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In the previous set of notes, we saw that one could derive expansion of Cayley graphs from three ingredients: non-concentration, product theorems, and quasirandomness. Quasirandomness was discussed in Notes 3. In the current set of notes, we discuss product theorems. Roughly speaking, these theorems assert that in certain circumstances, a finite subset ${A}$ of a group ${G}$ either exhibits expansion (in the sense that ${A^3}$, say, is significantly larger than ${A}$), or is somehow “close to” or “trapped” by a genuine group.

Theorem 1 (Product theorem in ${SL_d(k)}$) Let ${d \geq 2}$, let ${k}$ be a finite field, and let ${A}$ be a finite subset of ${G := SL_d(k)}$. Let ${\epsilon >0}$ be sufficiently small depending on ${d}$. Then at least one of the following statements holds:

• (Expansion) One has ${|A^3| \geq |A|^{1+\epsilon}}$.
• (Close to ${G}$) One has ${|A| \geq |G|^{1-O_d(\epsilon)}}$.
• (Trapping) ${A}$ is contained in a proper subgroup of ${G}$.

We will prove this theorem (which was proven first in the ${d=2,3}$ cases for fields ${F}$ of prime order by Helfgott, and then for ${d=2}$ and general ${F}$ by Dinai, and finally to general ${d}$ and ${F}$ independently by Pyber-Szabo and by Breuillard-Green-Tao) later in this notes. A more qualitative version of this proposition was also previously obtained by Hrushovski. There are also generalisations of the product theorem of importance to number theory, in which the field ${k}$ is replaced by a cyclic ring ${{\bf Z}/q{\bf Z}}$ (with ${q}$ not necessarily prime); this was achieved first for ${d=2}$ and ${q}$ square-free by Bourgain, Gamburd, and Sarnak, by Varju for general ${d}$ and ${q}$ square-free, and finally by this paper of Bourgain and Varju for arbitrary ${d}$ and ${q}$.

Exercise 1 (Girth bound) Assuming Theorem 1, show that whenever ${S}$ is a symmetric set of generators of ${SL_d(k)}$ for some finite field ${k}$ and some ${d\geq 2}$, then any element of ${SL_d(k)}$ can be expressed as the product of ${O_d( \log^{O_d(1)} |k| )}$ elements from ${S}$. (Equivalently, if we add the identity element to ${S}$, then ${S^m = SL_d(k)}$ for some ${m = O_d( \log^{O_d(1)} |k| )}$.) This is a special case of a conjecture of Babai and Seress, who conjectured that the bound should hold uniformly for all finite simple groups (in particular, the implied constants here should not actually depend on ${d}$. The methods used to handle the ${SL_d}$ case can handle other finite groups of Lie type of bounded rank, but at present we do not have bounds that are independent of the rank. On the other hand, a recent paper of Helfgott and Seress has almost resolved the conjecture for the permutation groups ${A_n}$.

A key tool to establish product theorems is an argument which is sometimes referred to as the pivot argument. To illustrate this argument, let us first discuss a much simpler (and older) theorem, essentially due to Freiman, which has a much weaker conclusion but is valid in any group ${G}$:

Theorem 2 (Baby product theorem) Let ${G}$ be a group, and let ${A}$ be a finite non-empty subset of ${G}$. Then one of the following statements hold:

• (Expansion) One has ${|A^{-1} A| \geq \frac{3}{2} |A|}$.
• (Close to a subgroup) ${A}$ is contained in a left-coset of a group ${H}$ with ${|H| < \frac{3}{2} |A|}$.

To prove this theorem, we suppose that the first conclusion does not hold, thus ${|A^{-1} A| <\frac{3}{2} |A|}$. Our task is then to place ${A}$ inside the left-coset of a fairly small group ${H}$.

To do this, we take a group element ${g \in G}$, and consider the intersection ${A\cap gA}$. A priori, the size of this set could range from anywhere from ${0}$ to ${|A|}$. However, we can use the hypothesis ${|A^{-1} A| < \frac{3}{2} |A|}$ to obtain an important dichotomy, reminiscent of the classical fact that two cosets ${gH, hH}$ of a subgroup ${H}$ of ${G}$ are either identical or disjoint:

Proposition 3 (Dichotomy) If ${g \in G}$, then exactly one of the following occurs:

• (Non-involved case) ${A \cap gA}$ is empty.
• (Involved case) ${|A \cap gA| > \frac{|A|}{2}}$.

Proof: Suppose we are not in the pivot case, so that ${A \cap gA}$ is non-empty. Let ${a}$ be an element of ${A \cap gA}$, then ${a}$ and ${g^{-1} a}$ both lie in ${A}$. The sets ${A^{-1} a}$ and ${A^{-1} g^{-1} a}$ then both lie in ${A^{-1} A}$. As these sets have cardinality ${|A|}$ and lie in ${A^{-1}A}$, which has cardinality less than ${\frac{3}{2}|A|}$, we conclude from the inclusion-exclusion formula that

$\displaystyle |A^{-1} a \cap A^{-1} g^{-1} a| > \frac{|A|}{2}.$

But the left-hand side is equal to ${|A \cap gA|}$, and the claim follows. $\Box$

The above proposition provides a clear separation between two types of elements ${g \in G}$: the “non-involved” elements, which have nothing to do with ${A}$ (in the sense that ${A \cap gA = \emptyset}$, and the “involved” elements, which have a lot to do with ${A}$ (in the sense that ${|A \cap gA| > |A|/2}$. The key point is that there is a significant “gap” between the non-involved and involved elements; there are no elements that are only “slightly involved”, in that ${A}$ and ${gA}$ intersect a little but not a lot. It is this gap that will allow us to upgrade approximate structure to exact structure. Namely,

Proposition 4 The set ${H}$ of involved elements is a finite group, and is equal to ${A A^{-1}}$.

Proof: It is clear that the identity element ${1}$ is involved, and that if ${g}$ is involved then so is ${g^{-1}}$ (since ${A \cap g^{-1} A = g^{-1}(A \cap gA)}$. Now suppose that ${g, h}$ are both involved. Then ${A \cap gA}$ and ${A\cap hA}$ have cardinality greater than ${|A|/2}$ and are both subsets of ${A}$, and so have non-empty intersection. In particular, ${gA \cap hA}$ is non-empty, and so ${A \cap g^{-1} hA}$ is non-empty. By Proposition 3, this makes ${g^{-1} h}$ involved. It is then clear that ${H}$ is a group.

If ${g \in A A^{-1}}$, then ${A \cap gA}$ is non-empty, and so from Proposition 3 ${g}$ is involved. Conversely, if ${g}$ is involved, then ${g \in A A^{-1}}$. Thus we have ${H = A A^{-1}}$ as claimed. In particular, ${H}$ is finite. $\Box$

Now we can quickly wrap up the proof of Theorem 2. By construction, ${A \cap gA| > |A|/2}$ for all ${g \in H}$,which by double counting shows that ${|H| < 2|A|}$. As ${H = A A^{-1}}$, we see that ${A}$ is contained in a right coset ${Hg}$ of ${H}$; setting ${H' := g^{-1} H g}$, we conclude that ${A}$ is contained in a left coset ${gH'}$ of ${H'}$. ${H'}$ is a conjugate of ${H}$, and so ${|H'| < 2|A|}$. If ${h \in H'}$, then ${A}$ and ${Ah}$ both lie in ${H'}$ and have cardinality ${|A|}$, so must overlap; and so ${h \in A A^{-1}}$. Thus ${A A^{-1} = H'}$, and so ${|H'| < \frac{3}{2} |A|}$, and Theorem 2 follows.

Exercise 2 Show that the constant ${3/2}$ in Theorem 2 cannot be replaced by any larger constant.

Exercise 3 Let ${A \subset G}$ be a finite non-empty set such that ${|A^2| < 2|A|}$. Show that ${AA^{-1}=A^{-1} A}$. (Hint: If ${ab^{-1} \in A A^{-1}}$, show that ${ab^{-1} = c^{-1} d}$ for some ${c,d \in A}$.)

Exercise 4 Let ${A \subset G}$ be a finite non-empty set such that ${|A^2| < \frac{3}{2} |A|}$. Show that there is a finite group ${H}$ with ${|H| < \frac{3}{2} |A|}$ and a group element ${g \in G}$ such that ${A \subset Hg \cap gH}$ and ${H = A A^{-1}}$.

Below the fold, we give further examples of the pivot argument in other group-like situations, including Theorem 2 and also the “sum-product theorem” of Bourgain-Katz-Tao and Bourgain-Glibichuk-Konyagin.