Journal de Théorie des Nombres de Bordeaux 00 (XXXX), 000–000

Small-sum pairs in abelian groups

par Reza Akhtar et Paul Larson

Résumé. Soit G un groupe abélien fini et A, B sous-ensembles de G tels que |A| = |B| = k et |A+A| = |A+B| = 2k −1. Pour sousensembles X, Y de G et c ∈ G, signifions par ν_{c}(X, Y) le nombre de couples (x, y) ∈ X × Y tels que c = x + y. Nous résolvons une question de Bihani et Jin en montrant qu’il existe g ∈ G tel que A = g + B si A + B et apériodique ou s’il existe a ∈ A et b ∈ B tels que ν_{a+b}(A, B) = ν_{a+a}(A, A) = 1. Nous caractérisons aussi les contre-exemples qui se présentent si ni l’une ni l’autre des ces hypothèses est satisfaite.

Abstract. Let G be an abelian group and A, B two subsets of equal size k such that A + B and A + A both have size 2k − 1. Answering a question of Bihani and Jin, we prove that if A + B is aperiodic or if there exist elements a ∈ A and b ∈ B such that a + b has a unique expression as an element of A + B and a + a has a unique expression as an element of A + A, then A is a translate of B. We also give an explicit description of the various counterexamples which arise when neither condition holds.

1. Introduction

Let G be an abelian group, written additively. If A and B are subsets of G, we write A + B to mean {a + b : a ∈ A, b ∈ B}, and similarly for A − B. We use A \ B to denote set difference. In their study [1] of sets of natural numbers with small upper Banach density, Bihani and Jin asked the following in the context of cyclic groups:

Question 1. Let G be an abelian group. Given a pair (A, B) of subsets of G such that |A| = |B| = k and |A + A| = |A + B| = 2k − 1, are A and B always a translates of each other; that is, does there exist h ∈ G such that B = A + h?

It is easily seen that this question cannot always be answered in the affirmative. To see this, let G be any abelian group of odd order 2k − 1 ≥ 5 and A, B any two subsets of G of order k. Since |A| + |B| > |G|, a rudimentary combinatorial result (cf. [7, Lemma 2.1] or Lemma 2.6 below) implies that A+B = G, regardless of whether or not A and B are translates of each other. It is far less obvious that all counterexamples are, in a suitable sense, based on this example.

The goal of this paper is to answer Question 1 in the affirmative under appropriate hypotheses and to provide explicit descriptions of counterexamples. We note that all our work is valid for arbitrary abelian groups G.

The key tool we use is the theory of critical pairs. A critical pair in an abelian group H is a pair (A, B) of subsets such that |A+B| = |A|+|B|−1; an essentially complete description of all such pairs was given by Vosper [6] for groups of prime order and by Kemperman [5] (see also [2], [3], [4]) for general abelian groups. We show how to use Kemperman’s results and Kneser’s Theorem to give an essentially complete answer to Question 1. We will refer to the exposition of Kemperman’s work in the paper of Grynkiewicz [2], as the constructions used there suit our purposes better than that of the original work [5]. For a detailed discussion of Kneser’s Theorem, we refer the reader to Chapter 4 of [7].

2. Preliminaries

Let H be an abelian group. A small-sum pair, or SS-pair for short, is a pair (X, Y) of subsets of H such that |X| = |Y | = k and |X + X| = |X + Y | = 2k − 1 for some k.

Given an abelian group H, subsets A, B ⊆ H, and c ∈ H, we define

ν_{c}(A, B) = |{(a, b) : a ∈ A, b ∈ B : a + b = c}|.

We say that the pair (A, B) has the unique expression property (UEP) if there exist a ∈ A and b ∈ B such that ν_{a+b}(A, B) = 1 and that (A, B) has the strong unique expression property (SUEP) if there exist a ∈ A and b ∈ B such that ν_{a+b}(A, B) = ν_{a+a}(A, A) = 1.

If P ⊆ H is a subgroup, we denote by ϕ_{P} : H → H/P the canonical quotient map.

The stabilizer of a subset A ⊆ H is the subgroup of H defined by:

Stab(A) = {h ∈ H : h + A = A}

A subset A ⊆ H is called periodic if Stab(A) ≠ {0}. For a nontrivial subgroup P, we say that A is P -periodic if A is a union of P -cosets; this is equivalent to the condition P ⊆ Stab(A). We say that A ⊆ H is P - subperiodic if there exists h ∈ H − A such that A ∪ {h} is P -periodic.

Definition 2.1. Let H be an abelian group and P ⊆ H a nontrivial subgroup. A subset A ⊆ H is said to have a quasi-periodic decomposition with respect to the quasi-period P if there exists a partition of A into two disjoint subsets A_{1} ∪ A_{0} such that A_{1} is either empty or P -periodic and A_{0} is a subset of a P -coset.

We refer to A_{1} as the periodic part of A and to A_{0} as the aperiodic part of A. We also say that A is quasi-periodic if A_{1} is nonempty. This notation (with the above meaning) will be used for quasi-periodic decompositions for the remainder of this article.

Lemma 2.2. Suppose H is finite, P is a subgroup of H and A ⊆ H is any nonempty subset. If A = A_{1} ∪ A_{0} is a quasi-periodic decomposition with respect to P and |A_{0}| < |P |, such a decomposition is unique.

Proof. With notation as above, suppose A has quasi-periodic decompositions A = A_{1} ∪A_{0} and A = A′_{1} ∪A′_{0} with respect to a common quasi-period P. Then A_{1},A′_{1} are unions of P -cosets, and each of A_{0}, A′_{0} is a proper subset of a single P -coset. It follows that A_{0} = A′_{0} and hence A_{1} = A′_{1}. □

Lemma 2.3. Suppose H is finite and A, B are subsets of H, and A = A_{1} ∪ A_{0} and B = B_{1} ∪ B_{0} are quasi-periodic decompositions with respect to a common quasi-period P. Suppose further that neither A nor B is P -periodic. If A = h + B, then A_{0} = h + B_{0} and A_{1} = h + B_{1}.

Proof. If A = h + B, then (h + B_{1}) ∪ (h + B_{0}) is also a quasi-periodic decomposition for A. By Lemma 2.2, it follows that A_{1} = h + B_{1} and A_{0} = h + B_{0}. □

The following results are elementary:

Lemma 2.4. Suppose H is an abelian group and A ⊆ H is a nonempty subset such that |A + A| = 2|A| − 1. Then A is not periodic.

Proof. Let k = |A|. If A is periodic, then there is a nontrivial subgroup P ⊆ H such that A is a disjoint union of P -cosets; in particular, |P | divides k. However, A + A is also a union of P -cosets, so |P | must divide 2k − 1, which forces P to be the trivial subgroup. □

Lemma 2.5. Let H be an abelian group. If A ⊆ H is P -periodic, then so is A + B for any B ⊆ H.

Finally, we mention a rudimentary but well-known result:

Lemma 2.6. [7, Lemma 2.1] Let H be a finite abelian group and A, B subsets of H such that |A| + |B| > |H|. Then A + B = H.

3. Kemperman pairs

In this section, we review the theory of critical pairs in abelian groups, following the exposition of Kemperman’s work in [2], and prove various results that will help in the solution of our problem.

Definition 3.1. Let H be an abelian group. A Kemperman pair in H is a pair (A, B) of subsets of H such that A + B is not periodic or (A, B) has UEP.

We now state Kemperman’s Theorem [5, Theorem 5.1 and p. 82] in the form given by Grynkiewicz [2].

Theorem 3.2. ([2], p. 563) Let G be an abelian group and (A, B) a Kemperman pair in G. Then |A + B| = |A| + |B| − 1 if and only if there exist quasi-periodic decompositions A = A_{1} ∪ A_{0} and B = B_{1} ∪ B_{0} with nonempty aperiodic parts and common quasi-period P such that (A_{0}, B_{0}) is a Kemperman pair and:

(1) ν_{c}(ϕ_{P} (A), ϕ_{P} (B)) = 1, where c = ϕ_{P} (A_{0}) + ϕ_{P} (B_{0});

(2) |ϕ_{P} (A) + ϕ_{P} (B)| = |ϕ_{P} (A)| + |ϕ_{P} (B)| − 1, and

(3) |A_{0} + B_{0}| = |A_{0}| + |B_{0}| − 1 and the pair (A_{0}, B_{0}) is of one of the

following (distinct) types: (a) |A_{0}| = 1 or |B_{0}| = 1.

(b) |A_{0}| ≥ 2, |B_{0}| ≥ 2, and A_{0}, B_{0} are arithmetic progressions with common difference d ∈ H such that d has order at least |A_{0}| + |B_{0}| − 1. In this case, ν_{c}(A_{0}, B_{0}) = 1 for exactly two values c ∈ A_{0} + B_{0}.

(c) |A_{0}| + |B_{0}| = |P | + 1, and there is exactly one element g ∈ G such that ν_{g}(A_{0}, B_{0}) = 1.

(d) A_{0} is aperiodic, B_{0} is of the form B_{0} = g − [(G \ A_{0}) ∩ (g′ + P)] for some g, g′ ∈ A_{0}, and ν_{c}(A_{0}, B_{0}) ≠ 1 for all c.

Remark.

We note (cf. [2], c.13 on p. 564) that if we select a different quasi-periodic decomposition for A or B, the type of the resulting pair of aperiodic parts does not change; hence we may speak of the type of (A, B) without reference to any choice of quasi-periodic decompositions.

If A = A_{1} ∪ A_{0} and B = B_{1} ∪ B_{0} are quasi-periodic decompositions of sets with respect to a common quasi-period P, we denote by (A + B)_{P} the P -periodic set (A_{1} + B) ∪ (A + B_{1}). Clearly A + B = (A + B)_{P} ∪ (A_{0} + B_{0}).

Lemma 3.3. Suppose (A, B) is a Kemperman SS-pair. With notation as in Theorem 3.2:

(1) A + B is the disjoint union of A_{0} + B_{0} and (A + B)_{P}. In particular, A + B is P -periodic if and only if A_{0} + B_{0} is P -periodic.

(2) If (A, B) has type (a) or (d), then A + B is not P -periodic. If (A, B) has type (c), then A + B is P -periodic.

Proof. For the first item, suppose A_{0} ⊆ g_{1} + P and B_{0} ⊆ g_{2} + P and let h_{1} = ϕ_{P} (g_{1}), h_{2} = ϕ_{P} (g_{2}). By Theorem 3.2, ν_{h_{1}+h_{2}}(ϕ_{P} (A), ϕ_{P} (B)) = 1.

Now if x ∈ (A + B)_{P} ∩ (A_{0} + B_{0}), then x = a + b, where a ∈ A, b ∈ B and a ∈ A_{1} or b ∈ B_{1}. Thus, ϕ(x) = h_{1} + h_{2} = ϕ(a) + ϕ(b), contradicting ν_{h_{1}+h_{2}}(ϕ_{P} (A), ϕ_{P} (B)) = 1. The second assertion now follows immediately. If (A, B) has type (a), then |A_{0}| = |B_{0}| = 1, so |A_{0} + B_{0}| = 1. Since P is nontrivial, A_{0} + B_{0} is a strict subset of a P -coset and so A_{0} + B_{0} (and hence also A + B) is not P -periodic. If (A, B) has type (d), then, as noted in [2, p. 563], A_{0} + B_{0} is P -subperiodic, so again A + B is not P -periodic. Finally, if (A, B) has type (c), then |A_{0} + B_{0}| = |A_{0}| + |B_{0}| − 1 = |P |, so A_{0} + B_{0} is a (full) P -coset; thus, A + B is P -periodic. □

The proof of the following is immediate: Lemma 3.4. Suppose A, B are subsets of an abelian group H and that A = A_{1} ∪ A_{0}, B = B_{1} ∪ B_{0} are quasi-periodic decompositions with respect to some common quasi-period P. If A + B is P -periodic, then |A + B| = |ϕ_{P} (A + B)||P |; otherwise, |A + B| = (|ϕ_{P} (A + B)| − 1)|P | + |A_{0} + B_{0}|. Lemma 3.5. Suppose (A, B) is a Kemperman SS-pair of subsets of an abelian group H with corresponding quasi-periodic decompositions A = A_{0}∪ A_{1} and B = B_{0} ∪ B_{1} with respect to some common quasi-period P. Then |A_{0}| = |B_{0}|, |ϕ_{P} (A)| = |ϕ_{P} (B)|, |A_{0} + B_{0}| = |A_{0}| + |B_{0}| − 1, |ϕ_{P} (A + B)| = |ϕ_{P} (A)|+|ϕ_{P} (B)|−1, |A_{0}+A_{0}| = 2|A_{0}|−1, and |ϕ_{P} (A+A)| = 2|ϕ_{P} (A)|−1. Proof. Let p = |P |, and define x and y by |A_{1}| = xp and |B_{1}| = yp. We have |A| = xp + |A_{0}|, |B| = yp + |B_{0}|; since A_{0} and B_{0} are nonempty, we have 1 ≤ |A_{0}| ≤ p, 1 ≤ |B_{0}| ≤ p. By properties of integer division, |A_{0}| = |B_{0}| = k and x = y, so |ϕ_{P} (A)| = |ϕ_{P} (A_{1})| + 1 = x + 1 = y + 1 = |ϕ_{P} (B_{1}) + 1| = |ϕ_{P} (B)|; this establishes the first two formulas. The next two formulas follow from Theorem 3.2. If A + B is not P -periodic, then |A + A| = |A + B| ≢ 0(mod |P |) so A + A is not P -periodic, either. It follows that A_{0} + A_{0} is a strict subset of a single P -coset, and also that |A_{0} + A_{0}| = |A_{0} + B_{0}| = |A_{0}| + |B_{0}| − 1 = 2|A_{0}| − 1. If A + B is P -periodic, then |A + A| = |A + B| ≡ 0(mod |P |), so A + A is likewise P -periodic. Letting c = |A + A|/p = |ϕ_{P} (A + A)|, we have cp = |A + A| = 2|A| − 1 = 2(xp + k) − 1. Thus, p(c − 2x) = 2k − 1. Since 1 ≤ k ≤ p, it follows that 2k − 1 = p, so c = 2x + 1; that is, |ϕ_{P} (A + A)| = 2|A_{1}|/p + 1 = 2(|ϕ_{P} (A)| − 1) + 1 = 2|ϕ_{P} (A)| − 1, as desired. □

Proposition 3.6. Suppose (A, B) is a Kemperman SS-pair with quasiperiodic decompositions A = A_{1} ∪ A_{0} and B = B_{1} ∪ B_{0} with respect to a common quasi-period P.

• If (A, B) has type (a) or (b), or (d) then A_{0} is a translate of B_{0}.

• If (A, B) has type (c), then A + B is P -periodic, |P | ≥ 3 is odd, and |A_{0}| = |B_{0}| ≥ 2. Furthermore, if (A, B) has SUEP, then there exist h_{1}, h_{2} ∈ H and a set S ⊆ P such that A_{0} = h_{1} + S, B_{0} = h_{2} + S, and h_{1} + h_{2} ∉ (A_{1} + B_{1}) ∪ (A_{1} + A_{1}); in particular, A_{0} is a translate of B_{0}.

Proof. Since all properties mentioned in the proposition remain invariant when either A or B is replaced by a translate, we assume throughout this proof that A_{0} and B_{0} are subsets of P.

If (A, B) is of type (a) or (b), then clearly A_{0} is a translate of B_{0}. If (A, B) has type (c), A + B is P -periodic by Lemma 3.3. Moreover, because |A_{0}| = |B_{0}|, it follows that |P | = 2|A_{0}| − 1 is odd. In particular, since P is nontrivial, |P | ≥ 3, and so |A_{0}| = |B_{0}| ≥ 2. Now suppose (A, B) has SUEP. By translating each appropriately, we may assume that ν_{0}(A, B) = ν_{0}(A, A) = 1. Then B_{0} ∩ −A_{0} = A_{0} ∩ −A_{0} = {0}; hence A_{0} = B_{0} ⊆ P.

Finally, suppose (A, B) has type (d). As remarked in [2, p. 563], a pair of type (d) satisfies A_{0} + B_{0} = (h_{0} + P) − {h_{0}} for some h_{0} ∈ H, so |A + B| ≡ −1(mod |P |). In particular, letting k = |A_{0}|, we have |P | = |A_{0} + B_{0}| + 1 = |A_{0}| + |B_{0}| = 2k. Since (A, B) is an SS-pair, A_{0} + A_{0} is a subset of P of order 2k − 1. Hence there exist x, y ∈ P such that A_{0} + A_{0} = P \ {x} and A_{0} + B_{0} = P \ {y}; setting B′_{0} = B_{0} + x − y, we have A_{0} + B′_{0} = P \ {x}. Thus, A_{0} and B′_{0} are subsets of P of order k, each disjoint from −A_{0} + x. This forces A_{0} = B′_{0}, and hence A_{0} is a translate of B_{0}. □

Lemma 3.7. Suppose (A, B) is a Kemperman SS-pair in H together with quasi-periodic decompositions A = A_{1} ∪ A_{0} and B = B_{1} ∪ B_{0} with respect to a common quasi-period P. If A + B is aperiodic or (A, B) has SUEP, then (ϕ_{P} (A), ϕ_{P} (B)) has SUEP.

Proof. For convenience, let C = ϕ_{P} (A) and D = ϕ_{P} (B). By translating A and B appropriately, we may assume that A_{0} ⊆ P, B_{0} ⊆ P, and hence that ϕ_{P} (A_{0}) = ϕ_{P} (B_{0}) = 0 ∈ C ∩ D. Suppose (C, D) does not have SUEP. Since ν_{0}(C, D) = 1 by statement 1 of Theorem 3.2, it follows that ν_{0}(C, C) > 1; thus, we may select 0 ≠ c ∈ C such that −c ∈ C. Writing c = ϕ_{P} (a) for some a ∈ A_{1}, observe that since ϕ(−a) = −c ≠ 0 ∈ C, −a + P ⊆ A_{1}, so −a ∈ A_{1} also. Next, choose p ∈ P, p ≠ 0. Since A_{1} is P -periodic, a − p, −a + p ∈ A_{1}, and so (a − p) + (−a + p) = 0 witnesses that ν_{0}(A, A) ≥ 2; hence (A, B) does not have SUEP. Moreover, A + A is the union of a P -periodic set with A_{0} + A_{0}. Since the latter is a subset of P and (a + P) + (−a + P) = P ⊆ A + A, it follows that A + A is P -periodic. Since |A + A| = |A + B| and A, B both have quasi-periodic decompositions with respect to P, A + B is also P -periodic. □

Finally, we need the following “reconstruction lemma”:

Lemma 3.8. Suppose (A, B) is a critical Kemperman pair in an abelian group H with |A| = |B| and that A = A_{0} ∪ A_{1}, B = B_{0} ∪ B_{1} are quasiperiodic decompositions with respect to a common quasi-period P. If A_{0} is a translate of B_{0} and ϕ_{P} (A) is a translate of ϕ_{P} (B) then A is a translate of B.

Proof. Note that the conclusion of Lemma 3.8 remains valid if we translate either A or B by any h ∈ H. Moreover, the assertion is automatically true if A_{1} and B_{1} are empty, so we assume henceforth that these are nonempty. Similarly, if A_{0} (and hence also B_{0}) is a full P -coset, the assertion is clear, so we assume otherwise. If ϕ_{P} (B) = ϕ_{P} (A) + h̄ for some h̄ ∈ H/P, then picking h ∈ G such that ϕ_{P} (h) = h̄, we may replace B by B −h and assume henceforth that ϕ_{P} (B) = ϕ_{P} (A). In particular, this means that there exist distinct cosets h_{1} +P, …, h_{s} +P of P in G and decompositions A = ∪_{i=1}^{s}C_{i} and B = ∪_{i=1}^{s}D_{i} where for every i = 1, …, s, C_{i} ⊆ h_{i} + P and D_{i} ⊆ h_{i} + P with C_{i} = h_{i} + P if and only if i ≠ i_{0} and D_{j} = h_{j} + P if and only if j ≠ j_{0}. With this notation, C_{i_{0}} = A_{0} and D_{j_{0}} = B_{0}. We claim that i_{0} = j_{0}.

Note that A+B is the union of a P -periodic set, together with A_{0}+B_{0} ⊆ (h_{i_{0}} + h_{j_{0}}) + P. If i_{0} ≠ j_{0}, then A_{1} + B_{1} ⊇ C_{j_{0}} + D_{i_{0}} = (h_{i_{0}} + h_{j_{0}}) + P ⊇ A_{0} + B_{0}, contradicting Lemma 3.3. Thus i_{0} = j_{0}, and so A_{1} = B_{1}. Since A_{0} = p + B_{0} for some p ∈ P, we have that A = p + B, as desired. □

4. Results

4.1. The Kemperman case. In this section, we answer our question in the case that our original SS-pair (A, B) is a Kemperman pair.

Theorem 4.1. Let G be an abelian group and (A, B) an SS-pair in G. If A + B is aperiodic or (A, B) has SUEP, then A = x + B for some x ∈ G.

Proof. We prove the theorem by induction on |A| + |B|, the theorem being trivially true when |A| = |B| = 1. Since the hypothesis clearly implies that (A, B) is a Kemperman pair, fix quasi-periodic decompositions A = A_{1}∪A_{0} and B = B_{1} ∪ B_{0} as in Theorem 3.2 with respect to some common quasiperiod P. By Proposition 3.6, A_{0} is a translate of B_{0}. If either of A_{1}, B_{1} is empty, the other is, too, so A = A_{0}, B = B_{0} and the claim is proven. Now suppose A_{1} and B_{1} are nonempty. Then (ϕ_{P} (A), ϕ_{P} (B)) is an SS-pair by Lemma 3.5, so Lemma 3.7 implies that the hypotheses of the theorem are satisfied by (ϕ_{P} (A), ϕ_{P} (B)). By induction, ϕ_{P} (A) is a translate of ϕ_{P} (B). Finally, Lemma 3.8 implies that A is a translate of B.

□

Corollary 4.2. Suppose G is a 2-group and (A, B) is an SS-pair of subsets of G. Then A is a translate of B.

Proof. Let k = |A| = |B|. First, note that A + B cannot be periodic, because |A + B| = 2k − 1, which is odd, so it cannot be the union of cosets of a nontrivial subgroup of G. Hence (A, B) is a Kemperman pair, and so A is translate of B by Theorem 4.1. □

4.2. The periodic case. In this section, we study the case in which (A, B) is an SS-pair with A + B periodic. If (A, B) is not a Kemperman pair, we cannot use Kemperman’s Theorem, so we use Kneser’s Theorem instead. We note that the results of this section also apply to the situation in which (A, B) has SUEP.

Theorem 4.3. (Kneser, [7, Theorem 4.2])

Let G be an abelian group and A, B finite nonempty subsets of G; set H = Stab(A + B). If |A + B| < |A| + |B|, then

|A + B| = |A + H| + |B + H| − |H|

In particular, if |A + B| < |A| + |B| − 1, then H is nontrivial and so A + B is periodic.

One easily constructs examples of SS-pairs (A, B) such that ν_{c}(A, B) > 1 for all c ∈ A + B. We describe how, starting with one such pair, one may construct SS-pairs (A, B′) where B′ is not a translate of A.

Lemma 4.4. Let G be an abelian group, (A, B) an SS-pair in G, and P = Stab(A + B). Let c_{1} + P, …, c_{k} + P be distinct cosets whose union is A + P and d_{1} + P, …, d_{l} + P distinct cosets whose union is B + P. Then:

• |P | is odd.

• |A + P | − |A| = |B + P | − |B| = \frac{|P | − 1}{2}. In particular,

|(c_{i} + P) ∩ A| ≥ \frac{|P | + 1}{2} and |(d_{j} + P) ∩ B| ≥ \frac{|P | + 1}{2}

for all i = 1, …, k and j = 1, …, l.

• Let (A′, B′) be any pair such that A′ ⊆ ∪_{i=1}^{k}(c′_{i}+P), B′ ⊆ ∪_{j=1}^{l}(d′_{j}+P),

where the c′_{i} and d′_{j} are elements of G such that |(c′_{i}+P)∩A′| ≥ \frac{|P | + 1}{2}

and |(d′_{j} +P)∩B′| ≥ \frac{|P | + 1}{2} for all i = 1, …, k and j = 1, …, l. Then

A′ + B′ = ∪_{i=1}^{k} ∪_{j=1}^{l} ((c′_{i} + d′_{j}) + P).

Proof. For convenience of notation, let C_{i} = c_{i} + P, D_{j} = d_{j} + P for i = 1, …, k, j = 1, …, l; hence A + B = ∪_{i=1}^{k} ∪_{j=1}^{l} C_{i} + D_{j}. Because |A| = |B| and (A, B) is critical, |A + B| = |A| + |B| − 1 = 2|A| − 1 is odd. Since A + B is a disjoint union of P -cosets, |P | must be odd. Now Kneser’s Theorem gives |A|+|B|−1 = |A+B| = |A+P |+|B +P |−|P |, or (|A+P |−|A|)+(|B+P |−|B|) = |P |−1. Hence |A+P | (respectively, |B+P |) is the smallest multiple of |P | greater than or equal to |A| (respectively |B|). However, since |A| = |B|, we must have |A + P | = |B + P |, so 2(|A + P | − |A|) = |P | − 1. Reduction modulo |P | gives 2|A| ≡ 1(mod |P |),

so since |P | is odd, |A| = |B| ≡ \frac{|P | + 1}{2}(mod |P |). Thus,

|A + P | − |A| = |B + P | − |B| = \frac{|P | − 1}{2}

and

|C_{i} ∩ A| = |C_{i}| − |C_{i} \ A| ≥ |C_{i}| − |(A + P) \ A| = |P | − \frac{|P | − 1}{2} = \frac{|P | + 1}{2}.

A similar argument shows that |D_{j} ∩ B| ≥ \frac{|P | + 1}{2}, establishing the first statement. For the second statement, note that for every i and j,

((c′_{i} + P) ∩ A′) + ((d′_{j} + P) ∩ B′) ⊆ (c′_{i} + d′_{j}) + P.

Moreover,

|(c′_{i} + P) ∩ A′| + |(d′_{j} + P) ∩ B′| ≥ |P | + 1 > |P |

so by Lemma 2.6, ((c′_{i} + P) ∩ A′) + ((d′_{j} + P) ∩ B′) = (c′_{i} + d′_{j}) + P. Thus, A′ + B′ = ∪_{i=1}^{k} ∪_{j=1}^{l} (c′_{i} + d′_{j}) + P. □

Proposition 4.5. If G is an abelian group and (A, B) a critical pair in G with P = Stab(A + B), then ϕ_{P} (A + B) is aperiodic and (ϕ_{P} (A), ϕ_{P} (B)) is a critical Kemperman pair of subsets of G/P. In particular, if (A, B) is an SS-pair in G, then (ϕ_{P} (A), ϕ_{P} (B)) is a Kemperman SS-pair in G/P and ϕ_{P} (A) is a translate of ϕ_{P} (B).

Proof. Aperiodicity is clear in view of the definition of stabilizer. Let A′ = A + P and B′ = B + P; then ϕ_{P} (A) = ϕ_{P} (A′), ϕ_{P} (B) = ϕ_{P} (B′) and A′ + B′ = (A + B) + P = A + B. By Kneser’s Theorem, |A + B| = |A′+B′| = |A′|+|B′|−|P |. Dividing all terms by |P |, we have |ϕ_{P} (A+B)| = |ϕ_{P} (A)|+|ϕ_{P} (B)|−1, and so (ϕ_{P} (A), ϕ_{P} (B)) is a critical Kemperman pair. This proves the first part of the Proposition.

Now assume that (A, B) is an SS-pair. We will prove that (ϕ_{P} (A), ϕ_{P} (B)) is an SS-pair. Write A′ = ∪_{i=1}^{k}c_{i} + P and B′ = ∪_{i=1}^{l}d_{j} + P (as disjoint

unions). By the second statement of Lemma 4.4, |A ∩ (c_{i} + P)| ≥ \frac{|P | + 1}{2},

|B ∩ (d_{j} + P)| ≥ \frac{|P | + 1}{2} for each i and j, so by the third statement, A + A is P -periodic; that is, P ⊆ Stab(A + A). Furthermore, A′ + A′ = (A + P) + (A + P) = (A + A) + P = A + A. Now Kneser’s Theorem implies

2|A|−1 = |A+A| = 2|A′|−|Stab(A+A)|, or |A′|−|A| = \frac{|Stab(A + A)| − 1}{2}.

Because |A′|−|A| = \frac{|P | − 1}{2} by Lemma 4.4, it follows that P = Stab(A+A). Finally, from |A′ + A′| = |A + A| = 2|A′| − |P | we get |ϕ_{P} (A) + ϕ_{P} (A)| = |ϕ_{P} (A + A)| = 2|ϕ_{P} (A)| − 1. This establishes that (ϕ_{P} (A), ϕ_{P} (B)) is an SS-pair; hence ϕ_{P} (A) is a translate of ϕ_{P} (B) by Theorem 4.1. □

Since Question 1 obviously has an affirmative answer if A and B have size 1 or 2, we assume henceforth without further mention that |A| ≥ 3, |B| ≥ 3. The next result shows that in general there are“many” SSpairs (A, B) such that A + B is periodic but A is not a translate of B. Since Proposition 4.5 implies that ϕ_{P} (A) is always a translate of ϕ_{P} (B), we assume ϕ_{P} (A) = ϕ_{P} (B) in the following to simplify the discussion.

Theorem 4.6. Suppose G is an abelian group and A ⊆ G is a subset such that |A + A| = 2|A| − 1 and P = Stab(A + A) ≠ {0}. Write A = ∪_{i=1}^{l}S_{i}, where each S_{i} is a subset of (the P -coset) c_{i} + P. Consider the collection B of all subsets B ⊆ G such that |A| = |B| = k, ϕ_{P} (A) = ϕ_{P} (B), and

|B ∩ (c_{i} + P)| ≥ \frac{|P | + 1}{2} for all i = 1, …, l. Then each (A, B) is an SSpair in G. Moreover, at most |P | elements of B are translates of A, and |B| > |P |, so there is at least one B ∈ B which is not a translate of A.

Proof. Since ϕ_{P} (A) = ϕ_{P} (B), any B ∈ B must satisfy B ⊆ A + P. Observe that by Lemma 2.4, A must be a strict subset of A + P. Clearly P ⊆ Stab(A + P) = P′. Moreover, we have A + A + P′ = (A + A) + P + P′ = (A + P) + (A + P) + P′ = (A + P) + (A + P) = (A + A) + P = A + A, so P′ ⊆ Stab(A + A) = P and hence Stab(A + P) = P. For any B ∈ B, Lemma 4.4 shows that A+B = A+A, so (A, B) is an SS-pair. If B = g+A for some g ∈ G, then (by definition) A + P = B + P = g + (A + P), so g ∈ Stab(A + P) = P. Now |P | is odd by Lemma 4.4, so |P | ≥ 3. Since A is a strict subset of A + P, there exists i_{0}, 1 ≤ i_{0} ≤ l such that S_{i_{0}} is a strict subset of c_{i_{0}} + P. If l ≥ 2, pick a nonzero element p ∈ P and consider B = ∪_{1≤i≤l:i≠i_{0}}S_{i} ∪ (S_{i_{0}} + p). Then B ∈ B, so (A, B) is an SS-pair, but B is not a translate of A. If l = 1, then A ⊆ c_{1} +P is a set of size n = |A| ≥ 3, and A + A = 2c_{1} + P has size 2n − 1, so |P | = 2n − 1. There are at most

|P | = 2n−1 translates of A within A+P = c_{1} +P and \binom{|P |}{|A|} = \binom{2n − 1}{n}

subsets of A + P of size n. Since n ≥ 3, \binom{2n − 1}{n} > 2n − 1, we may choose

B to be a subset of A + P of size n which is not a translate of A. □

Small-sum pairs in abelian groups 11

References

[1] P. Bihani and R. Jin. Kneser’s Theorem for Upper Banach Density. Journal de Théorie des Nombres de Bordeaux. 18 (2006), no. 2, 323-343.

[2] D. Grynkiewicz. Quasi-periodic decompositions and the Kemperman structure theorem. European Journal of Combinatorics. 26 (2005), 559-575.

[3] Y. O. Hamidoune. Subsets with small sums in abelian groups. I. The Vosper property. European Journal of Combinatorics. 18 (1997), no. 5, 541-556.

[4] Y. O. Hamidoune. Subsets with a small sum. II. The critical pair problem. European Journal of Combinatorics. 21 (2000), no. 2, 231-239.

[5] J. H. B. Kemperman. On small subsets of an abelian group. Acta Mathematica. 103 (1960), 63-88.

[6] A. G. Vosper. The critical pairs of subsets of a group of prime order. J. London Math. Soc. 31 (1956), 200-205 and 280-282.

[7] M. Nathanson. Additive Number Theory. Springer, 1996.

Reza Akhtar Department of Mathematics Miami University Oxford, OH 45056 USA

E-mail : reza@calico.mth.muohio.edu
URL: http://calico.mth.muohio.edu/reza/

Paul Larson Department of Mathematics Miami University Oxford, OH 45056 USA

E-mail : larsonpb@muohio.edu
URL: http://www.users.muohio.edu/larsonpb/