The Erdos-Turan problem in infinite groups

arXiv:0901.1649v1 [math.NT] 12 Jan 2009 THE ERDŐS-TURÁN PROBLEM IN INFINITE GROUPS SERGEI V. KONYAGIN AND VSEVOLOD F. LEV Dedicated to Mel Nathanson on the occasion of his 60th birthday Abstract. Let G be an infinite abelian group with |2G| = |G|. We show that if G is not the direct sum of a group of exponent 3 and the group of order 2, then G possesses a perfect additive basis; that is, there is a subset S ⊆ G such that every element of G is uniquely representable as a sum of two elements of S. Moreover, if G is the direct sum of a group of exponent 3 and the group of order 2, then it does not have a perfect additive basis; however, in this case there is a subset S ⊆ G such that every element of G has at most two representations (distinct under permuting the summands) as a sum of two elements of S. This solves completely the Erdős-Turán problem for infinite groups. It is also shown that if G is an abelian group of exponent 2, then there is a subset S ⊆ G such that every element of G has a representation as a sum of two elements of S, and the number of representations of non-zero elements is bounded by an absolute constant. 1. The background A subset of an abelian semigroup is called an additive basis of order 2, or basis for short, if every element of the semigroup is representable as a sum of two elements of the subset. We say that a basis is perfect if every element is represented uniquely, up to the order of summands. The representation function of a basis associates with each element the number of its (ordered) representations as a sum of two elements from the basis. If the semigroup can be embedded into an involution-free group, then for a basis to be perfect it is necessary and sufficient that its representation function is bounded by 2. A famous open conjecture of Erdős and Turán [ET41] is that every basis of the semigroup N0 of non-negative integers has unbounded representation function; that is, if S ⊆ N0 is a set such that each non-negative integer is representable as a sum of two elements of S, then there are integers with arbitrarily many representations. Most investigations related to the Erdős-Turán conjecture study representation functions of bases of N0 (see [NS07] for a survey) or consider the analogous problem The first author was supported by grants 08-01-00208 from the Russian Foundation for Basic Research and NSh-3233.2008.1 from the Program Supporting Leading Scientific Schools. 1 2 SERGEI V. KONYAGIN AND VSEVOLOD F. LEV for infinite abelian semigroups other than N0 , and also for infinite families of abelian semigroups (see [HH04]). In the present paper we are concerned with the latter line of research. There are several noticeable cases where bases with bounded representation functions are known to exist. As an example, Nathanson [N03] proved that the group of possesses a perfect basis. Ruzsa [R90] showed that if p is a prime with   integers 2 = −1, then the group Fp × Fp possesses a basis such that every group element p has at most 18 representations as a sum of two elements of this basis. (Here and below for a prime p we denote by Fp the finite field with p elements. To simplify the notation, we occasionally identify a field with its additive group.) As a corollary, Ruzsa derived a result [R90, Theorem 1] which easily implies that every finite cyclic group has a basis whose representation function is bounded by an absolute constant, independent of the order of the group. The approach of [R90] was further developed by Haddad and Helou [HH04] to show that for any finite field F of odd characteristic, the group F × F has a basis whose representation function does not exceed 18. In the case where F is a finite field of characteristic 2, a basis in F × F with a bounded representation function was constructed in [GDT91, Lemma 1], though the property we are interested in has never been identified explicitly to our knowledge. 2. The results For a subset C of an abelian group and an integer n ≥ 1 we write nC := {nc : c ∈ C}. Our main result is Theorem 1. Let G be an infinite abelian group with |2G| = |G|. (i) If G is not the direct sum of a group of exponent 3 and the group of order 2, then G has a perfect basis. (ii) If G is the direct sum of a group of exponent 3 and the group of order 2, then G does not have a perfect basis, but has a basis such that every element of G has at most two representations (distinct under permuting the summands) as a sum of two elements of the basis. Clearly, if G is an infinite abelian group with |2G| < |G|, then for any basis S of G (and indeed, for any subset S ⊆ G with |S| = |G|) there is an element of G having as many as |G| representations of the form 2s with s ∈ S. In particular, this applies to infinite abelian groups of exponent 2. Similarly, for no infinite family of groups of exponent 2 can one find bases with uniformly bounded representation functions, even THE ERDŐS-TURÁN PROBLEM 3 if the groups of the family are finite. We show that, nevertheless, efficient bases in such groups do exist if we exclude the zero element from consideration. Theorem 2. Each abelian group of exponent 2 possesses a basis such that every nonzero element of the group has at most 36 representations as a sum of two elements of this basis. Combined with the result of Haddad and Helou and the corollary of Ruzsa’s result, mentioned in Section 1, Theorems 1 and 2 readily yield Corollary 1. Let G be an abelian group. If G is either infinite with |2G| = |G|, or has prime exponent, then it possesses a basis with the representation function bounded by an absolute constant (independent of the group), except for the value of the function on the zero element in the case where G is of exponent 2. We notice that excluding the zero element for groups of exponent 2 is equivalent to disregarding representations with equal summands. To our present knowledge, a universal constant K may exist with the property that each abelian group possesses a basis such that every element of the group has at most K representations as a sum of two distinct elements of this basis. 3. The proofs In this section we use the word “basis” both in the above-defined and linearalgebraic meaning, adding the attribute linear in the latter case to avoid confusion. Our argument depends on the axiom of choice, which we assume for the rest of the paper. To handle infinite groups of exponents 2 and 3, we need the following lemma. Lemma 1. If G is an infinite abelian group of prime exponent p, then there exists an algebraically closed field F of characteristic p such that G ∼ = F × F. The proof uses several facts, well known from algebra and set theory; namely, (i) every vector space has a linear basis; (ii) an infinite set can be partitioned into two disjoint subsets of equal cardinality; (iii) an infinite vector space over a finite field has the same cardinality as any of its linear bases; (iv) the field of rational functions over a finite field in the variables, indexed by the elements of an infinite set, has the same cardinality as this set; (v) the algebraic closure of an infinite field has the same cardinality as the field itself. 4 SERGEI V. KONYAGIN AND VSEVOLOD F. LEV We notice that (i) follows easily from Zorn’s lemma, while (ii)–(v) are not difficult to derive from the basic set theory result saying that for any infinite cardinal m, a union of at most m sets, each of cardinality at most m, has cardinality at most m. Proof of Lemma 1. Considering G as a vector space over the field Fp , find a linear basis B of G and fix a partition B = B1 ∪ B2 , where B1 and B2 are disjoint subsets of equal cardinality. For i ∈ {1, 2} denote by Gi the group of functions from Bi to Fp with a finite support; thus, G1 ∼ = G2 , and since G is isomorphic to the group of functions from B to Fp with a finite support, we have G ∼ = G1 × G2 . Let F be the algebraic closure of the field of rational functions over Fp in the variables, indexed by the elements of G1 . By (iv) and (v), the cardinality of F is equal to the cardinality of G1 . From (iii) we conclude now that every linear basis of F has the same cardinality as B1 which, we recall, is a linear basis of G1 . Any bijection from B1 to a linear basis of F determines a group isomorphism between G1 and the additive group of F. As a  result, we have G1 ∼ = F, implying the assertion. = F, and hence also G2 ∼ For an abelian group G, an integer n ≥ 1, and subsets A, B, C ⊆ G we write Gn := {g ∈ G : ng = 0}, A ± B := {a ± b : a ∈ A, b ∈ B}, and A + B − C := {a + b − c : a ∈ A, b ∈ B, c ∈ C}. From G/Gn ∼ = nG we conclude that if |G| is infinite, then max{|Gn |, |nG|} = |G|. Yet another result used in the proof of Theorem 1 is Lemma 2. Let G be an abelian group such that 2G is infinite. If A, B ⊆ G satisfy max{|A|, |B|} < min{|2G|, |3G|}, then there exists an element s ∈ G with 2s ∈ / A and 3s ∈ / B. Proof. Suppose for a contradiction that for every s ∈ G we have either 2s ∈ A, or 3s ∈ B. Without loss of generality we assume B ⊆ 3G, and we find then a subset U ⊆ G with |U| = |B| and B = {3u : u ∈ U}. Fix w ∈ G with 3w ∈ / B. For any g ∈ G3 we have 3(w + g) = 3w ∈ / B, whence 2(w+g) ∈ A and therefore 2g ∈ −2w+A. Now if s ∈ G satisfies 3s ∈ B, then s = u+g with some u ∈ U and g ∈ G3 , implying 2s = 2u + 2g ∈ 2u − 2w + A ⊆ −2w + 2U + A. It follows that for every s ∈ G we have either 2s ∈ −2w + 2U + A, or 2s ∈ A; this, however, is impossible as | − 2w + 2U + A| ≤ |U||A| < |2G| and |A| < |2G| by the assumptions.  THE ERDŐS-TURÁN PROBLEM 5 Eventually, we are ready to prove Theorem 1. Proof of Theorem 1. We split the proof into three parts. 1. First, suppose that G is of exponent 3. By Lemma 1, we can assume G = F × F, where F is an algebraically closed field of characteristic 3. Set S := {(x, x2 ) : x ∈ F}. For each pair (u, v) ∈ G, the number of representations of (u, v) as a sum of two elements of S is the number of solutions of the equation x2 + (u − x)2 = v; x ∈ F, which is either 1, or 2. The assertion follows. We remark that this argument above actually goes through for any odd prime exponent; however, only the case of exponent 3 is not covered by the proof below. 2. Now suppose that G = F ⊕ {0, h}, where F is of exponent 3 and h has order 2. As shown above, F has a perfect basis S, and it is immediate that S ∪ (h + S) is a basis of G such that every element of G has at most two representations (distinct under permuting the summands) as a sum of two elements of this basis. Assuming, on the other hand, that G possesses a perfect basis, we write this basis as T = T0 ∪(h+T1 ) with T0 , T1 ⊆ F . Shifting T appropriately, we assume furthermore that 0 ∈ T0 . To obtain a contradiction we observe that the unique representation of h as a sum of two elements of T has the form h = t0 + (h + t1 ) with t0 ∈ T0 and t1 ∈ T1 ; hence, 2(t1 + h) = t0 + 0 gives two representations of t0 as a sum of two elements of T. 3. Turning to the general case, we denote by µ the initial ordinal of the cardinal |G| and consider a well-ordering G = {gι : ι < µ}. Notice, that µ is a limit ordinal, and hence the successor of any ordinal, smaller than µ, is also smaller than µ. We set S0 := ∅ and construct a chain of subsets Sι , for each ordinal ι ≤ µ, so that – – – – Sι ⊆ Sρ whenever ι < ρ ≤ µ; if ι is a finite ordinal, then Sι is finite, and if ι ≤ µ is infinite, then |Sι | ≤ |ι|; gι ∈ Sρ + Sρ whenever ι < ρ ≤ µ; for any ordinal ι ≤ µ and element g ∈ G there is at most one representation of g as a sum of two elements of Sι . The proof is then completed by observing that Sµ is a perfect basis of G; hence, it suffices to show that the subsets Sι can be constructed. We use transfinite recursion, assuming that ν ≤ µ and that Sι has already been found for each ordinal ι < ν, and constructing Sν . If ν is a limit ordinal, then we put Sν := ∪ι<ν Sι . If ν is a successor ordinal and gν−1 ∈ Sν−1 + Sν−1 , then we put Sν := 6 SERGEI V. KONYAGIN AND VSEVOLOD F. LEV Sν−1 . In the remaining case where ν is a successor ordinal and gν−1 ∈ / Sν−1 + Sν−1 we put Sν := Sν−1 ∪ {s, t}, where s, t ∈ G with s + t = gν−1 are chosen to satisfy the following conditions: (a) (b) (c) (d) (e) if |3G| < |G|, then s, t ∈ / Sν−1 + 3G; s, t ∈ / Sν−1 + Sν−1 − Sν−1 ; 2s, 2t ∈ / Sν−1 + Sν−1 ; s−t∈ / Sν−1 − Sν−1 ; 2s − t, 2t − s ∈ / Sν−1 . Condition (a) is of technical nature and its exact purpose will be clarified in the following paragraph, while the last four conditions ensure that the unique representation property of Sν−1 is inherited by Sν . Thus, to complete the proof it suffices to show that s and t := gν−1 − s satisfying (a)–(e) can be found. To this end we first observe that if Sν−1 is infinite, then condition (e) excludes at most |Sν−1 | ≤ |ν − 1| < |µ| = |G| options for 3s, and similarly (a)–(d) together exclude fewer, than |G| options for 2s. Clearly, this conclusion remains valid if Sν−1 is finite. Therefore, in view of Lemma 2 we can assume that |3G| < |G|. Consequently, securing (a) at each step of the construction, we have ensured that all elements of Sν−1 fall into distinct cosets of 3G, and in particular each of gν−1 + Sν−1 and 2gν−1 − 2Sν−1 contains at most one element from 3G. Since (e) can be re-written as 3s ∈ / (gν−1 + Sν−1 ) ∪ (2gν−1 − Sν−1 ), if |3G| ≥ 3, then there exists g ∈ G such that every s ∈ g + G3 satisfies (e). As remarked above, (a)–(d) reduce to forbidding fewer than |G| values for 2s; that is, forbidding fewer than |G| cosets of G2 for s. Since |g+G3 | = |G| in view of |3G| < |G|, and every G2 -coset intersects g + G3 by at most one element, there exists s ∈ g + G3 with an admissible value of 2s, proving the assertion. Suppose, therefore, that |3G| < 3. If |3G| = 1, then G is of exponent 3, the case which has been addressed above. If |3G| = 2, then the identity g = −2g + 3g shows that G = G3 + 3G, and the sum is direct as if g ∈ G3 ∩ 3G, then 2g = 0 (as g ∈ 3G and |3G| = 2) and 3g = 0 (as g ∈ G3 ), implying g = 0. Consequently, G is the direct sum of a group of exponent 3 and the group of order 2. This completes the proof.  Finally, we prove Theorem 2. As indicated in the introduction, the construction employed in the proof is adopted from [GDT91], where it is used (in the finitedimensional case) to find small codes with covering radius 2. THE ERDŐS-TURÁN PROBLEM 7 Proof of Theorem 2. In view of Lemma 1, it suffices to show that if the field F of characteristic 2 is either finite or algebraically closed, then the group F ×F has a basis with the representation function bounded by 18. Clearly, we can assume |F| > 2. We fix d1 , d2, d3 ∈ F× with d1 + d2 + d3 = 0, write Si := {(x, di /x) : x ∈ F× }; i ∈ {1, 2, 3}, and put S = S1 ∪ S2 ∪ S3 . For (u, v) ∈ F × F let r(u, v) denote the number of representations of (u, v) as a sum of two elements of S, and for i, j ∈ {1, 2, 3} denote by rij (u, v) the number of representations of (u, v) as a sum of an element of Si and an element of Sj . Since the sets S1 , S2 , and S3 are pairwise disjoint, we have r(u, v) = 3 X rij (u, v) i,j=1 and furthermore, rij (u, v) = {x ∈ F \ {0, u} : di /x + dj /(x + u) = v} ; i, j ∈ {1, 2, 3}. The equation di /x + dj /(x + u) = v can be re-written as vx2 + (uv + di + dj )x + di u = 0 (1) and since it has a non-zero coefficient unless (u, v) 6= (0, 0), we have rij (u, v) ≤ 2, except if u = v = 0. It follows that r(u, v) ≤ 18 and to achieve our goal it suffices to show that for any (u, v) 6= (0, 0) there are i, j ∈ {1, 2, 3} with rij (u, v) > 0. We consider three cases. If u = 0 and v 6= 0, then r12 (u, v) is the number of solutions of d1 /x + d2 /x = v, which is 1. If u 6= 0 and v = 0, then r12 (u, v) is the number of solutions of d1 /x = d2 /(x + u); this leads to a non-degenerate linear equation, the solution of which is distinct from both 0 and u. Finally, suppose that u 6= 0 and v 6= 0. In this case for j = i equation (1) takes the form vx2 + uvx + di u = 0, (2) and for rii (u, v) to be non-zero it is necessary and sufficient that (2) has a solution (which automatically is then distinct from 0 and u). If F is algebraically closed, then we are done; suppose, therefore, that F is finite. Since (2) can be re-written as (x/u)2 + (x/u) = di /(uv), it has a solution if and only if di /(uv) belongs to the image of the linear transformation x 7→ x + x2 of the field F considered as a vector space over F2 . The kernel of this 8 SERGEI V. KONYAGIN AND VSEVOLOD F. LEV transformation is a subspace of dimension 1; hence its image is a subspace of F of co-dimension 1. (This image actually is the set of all the elements of F with zero trace, but we do not use this fact.) Consequently, d1 /(uv) + d2 /(uv) + d3 /(uv) = 0 implies that at least one of d1 /(uv), d2 /(uv), and d3 /(uv) is an element of the image.  References [ET41] P. Erdős and P. Turán, On a problem of Sidon in additive number theory, and on some related problems, J. London Math. Soc. 16 (1941), 212–215. [GDT91] E. Gabidulin, A. Davydov, and L. Tombak, Linear codes with covering radius 2 and other new covering codes, IEEE Trans. Inform. Theory 37 (1) (1991), 219–224. [HH04] L. Haddad and C. Helou, Bases in some additive groups and the Erdős-Turán conjecture, J. Combin. Theory Ser. A 108 (1) (2004), 147–153. [N03] M. Nathanson, Unique representation bases for the integers, Acta Arith. 108 (1) (2003), 1–8. [NS07] J. Nešetřil and O. Serra, On a conjecture of Erdős and Turán for additive bases, In: Bib. de la Revista Matematica Iberoamericana, Proceedings of the Segundas Jornadas de Theoria de Numeros (Madrid 2007), 1–12. [R90] I.Z. Ruzsa, A just basis, Monatsh. Math. 109 (2) (1990), 145–151. E-mail address: [email protected] Steklov Mathematical Institute, 8 Gubkina St, Moscow 119991, Russia E-mail address: [email protected] Department of Mathematics, The University of Haifa at Oranim, Tivon 36006, Israel