\_{i,j}^k \neq 0$ only if $k \in S$. It is called *simple* when every fusion subring is given by the subset $\{b\_1\}$ or $B$. Let $G$ be a finite group, then the Grothendieck ring of $Rep(G)$ is simple (as fusion ring) if and only if $G$ is simple.


**Question**: Is there a noncommutative simple fusion ring? 




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*Investigation* 


According to [this post](https://mathoverflow.net/q/348527/34538), a noncommutative fusion ring must be of rank at least $5$, and at rank exactly $5$, we can assume: $2^\*=3$, $4^\*=4$ and $5^\*=5$. A brute-force computation shows that at multiplicity at most three, there is a unique *simple* example (upto equiv.):  

$$\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix} , \
\begin{smallmatrix}0&1&0&0&0\\0&0&1&1&0\\1&0&0&0&1\\0&0&1&0&1\\0&1&0&1&1\end{smallmatrix} , \
\begin{smallmatrix}0&0&1&0&0\\1&0&0&0&1\\0&1&0&1&0\\0&1&0&0&1\\0&0&1&1&1\end{smallmatrix} , \
\begin{smallmatrix}0&0&0&1&0\\0&0&1&0&1\\0&1&0&0&1\\1&0&0&1&1\\0&1&1&1&1\end{smallmatrix} , \
\begin{smallmatrix}0&0&0&0&1\\0&1&0&1&1\\0&0&1&1&1\\0&1&1&1&1\\1&1&1&1&2\end{smallmatrix}$$ but it is commutative. Note that $[d(b\_1),d(b\_2),d(b\_3),d(b\_4),d(b\_5)] =[1,\alpha,\alpha,\beta,\gamma]$ with: 


* $\alpha =1+2cos(2\pi/7) \simeq 2.246979603\dots$,
* $\beta =1-2cos(6\pi/7) \simeq 2.801937735\dots$,
* $\gamma =\alpha+\beta-1 \simeq 4.048917339\dots$,
* so that: $gdim:=\sum\_i d(b\_i)^2 \simeq 36.65039990\dots$,


moreover, it is of Frobenius type and satisfies Schur Product Property.



Is there a noncommutative simple fusion ring?
The fusion ring of one of the even parts of Extended Haagerup will work (see [Appendix A of our paper](https://arxiv.org/abs/0909.4099)). It has rank 8, and you can see that $AB \neq BA$. Surely there's less complicated examples though.


