In this paper, we study product-free subsets of the free semigroup over a finite alphabet A. We prove that the maximum density of a product-free subset of the free semigroup over A, with respect to the natural measure that assigns a weight of |A|−n to each word of length n, is precisely 1∕2

Leader, I

Letzter, S

Narayanan, B

Walters, M

English

UCL Discovery

PRODUCT-FREE SETS IN THE FREE SEMIGROUPIMRE LEADER, SHOHAM LETZTER, BHARGAV NARAYANAN, AND MARK WALTERSAbstract. In this paper, we study product-free subsets of the free semigroupover a finite alphabet A. We prove that the maximum density of a product-freesubset of the free semigroup over A, with respect to the natural measure thatassigns a weight of |A|−n to each word of length n, is precisely 1/2.1. IntroductionA subset S of a semigroup is said to be product-free if there do not exist x, y, z ∈ S(not necessarily distinct) such that x • y = z; it is customary to call S sum-freewhen the underlying semigroup is abelian.It is a well known fact (and an easy exercise) that any sum-free subset of theintegers has upper density at most 1/2. Sum-free subsets of the integers, and ofabelian groups in general, have been studied by very many researchers over thelast fifty years. For example, from the work of Green and Ruzsa [4], there is nowa complete picture of how large a sum-free set we can find in any finite abeliangroup. We refer the reader to the surveys of Tao and Vu [7] and Kedlaya [5] formore information on these questions.Product-free subsets of finite non-abelian groups were first investigated by Babaiand Sós [1]. Following foundational work by Gowers [3] demonstrating so-called‘product-mixing’ phenomena in groups with no low-dimensional representations,there has been a great deal of recent work in the non-abelian setting; for instance,in a recent breakthrough, Eberhard [2] determined how large a product-free subsetof the alternating group can be.In light of these developments, it is natural to ask what one can say about product-free sets in infinite non-abelian structures, a setting in which our knowledge is a bitmore limited. Perhaps the first natural place to look among infinite non-abelianDate: 6 December 2018.2010 Mathematics Subject Classification. Primary 20M05; Secondary 05D05.1structures is among those that are free, so here, we shall investigate how largeproduct-free subsets of the free semigroup can be.2. Our resultsLet A be a finite set. We write F = FA for the free semigroup over A; in otherwords, F is the set of all finite words over the alphabet A equipped with theassociative operation of concatenation. While we state and prove our results forfinite alphabets of all possible sizes for the sake of completeness, the reader willlose nothing by supposing that A is a two-element set in what follows; indeed, thiscase captures all the difficulties inherent in the questions we study.Recall that a set S ⊂ F is product-free if, writing • for the operation of concate-nation, there do not exist words x, y, z ∈ S (not necessarily distinct) such thatx • y = z. There is an obvious example of a ‘large’ subset of F that is product-free:when A = {a, b} for instance, the set of words which contain an odd number ofoccurrences of the symbol a (or b, for that matter) is easily seen to be a product-freeset that contains, roughly, half the words from F . Our aim in this paper is to provethat these sets are, in a precise sense, the largest product-free subsets of F . Weremark in passing that there are several other product-free sets that are ‘equallylarge’: for any nonempty subset Γ ⊂ A, the odd-occurrence set OΓ ⊂ F generatedby Γ, namely the set of words in which the total number of occurrences of symbolsfrom Γ is odd, is easily seen to be a product-free set; in the case where A = {a, b},our earlier example corresponds to taking Γ = {a}, and taking Γ = {a, b} gives usthe set of all words of odd length, for example.To formally state our results, we need a way to measure the size of a set S ⊂ F .For an integer n ∈ N, the layer F(n) ⊂ F is the set of words of length n, and theball F≤(n) ⊂ F is the set of words of length at most n. As a first attempt, onemight define the density of a set S ⊂ F via its densities in balls, namely as thequantitylim supn→∞|S ∩ F≤(n)||F≤(n)|.However, a little thought should convince the reader that the counting measure issomewhat ill-suited for our purposes. Indeed, when |A| > 1, almost all the words inF≤(n) are long since |F(n)| ≥ |F≤(n)|/2. Consequently, we may find product-freesets that are intuitively small, and yet have density arbitrarily close to 1 in the2above sense; for example, for any sufficiently large c ∈ N, the set⋃n≥c(F≤(3n + c) \ F≤(3n))is product-free and has density at least 1−1/c in the above sense, provided |A| > 1.A more natural approach is to assign a weight of |A|−n to each word of F(n),thereby ensuring that the layers F(n) have the same total weight for all n ∈ N.To this end, for a subset S ⊂ F and an integer n ∈ N, we define the density ofS in the layer F(n) by dS(n) = |S ∩ F(n)|/|F(n)|. With this definition in place,most standard notions of density may now be carried over: we define the upperasymptotic density of S byd̄(S) = lim supn→∞∑ni=1 dS(i)n,and the upper Banach density of S byd∗(S) = lim supn−m→∞∑ni=m dS(i)n−m+ 1.Of course, the latter is a weaker notion of density than the former; indeed, it isclear that d̄(S) ≤ d∗(S) for any S ⊂ F .It is easy to see that any odd-occurrence set has both an upper asymptoticdensity and an upper Banach density of 1/2. Our aim in this note is to show thatproduct-free sets cannot be any larger; our main result is as follows.Theorem 2.1. Let A be a finite set. If S ⊂ FA is product-free, then d∗(S) ≤ 1/2.Let us mention that product-free sets in cancellative semigroups have beenstudied by  Luczak and Schoen [6]; while their results are sharp for such semigroupsin general, these results do not give us any effective bounds on the size of aproduct-free subset of F .Before we turn to the proof of Theorem 2.1, it is worth pointing out that thereis a simple argument that allows us to bound the upper Banach density of aproduct-free subset of F away from 1. Indeed, suppose that S ⊂ F is product-free.We then havedS(m)dS(n) + dS(m+ n) ≤ 1for any m,n ∈ N since the sets S ∩ F(m+ n) and (S ∩ F(m)) • (S ∩ F(n)) mustbe disjoint. By averaging this inequality, with m fixed and with n ranging oversuitably long intervals, we learn that (dS(m) + 1)d∗(S) ≤ 1 for each m ∈ N. It3is now immediate that d∗(S) ≤ φ, where φ = (√5 − 1)/2 ≈ 0.618 is the uniquepositive solution to the equation x2 + x = 1.We shall have to work somewhat harder to prove Theorem 2.1, which improvesthis bound of φ for the upper Banach density to the optimal bound of 1/2. Theproof of Theorem 2.1 is given in Section 3. We conclude this note with a discussionof some open problems in Section 4.3. Proof of the main resultWe begin by fixing our finite alphabet A. In the sequel, F will always mean FA,the free semigroup over this fixed alphabet A.It will be helpful to establish some notation. For a pair of words x,w ∈ F , wesay that x is a prefix of w if w = x • y for some y ∈ F , and that x is a suffix of wif w = y • x for some y ∈ F . For a pair of sets S1, S2 ⊂ F , we write S1 • S2 for their(Minkowski) product; in other words,S1 • S2 = {w1 • w2 : w1 ∈ S1, w2 ∈ S2}.For a set S ⊂ F and an integer n ∈ N, we set S(n) = S ∩ F(n). One of the keyideas in the proof of Theorem 2.1 is the following definition. For any sequence ofpositive integers `1 < `2 < · · · < `k < n, we defineS(n; `1, `2, . . . , `k) = {w ∈ S(n) : w has no prefix in S(`1) ∪ S(`2) ∪ · · · ∪ S(`k)};in other words,S(n; `1, `2, . . . , `k) = S(n) \(k⋃i=1S(`i) • F(n− `i)).Let us note, for any S ⊂ F , that the sets S(n;m) and S(m) • F(n−m) are disjointfor any pair of positive integers m < n. Recall that dS(n) = |S(n)||F(n)|−1; weanalogously definedS(n; `1, `2, . . . , `k) =|S(n; `1, `2, . . . , `k)||F(n)|.When the set S in question is clear, we write d(n) and d(n; `1, `2, . . . , `k) fordS(n) and dS(n; `1, `2, . . . , `k), respectively. Recall that for any product-free setS ⊂ F and any m,n ∈ N, we haved(m)d(n) + d(m+ n) ≤ 1.4We start by proving a generalisation of this fact.Proposition 3.1. If S ⊂ F is product-free, then for any sequence of positiveintegers `1 < `2 < · · · < `k < n, we haved(`1)d(n− `1) + d(`2; `1)d(n− `2) + · · ·+ d(`k; `1, `2, . . . , `k−1)d(n− `k) + d(n)≤ d(`1) + d(`2; `1) + · · ·+ d(`k; `1, `2, . . . , `k−1) + d(n; `1, `2, . . . , `k) ≤ 1.Proof. First, consider the productsS(`1) • S(n− `1), S(`2; `1) • S(n− `2), . . . , S(`k; `1, `2, . . . , `k−1) • S(n− `k).These subsets of F(n) are by definition pairwise disjoint. Let L′ be the unionof these k sets. Since S is product-free, L′ and S(n) are disjoint as well. LetL = L′ ∪ S(n); clearly, the density of L in F(n) isd(`1)d(n− `1) + d(`2; `1)d(n− `2) + · · ·+ d(`k; `1, `2, . . . , `k−1)d(n− `k) + d(n).Next, consider the Minkowski productsS(`1) • F(n− `1), S(`2; `1) • F(n− `2), . . . , S(`k; `1, `2, . . . , `k−1) • F(n− `k).These subsets of F(n) are again pairwise disjoint by definition; let R′ denotetheir union. Note that R′ and S(n; `1, `2, . . . , `k) are disjoint. Let R = R′ ∪S(n; `1, `2, . . . , `k); it is easy to see that the density of R in F(n) isd(`1) + d(`2; `1) + · · ·+ d(`k; `1, `2, . . . , `k−1) + d(n; `1, `2, . . . , `k)and that this quantity is therefore at most 1.To finish the proof, it suffices to show that L ⊂ R. It is easy to see that L′ ⊂ R′.It remains to show that S(n) is a subset of R′∪S(n; `1, `2, . . . , `k). To see this, notethat any word from S(n) which has a prefix in S(`1) ∪ S(`2) ∪ · · · ∪ S(`k) belongsto R′. In other words, S(n) \ S(n; `1, `2, . . . , `k) ⊂ R′; the result follows. With the above observation in hand, we are now ready to prove Theorem 2.1.Proof of Theorem 2.1. We prove by contradiction that the upper Banach densityof a product-free set is at most 1/2.Suppose that S ⊂ F is product-free and that d∗(S) > 1/2 + ε for some ε > 0.We then claim that we may find an increasing sequence of positive integers (`k)k∈Nsuch thatd(`1) + d(`2; `1) + · · ·+ d(`k; `1, `2, . . . , `k−1) ≥12+14+ · · ·+ 12k= 1− 12k5for each k ∈ N.We construct this sequence inductively. Since d∗(S) > 1/2, it is clear that wemay find `1 ∈ N such that d(`1) ≥ 1/2. Having found `1 < `2 < · · · < `k asrequired, we choose `k+1 as follows. Since d∗(S) > 1/2 + ε, there exist arbitrarilylong intervals I ⊂ N that satisfy∑n∈I d(n)|I|>12+ ε.Choose such an interval I whose length is sufficiently larger than `k; we may assume(by passing to a sub-interval if necessary) that min I > `k, and that |I| > (`k + 1)/ε.We claim that it is possible to choose `k+1 from I; in other words, we claim thatthere exists an n ∈ I such thatd(`1) + d(`2; `1) + · · · + d(`k; `1, `2, . . . , `k−1) + d(n; `1, `2, . . . , `k) ≥ 1 −12k+1.We prove this claim by contradiction. Suppose that there is no such n ∈ I. Then,by Proposition 3.1, we haved(`1)d(n− `1) + d(`2; `1)d(n− `2) + · · ·+ d(`k; `1, `2, . . . , `k−1)d(n− `k) + d(n)≤ d(`1) + d(`2; `1) + · · ·+ d(`k; `1, `2, . . . , `k−1) + d(n; `1, `2, . . . , `k) < 1−12k+1for each n ∈ I. By summing the above bound over all n ∈ I, we get∑n∈I′d(n)(1 + d(`1) + d(`2; `1) + · · ·+ d(`k; `1, `2, . . . , `k−1)) < |I|(1− 12k+1),where I ′ ⊂ I is the set of n ∈ I with n+ `k < max I; indeed, to see this, note thatfor every 1 ≤ i ≤ k and every n′ ∈ I ′, there is an n ∈ I such that d(n′) = d(n− `i).This implies, by the inductive hypothesis, that∑n∈I′d(n)(2− 12k)< |I|(1− 12k+1),or equivalently,∑n∈I′ d(n) < |I|/2. Therefore, we have∑n∈Id(n) ≤∑n∈I′d(n) + `k + 1 <|I|2+ `k + 1,which contradicts the fact that∑n∈I d(n) > |I|/2 + ε|I|, provided |I| > (`k + 1)/ε.6We now finish the proof of the proposition by showing that the existence of thissequence (`k)k∈N contradicts our initial assumption that d∗(S) > 1/2 + ε. Fix ak ∈ N large enough to ensure that2k2k+1 − 1<1 + ε2and consider any interval I ⊂ N with |I| > 4(`k + 1)/ε. We know from Proposi-tion 3.1 thatd(`1)d(n− `1) + d(`2; `1)d(n− `2) + · · ·+ d(`k; `1, `2, . . . , `k−1)d(n− `k) + d(n) ≤ 1for each n ∈ N with n > `k; as earlier, we sum the above bound over such n ∈ I toget ∑n∈I′d(n)(2− 12k)≤ |I|,where I ′ is the set of n ∈ I with n > `k and n+ `k < max I. Therefore,∑n∈I d(n)|I|≤ 2k(2k+1 − 1)+2(`k + 1)|I|<12+ ε,which is a contradiction; this proves the claimed upper bound in Theorem 2.1. 4. ConclusionA common line of enquiry in the study of product-free sets is to ask for ‘asym-metric’ versions of results bounding the upper density of product-free sets. Inthis spirit, it is natural to ask whether an analogue of Theorem 2.1 continues tohold when one wishes to solve the equation x • y = z with x, y and z in specifiedsubsets of F . More precisely, if X, Y, Z ⊂ F are such that there are no solutionsto x • y = z with x ∈ X, y ∈ Y and z ∈ Z, one might ask if one of X, Y or Z hasan upper asymptotic density of at most 1/2. However, it is possible to constructfor any ε > 0, three sets X, Y, Z ⊂ F , each of upper asymptotic density at leastφ − ε, where φ = (√5 − 1)/2, such that there are no solutions to x • y = z withx ∈ X, y ∈ Y and z ∈ Z; of course, the simple argument given in the introductionshows that one cannot do any better than this. Pick a suitably large n ∈ N andchoose any set W ⊂ F(n) such that ||W |/|F(n)| − φ| < ε/3. Now take X to bethe set of all words with a prefix in W , Y to be the set of all words with a suffix inW , and Z to be the set F \ (X • Y ). Clearly, there are no solutions to x • y = zwith x ∈ X, y ∈ Y and z ∈ Z; it is also not hard to check that each of X, Y andZ has an upper asymptotic density at least φ− ε.7Next, it would be interesting to understand what product-free sets of maximaldensity look like. As we saw earlier, several non-isomorphic extremal constructionsare furnished by the family of odd-occurrence sets. We suspect that these mightbe the only constructions of maximal density, and make the following conjecture.Conjecture 4.1. Let A be a finite set. If S ⊂ FA is product-free and d∗(S) = 1/2,then S ⊂ OΓ for some nonempty subset Γ ⊂ A.Finally, another natural direction is to study product-free subsets of the freegroup FA over a finite alphabet A. Similarly to the situation in this paper, the mostnatural measure to consider in the case of the free group FA would be the one thatassigns a weight of |A|(|A| − 1)−(n−1) to each irreducible word of length n. Thedifferent notions of density defined here for the free semigroup then have analogousdefinitions in the free group, and we believe that an analogue of Theorem 2.1 shouldhold in the free group as well; concretely, we conjecture the following.Conjecture 4.2. For any finite alphabet A, no product-free subset of the free groupFA has upper Banach density exceeding 1/2.Note that, in the proof of Theorem 2.1, we rely crucially on the fact that thereis exactly one way to write a word of length m+ n as the concatenation of a wordof length m with a word of length n; of course, we lose this property when workingwith free groups, so we believe that some new ideas will be required to understandproduct-free sets in free groups.AcknowledgementsThe second author would like to acknowledge the support of Dr. Max Rössler,the Walter Haefner Foundation, and the ETH Zurich Foundation. The third authorwishes to acknowledge support from NSF grant DMS-1800521.References1. L. Babai and V. T. Sós, Sidon sets in groups and induced subgraphs of Cayleygraphs, European J. Combin. 6 (1985), 101–114.2. S. Eberhard, Product mixing in the alternating group, Discrete Analysis (2016:2),19 pp.3. W. T. Gowers, Quasirandom groups, Combin. Probab. Comput. 17 (2008),363–387.84. B. Green and I. Z. Ruzsa, Sum-free sets in abelian groups, Israel J. Math. 147(2005), 157–188.5. K. S. Kedlaya, Product-free subsets of groups, then and now, Communicatingmathematics, Contemp. Math., vol. 479, Amer. Math. Soc., Providence, RI,2009, pp. 169–177.6. T.  Luczak and T. Schoen, Sum-free subsets of right cancellative semigroups,European J. Combin. 22 (2001), 999–1002.7. T. Tao and V. Vu, Sumfree sets in groups: a survey, J. Comb. 8 (2017), 541–552.Department of Pure Mathematics and Mathematical Statistics, University ofCambridge, Wilberforce Road, Cambridge CB3 0WB, UKEmail address: i.leader@dpmms.cam.ac.ukETH Institute for Theoretical Studies, 8092 Zurich, SwitzerlandEmail address: shoham.letzter@math.ethz.chDepartment of Mathematics, Rutgers University, Piscataway, NJ 08854, USAEmail address: narayanan@math.rutgers.eduSchool of Mathematical Sciences, Queen Mary, University of London, LondonE1 4NS, UKEmail address: m.walters@qmul.ac.uk9

Product-free sets in the free semigroup

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Product-free sets in the free semigroup

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