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To summarise, if we let be the set of all elements of that lie in the support of , then (4) holds for all . This severely limits the size of to only be of polynomial size, rather than exponential size:

Proposition 7 Let be a subset of the support of (thus, consists of words in of length ) such that the law (4) holds for all . Then .

The proof of this proposition is laid out in the exercise below.

Exercise 5 Let be a free group generated by two generators . Let be the set of all words of length at most in .

• (i) Show that if commute, then lie in the same cyclic group, thus for some and .
• (ii) Show that if , there are at most elements of that commute with .
• (iii) Show that if , there are at most elements of with .
• (iv) Prove Proposition 7.

Now we can conclude the proof of Theorem 3:

Exercise 6 Let be a free group generated by two generators .

• (i) Show that for some absolute constant . (For much more precise information on , see this paper of Kesten.)
• (ii) Conclude the proof of Theorem 3.

— 2. Random generators expand —

We now prove Theorem 4. Let be the free group on two formal generators , and let be the generator of the random walk. For any word and any in a group , let be the element of formed by substituting for respectively in the word ; thus can be viewed as a map for any group . Observe that if is drawn randomly using the distribution , and , then is distributed according to the law , where . Applying Corollary 2, it suffices to show that whenever is a large prime and are chosen uniformly and independently at random from , that with probability , one has

for some absolute constant , where ranges over all Borel subgroups of and is drawn from the law for some even natural number .

Let denote the words in of length at most . We may use the law (4) to obtain good bound on the supremum in (5) assuming a certain non-degeneracy property of the word evaluations :

Exercise 7 Let be a natural number, and suppose that is such that for . Show that

for some absolute constant , where is drawn from the law . (Hint: use (4) and the hypothesis to lift the problem up to , at which point one can use Proposition 7 and Exercise 6.)

In view of this exercise, it suffices to show that with probability , one has for all for some comparable to a small multiple of . As has elements, it thus suffices by the union bound to show that

for some absolute constant , and any of length less than for some sufficiently small absolute constant .

Let us now fix a non-identity word of length less than , and consider as a function from to for an arbitrary field . We can identify with the set . A routine induction then shows that the expression is then a polynomial in the eight variables of degree and coefficients which are integers of size . Let us then make the additional restriction to the case , in which case we can write and . Then is now a rational function of whose numerator is a polynomial of degree and coefficients of size , and the denominator is a monomial of of degree .

We then specialise this rational function to the field . It is conceivable that when one does so, the rational function collapses to the constant polynomial , thus for all with . (For instance, this would be the case if , by Lagrange’s theorem, if it were not for the fact that is far too large here.) But suppose that this rational function does not collapse to the constant rational function. Applying the Schwarz-Zippel lemma (Exercise 23 from Notes 5), we then see that the set of pairs with and is at most ; adding in the and cases, one still obtains a bound of , which is acceptable since and . Thus, the only remaining case to consider is when the rational function is identically on with .

Now we perform another “Lefschetz principle” maneuvre to change the underlying field. Recall that the denominator of rational function is monomial in , and the numerator has coefficients of size . If is less than for a sufficiently small , we conclude in particular (for large enough) that the coefficients all have magnitude less than . As such, the only way that this function can be identically on is if it is identically on for all with , and hence for or also by taking Zariski closures.

On the other hand, we know that for some choices of , e.g. , contains a copy of the free group on two generators (see e.g. Exercise 23 of Notes 2). As such, it is not possible for any non-identity word to be identically trivial on . Thus this case cannot actually occur, completing the proof of (6) and hence of Theorem 4.

Remark 5 We see from the above argument that the existence of subgroups of an algebraic group with good “independence” properties – such as that of generating a free group – can be useful in studying the expansion properties of that algebraic group, even if the field of interest in the latter is distinct from that of the former. For more complicated algebraic groups than , in which laws such as (4) are not always available, it turns out to be useful to place further properties on the subgroup , for instance by requiring that all non-abelian subgroups of that group be Zariski dense (a property which has been called strong density), as this turns out to be useful for preventing random walks from concentrating in proper algebraic subgroups. See this paper of Breuillard, Guralnick, Green and Tao for constructions of strongly dense free subgroups of algebraic groups and further discussion.