Sets with few differences in abelian groups

Let $(G, +)$ be an abelian group. In 2004, Eliahou and Kervaire found an explicit formula for the smallest possible cardinality of the sumset $A+A$, where $A \subseteq G$ has fixed cardinality $r$. We consider instead the smallest possible cardinality of the difference set $A-A$, which is always greater than or equal to the smallest possible cardinality of $A+A$ and can be strictly greater. We conjecture a formula for this quantity and prove the conjecture in the case that $G$ is a cyclic group or a vector space over a finite field. This resolves a conjecture of Bajnok and Matzke on signed sumsets.


Introduction
Let G be a finite abelian group of order N written with additive notation. Given subsets A, B ⊆ G, the sumset of A and B is defined as and the difference set of A and B is defined as Let −A denote the difference set {0} − A = {−a | a ∈ A}. Given integers r and s with 1 ≤ r, s ≤ N, define µ G (r, s) = min{|A + B| | A, B ⊆ G, |A| = r, |B| = s} (1) We remark that taking B = A in (1) yields µ G (r, r) ≤ ρ + G (r) and taking B = −A yields µ G (r, r) ≤ ρ − G (r). The functions µ G (r, s) and ρ + G (r) have held considerable interest for over 200 years. In 1813, Cauchy [4] proved the following classical result, which was rediscovered by Davenport [5] in 1935.
In 2004, Eliahou and Kervaire [7] used a classical result of Kneser [8] to compute µ G (r, s) and ρ + G (r) for all finite abelian groups G. Theorem 2 (Eliahou and Kervaire, [7,Theorem 2,Proposition 7]). Let G be a finite abelian group of order N. Then for 1 ≤ r, s ≤ N, where D(N) denotes the set of positive divisors of N. Furthermore, we have ρ + G (r) = µ G (r, r). Remark 1. By Theorem 2, the quantities µ G (r, s) and ρ + G (r) depend on N, r, and s, but not the group structure of G.
However, there is no known explicit formula for ρ − G (r). In [1,2], Bajnok and Matzke considered an h-fold variant of this problem. A small adaptation of their proofs yields the following upper bound for ρ − G (r), which we conjecture holds with equality. Theorem 3 (cf. [1,Theorem 5]). Let G be a finite abelian group of order N. Let e = exp G be the exponent of G; that is, the least common multiple of the orders of the elements of G.
The goal of this paper is to prove two important special cases of Conjecture 1.
First, consider the case that G = Z/NZ is a finite cyclic group. In this case, we have e = exp G = N, so D(N, e, r) = D(N) for 1 ≤ r ≤ N. Thus, the statement of Conjecture 1 becomes Theorem 4 below.
Theorem 4 (cf. [1,Theorem 4]). Let G = Z/NZ. Then Second, consider the case that G = (Z/pZ) d where p is prime and d ≥ 0. Then Theorem 5 below, which is the main result of this paper, computes ρ − G (r) for 1 ≤ r ≤ p d . We will verify in Section 4 that Theorem 5 agrees with the prediction given by Conjecture 1.
Theorem 5. Let G = (Z/pZ) d where p is prime and d ≥ 0. Let t and r be integers with 0 ≤ t ≤ d and p t < r ≤ p t+1 . Then As a consequence of Theorem 5, we obtain the following result, which appears as Conjecture 18 in [2]. We use the notation ρ ± (G, m, r) defined in [2].
Theorem 6 ([2, Conjecture 18]). Let p > 2 be a prime number, and let c and v be integers with 0 ≤ c ≤ p − 1 and In Section 2, we will prove Theorem 4. In Section 3, we will prove Theorem 3. In Sections 4 to 7, we will prove Theorem 5. Finally, in Section 8, we will prove Theorem 6.

The cyclic case
The goal of this section is to prove Theorem 4, which computes ρ − G (r) in the case that G is a finite cyclic group. The proof closely follows that of [1,Theorem 4], though it should be noted that Theorem 4 does not follow directly from [1, Theorem 4] due to differences in the definitions of 2 ± A and A − A.
Theorem 4 (cf. [1,Theorem 4]). Let G = Z/NZ. Then Proof of Theorem 4. By Theorem 2, we have It suffices to show that for each d ∈ D(N). For this, we will construct a set A ⊆ G with |A| ≥ r and Let H be the subgroup of G of order d, and let x be a generator for G/H. Take A to be the "coset arithmetic progression" We compute so |A| = d⌈r/d⌉ ≥ r and as desired.
Remark 3. By comparing the expressions in Theorem 2 and Theorem 4, we see that 3. An upper bound on ρ − G (r) We shall now restate and prove Theorem 3. The proof very closely follows that of [1,Theorem 5].  Proof. It suffices to show that for each d ∈ D(N, e, r). For this, we will construct a set A ⊆ G with |A| ≥ r and By the structure theorem for finitely generated abelian groups, the group G is isomorphic to a direct product H × (Z/eZ) for some abelian group H with |H| = N/e. Since d 1 ∈ D(N/e), we can find a subgroup 4. An outline of the proof of Theorem 5 Sections 4 to 7 of this paper will contain the proof of Theorem 5, which will proceed in four steps: (1) We will show that the bound given in Theorem 5 is achieved. That is, we will show that (2) We will show that for G = (Z/pZ) d , the quantity ρ − G (r) only depends on r and p and not d, as long as d is large enough that (3) By applying the Cauchy-Davenport Theorem (Theorem 1) repeatedly, we will prove Theorem 5 in the case that r ≤ p 2 . (4) We will conclude the proof of the theorem by induction on r. We start with the following result, which is step (1) above. Lemma 1. With the notation of Theorem 5, we have Proof. Using the notation of Theorem 3, we have N = |G| = p d and e = exp G = p, so By Theorem 3, we have min d∈D(N,e,r) as desired.
Remark 4. The proof of Lemma 1 given above shows that Theorem 5 agrees with the prediction given by Conjecture 1.
Remark 5. Here is an explicit example of a subset A ⊆ G achieving the bound of Lemma 1. Put a total order < on Z/pZ by identifying it with {0, 1, . . . , p − 1} in the usual way. Then, recall that (Z/pZ) d is totally ordered by the lexicographic order, which is defined as follows: we say that x = (x 1 , . . . , x d ) precedes y = (y 1 , . . . , y d ) in the lexicographic order if for some i we have x i < y i and x j = y j for j < i. Let A be the set of the smallest r elements of (Z/pZ) d in the lexicographic order. Then one can easily verify that which provides an alternative constructive proof of Lemma 1. It is worth noting that by [6, Proposition 3.1], the same set A satisfies |A + A| = ρ + G (r).

Independence of dimension
The following result is step (2) in the proof of Theorem 5.
Lemma 2. Let p be a prime and let lines containing 0 (that is, vector subspaces of dimension 1) in G. On the other hand, there are only Considering H as a vector space of dimension d 2 = d 1 − 1 over F p , fix an F p -linear transformation π : G → H whose kernel is the line ℓ. Such a transformation π exists because We claim that the restriction π| A is an injection. To show this, take x, y ∈ A with π(x) = π(y); we will show that x = y. Since π is linear, we have π(x − y) = 0, so x − y ∈ ker π = ℓ. Therefore, we have In this section, we show that the statement of Theorem 5 holds when r ≤ p 2 , which is step (3) in the proof of Theorem 5.
Lemma 3. Let p be a prime and let d be a nonnegative integer. Let G be the group (Z/pZ) d . Then The following lemma will be instrumental in the proof of Lemma 3.
Proof. We defer the proof to Appendix A.
Proof of Lemma 3. By Lemma 1, we have so it remains to show that (4) ρ − G (r) ≥ p t min 2 If r ≤ p, then this follows directly from Lemma 2 and the Cauchy-Davenport Theorem. Thus, we may assume r > p.
By Lemma 2, we may assume that d = 2, so G = (Z/pZ) 2 . If p = 2, then the theorem follows easily from enumerating all possible values of r and all sets A ⊆ G, so assume that p > 2. Let Since r ≥ r ′ , replacing r with r ′ cannot increase the left-hand side of (4), and it is easy to check that this replacement leaves the right-hand side unchanged. Therefore, we may assume that r = np + 1 where 1 ≤ n ≤ (p − 1)/2. Take a subset A ⊂ G with |A| = r; we will show that |A − A| ≥ (2n + 1)p = p t min 2 r p t − 1, p .
Identify G with the two-dimensional vector space F 2 p over the field F p . We will now count the two-element subsets of A in two ways. By definition, the number of two-element subsets of A is the binomial coefficient np+1 2 . On the other hand, every two-element subset of A is contained in a unique line (that is, affine subspace of G of dimension 1), so we can count these subsets according to the lines containing them. This yields where the sum is over all lines ℓ ⊂ G. Every line in G is parallel to exactly one line ℓ ′ ⊂ G containing 0, so (5) can be rewritten as where the outer sum is over all lines ℓ ′ ⊂ G containing 0, and the inner sum is over all lines ℓ ⊂ G parallel to ℓ ′ . Since there are exactly p + 1 lines in G containing 0, there is a particular line ℓ 0 ⊂ G containing 0 such that We may assume, by applying an F p -linear change of coordinates, that ℓ 0 is the line {(0, y) | y ∈ F p } ⊂ F 2 p = G. For any x ∈ F p , define the line ℓ x = {(x, y) | y ∈ F p }. Then, the lines in G parallel to ℓ 0 are exactly the lines ℓ we have m ≥ ⌈(np + 1)/p⌉ = n + 1. We consider three cases, depending on whether m ≥ (p + 1)/2, or m = n + 1, or n + 2 ≤ m ≤ (p − 1)/2. Case 1 (m ≥ (p + 1)/2): Take x ∈ F p such that |A ∩ ℓ x | = m. Since ℓ x is a translate of ℓ 0 , which is isomorphic as a group to Z/pZ, the Cauchy-Davenport Theorem applies to the difference ( (Essentially, we are applying the Cauchy-Davenport Theorem only to the second coordinates of the elements of A ∩ ℓ x , which lie in Z/pZ.) That is, the line ℓ 0 is a subset of A − A. Now, take any line ℓ ′ ⊂ G containing 0. There is a line ℓ parallel to ℓ ′ such that |A∩ℓ| ≥ ⌈(np+1)/p⌉ = n+1. Since ℓ is a translate of ℓ ′ , which is isomorphic as a group to Z/pZ, the Cauchy-Davenport Theorem again applies to the difference ( over all lines ℓ ′ ⊂ G containing 0, we conclude ≥ 1 + (p − 1) + p · ((2n + 1) − 1) = (2n + 1)p which is the desired inequality.
Case 2 (m = n + 1): Let S = {x ∈ F p | |A ∩ ℓ x | = n + 1} and let s = |S|. For each x ∈ F p \ S we have |A ∩ ℓ x | ≤ n, so Simplifying this inequality and using the bound n ≤ (p − 1)/2, we obtain Thus s ≥ (p + 1)/2, so by the Cauchy-Davenport Theorem, we have |S − S| ≥ min{2s − 1, p} = p, so S − S = F p . Now, take any x ∈ F p . Since x ∈ S − S, there is y ∈ F p such that y, x + y ∈ S. By the Cauchy-Davenport Theorem again, we have Summing over all x ∈ F p yields By definition, we have p ≥ λ 1 ≥ · · · ≥ λ m > 0 and p ≥ µ 1 ≥ · · · ≥ µ p ≥ 0. We have We claim that M i+j−1 ⊇ Λ i − Λ j for 1 ≤ i, j ≤ m. To show this, take x 1 ∈ Λ i and x 2 ∈ Λ j ; we will show that x 1 − x 2 ∈ M i+j−1 . By the Cauchy-Davenport Theorem, we have where the last equality follows from the bound i, j ≤ m ≤ (p − 1)/2. That is, we have x 1 − x 2 ∈ M i+j−1 , as desired.
By the Cauchy-Davenport Theorem again, we conclude Therefore, the conditions of Lemma 4 are satisfied, so µ k ≥ (2n + 1)p as desired.

Completing the proof of Theorem 5
Before proceeding to the proof of Theorem 5, we prove a general lemma about sets in vector spaces over finite fields.
Lemma 5. Let p be a prime and let m be an integer. Let G be a vector space over the field F p of dimension d ≥ 3, and let S be a subset of G such that |S ∩ H| ≥ mp d−2 for each vector hyperplane H (that is, vector subspace of dimension Proof of Lemma 5. Assume for the sake of contradiction that |S| < mp d−1 . We first claim that there is a (d − 2)-dimensional vector sub- To show this, take a (d − 2)dimensional vector subspace V ⊂ G uniformly at random. It is clear that V has p d−2 − 1 nonzero elements, that G has p d − 1 nonzero elements, and that each nonzero element of G is in V with equal probability. Therefore, the probability that x ∈ V for a fixed x ∈ G \ {0} is Since mp d−3 is an integer, we conclude that there is a particular Finally, consider the integer N defined by the sum

|S ∩ H|
where H ranges over all vector hyperplanes with V 0 ⊂ H ⊂ G. Such hyperplanes H are in bijection with lines through the origin in the twodimensional quotient space G/V 0 , so there are p+1 of them. Therefore, by the assumption of the theorem, we have On the other hand, the sum defining N counts every element of S \ V 0 once and every element of S ∩ V 0 exactly p + 1 times, so Therefore, we have which contradicts our assumption that |S| < mp d−1 .
We are now ready to restate and prove Theorem 5.
Theorem 5. Let G = (Z/pZ) d where p is prime and d ≥ 0. Let t and r be integers with 0 ≤ t ≤ d and p t < r ≤ p t+1 . Then Proof. We proceed by induction on r. If t < 2, then the result follows from Lemma 3, so we may assume t ≥ 2. By Lemma 2, we may also assume that d = t + 1. Let m = min{2 ⌈r/p t ⌉ − 1, p}. We wish to show that ρ − G (r) = mp t . By Lemma 1, we have ρ − G (r) ≤ mp t , so it remains to show that ρ − G (r) ≥ mp t . Let A be a subset of G with |A| = r; we will show that |A − A| ≥ mp t .
Consider G as a vector space of dimension d = t + 1 ≥ 3 over F p . By Lemma 5 applied to S = A−A, it suffices to show that |(A−A)∩H| ≥ mp t−1 for each vector hyperplane H ⊂ G. For this, note that there are exactly p distinct translates H + x, where x ∈ G, and that the entire space G is the disjoint union of these p translates. Therefore, there exists x 0 ∈ G such that |A ∩ (H + x 0 )| ≥ ⌈r/p⌉. By the inductive hypothesis, H (⌈r/p⌉) = mp t−1 as desired.

Applications to signed sumsets
In this section, we prove Theorem 6. In particular, we will show that it is a consequence of the following more general result. The notations ρ ± (G, m, r) and r ± A used in this section are defined in [2]. Lemma 6. Let G be a finite abelian group of order N. Then Proof. Let A ⊆ G be a subset with |A| = m. We will show that We consider two cases, depending on whether or not A ∩ (−A) = ∅. Case 1 (A ∩ (−A) = ∅): Choose x ∈ A ∩ (−A). By definition, the signed sumset 2 ± A contains 0 = x+ (−x) and it contains the difference of any two distinct elements of A. Therefore, we have A − A ⊆ 2 ± A. It follows that Proof. (a) By Lemma 6 and Theorem 5, we have The reverse inequality ρ ± ((Z/pZ) 2 , m, 2) ≤ (2c + 1)p follows from [1, Theorem 5]. (b) By Lemma 6 and Theorem 5, we have The reverse inequality ρ ± ((Z/pZ) 2 , m, 2) ≤ p 2 − 1 follows from [1,Proposition 8].
A. Proof of Lemma 4 In this appendix, we prove Lemma 4. The following lemma is essential to our proof of Lemma 4.