It is mentioned in here (last paragraph of the first page) that Dixon proved the following result: the probability that two elements of a finite nonabelian simple group commute is at most $\frac{1}{12}$. I cannot seem to find this proof anywhere online. Do you know a proof of this fact? Or do you have a reference for it? (I am actually more interested in the original proof)
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2$\begingroup$ In fact, for any $\epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) \geq \epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345 $\endgroup$– Ian AgolSep 19, 2018 at 4:07

$\begingroup$ In fact, that can be proven without the use of any representation theory. Let $c(G)$ denote the number of conjugacy classes of the finite group $G$, so that $cp(G) \geq \epsilon$ means $\frac{c(G)}{G} \geq \epsilon$. Then if $k$ is the size of the smallest nonidentity conjugacy class of $G$, comparing that with the average size of one of them gives us $k \leq \frac{G1}{c(G)1}$. Since $c(G) \geq 5$ for any nonabelian finite simple $G$ (in fact, $c(G) \leq 4$ implies $G \leq 12$), we get $k < \frac{5G}{4c(G)} \leq \frac{5}{4\epsilon}$. Then remember $G$ acts faithfully on $k$ points. $\endgroup$– DavidLHardenJan 19, 2019 at 22:03

$\begingroup$ Clarifying: I meant that $c(G) \leq 4$ and $G$ finite imply $G \leq 12$. $\endgroup$– DavidLHardenJan 19, 2019 at 22:04
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You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

3$\begingroup$ Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $G$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $I$ (on consideration of determinant), contrary to simplicity. $\endgroup$ Sep 19, 2018 at 10:48