Lower bound for the order of a simple group with a given class number

Question

Every simple group below are assumed non-abelian.

Let us call the class number $k(G)$ of a finite group $G$ the number of its conjugacy classes (also, the number of its irreducible complex representations, up to equivalence). Here is the computation of the order and the class number of the first finite simple groups:

gap> it:=SimpleGroupsIterator(10,1000);
<iterator>
gap> for G in it do Print([G,Order(G),NrConjugacyClasses(G)]); od;
[ A5, 60, 5 ][ PSL(2,7), 168, 6 ][ A6, 360, 7 ][ PSL(2,8), 504, 9 ][ PSL(2,11), 660, 8 ]  

The points of the following picture corresponds to the $161$ finite simple groups of order less than $10^8$, where $x$-axis is the class number and $y$-axis the order.

We observe that these points are bounded below by the curve interpolating the points for the groups $\mathrm{PSL}(2,2^n)$. We checked that $k(\mathrm{PSL}(2,2^n)) = 2^n+1$, for $n \le 50$, so that the equality should be known true in general. Now $$|\mathrm{PSL(2,2^n)}| = 2^n (2^{2n}-1) = (r-1)((r-1)^2-1)) = (r-2)(r-1)r,$$ with $r = 2^n+1$, which leads to:

Question: Let $G$ be a finite simple group of class number $r$. Is it true that $|G| \ge (r-2)(r-1)r$?

If so, do you expect the existence of a proof without CFSG?
(because motivation: extend such a result to the simple integral fusion rings)

gap> for r in [5..15] do Print([r,(r-2)*(r-1)*r]); od;
[ 5, 60 ][ 6, 120 ][ 7, 210 ][ 8, 336 ][ 9, 504 ][ 10, 720 ][ 11, 990 ][ 12, 1320 ][ 13, 1716 ][ 14, 2184 ][ 15, 2730 ]  

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Bonus

We checked that $k(\mathrm{PSL}(2,3^n)) = (5 + 3^n) / 2$ for $n \le 40$ (see A289521), moreover, $$|\mathrm{PSL}(2,3^n)| = \frac{1}{2}3^n (3^{2n}-1) = \frac{1}{2}(2r-5)((2r-5)^2-1)) = \frac{1}{2}(2r-6)(2r-5)(2r-4),$$ with $r = \frac{1}{2}(5 + 3^n)$. These groups seems to be in the main curve in the middle. Then:

Augmented question: Let $G$ be a finite simple group of class number $r$. Is it true that $$|G| \ge (r-2)(r-1)r,$$ and that the equality holds iff $G = \mathrm{PSL}(2,2^n)$, otherwise that $G=\mathrm{PSU}(3,3)$ xor $$|G| \ge \frac{1}{2}(2r-6)(2r-5)(2r-4),$$ and that the equality holds iff $G = \mathrm{PSL}(2,q)$ with $q$ a odd prime power?

Remark: It is checked by GAP for $|G| < 15000000$.

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gap> for r in [5..15] do Print([r,(2*r-6)*(2*r-5)*(2*r-4)/2]); od;
[ 5, 60 ][ 6, 168 ][ 7, 360 ][ 8, 660 ][ 9, 1092 ][ 10, 1680 ][ 11, 2448 ][ 12, 3420 ][ 13, 4620 ][ 14, 6072 ]
[ 15, 7800 ]

Answer 1

Apart from finitely many examples, such results would follow from the work of Liebeck and Shalev (Proc. London Math. Soc. 2005). In Corollary 5.2 of that paper they show that for some constant $c >0$ one has $$ |G| \ge c k(G)^4, $$ except for the finite simple groups in the families $L_2(q)$, $L_3(q)$ and $U_3(q)$. Naturally in the remaining families one can compute $k(G)$ directly.

Answer 2

It has been proved by J. Fulman and R. Guralnick that is $G$ is ``almost simple", (and $F(G) = 1$), then $k(G) < |G|^{0.41}.$ This yields $|G| > k(G)^{2.436}.$ This result needs the classification of finite simple groups. I do not know how much it can be improved by restricting to simple groups.

I would not expect your question to be answered without the classification of finite simple groups. It would seem to be necessary, for example, to at least know that there are only finitely many sporadic simple groups.