Double log ratio of number of conjugacy classes to order of group in infinite simple group families (or their corresponding Schur covers, automorphism groups)

Question

For a group G, denote by c(G) the number of conjugacy classes in G.

If $S_n$ denotes the symmetric group on n letters, then:

$$\lim_{n \to \infty} \frac{\log \log c(S_n)}{\log \log |S_n|} = \frac{1}{2}$$

[NOTE: Assume all logs are to the same base. It doesn't matter what the base is, as long as we use the same one in the numerator and denominator.]

[Proof that I know of: $c(S_n)$ equals the number of unordered integer partitions of n, and there exist asymptotic formulas for that which show it to be exponential in $\sqrt{n}$. $|S_n|$ is exponential in $n \log n$. Taking double logs, we get roughly $(1/2)\log n$ and $\log n + \log(\log n)$ respectively, and the quotient goes to $1/2$.]

Obviously, the result also holds for alternating groups instead of symmetric groups.

Similarly, if $GL(n,q)$ denotes the general linear group of degree n over a field of q elements, then for fixed q:

$$\lim_{n \to \infty} \frac{\log \log c(GL(n,q))}{\log \log |GL(n,q)|} = \frac{1}{2}$$

[Proof that I know of: the number of conjugacy classes can be expressed as a polynomial in q of degree n, and the group order is a polynomial in $q$ of degree $n^2$, so we get the result.]

The result holds if we replace the general linear group by the special linear group, projective general linear group, or projective special linear group (which is the Chevalley group of type A).

My question:

1. Does the result also hold for the groups in the other three infinite Chevalley families B, C, D, and in the other infinite families of simple groups?
2. If the answer to 1 is yes, is there some deeper reason why the result holds both in the symmetric/alternating group case and in the linear groups case? I imagine that the linear groups cases can be tied together using some general Lie group or algebraic group principles. If so, what are they? How do the alternating groups fit into the picture? May be something to do with Iwahori-Hecke algebras or the field of one element? All explanations are welcome.

Answer 1

(1) is true, and follows from a paper of Liebeck and Shalev ("Character degrees and random walks in finite groups of Lie type" on Liebeck's webpage). They count representations according to their degrees rather than conjugacy classes, but it follows from their results that if $\Phi$ is a (reduced, irreducible) root system and G is a corresponding algebraic group (possibly twisted), then the number of conjugacy classes of $G(F_q)$ is roughly $q^{rk \Phi}$ (the lower bound is easy, just take conjugacy classes of elements in the torus). The limit of $$ \frac{\log \log q ^{rk \Phi}}{\log \log |G(\mathbb{F}_q)|}$$ as the rank of $\Phi$ tends to infinity is 1/2 also for types B,C, and D.