# Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?

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 $\vert G \vert < 10^4$).

• No. It is of mult. $\le 3$ iff it is in the family PSL2, and there mult. $>1$. See this post: mathoverflow.net/q/382380/34538 Feb 10, 2021 at 22:12

There may be some clever trick to do this by elementary arguments, but I don't see it at the moment. With CFSG, the following argument shows that it is likely to be rare to have a Grothendieck ring with multiplicity $1$, if it happens at all. Let $b$ the the largest character degree of the non-Abelian finite simple group $G$, and suppose that $G$ has $k$ conjugacy classes. It is conjectured that $|G| <b^{3}$, and it has been proved by Cossey,Halasi,Maroti and Nguyen (using CFSG) that $|G| <b^{4}$. It has been proved by Fulman and Guralnick (again using CFSG) that $k < |G|^{0.41}$.
Your last inequality ( together with Cauchy_Schwarz) gives $b^{2} < \sqrt{k}\sqrt{|G|}$, so using the Fulman-Guralnick result, $b < |G|^{0.3525}$. Certainly $kb^{2} > |G|$, so we must have $k > |G|^{0.295}$. This last inequality can be achieved ( eg in ${\rm SL}(2,2^{n})$), but it seems likely to be relatively rare ( though precise checking may be painful).
Later edit: In fact, it is interesting to note that when $G = {\rm SL}(2,2^{n}) (n \geq 2)$, if we let $\chi$ denote the Steinberg character (which has degree $2^{n}$, whereas the largest irreducible character degree is $2^{n}+1$), we always .have $\chi \chi = \sum_{ \mu \in {\rm Irr}(G)} \mu$. This is because the projective cover of the (characteristic $2$) trivial module has dimension $2^{2n}-2^{n}$, and occurs as a summand of ${\rm St} \otimes {\rm St}.$
• Is the conjecture $\vert G \vert <b^3$ optimal? Because by your answer, multiplicity one implies $\vert G \vert > b^{2.8368}$. Mar 25, 2015 at 12:45
• It is not far from optimal in general ( again, ${\rm SL}(2,2^{n})$ is not far from achieving it). Mar 25, 2015 at 12:53
• @GeoffRobinson: Your inequalities $k b^2 > |G|$ and $k |G| > b^4$ imply $k > b$ (after multiplying the first inequality by $k$). This is (conjecturally) stronger than your conclusion $k > |G|^{0.295}$. Can the stronger inequality ever be satisfied by a simple group? (It seems that $\mathrm{SL}(2,2^n)$ doesn't quite make it.) Mar 25, 2015 at 17:53