Bounds on the number of elements of a given order in a finite simple group Let $G$ be a finite simple group (nonabelian, for simplicity). Let $n(G,k)$ denote the number of elements of $G$ of order $k$. I'm interested in results which bound quantities related to $n(G,k)$.
For example, are there bounds for $\frac{n(G,k)}{|G|}$? For $\frac{n(G,k)}{k}$? For some other reasonable normalization of $n(G,k)$?
 A: For $w$ a word in a free group $F_d$, we can consider the word map $\bar w : G^d \to G$. Let $P_w(G)$ denote the proportion of $d$-tuples $x \in G^d$ such that $\bar w (x) = 1$. Your notation is related by
$$P_{x^k}(G) = \sum_{\ell \mid k} n(G, \ell) / |G|.$$
For any fixed nontrivial word $w$ we have $P_w(G) \to 0$ as $|G| \to \infty$ for nonabelian finite simple groups $G$. [1]
[1] Dixon, John D.; Pyber, László; Seress, Ákos; Shalev, Aner, Residual properties of free groups and probabilistic methods, J. Reine Angew. Math. 556, 159-172 (2003). ZBL1027.20013.
A: Here is a different, reasonably elementary, proof that  for any $\varepsilon > 0,$ there are only finitely many non-Abelian simple groups $G$ with $n(G,2) > \varepsilon |G|.$
The number of involutions of such a $G$ is $\sum_{\chi \neq 1} \nu(\chi) \chi(1)$ where $\nu$ is the Frobenius-Schur
indicator (which always takes value $0 , 1 $ or $-1$).
This is certainly less than $\frac{|G|}{d},$ where $d$ is the minimum degree of a non-trivial complex irreducible character of $G$.
By Jordan's theorem on complex linear groups, for any positive integer  $n$, there are only finitely many non-Abelian simple groups which have a non-trivial irreducible character of degree $n$, so the claimed result follows.
