# In what sense is SL(2,q) “very far from abelian”?

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:

• I think one reason is that its derived subgroup is itself, while an abelian group has trivial derived subgroup. This however is irrelevant to the theorem. – WhatsUp Sep 28 at 0:31
• I don't have an answer, but one way to figure out why the theorem has to do with the "abelian-ness" of $SL(2,q)$ is to take "more abelian" groups (abelian, nilpotent, solvable) and see how you would have to modify the theorem for them. It might also help to try to note how the "abelian-ness" of these groups changes the calculation. – Jon Aycock Sep 28 at 1:25
• But the author says what was meant by "very far from abelian" in the following sentence. – Buster Sep 28 at 12:03
• The following sentence is: "Note that the conclusion of the theorem in particular implies that abc is supported over the entire group." – Joseph O'Rourke Sep 28 at 12:50
• I'd really like to see an answer to the title question (not so much the more precise version in the question statement) which starts from the premise that any free product of two nontrivial groups is "far from abelian", so that $PSL_2(\mathbb Z) \cong C_2 \ast C_3$ is "far from abelian", and deduces that $SL_2(\mathbb F_q)$ must also be "far from abelian". – Tim Campion Sep 29 at 3:37

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 $${\rm cp}(G) \leq \frac{1}{\mu_{d}(G)^{2}}.$$

One can use the smallest dimension of nontrivial representations over $$\mathbb C$$ (also called the quasirandom degree) to measure how far a group is from being abelian. The representations of abelian groups are all one-dimensional. A group with a non-trivial abelian quotient has a nontrivial character of dimension $$1$$. In other words, the quasirandom degree of $$G$$ is at least $$2$$ if and only if $$G$$ is a perfect group.

$$SL(2,q)$$ has quasirandom degree $$\frac{q-1}2$$, roughly the $$1/3$$-th power of the group size, so it is "very far from abelian" in this sense.

The notion of quasirandom degree is used in combinatorics to produce random-like structures, e.g. high-expansion graphs.

See the paper by Timothy Gowers :

Having proved this theorem, we step back and look at what we have done from a more abstract point of view. The property of $$PSL_2(q)$$ that makes it suitable for results of this kind is that it has no non-trivial irreducible representations of low dimension. This property has been used in a similar way before: it is an important ingredient in the famous construction of Ramanujan graphs by Lubotzky, Phillips and Sarnak, and it has recently been used by Bourgain and Gamburd to show that certain Cayley graphs are expanders.

• But ${\rm SL}(2,q)$ has a representation of degree $2$ (and ${\rm PSL}(2,q)$ one of degree $3$) over the field of order $q$. – Derek Holt Sep 28 at 8:55
• The representations are taken as on $\mathbb{C}$. – LeechLattice Sep 28 at 9:12
• There are groups with structure $C_q \rtimes C_{q-1}$ (for prime powers $q$) in which the smallest degree faithful representation over any field is $q-1$, which is about the square root of the group order. But these groups are metabelian (in fact metacyclic) - not exactly what you would call far from abelian. – Derek Holt Sep 28 at 9:48
• Yes, I understand your definition of quasirandom degree, but I am querying whether this is really a good measure of how far the group is from being abelian. ${\rm PSL}(2,q)$ has high quasirandom degree, but it is one for ${\rm PGL}(2,q)$ when $q$ is odd. The commuting probability of a pair of elements seems better. – Derek Holt Sep 28 at 12:28
• @DerekHolt, it is no doubt true that commuting probability is the best measure of abelian-ness in many contexts. Nonetheless, for the results cited in the question, the minimum degree of a non-trivial complex rep (a.k.a. quasirandomness) is exactly the right measure... The absence of small non-trivial reps is crucial to both of the proofs cited there. – Nick Gill Oct 2 at 10:12

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.