A COURSE IN ARITHMETIC COMBINATORICS MICHAEL BJÖRKLUND CONTENTS 1. Impact functions and Plünnecke’s Theorem 1 2. Small product sets 6 3. Freiman’s Theorem for finite groups (Lecture by Hegarty) 9 4. Roth’s Theorem (Lecture by Roginskaya) 9 5. Schur’s Theorem 9 6. van der Waerden’s Theorem 11 7. Quasi-randomness and Gowers’ Theorem 11 8. A Sum-Product Theorem (Lecture notes by Hegarty) 15 9. Home work assignments 20 References 21 1. IMPACT FUNCTIONS AND PLÜNNECKE’S THEOREM Let G be a finite group and denote by mG the normalized counting (Haar) measure on G, i.e. B mG(B) j j , for all B G. Æ G ½ j j Given non-empty subsets A,B G, we define the product set AB of A and B by ½ AB ©ab : a A, b Bª. Æ 2 2 We say that a set A G is a basis of order k if ½ Ak A A A G. Æ ¢ ¢¢¢ Æ Note that a proper subgroup of G can never be a basis of any order. Given a basis A of order k in G, we wish to understand in this lecture the rough behavior of the associated impact function ½ A which is defined by n o ½ A(t) min mG(AB): mG(B) t for 0 t 1. Æ ¸ · · 1.1. The Erdös-Raikov Inequality. Suppose that A G contains at least two elements. We ½ may assume that one of these elements is the identity and we note that mG(AB) mG(B aB) mG(B) mG(aB \ B) for all a A. ¸ [ Æ Å 2 Define the ejectivity function ³A by n ª ³A(t) inf "A(B): B G such that mG(B) t , Æ ½ ¸ where "A(B) sup mG(aB \ B), Æ a A 2 1 2 MICHAEL BJÖRKLUND so that mG(AB) mG(B) ³A(mG(B)). ¸ Å In particular, we have ½ A(t) t ³A(t), for all 0 t 1. ¸ Å · · The following basic result (see the papers [1] and [5])) is sometimes referred to as the Erdös- Raikov Inequality. Theorem 1.1. Let G be a finite group and suppose A G is a basis of order k in G. Then ½ 1 ³A(t) t (1 t) for all 0 t 1. ¸ k ¢ ¢ ¡ · · To prove this theorem, we shall need two lemmata. Lemma 1.1. For any A G and for every positive integer n, we have ½ ³An (t) n ³A(t) · ¢ for all 0 t 1. · · Proof. First note that if X,Y and Z are subsets of G, then mG(X \ Y ) mG(X \ Z) mG(Z \ Y ). · Å In particular, if we define the function »(x) mG(xB \ B), x G, Æ 2 then »(xy) »(x) »(y) for all x, y G. · Å 2 For any x1,..., xn, we have mG(x1 xnB \ B) »(x1 ... xn) »(x1) ... »(xn) mG(x1B \ B) ... mG(xnB \ B) ¢¢¢ Æ · Å Å Æ Å which readily implies the lemma. Lemma 1.2. For every finite group G, we have ³G(t) t (1 t), ¸ ¢ ¡ for all 0 t 1. · · Proof. One readily verifies that Z 2 mG(B xB) dmG(x) mG(B) G \ Æ for every B G and thus there exists at least one x G such that ½ 2 2 mG(B xB) mG(B) , \ · and thus mG(xB \ B) mG(B) mG(B xB) mG(B) (1 mG(B)), Æ ¡ \ ¸ ¢ ¡ which finishes the proof. To prove Theorem 1.1: Let A G be a basis of order k. Then, by the lemmata above, we have ½ 1 1 ³A(t) ³G(t) t (1 t) ¸ k ¢ ¸ k ¢ ¢ ¡ for all 0 t 1. · · A COURSE IN ARITHMETIC COMBINATORICS 3 1.2. Plünnecke’s Inequality. In the case when G is a finite abelian group, Plünnecke [4] was able to provide a better lower bound on the impact function for a basis of order k. In this subsection, we shall outline the recent argument by Petridis [3] to obtain this celebrated bound. Theorem 1.2. Let G be a finite abelian group and suppose A and B are non-empty subsets of G. Let K 1 be a constant such that ¸ mG(AB) K mG(B). · ¢ Then there exists a non-empty subset B0 B such that ½ k k mG(A B0) K mG(B0), for all k. · ¢ In particular, if A G is a basis of order k, then ½ 1 1 mG(AB) mG(B) ¡ k for all B G, ¸ ½ 1 1 or equivalently, ½ A(t) t ¡ k for all 0 t 1. ¸ · · We shall need the following lemma. Lemma 1.3 (Petridis, [3]). Let G be a finite abelian group and suppose A and B are non-empty subsets of G. Let B0 B be a non-empty subset such that ½ mG(AB0) n mG(AB00) o min : B00 B . mG(B0) Æ mG(B00) ; 6Æ ½ Then, for every F G, we have ½ mG(AB0) mG(F AB0) mG(FB0) . (1.1) · ¢ mG(B0) Proof. Note that inequality (1.1) trivially holds whenever the set F consists of a single point. Our argument now goes as follows. Fix a finite set F G for which (1.1) holds and pick g G\F. ½ 2 We shall prove that (1.1) then holds for the set F0 F {g}. Æ [ Since G is abelian, we have the inclusion 1 1 A(B0 g¡ FB0) AB0 g¡ F AB0, \ ⊆ \ and thus, ³ ´ F0 AB0 F AB0 gAB0 \ F AB0 Æ [ ³ ¡ 1 ¢´ F AB0 g AB0 \ AB0 g¡ F AB0 Æ [ \ ³ 1 ´ F AB0 g AB0 \ A(B0 g¡ FB0) . ⊆ [ \ 1 Since B0 g¡ FB0 B0 B, we have \ ½ ½ 1 ¢ ¡ 1 ¢ mG(AB0) mG(A(B0 g¡ FB0) mG B0 g¡ FB0 , \ ¸ \ ¢ mG(B0) and thus ¡ ¢ 1 mG F0 AB0 mG(F AB0) ¹(AB0) mG(A(B0 g¡ FB0)) · Å ¡ \ 1 mG(AB0) mG(F AB0) mG(AB0) mG(B0 g¡ FB0) · Å ¡ \ ¢ mG(B0) ³ 1 ´ mG(AB0) mG(F AB0) mG(B0) mG(B0 g¡ FB0) . · Å ¡ \ ¢ mG(B0) 4 MICHAEL BJÖRKLUND Since (1.1) is assumed to hold for the set F, we conclude that ¡ ¢ ³ 1 ´ mG(AB0) mG F0 AB0 mG(FB0) mG(B0) mG(B0 g¡ FB0) . · Å ¡ \ ¢ mG(B0) Note that 1 mG(FB0) mG(B0) mG(B0 g¡ FB0) mG(F0B), Å ¡ \ Æ which finishes the proof. Proof of Theorem 1.2. Let G be a finite abelian group and suppose A and B are non-empty sub- sets of G. Fix a non-empty subset B0 B such that ½ mG(AB0) n mG(AB00) o Ko min : B00 B K. Æ mG(B0) Æ mG(B00) ; 6Æ ½ · We wish to prove that k k mG(A B) K mG(B), for all k. · o ¢ Clearly this inequality holds for k 1. Assume that Æ k 1 k 1 mG(A ¡ B) K ¡ mG(B), for some k 2. · o ¢ ¸ k 1 Then, by Lemma 1.3 applied to the set F A ¡ , we have Æ k k 1 k mG(A B0) Ko mG(A ¡ B0) K mG(B0), · ¢ · o ¢ which finishes the proof of Theorem 1.2. In particular, if we apply Theorem 1.2 to the sets A B for some non-empty subset B G, Æ ½ then we can find a non-empty subset B0 B such that ½ k k k k mG(B ) mG(B B0) K mG(B0) K mG(B), · · ¢ · ¢ 2 whenever mG(B ) K mG(B). We have thus proved: · ¢ Corollary 1.1. Let G be a finite abelian group and suppose 2 mG(B ) K mG(B) · ¢ k k for some constant K. Then mG(B ) K mG(B) for all k. · ¢ 1.3. Ruzsa’s Triangle Inequality. Proposition 1.1. Let G be a finite group and suppose that A and B are non-empty subsets of G. Let K 1 be a constant such that AB K B . Then, ¸ j j · ¢ j j k l k l A A¡ K Å B , for all k, l 0. j j · ¢ j j ¸ Lemma 1.4. 1.4. Failure of Plünnecke’s inequality for non-abelian groups. We shall now see that if we drop the assumption that G is an abelian group, then counterexamples to the conclusion of Corollary 1.1 are easy to come by. We say that a subgroup H G is locally malnormal if Ç 1 xHx¡ H {e} for some x G \ H. \ Æ 2 Note that abelian groups do not admit non-trivial locally malnormal subgroups. Proposition 1.2. Let G be a finite group which contains a proper locally malnormal subgroup 1 H G with at least 33 elements. Then, for every x G \ H such that xHx¡ H {e}, we have Ç 2 \ Æ 2 3 mG(B ) 3 mG(B) and mG(B ) 27 mG(Bx), x · ¢ x È ¢ where Bx {x} H. Æ [ A COURSE IN ARITHMETIC COMBINATORICS 5 Proof. We first note that B2 H Hx xH {x2} and B3 HxH, x Æ [ [ [ x ⊃ for every x G, so in particular 2 2 3 mG(B ) 3 mG(Bx) and mG(B ) mG(HxH). x · ¢ x ¸ 1 Since H is a locally malnormal subgroup, there exists x G such that xHx¡ H {e} and thus 2 \ Æ the identity HxH H 2 holds. Indeed, one readily checks that the map q : H H HxH given j j Æ j j £ ! by q(s, t) sxt (s, t) H H, Æ 2 £ 3 is bijective. In particular, if mG(B ) 27 mG(Bx), then x · ¢ 27 ( H 1) H 2, ¢ j j Å ¸ j j which does not have a solution if H 33.