8
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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.

enter image description here

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 ]  

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 $|G| < 15000000$.


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 ]
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  • $\begingroup$ I'm not sure "rank" is the right word. For instance the groups $\mathrm{PSL}(2,q)$ are usually considered to have bounded rank, when $q$ varies. $\endgroup$ – YCor Dec 24 '19 at 9:15
  • $\begingroup$ @YCor What is called "rank" here is not the usual one in group theory (i.e. smallest cardinality of a generating set for $G$) but the dimension of the Grothendieck ring of $Rep(G)$; this word has this sense in this framework. Of course these two notions are not equivalent. What word should be used in group theory? $\endgroup$ – Sebastien Palcoux Dec 24 '19 at 13:39
  • 1
    $\begingroup$ What you are calling the rank is just the number of conjugacy classes, $k(G)$. $\endgroup$ – Geoff Robinson Dec 24 '19 at 16:36
  • $\begingroup$ @GeoffRobinson: Yes, thanks! I will improved the post. $\endgroup$ – Sebastien Palcoux Dec 24 '19 at 17:10
  • $\begingroup$ @YCor: I have improved the post. $\endgroup$ – Sebastien Palcoux Dec 24 '19 at 20:17
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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.

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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.

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  • $\begingroup$ Why can't we expect to prove without CFSG that the average size of a conjugacy class of a non-abelian finite simple group $G$ is greater than $|G|^{2/3}-4$ (by an elementary argument using the definition of a simple group)? A proof could start as follows: let $g$ be a non-trivial element of a non-abelian finite simple group $G$. If $|Cl(g)| < |G|^{2/3}$ and $ord(g) \le |G|^{1/3}$, then $|\langle g \rangle^G| < |G|$, so that $\langle g \rangle^G$ is a proper non-trivial normal subgroup, contradiction. So "it suffices" to show that the elements of order less than $|G|^{1/3}$ "really" dominate. $\endgroup$ – Sebastien Palcoux Dec 26 '19 at 4:37
  • $\begingroup$ The argument you give here seems to prove that $G$ does not just consist of conjugates of powers of $g$, but doesn't seem to prove that the conjugates of $g$ don't generate $G$. $\endgroup$ – Geoff Robinson Dec 26 '19 at 9:53
  • $\begingroup$ You are right, $Cl(\langle g \rangle):=\{Cl(h) \ | \ h \in \langle g \rangle \} = \cup_n Cl(g^n)$ could be not a group, i.e. a strict subset of $\langle g \rangle^G$. $\endgroup$ – Sebastien Palcoux Dec 27 '19 at 0:54

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