On the generalized restricted sumsets in abelian groups
aa r X i v : . [ m a t h . N T ] S e p ON THE GENERALIZED RESTRICTED SUMSETS IN ABELIANGROUPS
SHANSHAN DU AND HAO PAN
Abstract.
Suppose that A , B and S are non-empty subsets of a finite abeliangroup G . Then the generalized restricted sumset A S + B := { a + b : a ∈ A, b ∈ B, a − b S } contains at least min {| A | + | B | − | S | , p ( G ) } elements, where p ( G ) is the least prime factor of | G | . Further, we also have | A S + B | ≥ min {| A | + | B | − | S | − , p ( G ) } , provided that both | A | and | B | are large with respect to | S | . Introduction
Suppose that G is a finite abelian group and A, B are two non-empty subsets of G . Define the sumset A + B := { a + b : a ∈ A, b ∈ B } . For a positive integer n , let Z n denote the cyclic group of order n . If p is primeand ∅ 6 = A, B ⊆ Z p , then classical Cauchy-Davenport theorem (cf. [12, Theorem2.2]) says that | A + B | ≥ min {| A | + | B | − , p } . (1.1)For a finite abelian group G , let p ( G ) denote the least prime factor of | G | . Inview of the well-known Kneser theorem, we have the following extension of theCauchy-Davenport theorem in abelian groups: | A + B | ≥ min {| A | + | B | − , p ( G ) } , (1.2)where A, B are two non-empty subsets of G .On the other hand, Erd˝os and Heilbronn [7, 8] considered the restricted sumset A ∔ B := { a + b : a ∈ A, b ∈ B, a = b } . They conjectured that if p is prime and A, B are two non-empty subsets Z p , then | A ∔ A | ≥ min { | A | − , p } . (1.3) Mathematics Subject Classification.
Primary 11P70; Secondary 11B13.
Key words and phrases.
Restricted sumsets.
With help of the exterior algebra, Dias da Silva and Hamidoune [5] confirmed theconjecture of Erd˝os and Heilbronn. Subsequently, using the polynomial method,Alon, Nathanson and Ruzsa [2, 3] gave a simple proof of the Erd˝os-Heilbronnconjecture. In fact, they obtained that | A ∔ B | ≥ min {| A | + | B | − , p } , (1.4)provided | A | 6 = | B | .In [9, 10], K´arolyi considered the Erd˝os-Heilbronn problem in abelian groups.He proved that | A ∔ A | ≥ min { | A | − , p ( G ) } , (1.5)where A is a non-empty subset of a finite abelian group G . In [4], Balister andWheeler showed that for ∅ 6 = A, B ⊆ G , | A ∔ B | ≥ min {| A | + | B | − , p ( G ) } . (1.6)They even proved that (1.6) is still valid when G is a finite group (not necessarilycommutative). The key ingredient of the proofs of K´arolyi and Balister-Wheeleris an inductive step, i.e., to prove (1.5) and (1.6) under the hypothesis that theyare valid for H and G/H , where H is a subgroup of G .In this paper, we shall consider the generalized restricted sumset A S + B := { a + b : a ∈ A, b ∈ B, a − b S } , where A, B, S are the non-empty subsets of G . When G = Z p with p is prime, Panand Sun proved that | A S + B | ≥ min {| A | + | B | − | S | − , p } , (1.7)provided that | S | < p . Notice that the restriction | S | < p is necessary, since if S = Z p , then A S + B is always empty.It is natural to find a generalization of (1.7) for abelian groups. However, as weshall see later, the inductive step will become much more complicated when | S | islarge. Here we can establish the following weak type extension of (1.7) for abeliangroups. Theorem 1.1.
Let G be a finite abelian group. Suppose that A , B and S arenon-empty subsets of G . Then | A S + B | ≥ min {| A | + | B | − | S | , p ( G ) } . (1.8)In fact, essentially our proof of Theorem 1.1 doesn’t depend on the fact that G is abelian. In Section 2, we shall also give a brief explanation how to extend (1.8)to general finite groups. N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 3
Although Theorem 1.1 holds unconditionally, we can get the better lower boundof | A S + B | under some additional assumptions. For example, when G = Z p α , (1.8)can be improved to [14, Remark 1.3] | A S + B | ≥ min {| A | + | B | − | S | − , p } . (1.9)On the other hand, if | A | , | B | are large with respect to | S | , we can show that (1.7)is also valid for any finite abelian groups. Theorem 1.2.
Let A , B and S be non-empty subsets of an abelian group G .Suppose that min {| A | , | B |} ≥ | S | − | S | − . Then | A S + B | ≥ min {| A | + | B | − | S | − , p ( G ) } . (1.10)2. Proof of Theorem 1.1
Lemma 2.1.
Let p ≥ be a prime and α ≥ . Suppose that A, B, S are non-emptysubsets of F p α with | S | < p , where F p α is the finite field with p α elements. Thenfor any γ ∈ F p α \ { , − } , the cardinality of the restricted sumset { a + b : a ∈ A, b ∈ B, a − γb S } is at least min {| A | + | B | − | S | − , p } . Proof.
This is [13, Corollary 2]. (cid:3)
Lemma 2.2.
Let p be a prime and α ≥ . Suppose that A = { a , . . . , a m } , B = { b , . . . , b n } and S are non-empty subsets of F p α with | S | < p . Let h = | S | and suppose that m ≥ h + 3 . If m + n − h − ≤ p ,then the set ( A S + B ) \ { a + b , . . . , a + b n } contains the distinct elements a i + b j , a i + b j , . . . , a i m − h − + b j m − h − such that for each ≤ k ≤ m − h − i k ∈ { , , . . . , h + 2 , k + h + 2 } . Proof.
Let X k = ( { a , a , . . . , a h +2 , a k + h +2 } S + B ) \ { a + b , . . . , a + b n } SHANSHAN DU AND HAO PAN for each 1 ≤ k ≤ m − h −
2. Clearly for any ∅ 6 = k ⊆ { , , . . . , m − h − } , usingLemma 2.1, we have [ k ∈ I X k = (cid:18)(cid:18) { a , a , . . . , a h +2 } ∪ [ k ∈ I { a k + h +2 } (cid:19) S + B (cid:19) \ { a + b , . . . , a + b n }≤ (cid:0) h + 2 + | I | ) + | B | − | S | − (cid:1) − n = | I | . With help of the well-known Hall theorem, we may choose distinct x , x , . . . , x m − h − such that x k ∈ X k for each k = 1 , . . . , m − h −
2, i.e., x k = a i k + b j k with i k ∈ { , . . . , h + 2 , k + h + 2 } . (cid:3) Proof of Theorem 1.1.
Assume that our assertion is true for any proper subgroupof G . Suppose that | A | + | B |− | S | > p ( G ). Then we may choose non-empty A ′ ⊆ A and B ′ ⊆ B such that | A ′ | + | B ′ | − | S | = p ( G ). Clearly | A S + B | ≥ | A ′ S + B ′ | . Sowe may assume that | A | + | B | − | S | ≤ p ( G ). On the other hand, trivially we have | A S + B | ≥ max {| A | − | S | , | B | − | S |} . So | A | + | B | − | S | > | B | − | S | will imply | A | > | S | , i.e., | A S + B | ≥ | A | − | S | ≥ | S | + 1 . Hence we always assume that | S | < p ( G ).Let H be a proper subgroup of G such that [ G : H ] = p ( G ). Write A = m [ i =1 ( a i + A i ) , B = n [ i =1 ( b i + B i ) , S = h [ i =1 ( s i + S i ) , where A i , B i , S i are non-empty subsets of H and a i − a j , b i − b j , s i − s j H for anydistinct i, j . Noting that | A S + B | = | ( − B ) ( − S ) + ( − A ) | , without loss of generality, we may assume that n ≥ m . Furthermore, assume that |A | ≥ |A | ≥ · · · ≥ |A m | . For each a ∈ G , let ¯ a denote the coset a + H . Let¯ A = { ¯ a , . . . , ¯ a m } , ¯ B = { ¯ b , . . . , ¯ b n } , ¯ S = { ¯ s , . . . , ¯ s h } . In view of Lemma 3.1, | ¯ A ¯ S + ¯ B | ≥ | ¯ A | + | ¯ B | − | ¯ S | − m + n − h − . Let 1 ≤ µ < µ < · · · < µ r ≤ n be all integers such that¯ a − ¯ b µ , ¯ a − ¯ b µ , . . . , ¯ a − ¯ b µ r ∈ ¯ S. N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 5
Without loss of generality, we may assume that¯ a − ¯ b µ k = ¯ s k for each 1 ≤ k ≤ r . Then by the induction hypothesis, | ( a + A ) s k + S k + ( b µ k + B µ k ) | = |A S ∗ k + B µ k | ≥ |A | + |B µ k | − |S k | , where S ∗ k = ( b µ k − a ) + s k + S k .Let T = ( ¯ A ¯ S + ¯ B ) \ { ¯ a + ¯ b , . . . , ¯ a + ¯ b n } and let τ = | T | . It follows from Lemma 2.1 that τ = | T | ≥ | ¯ A ¯ S + ¯ B | − n ≥ m − h − . We may assume that ¯ a γ + ¯ b η , ¯ a γ + ¯ b η . . . , ¯ a γ τ + ¯ b η τ are distinct elements of T , where 2 ≤ γ , . . . , γ τ ≤ m , 1 ≤ η , . . . , η τ ≤ m and¯ a γ − ¯ b η , ¯ a γ − ¯ b η . . . , ¯ a γ τ − ¯ b η τ ¯ S. Further, in view of Lemma 2.2, if m ≥ h + 3, we may assume γ j ∈ { , , . . . , h + 2 , j + h + 2 } for each 1 ≤ j ≤ m − h −
2. Clearly (cid:18) r [ k =1 (cid:0) ( a + A ) s k + S k + ( b µ k + B µ k ) (cid:1)(cid:19) ∪ (cid:18) [ ≤ ν ≤ nν µ ,...,µ r } (cid:0) ( a + A ) + ( b ν + B µ ) (cid:1)(cid:19) ∪ (cid:18) τ [ j =1 (cid:0) ( a γ j + A γ j ) + ( b η j + B η j ) (cid:1)(cid:19) SHANSHAN DU AND HAO PAN forms a disjoint union of a subset of A S + B . Therefore | A S + B | ≥ r X k =1 |A S ∗ k + B µ k | + X ≤ ν ≤ nν µ ,...,µ r } |A + B ν | + τ X j =1 |A γ i + B η j |≥ r X k =1 ( |A | + |B µ k | − |S k | ) + X ≤ ν ≤ nν µ ,...,µ r } ( |A | + |B ν | −
1) + τ X j =1 |A γ j |≥ n |A | + | B | + τ X j =1 |A γ j | − r X k =1 |S k | − ( n − r ) . (2.1)Evidently 3 r X k =1 |S k | − r ≤ h X k =1 |S k | − h = 3 | S | − h. Let Ψ = n ( |A | −
1) + | B | + τ X j =1 |A γ j | . Clearly (1.8) is true if we can show thatΨ ≥ | A | + | B | − h. We need to consider the following four cases:(I) |A | = 1.Note that now |A S ∗ k + B µ k | is greater than or equal to |B µ k | − |S k | , rather than1 + |B µ k | − |S k | . In view of (2.1), | A S + B | ≥ r X k =1 ( |B µ k | − |S k | ) + X ≤ ν ≤ nν µ ,...,µ r } |B ν | + τ X j =1 ≥| B | + ( m − h − − r X k =1 ( |S k | − ≥ | A | + | B | − | S | − . (II) |A | ≥ n ≥ h + 4.Suppose that m ≥ h + 3. Recall that for each j = 1 , . . . , m − h − γ j ∈{ , h + 2 , j + h + 2 } , i.e., γ j ≥ j + h + 2. So τ X j =1 |A γ k | ≥ m − h − X j =1 |A γ j | ≥ m X k = h +3 |A k | . N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 7
It follows thatΨ = n ( |A | −
1) + | B | + τ X j =1 |A γ k |≥ ( h + 2) |A | − ( h + 2) + m X k = h +3 |A k | + | B | + ( n − h − |A | − ≥| A | + | B | − h. (2.2)And if m ≤ h + 2, we also haveΨ ≥ n ( |A | −
1) + | B | ≥ m |A | − m + | B | + ( n − m ) ≥ | A | + | B | − h. (III) |A | ≥ n ≤ h + 3 and m ≥ h + 2.Suppose that | ¯ A ¯ S + ¯ B | ≥ m + n − h − , i.e., τ ≥ m − h −
1. ThenΨ ≥ m ( |A | −
1) + | B | + τ X j =1 |A γ k |≥ (( m − |A | + 2) − m + | B | + ( |A γ | + ( τ − ≥ | A | + | B | − h. (2.3)So we may assume that | ¯ A ¯ S + ¯ B | = m + n − h −
2, i.e., τ ≥ m − h −
2. Let J = { ¯ a j : |A j | ≤ |A | − , ≤ j ≤ m } . If | J | ≥
2, then we get m |A | ≥ | A | + 2. ThusΨ ≥ m |A | + | B | + τ − m ≥ | A | + | B | − h. Therefore we may assume that | J | ≤ ≤ j ≤ m and 1 ≤ k ≤ n such that¯ a j + ¯ b k ¯ A ¯ S + ¯ B and |A j | = |A | . (2.4)Then we can exchange a + A and a j + A j , i.e., set ¯ a j as the new ¯ a . Since¯ a + ¯ b k ¯ A ¯ S + ¯ B now , we get τ = (cid:12)(cid:12) ( ¯ A ¯ S + ¯ B ) \ { ¯ a + ¯ b , . . . , ¯ a + ¯ b n } (cid:12)(cid:12) ≥ ( m + n − h − − ( n −
1) = m − h − . By (2.3), we also haveΨ ≥ m ( |A | −
1) + | B | + τ X j =1 |A γ j | ≥ | A | + | B | − h. SHANSHAN DU AND HAO PAN
We still need to find j, k satisfying the assumption (2.4). Clearly we may assumethat | ¯ A ¯ S + ¯ B | < p ( G ), otherwise we immediately get | A S + B | ≥ p ( G ). Supposethat m + n − ≤ p ( G ). Then | ¯ A + ¯ B | − | ¯ A ¯ S + ¯ B | ≥ ( m + n − − ( m + n − h −
2) = h + 1 . Since | J | ≤
1, we have | J + ¯ S | ≤ h . So there exist 1 ≤ j ≤ m and 1 ≤ k ≤ n suchthat ¯ a j + ¯ b k ¯ A ¯ S + ¯ B and ¯ a j J , i.e., |A j | = |A | .Suppose that m + n − > p ( G ), i.e., ( m −
1) + n − ≥ p ( G ). Then( ¯ A \ J ) + ¯ B = G/H ) ¯ A ¯ S + ¯ B. We also can find ¯ a j + ¯ b k ¯ A ¯ S + ¯ B with ¯ a j J .(IV) |A | ≥ n ≤ h + 3 and m ≤ h + 1.Suppose that τ ≥
1. From (2.3), we know thatΨ ≥ (cid:0) ( m − |A | + 2 (cid:1) + | B | + (cid:0) ( τ −
1) + A γ (cid:1) − m ≥| A | + | B | − h + 1 . And if there exists 1 ≤ j ≤ m such that |A j | ≤ |A | −
1, then we also haveΨ ≥ m |A | + | B | − m ≥ ( | A | + 1 (cid:1) + | B | − ( h + 1) = | A | + | B | − h. Furthermore, we getΨ ≥ n ( |A | −
1) + | B | ≥ m ( |A | −
1) + | B | + ( n − m ) ≥ | A | + | B | − h provided that n ≤ h or n > m .Below we assume that τ = 0 , |A | = · · · = |A m | , n = m = h + 1 . This is the most difficult part of the proof of Theorem 1.1. Clearly we may set¯ a = ¯ a for any ¯ a ∈ ¯ A . Assume that | ¯ A ¯ S + ¯ B | < p ( G ). Note that τ = 0 implies¯ A ¯ S + ¯ B ⊆ { ¯ a + ¯ b , . . . , ¯ a + ¯ b m } , On the other hand, since | ¯ A ¯ S + ¯ B | < min { m − , p ( G ) } ≤ | ¯ A + ¯ B | , we may choose ¯ a = ¯ a ∈ ¯ A and ¯ b µ ∈ ¯ B such that ¯ a + ¯ b µ ¯ A ¯ S + ¯ B . So¯ A ¯ S + ¯ B ⊆ { ¯ a + ¯ b k : 1 ≤ k ≤ m, k = µ } . N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 9
But | ¯ A ¯ S + ¯ B | ≥ m − h − h = m − . We must have ¯ A ¯ S + ¯ B = { ¯ a + ¯ b k : 1 ≤ k ≤ m, k = µ } , i.e., | ¯ A ¯ S + ¯ B | = m − a ∈ ¯ A as ¯ a . Assume that¯ a − ¯ b µ k = ¯ s λ k , ≤ k ≤ r and ¯ a − ¯ b ν ¯ S, ν
6∈ { µ , . . . , µ r } . Without loss of generality, assume that ¯ a + ¯ b µ ¯ A ¯ S + ¯ B , i.e., ¯ A ¯ S + ¯ B = { ¯ a + ¯ b k : k = µ } . Letting S ∗ λ k = ( b k − a k ) + s λ k + S λ k , we get | A S + B | ≥ r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν |≥ r X k =1 ( |A | + |B µ k | − |S λ k | ) + X ≤ ν ≤ mν µ ,...,µ r } ( |A | + |B ν | − m ( |A | −
1) + m X k =1 |B k | − r X k =1 (3 |S λ k | − . (2.5)Hence | A S + B | ≥ | A | + | B | − m − | S | + h + δ = | A | + | B | − | S | − δ, (2.6)where δ ∈ { , } and δ = 1 if none of the following conditions is true:(a) r = h ;(b) |A S ∗ λk + B µ k | = |A | + |B µ k | − |S λ k | for each 1 ≤ k ≤ h ;(c) |A + B ν | = |A | + |B ν | − ν
6∈ { µ , . . . , µ h } .Suppose that there exists some ¯ a ∈ ¯ A such that δ = 1 when ¯ a = ¯ a , i.e., at leastone of (a)-(c) fails. Evidently by (2.6), we immediately get (1.8). Assume that forany ¯ a ∈ ¯ A , we always have δ = 0 if ¯ a = ¯ a , i.e., (a)-(c) all hold.For each 2 ≤ i ≤ m , it is impossible that¯ a i − ¯ b ν ¯ S, where 1 ≤ ν ≤ m is the unique one satisfying ¯ a − ¯ b ν ¯ S . Otherwise, by (a), wealso can get { ¯ b k : 1 ≤ k ≤ m, k = ν } = ¯ a i − ¯ S. It follows that ¯ a i − ¯ a + ¯ S = ¯ S, i.e., ¯ S = G/H . This contradicts our assumption h < m ≤ p ( G ).Below we need to the following inverse theorem of Kar´olyi [11, Theorem 4]. Lemma 2.3.
Let A and B be non-empty subsets of a finite group G . Suppose that | A + B | = | A | + | B | − ≤ p ( G ) − . Then the following one holds:(1) | A | = 1 or | B | = 1 ;(2) A and B are two arithmetic progressions with the common difference, i.e., A = { a, a + q, . . . , a + ( k − q } and B = { b, q + b, . . . , ( l − q + b } ;(3) A ⊆ a + K and B ⊆ K + b , where K is a subgroup of G of order p ( G ) . Since |A | + |B ν | > p ( H ) implies | A S + B | ≥ |A + B ν | ≥ p ( H ), we may assumethat |A | + |B ν | ≤ p ( H ) . (2.7)By (c), we get |A + B ν | = |A | + |B ν | − . (2.8)By Lemma 2.3 and (2.8), we must have |B ν | = 1, or A and B ν are the arithmeticprogressions with the common difference, or A = α + K and B ν = β + K where K is a subgroup of H of order p ( H ).For each 2 ≤ i ≤ m , since ¯ a i − ¯ b ν ∈ ¯ S , we may assume that ¯ a i − ¯ b ν = ¯ s κ i for some1 ≤ κ i ≤ h . And there exists a unique 1 ≤ υ i ≤ m such that ¯ a i + ¯ b υ i ¯ A ¯ S + ¯ B .Note that (a)-(c) are still valid even if we exchange a + A and a i + A i . It followsfrom (b) and (c) that |A i S ∗ κi + B ν | = |A i | + |B ν | − |S ∗ κ i | (2.9)and |A i | + B υ i | = |A i | + |B υ i | − , (2.10)where S ∗ κ i = s κ i + b m − a + S κ i .According to Lemma 2.3, there are four sub-cases:(i) |B ν | = 1.Trivially |A i S ∗ κi + B ν | = |A i | − |S κ i | > |A | + |B ν | − |S κ i | , which evidently contradicts (2.9).(ii) There exist the subgroups K , K of H of order p ( H ) and α, β ∈ H such that A i ⊆ α + K and B ν ⊆ β + K . N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 11
Since | K | = | K | = p ( H ), either K = K or | K ∩ K | = 1. If | K ∩ K | = 1,it is easy to see that |A i + B ν | = | ( A i − α ) + ( B ν − β ) | = |A i | · |B ν | . Similarly, |A i − B ν | = |A i | · |B ν | . So |{ ( a, b ) : a − b ∈ S ∗ κ i , a ∈ A i , b ∈ B ν }| ≤ |S ∗ κ i | . Thus |A i S ∗ κi + B ν | ≥|A i | · |B ν | − |S κ i |≥|A i | + |B ν | − |S κ i | − > |A i | + |B ν | − |S κ i | . Suppose that K = K . Then letting S ◦ κ i = ( β − α + S ∗ κ i ) ∩ K and applyingLemma 2.1 , we get |A i S ∗ κi + B ν | = | ( A − α ) S ◦ κi + ( B m − β ) | ≥ |A i | + |B ν | − |S ◦ κ i | − . From (2.9), we obtain that |S κ i | = 1 . (iii) Either A i is an arithmetic progression and B ν ⊆ β + K where K is a subgroupof H , or A i = α + K and B ν is an arithmetic progression.We only need to consider the first possibility, i.e., A i = { a, a + q, . . . , a +( d − q } and B ν = β + K . If q ∈ K , then A i ∈ a + K . By the discussion in Case (ii), wecan get that |S κ i | = 1.Suppose that q K . Since |A i | = d < p ( H ) by (2.7), a + K, . . . , a + ( d − q + K must be disjoint cosets of K . Otherwise, we can get d ≥ [ h q, K i : K ] ≥ p ( H ),where h q, K i denotes the subgroup generated by q and K . Write S ∗ κ i = h ∗ [ j =1 ( t j + T j ) , where T j ⊆ K and t j − t k K if j = k . Assume that a + µ ∗ k q − t k ∈ K for1 ≤ k ≤ r ∗ . Then |A i S ∗ κi + B ν | ≥ r ∗ X k =1 ( |B ν | − |T k | ) + | B ν | · ( d − r ∗ )= d |B ν | − r ∗ X k =1 |T k | ≥ |A i | + |B ν | − |S κ i | − . (iv) Both A i and B ν are arithmetic progressions.Write A i = { a + kq : 0 ≤ k ≤ d − } , B ν = { a + kq : 0 ≤ k ≤ d − } . Let h q i denote the subgroup generated by q . If q
6∈ h q i , then B ν ⊆ a + h q i and a + h q i , . . . , a + ( d − q + h q i are disjoint cosets of h q i . According to thediscussion in Case (iii), we can get |A i S ∗ κi + B ν | ≥ |A i | + |B ν | − |S κ i | −
1. Similarly,the same result can be deduced from the assumption q
6∈ h q i , too.Suppose that h q i = h q i . Let K be a maximal proper subgroup of h q i , i.e.,[ h q i : K ] is prime. Let A ◦ = A i − a , B ◦ = B ν − a , S ◦ = ( a − a + S ∗ κ i ) ∩ K. Clearly |A i S ∗ κi + B ν | = |A ◦ S ◦ + B ◦ | . Let b A ◦ = { a + K : a ∈ A ◦ } , b B ◦ = { b + K : b ∈ B ◦ } , b S ◦ = { s + K : s ∈ S ◦ } . Since q , q K , we have | b A ◦ | = |A ◦ | , | b B ◦ | = |B ◦ | and | b S ◦ | ≤ |S ◦ | . Since [ h q i : K ]is prime, by Lemma 2.1, we get |A i S ∗ κi + B ν | ≥ | b A ◦ b S ◦ + b B ◦ | ≥ | b A ◦ | + | b B ◦ | − | b S ◦ | − ≥ |A i | + |B ν | − |S κ i | − , which implies |S κ i | = 1 by (2.9).Now we have deduced that either |A i S ∗ κi + B ν | ≥ |A i | + |B ν | − |S κ i | − > |A i | + |B ν | − |S κ i | which leads to a contradiction to (2.9), or |A i S ∗ κi + B ν | = |A i | + |B ν | − |S κ i | − |S κ i | = 1 from (2.9). Since ¯ a i − ¯ b ν ∈ ¯ S for each 2 ≤ i ≤ m , we get { ¯ a , . . . , ¯ a m } = ¯ b ν + ¯ S . Hence we must have |S k | = 1for each 1 ≤ k ≤ h .If m = h + 1 < p ( G ), then | ¯ A + ¯ B | ≥ min { m − , p ( G ) } > m. So there exist 2 ≤ j ≤ m and 1 ≤ k ≤ m such that¯ a j + ¯ b k ∈ ( ¯ A + ¯ B ) \ { ¯ a + ¯ b , . . . , ¯ a + ¯ b m } . Assume that ¯ a j − ¯ b k = ¯ s λ . Let S ∗ λ = s λ + b k − a j + S λ . In view of (2.5), we have | A S + B | ≥|A j S ∗ λ + B k | + r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν |≥|A j S ∗ λ + B k | + | A | + | B | − | S | − . N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 13
Since |A j | ≥ |S ∗ λ | = 1, A j S ∗ λ + B k = ∅ , i.e., | A S + B | ≥ | A | + | B | − | S | .Suppose that m = p ( G ). Then we also have A S ∗ λ + B µ = ∅ . Recall that¯ A ¯ S + ¯ B = { ¯ a + ¯ b k : 1 ≤ k ≤ m, k = µ } . We get | A S + B | ≥|A S ∗ λ + B µ | + ( m − ≥ p ( G ) . (cid:3) Let us briefly explain how to extend Theorem 1.1 to finite non-commutativegroups. Suppose that G is a finite group and H is a non-trivial subgroup of G .Note that for a, b ∈ G and A , B ⊆ H ,( a + A ) + ( b + B ) = ( a + b ) + (cid:0) ψ b ( A ) + B (cid:1) , where ψ b : x
7→ − b + x + b is an inner automorphism of G . So we have to studythe restricted sumset A σ,S + B := { a + b : a ∈ A, b ∈ B, a − σ ( b ) S } , where σ is an automorphism of G . If a − b ∈ s + H and S ⊆ H , then we have( a + A ) σ,s + S + ( b + B ) = ( a + b ) + ( ψ b ( A ) ψ b σ, S ∗ + B ) , where S ∗ = ψ σ ( b )+ b (cid:0) ( ψ b − a ) + s + S (cid:1) .On the other hand, in order to use the induction process, Balister and Wheelerproved the following lemma. Lemma 2.4 ([4, Theorem 3.2]) . Suppose that G is a finite group of odd order and σ is an automorphism of G . Then there exists a proper normal subgroup H of G satisfying that(1) σ ( H ) = H .(2) G/H is isomorphic to the additive group of some finite field F p α .(3) Let χ denote the isomorphism from G/H to the additive group of F p α .Then there exists some γ ∈ F p α \{ } such that χ ( σ (¯ a )) = γ · χ (¯ a ) for each ¯ a ∈ G/H . With help of Lemmas 2.1, 2.3 and 2.4, we can obtain that the following gener-alization of Theorem 1.1 for general finite groups.
Theorem 2.1.
Suppose that A , B and S are non-empty subsets of a finite group G . Let σ be an automorphism of G with odd order. Then | A σ,S + B | ≥ min {| A | + | B | − | S | , p ( G ) } . (2.11) Proof of Theorem 1.2 for Z p α Lemma 3.1.
Let p be a prime and α ≥ . Suppose that A, B, S are non-emptysubsets of Z p α . Then | A S + B | ≥ min {| A | + | B | − | S | − , p } . Proof.
See [14, Remark 1.3]. (cid:3)
Lemma 3.2 ([6, Lemma 2.1]) . Suppose that A = { a , . . . , a m } and B = { b , . . . , b n } are non-empty subsets of a finite abelian group G . If m + n − ≤ p ( G ) , then theset ( A + B ) \ { a + b , . . . , a + b n } contains the distinct elements a + b j , a + b j , . . . , a i m + b j m . Lemma 3.3.
Suppose that A = { a , . . . , a m } , B = { b , . . . , b n } and S are non-empty subsets of a finite abelian group G . Let h = | S | and suppose that m ≥ h +1 .If m + n − h ≤ p ( G ) and h < p ( G ) , then the set ( A S + B ) \ { a + b , . . . , a + b n } contains the distinct elements a i + b j , a i + b j , . . . , a i m − h + b j m − h such that for each ≤ k ≤ m − hi k ∈ { , , . . . , h, k + 3 h } . Proof.
This lemma immediately follows from Theorem 1.1 by using the same dis-cussions in the proof of Lemma 3.1. (cid:3)
Proposition 3.1.
Let
A, B, S be non-empty subsets of Z p α . Suppose that min {| A | , | B |} ≥ | S | − . Then | A S + B | ≥ min {| A | + | B | − | S | − , p } . (3.1) Proof.
We use an induction on α . Assume that the assertion of Proposition 3.1 istrue for Z p α − . In view of (1.6), we alway assume that | S | ≥
2. Note that | A S + B | ≥ | A | − | S | ≥ | S | − | S | − ≥ | S | when | S | ≥
2. So we only need to consider that case | S | < p .Let H be the subgroup of Z p α of order p . For x ∈ Z p α , let ¯ x denote the coset x + H , and let ¯ X := { ¯ x : x ∈ X } for X ⊆ Z p α . Assume that¯ A = { ¯ a , . . . , ¯ a m } , ¯ B = { ¯ b , . . . , ¯ b n } , ¯ S = { ¯ s , . . . , ¯ s h } . By exchanging A and B , we may assume that m ≤ n . Write A = m [ i =1 ( a i + A i ) , B = n [ i =1 ( b i + B i ) , S = h [ i =1 ( s i + S i ) , N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 15 where those A i , B i , S i ⊆ H . Moreover, assume that |A | ≥ |A | ≥ · · · ≥ |A m | , |B | ≥ |B | ≥ · · · ≥ |B n | . Suppose that m = n = 1. Let T = ( b − a ) + s i + S i if ¯ a − ¯ b = ¯ s i for some1 ≤ i ≤ h , and let T = ∅ if ¯ a − ¯ b ¯ S . Then | A S + B | = |A T + B | ≥ min {|A | + |B | − |T | − , p }≥ min {| A | + | B | − | S | − , p } . Below we always assume that either m >
1, or n > m .(I) |A | ≥ a − ¯ b µ k = ¯ s λ k for each 1 ≤ k ≤ r , and ¯ a − ¯ b ν ¯ S if 1 ≤ ν ≤ n and ν
6∈ { µ , . . . , µ r } . Let τ = | ( ¯ A ¯ S + ¯ B ) \ { ¯ a + ¯ b , . . . , ¯ a + ¯ b n }| . By Lemma 3.1,we have τ ≥ max { m − h − , } . Furthermore, when m ≥ h + 2, we may assumethat ¯ a γ + ¯ b η , . . . , ¯ a γ m − h + ¯ b η m − h − are distinct elements of ( ¯ A ¯ S + ¯ B ) \ (¯ a + ¯ B )with γ j ≤ j + 2 h + 1 for each 1 ≤ j ≤ m − h − S ∗ λ k = ( b µ k − a k ) + s λ k + S λ k , we have | A S + B | ≥ r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν | + τ X j =1 |A γ j + B η j | . We may assume that those |A S ∗ λk + B µ k | , |A + B ν | , |A γ j + B η j | < p, otherwise we immediately get | A S + B | ≥ p . Thus | A S + B | ≥ r X k =1 ( |A | + |B µ k | − |S λ k | −
2) + X ≤ ν ≤ mν µ ,...,µ r } ( |A | + |B ν | −
1) + τ X j =1 |A γ j |≥ n |A | + | B | + m − h X j =1 |A γ j | − r X k =1 ( |S λ k | + 1) − n. (3.2)When m ≥ h + 2, we obtain that | A S + B | ≥ (2 h + 1) |A | + m X i =2 h +2 |A i | + | B | + ( n − h − |A | − n − h X k =1 ( |S k | + 1) ≥| A | + | B | − | S | − (cid:0) ( n − h − |A | − n − h + 2 (cid:1) . (3.3) While if m ≤ h + 1, we also have | A S + B | ≥ n |A | + | B | − r X k =1 ( |S λ k | + 1) − n ≥ m |A | + | B | + ( n − m ) |A | − h X k =1 ( |S k | + 1) − n ≥| A | + | B | − | S | − (cid:0) ( n − m ) |A | − n − h + 2 (cid:1) . (3.4)If n ≥ h , then( n − h − |A | − n − h + 2 ≥ n − h − − n − h + 2 ≥ . Below assume that n ≤ h −
1. Note that the function ( x − h − | A | /x − x isincreasing on (0 , p (2 h + 1) | A | ] and is decreasing on [ p (2 h + 1) | A | , + ∞ ). Since m |A | ≥ | A | ,( n − h − |A | − n ≥ ( n − h − · | A | m − n ≥ ( n − h − · | A | n − n ≥ min (cid:26) | A | h + 2 − (2 h + 2) , (3 h − | A | h − − (5 h − (cid:27) . (i) | S | ≥ h + 1.Suppose that m ≥ h + 2. Since | A | ≥ | S | − > h ( h + 2) , we have | A | h + 2 − (2 h + 2) ≥ h − (2 h −
2) = h − . And it is easy to check(3 h − | A | h − − (5 h − ≥ (3 h − · h ( h + 2)5 h − − h + 1 ≥ h − h ≥
1. Hence we have | A S + B | ≥ | A | + | B | − | S | − | S | > h .Suppose that m ≤ h + 1 and m < n . Then( n − m ) |A | − n ≥ n − mm · | A | − n ≥ | A | m − m − ≥ | A | h + 1 − (2 h + 1) − ≥ h ( h + 2)2 h + 1 − h − ≥ h − . From (3.4), we also can get | A S + B | ≥ | A | + | B | − | S | − N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 17
Suppose that m = n ≤ h + 1. Since | ¯ A + ¯ B | ≥ m + 1, there exist 1 ≤ γ, η ≤ h such that ¯ a γ +¯ b η
6∈ { ¯ a +¯ b , . . . , ¯ a +¯ b m } . Let T = ( b η − a γ ) + s i + S i if ¯ a γ − ¯ b η = ¯ s i ,and let T = ∅ if ¯ a γ − ¯ b η ¯ S . Clearly |T | ≤ max ≤ i ≤ h |S i | ≤ | S | − h + 1 . Then | A S + B | ≥ r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν | + |A γ T + B η |≥ m |A | + | B | − | S | − h − n + ( |A γ | − |T | ) ≥| A | + | B | − | S | − |A | − | S | − h ) . Note that | A | ≥ | S | ( h + 1) − | S | (2 h + 1) + 3 | S | − ≥ ( | S | + 2 h ) · (2 h + 1) . We obtain that |A | ≥ | A | m ≥ | A | h + 1 ≥ | S | + 2 h, i.e., | A S + B | ≥ | A | + | B | − | S | − | S | = h .Now |S i | = 1 for each 1 ≤ i ≤ h . We shall use another way to give the lowerbound of | A S + B | Since h ≥ | B | ≥ h ( h − | A S + B | ≥ n ( |A | −
1) + | B | + ( m − h ) − h − | S |≥ n + m + | B | − h ≥ m + n + h. (3.5)So we may assume that m + n − ≤ p . According to Lemma 3.2, assume that¯ a + ¯ b , . . . , ¯ a + ¯ b n , ¯ a + ¯ b υ , . . . , ¯ a m + ¯ b υ m are distinct elements of ¯ A + ¯ B . For2 ≤ j ≤ m , let T j = ( ( b υ j − a j ) + s i + S i , if ¯ a j − ¯ b υ j = ¯ s i for some 1 ≤ i ≤ h, ∅ , if ¯ a j − ¯ b υ j ¯ S. Then | A S + B | ≥ r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν | + m X j =2 |A j T j + B υ j |≥ r X k =1 ( |A | + |B µ k | −
3) + X ≤ ν ≤ mν µ ,...,µ r } ( |A | + |B ν | −
1) + m X j =2 ( |A j | − ≥| A | + | B | + ( n − |A | − h − n − ( m − ≥| A | + | B | − | S | − (cid:0) ( n − |A | − n − h + 3 (cid:1) . (3.6)Recalling that 2 ≤ n ≤ h −
1, we have( n − |A | − n ≥ ( n − · (cid:24) | A | n (cid:25) − n ≥ min (cid:26) (2 − | A | − · , (5 h − · (cid:24) | A | h − (cid:25) − h − (cid:27) , where ⌈ x ⌉ denotes the least integer not less than x . It is easy to verify that | A | − ≥ (3 h − − ≥ h − h ≥
2, and(5 h − | A | h − − h − ≥ (5 h − · (6 h − h − − h − ≥ h − h ≥
2. When h = 2, we also have(5 h − · (cid:24) | A | h − (cid:25) − h −
1) = 8 · (cid:24) (cid:25) −
18 = 6 > − . So by (3.6), we get the desired result.(II) |A | = 1.Suppose that |B | ≥
2. Since |A | = · · · |A m | = 1, we have m = | A | ≥ | S | − ≥ h. Assume that 1 ≤ ˆ µ , . . . , ˆ µ ˆ r are all integers such that ¯ a ˆ µ k − ¯ b = ¯ s ˆ λ k ∈ ¯ S for1 ≤ k ≤ ˆ r . And assume that ¯ a ˆ γ + ¯ b ˆ η , . . . , ¯ a ˆ γ n − h − + ¯ b ˆ η n − h − are distinct elementsof ( ¯ A ¯ S + ¯ B ) \ { ¯ a + ¯ b , . . . , ¯ a m + ¯ b } with ˆ η j ≤ j + 2 h + 1 for 1 ≤ j ≤ n − h − N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 19
Letting S ∗ ˆ λ k = ( b − a ˆ µ k ) + s ˆ λ k + S ˆ λ k , we have | A S + B | ≥ ˆ r X k =1 |A ˆ µ k S ∗ ˆ λk + B | + X ≤ ˆ ν ≤ m ˆ ν ˆ µ ,..., ˆ µ ˆ r } |A ˆ ν + B | + n − h − X j =1 |A ˆ γ j + B ˆ η j |≥ ˆ r X k =1 ( |B | − |S ∗ ˆ λ k | ) + X ≤ ˆ ν ≤ m ˆ ν ˆ µ ,..., ˆ µ ˆ r } |B | + n − h − X j =1 |B ˆ η j |≥ (2 h + 1) |B | + n − h − X j =1 |B ˆ η j | − | S | ≥ | B | + ( m − h − |B | − | S |≥| B | + | A | + 2( m − h − − | S | − m ≥ | A | + | B | − | S | − . (3.7)Finally, assume that |B | = 1. Then by the induction hypothesis, | A S + B | ≥ | ¯ A ¯ S + ¯ B | ≥ | ¯ A | + | ¯ B | − | ¯ S | − ≥ | A | + | B | − | S | − . (cid:3) Proof of Theorem 1.2
By Proposition 3.1, there is nothing to do if G = Z p α . Suppose that G is nota cyclic group of prime power order. The case | S | = 1 easily follows from (1.6).Note that | A S + B | ≥ | A | − | S | ≥ | S | − | S | − | S | + 1)( | S | − ≥ | S | whenever | S | ≥
2. We always assume that 2 ≤ | S | < p ( G ).We use an induction on | G | . Assume that Theorem 1.2 holds for any abeliangroup whose order is less than | G | . For a subgroup H of G , define X A,H = { a + H ∈ G/H : A ∩ ( a + H ) = ∅} . We claim that | X A,H | ≥ p | A | for some subgroup H ⊆ G of prime order. Since G is not a cyclic group of primepower order, we may write G = K ⊕ K where | K | , | K | >
1. Assume that | X A,K | < p | A | . By the pigeonhole principle, there exists a coset a + K suchthat A ∩ ( a + K ) > √ A . Assume that a + K = { b , . . . , b k } . It is easy tosee that b + K , . . . , b k + K are distinct cosets of K . We get | X A,K | > p | A | .So max {| X A,K | , | X A,K |} ≥ p | A | . Assume that | X A,K | ≥ p | A | and let H be asubgroup of K with | H | is prime. Clearly we also have | X A,H | ≥ p | A | . Sincemin {| A | , | B |} ≥ | S | − | S | − , we may assume that.max {| X A,H | , | X B,H |} ≥ p | S | − | S | − . Assume that¯ A = { ¯ a , . . . , ¯ a m } , ¯ B = { ¯ b , . . . , ¯ b n } , ¯ S = { ¯ s , . . . , ¯ s h } , where ¯ a = a + H and ¯ A = X A,H . Further, without loss of generality, assume that n ≥ m . According to our choice of H , we know that n ≥ p | S | − | S | −
3. Write A = m [ i =1 ( a i + A i ) , B = n [ i =1 ( b i + B i ) , S = h [ i =1 ( s i + S i ) , where |A | ≥ |A | ≥ · · · ≥ |A m | , |B | ≥ |B | ≥ · · · ≥ |B n | . If m = n = 1, then for some 1 ≤ i ≤ h , | A S + B | ≥ |A | + |B | − |S i | − ≥ | A | + | B | − | S | − . So below assume that either m ≥ m < n .(I) |A | ≥ | A S + B | ≥ r X k =1 |A S ∗ λk + B µ k | + X ≤ ν ≤ mν µ ,...,µ r } |A + B ν | + τ X j =1 |A γ j + B η j |≥ r X k =1 ( |A | + |B µ k | − |S λ k | −
2) + X ≤ ν ≤ mν µ ,...,µ r } ( |A | + |B ν | −
1) + τ X j =1 |A γ j |≥ n |A | + | B | + τ X j =1 |A γ j | − r X k =1 ( |S k | + 1) − n, (4.1)where we have assumed that those |A S ∗ λk + B µ k | , |A + B ν | , |A γ j + B η j | < p ( G ) . Suppose that m ≥ h + 1. Then | A S + B | ≥ | A | + | B | − | S | − (cid:0) ( n − h ) |A | − n − h + 2 (cid:1) , (4.2)since τ ≥ m − h now. And if m ≤ h , we also have | A S + B | ≥| A | + | B | − | S | − (cid:0) ( n − m ) |A | − n − h + 2 (cid:1) . (4.3) N THE GENERALIZED RESTRICTED SUMSETS IN ABELIAN GROUPS 21 If n ≥ h −
2, then | A S + B | ≥ | A | + | B | + 2( n − h ) − | S | − h − n ≥ | A | + | B | − | S | − . Below assume that n ≤ h − | S | ≥ h + 1.By (4.2), we have | A S + B | ≥ | A | + | B | − | S | − (cid:18) n − hn · | A | − n − h + 2 (cid:19) . Clearly the function ( x − h ) | A | /x + x is increasing on (0 , p h | A | ] and is decreasingon [ p h | A | , + ∞ ). Note that | A | ≥ | S | − | S | − ≥ h + 13 h + 1 . It is easy to check that p h | A | ≥ p h (9 h + 13 h + 1) ≥ max {√ h + 13 h + 1 , h − } . If h ≥
2, then √ h + 13 h + 1 ≤ h −
3. So n − hn · | A | − n ≥ ( √ h + 13 h + 1 − h ) · | A |√ h + 13 h + 1 − √ h + 13 h + 1 ≥ ( √ h + 13 h + 1 − h ) · √ h + 13 h + 1 − √ h + 13 h + 1= h − (cid:0) h + 12 h + 3 − (3 h + 1) √ h + 13 h + 1 (cid:1) ≥ h − , where the last step follows from9 h + 12 h + 33 h + 1 = 3 h + 3 > √ h + 13 h + 1 . And if h = 1, then n − hn · | A | − n ≥ (7 h − · | A | h − − (7 h − ≥ · − > h − . Thus we get (1.10).(ii) | S | = h .By (3.5), we may assume that m + n − ≤ p ( G ). In view of (3.6), we get | A S + B | ≥| A | + | B | + ( n − |A | − h − n − ( m − ≥| A | + | B | − | S | − (cid:18) ( n − · (cid:24) | A | n (cid:25) − n − h + 3 (cid:19) . For 2 ≤ n ≤ h −
3, we have( n − · (cid:24) | A | n (cid:25) − n ≥ min (cid:26)(cid:24) | A | (cid:25) − , (7 h − · (cid:24) | A | h − (cid:25) − (7 h − (cid:27) . Since | A | ≥ h − h −
3, it is not difficult to verify that | A | / − ≥ h − h ≥
2, and (7 h − | A | h − − (7 h − ≥ h − h ≥
3. And if h = 2, then(7 h − · (cid:24) | A | h − (cid:25) − (7 h − ≥ · (cid:24) (cid:25) −
11 = 9 ≥ h − . So we have | A S + B | ≥ | A | + | B | − | S | − |A | = 1.If |A | = |B | = 1, then our assertion immediately follows from the inductionhypothesis on G/H . Suppose that |A | = 1 and |B | ≥
2. Then m = | A | ≥ | S | − | S | − ≥ h − . Assume that ¯ a ˆ µ k − ¯ b = ¯ s ˆ λ k ∈ ¯ S for 1 ≤ k ≤ ˆ r , and ¯ a ˆ γ + ¯ b ˆ η , . . . , ¯ a ˆ γ n − h + ¯ b ˆ η n − h are distinct elements of ( ¯ A ¯ S + ¯ B ) \ { ¯ A + ¯ b } with ˆ η j ≤ j + 3 h for 1 ≤ j ≤ n − h .Let S ∗ ˆ λ k = ( b − a ˆ µ k ) + s ˆ λ k + S ˆ λ k . We obtain that | A S + B | ≥ ˆ r X k =1 |A ˆ µ k S ∗ ˆ λk + B | + X ≤ ˆ ν ≤ m ˆ ν ˆ µ ,..., ˆ µ ˆ r } |A ˆ ν + B | + n − h X j =1 |A ˆ γ j + B ˆ η j |≥ ˆ r X k =1 ( |B | − |S ∗ ˆ λ k | ) + X ≤ ˆ ν ≤ m ˆ ν ˆ µ ,..., ˆ µ ˆ r } |B | + n − h X j =1 |B ˆ η j |≥ h |B | + n − h X j =1 |B ˆ η j | − | S | ≥ | B | + ( m − h ) |B | − | S |≥| B | + | A | + 2( m − h ) − | S | − m ≥ | A | + | B | − | S | − . (4.4) References [1] N. Alon,
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