TITLE: In what sense is SL(2,q) "very far from abelian"?
QUESTION [25 upvotes]: I am far from an expert in this area.
I'd be grateful if someone could explain in what sense $\mathop{SL}(2,q)$ is
"very far from abelian," to quote
Emanuele Viola?
Why does Theorem 1 (below) justify this "far from abelian" claim?

          


          

(Snapshot from blog here.)


Earlier related MO questions:

Measures of non-abelian-ness

How nearly abelian are nilpotent groups?
.

REPLY [26 votes]: Another measure of how far a finite group $G$ is from commutative is the "commuting probability", which goes back at least as far as W.H. Gustafson, and certainly predates the work of Gowers. This is just the probability that a pair of elements of $G$ commute, where the uniform distribution is put on $G \times G$. This turns out to be $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$.
It has been noted by several authors, including in recent years P. Lescot, and long ago, (implicitly) by E. Wigner, that this is somewhat related to the smallest degree $d$ of a non-linear complex irreducible character of $G$, though as Derek Holt's example in comments illustrates, the influence can wane if $G$ is far from perfect and has many linear characters. However, when $G$ is perfect, we have $1 +(k(G)-1)d^{2} < |G|$, so that $k(G) < \frac{|G|}{d^{2}}+1,$ and the commuting probability of $G$ is bounded above by something only slightly larger than $\frac{1}{d^{2}}.$
As noted in the question the smallest non-linear complex irreducible character degree of ${\rm SL}(2,q)$ is $\frac{q-1}{2}$ when $q$ is odd, so leading to a upper bound for the commuting probability of something close to $\frac{4}{q^{2}}$ for ${\rm SL}(2,q).$
Another approach to this in the case $G = {\rm PSL}(2,q)$ is to note that ${\rm PSL}(2,q)$ always has at most $q+1$ complex irreducible characters (equality is achieved when $q$ is even). In this case, we have $|G| = \frac{q(q-1)(q+1)}{2}$ if $q$ is odd, and the commuting probability for $G$ is less than $\frac{2}{q(q-1)}$ when $q$ is odd
(and is equal to $\frac{1}{q(q-1)}$ when $q$ is even). The same inequalities hold for ${\rm SL}(2,q)$.
Note that this gives that the commuting probability of $G = {\rm PSL}(2,q)$ is bounded above by something like $c|G|^{\frac{-2}{3}}$ for a small fixed constant $c.$ On the other hand, Bob Guralnick and I proved (using the classification of finite simple groups) that for any finite group $G$ with $F(G) = 1$, the commuting probability of $G$ is at most $|G|^{-\frac{1}{2}}$, so the bound which holds for ${\rm PSL}(2,q)$ is significantly smaller than the general bound we obtained.
Later edit: To be more precise, the arithmetic mean (say $\mu_{d}(G)$) of the complex irreducible character degrees of $G$ is quite strongly related to the commuting probability ${\rm cp}(G)$ of $G$. The Cauchy-Schwartz inequality gives $\sum_{\chi \in {\rm Irr}(G)} \chi(1) \leq \sqrt{k(G)|G|}$ so that 
$\mu_{d}(G){\rm cp}(G) \leq \sqrt{{\rm cp}(G)}$ and hence 
${\rm cp}(G) \leq \frac{1}{\mu_{d}(G)^{2}}.$

REPLY [9 votes]: I am going to piggyback off Bullet51's answer and mention product-free sets. In particular, one might want to see a failure of Theorem 1 (from the question) for a finite abelian group. 
As is explained in the introduction to Gowers' paper on quasirandom groups, if $G$ is a finite abelian group then there is some set $X\subseteq G$ of "constant density", namely $|X|\geq |G|/3$, such that $X$ is "product-free" (or "sum-free" for abelian groups). Specifically, if $x_1,x_2\in X$ then $x_1+x_2\not\in X$. 
So, in the context of Theorem 1, take $A=B=X$ and $C=-X$. Let $g=0$. Then we have $Pr[a+b+c=g]=0$, and so 
$$
|Pr[a+b+c=g]-1/|G||=1/|G|
$$
(which of course is not less than $1/|G|^{1+\Omega(1)}$). 
There are probably more sophisticated ways of handling this, but the upshot is that abelian groups have very large product-free sets, while "quasirandom" groups like $SL(2,q)$ do not. Gowers shows that if $G$ is a finite group with no nontrivial representations of degree less than $k$, then any set of size greater than $|G|/k^{1/3}$ is not product-free. 
For $G=SL(2,q)$ we can take $k=(q-1)/2$. So there is some absolute constant $\epsilon>0$ such that if $G=SL(2,q)$ then $G$ has no product-free set of size greater than $|G|^{1-\epsilon}$ (in contrast to abelian groups, which have product-free sets of "linear" size, namely, $|G|/3$).
By the way, in this paper Green and Ruzsa determine the maximal densities of sum-free subsets of finite abelian groups.