aa r X i v : . [ m a t h . N T ] M a y Asymptotically tight bounds on subset sums
Simon GriffithsOctober 25, 2018
Abstract
For a subset A of a finite abelian group G we define Σ( A ) = { P a ∈ B a : B ⊂ A } . In the case that Σ( A ) has trivial stabiliser,one may deduce that the size of Σ( A ) is at least quadratic in | A | ;the bound | Σ( A ) | ≥ | A | /
64 has recently been obtained by De Vos,Goddyn, Mohar and ˇS´amal [2]. We improve this bound to the asymp-totically best possible result | Σ( A ) | ≥ (1 / − o (1)) | A | .We also study a related problem in which A is any subset of Z n with all elements of A coprime to n ; it has recently been shown,by Vu [6], that if such a set A has the property Σ( A ) = Z n then | A | = O ( √ n ). This bound was improved to | A | ≤ √ n by De Vos,Goddyn, Mohar and ˇS´amal [2], we further improve the bound to theasymptotically best possible result | A | ≤ (2 + o (1)) √ n . For a subset A of a finite abelian group G we define,Σ( A ) = n X a ∈ B a : B ⊂ A o the set of all elements which may be expressed as a sum of elements of A (with repetition not allowed). For a subset S ⊂ G the stabiliser Stab( S )of S is the set of elements g ∈ G such that S + g = S ; the stabiliser is asubgroup of G . We say that S has trivial stabiliser if Stab( S ) = { } . Arecent result of De Vos, Goddyn, Mohar and ˇS´amal [2] shows that if Σ( A )has trivial stabiliser then its size is at least quadratic in the size of A .1 heorem 1.1. [2] Let A ⊂ G \ { } , and suppose that Σ( A ) has trivialstabiliser, then | Σ( A ) | ≥ | A | / /
64 replacedby 1 / − o (1); where o (1) denotes a function which converges to 0 as | A | → ∞ . Our aim is to improve this further by showing the result holdswith the constant replaced by 1 / − o (1). This result is asymptotically bestpossible as seen by considering A = {− n, − ( n − , ..., n − , n } ⊂ Z N , with N large. Theorem 1.2.
Let A ⊂ G \ { } , and suppose Σ( A ) has trivial stabiliser,then | Σ( A ) | ≥ (1 / − o (1)) | A | We follow [2] in deducing related results for the case A has non-trivialstabiliser. Theorem 1.3.
Let A ⊂ G , and let H be the stabiliser of Σ( A ) then, | Σ( A ) | ≥ (1 / − o (1)) | A \ H | where o (1) denotes a function which converges to as | A \ H | / | H | → ∞ .Remark. As the o (1) term of the above statement converges to zero as | A \ H | / | H | → ∞ , rather than as | A \ H | → ∞ , our result is only animprovement on the result of [2] in the case that | A \ H | is large relative to | H | .Erd˝os and Heilbronn [3] proved that if A ⊂ Z p , the integers modulo p (with p prime), and | A | ≥ √ p then Σ( A ) = Z p – the connection ofthis result to the current discussion is that this result is proved by givinga quadratic lower bound on | Σ( A ) | for A ⊂ Z p . They conjectured that theconstant 3 √ Z n , for n composite, one must put extra conditionson the set A . The following theorem of Vu [6] shows one way in which thiscan be done is to demand that A is contained in Z ∗ n , the elements coprimeto n . Theorem 1.4. (Vu [6]) There is a constant c such that every subset A ⊂ Z ∗ n ⊂ Z n with | A | ≥ c √ n satisfies Σ( A ) = Z n c = 8. We improve this further, by replacingthe constant by 2 + o (1). Theorem 1.5.
Let A ⊂ Z ∗ n ⊂ Z n be such that Σ( A ) = Z n , then | A | ≤ (2 + o (1)) √ n Remark.
This result is asymptotically best possible, consider for examplethe case n = p , where p is a large prime, and consider A = {− ( p − , ..., − , , ..., p − } . All elements of A are coprime to n , and the element p ( p − / A ) so that Σ( A ) = Z n , while | A | = 2 p − − o (1)) √ n .The results Theorem 1.2 and Theorem 1.3 will be deduced from Theorem1.6 which we state below. Theorem 1.6 will state that if A ⊂ G \ { } is such that Σ( A ) has trivial stabiliser, and has the extra property that A ∩ ( − A ) = φ then Σ( A ) ≥ (1 / − o (1)) | A | . We now state this resultformally.Let n = 1, and for k ≥
10 let n k = 2 k . We define also a sequence ofreal numbers α k as follows: α = 1 /
64 and for k ≥
10 we define α k = min n α k − , − k − o It is clear that the sequence α k is increasing and converges to 1 / k → ∞ . Theorem 1.6.
Let A ⊂ G \ { } and let k ≥ . Suppose that A ∩ ( − A ) = φ ,that Σ( A ) has trivial stabiliser and that | A | ≥ n k , then | Σ( A ) | ≥ α k | A | The layout of the article is as follows. We shall conclude the introductionby introducing definitions and observations that will be used throughout thearticle. In Section 2 we assume Theorem 1.6 and deduce Theorems 1.2 and1.3. In Section 3 we establish preliminary lemmas required for the proofof Theorem 1.6, we then prove Theorem 1.6 in Section 4. We then turn toTheorem 1.5, which is proved in Section 5.Given a set S ⊂ G and an element c ∈ G we write S + c for the set { s + c : s ∈ S } . We define λ S ( c ), or simply λ ( c ), to be the number ofelements in S + c which are not in S , ie. λ ( c ) = | ( S + c ) \ S | . The followingproperties of λ are elementary. 3i) λ (0) = 0(ii) λ ( − c ) = λ ( c ) for all c ∈ G (iii) λ ( b + c ) ≤ λ ( b ) + λ ( c ) for all b, c ∈ G (subadditivity of λ )For sets A , ..., A r ⊂ G , we define their sumset,Σ ri =1 A i = { a + ... + a r : a i ∈ A i for i = 1 , ..., r } We shall also use the notation rA to denote P ri =1 A = { a + ... + a r : a i ∈ A } .The key sumset inequality we shall use is Kneser’s addition theorem. Theorem 1.7. (Kneser [4]) Let A , ..., A r ⊂ G and let H be the stabiliserof P ri =1 A i , then, (cid:12)(cid:12) r X i =1 A i (cid:12)(cid:12) ≥ r X i =1 | A i | − ( r − | H | Let B ⊂ A ⊂ G . We may express Σ( A ) as Σ( A ) = Σ( B ) + Σ( A \ B ). Itis then easily observed that,Stab(Σ( B )) ⊂ Stab(Σ( A ))In particular if Σ( A ) has trivial stabiliser then so does Σ( B ). These obser-vations shall be used throughout. In this section we assume Theorem 1.6 and prove Theorem 1.2 and Theorem1.3. Let ( n k ) k ≥ and ( α k ) k ≥ be the sequences defined above, for a naturalnumber n let k ( n ) be the largest integer k such that n ≥ n k . We define, f ( n ) = α k ( n ) / − /n It is clear that f ( n ) satisfies f ( n ) → / n → ∞ , ie. f is of the form1 / − o (1). Proof of Theorem 1.2.
Let A ⊂ G \ { } be such that Σ( A ) has trivial sta-biliser, let n = | A | . We prove Theorem 1.2 by showing | Σ( A ) | ≥ f ( n ) n .4et m = ⌊ n/ ⌋ , it is possible to partition A into two subsets A , A withcardinalities m and n − m respectively, with the property A i ∩ ( − A i ) = φ for i = 1 ,
2. Let k = k ( n ), by the definition of k ( n ) we have that m ≥ n k , andso applying Theorem 1.6 to A , A individually we obtain, | Σ( A ) | ≥ α k m and | Σ( A ) | ≥ α k ( n − m ) . We have that, Σ( A ) = Σ( A )+Σ( A ), and so ap-plying Kneser’s theorem (and using the fact that Σ( A ) has trivial stabiliser)we obtain, | Σ( A ) | ≥ | Σ( A ) | + | Σ( A ) | − ≥ α k m + α k ( n − m ) − m + ( n − m ) ≥ n / | Σ( A ) | ≥ α k n − f ( n ) n Let A ⊂ G \ { } , and suppose Σ( A ) has trivial stabiliser, Theorem 1.2gives that | Σ( A ) | ≥ f ( | A | ) | A | for some function f ( n ) → / n → ∞ .Of course by setting f ′ ( n ) = f ( n ) − /n , we have | Σ( A ) | ≥ f ′ ( | A | ) | A | (1)for a function f ′ ( n ) → / n → ∞ . Let m ( n ) = ⌊√ n ⌋ , we define, f ( n ) = (1 − m ( n ) − ) f ′ ( m ( n ))It is clear that f ( n ) → / n → ∞ , ie. f ( n ) is of the form 1 / − o (1) Proof of Theorem 1.3.
Let m ( n ) and f ( n ) be as defined as above. Let A ⊂ G , let H be the stabiliser of Σ( A ), and let n = | A \ H | / | H | . We proveTheorem 1.3 by demonstrating that, | Σ( A ) | ≥ f ( n ) | A \ H | We work in
G/H , an element Q of G/H is a coset of H . We let h = | H | and define a sequence A , ..., A h of subsets of G/H by A i = { Q ∈ G/H : | A ∩ Q | ≥ i } Note that the coset Q appears in exactly | A ∩ Q | of the sets A i , this implies, h X i =1 | A i | = | A | A ′ i for A i \ { H } we have, h X i =1 | A ′ i | = | A \ H | The key observation is that Σ( A ) consists exactly of those elements whichlie in cosets in Σ( A ′ ) + ... + Σ( A ′ h ). So that, | Σ( A ) | = h | Σ( A ′ ) + ... + Σ( A ′ h ) | We now put a lower bound on | Σ( A ′ ) + ... + Σ( A ′ h ) | . The sets A ′ , ..., A ′ h are decreasing in size, let j be maximal such that | A ′ j | ≥ m ( n ). Now P i>j | A ′ i | < hm ( n ) so that we have,Σ ji =1 | A ′ i | ≥ | A \ H | − hm ( n ) ≥ h ( n − m ( n )) ≥ hn (1 − m ( n ) − )For all i ≤ j we have | A ′ i | ≥ m ( n ) so that from (1), and the fact that f ′ isincreasing, we have, | Σ( A ′ i ) | ≥ f ′ ( | A ′ i | ) | A ′ i | ≥ f ′ ( m ( n )) | A ′ i | Since H is the stabiliser of Σ( A ) we must have that Σ( A ′ ) + ... + Σ( A ′ j ) hastrivial stabiliser in G/H , and so by Kneser’s theorem, | Σ( A ′ )+ ... +Σ( A ′ j ) | ≥| Σ( A ′ ) | + ... + | Σ( A ′ j ) | − ( j − | Σ( A ′ ) + ... + Σ( A ′ j ) | ≥ j X i =1 (cid:0) f ′ ( m ( n )) | A ′ i | (cid:1) − ( j − ≥ f ′ ( m ( n )) j X i =1 | A ′ i | We now apply Cauchy-Schwarz to obtain, | Σ( A ′ ) + ... + Σ( A ′ j ) | ≥ f ′ ( m ( n )) j (cid:16) j X i =1 | A ′ i | (cid:17) ≥ f ′ ( m ( n )) h (cid:0) hn (1 − m ( n ) − ) (cid:1) and so, | Σ( A ′ ) + ... + Σ( A ′ j ) | ≥ (1 − m ( n ) − ) f ′ ( m ( n )) hn We now simply note that | Σ( A ′ ) + ... + Σ( A ′ h ) | ≥ | Σ( A ′ ) + ... + Σ( A ′ j ) | andrecall that, | Σ( A ) | = h | Σ( A ′ ) + ... + Σ( A ′ h ) | to obtain, | Σ( A ) | ≥ (1 − m ( n ) − ) f ′ ( m ( n )) h n = f ( n ) | A \ H | Some preliminary lemmas
In this section we prove the key lemmas which are central to our proof ofTheorem 1.6 in Section 4. That proof will work by building up a set B ⊂ A with | Σ( B ) | large. During this process we shall inspect the current choice of B and let S = Σ( B ), we shall then attempt to find an element c ∈ C = A \ B with λ S ( c ) relatively large, and then add c to B , this leads to set with sizeone more than | B | , with a significantly larger set of sums. We shall recallresults proved previously (Lemma 3.1 [3] and Lemma 3.2 [2]) which enableone to put a lower bound on max c ∈ C λ ( c ). More importantly we strengthenLemma 3.2 to Lemma 3.4, a result which is best possible, up to an errorterm; furthermore for our applications of the lemma the error term is smallasymptotically.For a subset Q ⊂ G , we define def Q ( S ) = min {| S ∩ Q | , | Q \ S |} . Let ρ ( d ) be the number of representations of d as a difference of two elementsof S , ie. ρ ( d ) = |{ ( x, y ) ∈ S : x − y = d }| = | S ∩ ( S + d ) | . The equality | ( S + d ) ∩ S | + | ( S + d ) \ S | = | S | implies that λ ( d ) = | S | − ρ ( d ). A key factwill be that, X d ∈ G ρ ( d ) = | S | this follows from the fact that each pair of elements ( x, y ) of S have a uniquedifference x − y , so is counted exactly once by the sum. In the followinglemmas, H denotes a finite abelian group. The following two lemmas wereoriginally proved in [3] and [2] respectively, we give proofs so that the readermay become familiar with ideas which we shall use in the more involvedproof of Lemma 3.4. Lemma 3.1. [3] Let
C, S ⊂ H be such that def H ( S ) ≤ | C | / , then | C | X c ∈ C λ ( c ) ≥ def H ( S )2 In particular λ ( c ) ≥ def H ( S ) / for some c ∈ C .Proof. Exchanging S for H \ S if required we may assume def H ( S ) = | S | ≤| H | /
2, this is allowed as λ S ( c ) = λ H \ S ( c ) for all c ∈ C . From the facts λ ( c ) = | S | − ρ ( c ) and P d ∈ H ρ ( d ) = | S | we deduce that, X c ∈ C λ ( c ) = | C || S | − X c ∈ C ρ ( c ) ≥ | C || S | − | S |
7e have also that | C | ≥ H ( S ) = 2 | S | so that | C | − | S | ≥ | C | /
2, itfollows that, X c ∈ C λ ( c ) ≥ | S | ( | C | − | S | ) ≥ | C || S | | C | def H ( S )2 Lemma 3.2. [2] Let
C, S ⊂ H be such that H = h C i and def H ( S ) ≥ | C | / then there exists c ∈ C with λ ( c ) ≥ | C | / Sketch proof.
Again we may assume def H ( S ) = | S | ≤ | H | /
2. We cannotproceed as in the previous proof, because we obtain no information in thecase | C | ≤ | S | . Instead we consider not only the elements of C but allelements which can be expressed as a sum of a small number of elements of C , we then obtain the required result from the subadditivity of λ . Supposefor contradiction that λ ( c ) < | C | / c ∈ C , and suppose d may beexpressed as a sum of k elements of C , then by the subadditivity of λ wehave λ ( d ) < k | C | /
8. Using the notation rA for the set { a + ... + a r : a i ∈ A } ,we have that the set of elements that may be expressed as a sum of at most r elements of C is the set rC ∗ where C ∗ = C ∪ { } . We may assume forall d ∈ rC ∗ that λ ( d ) < r | C | / ρ ( d ) > | S | − r | C | /
8. An applicationof Kneser (see [2] for details) now shows that for r = ⌊ | S | / | C |⌋ we have | rC ∗ | ≥ | S | . For this value of r we have ρ ( d ) > | S | − r | C | / ≥ | S | −| S | / | S | / d ∈ rC ∗ , so that, X d ∈ rC ∗ ρ ( d ) > | rC ∗ | | S | ≥ | S | This contradicts the equality P d ∈ H ρ ( d ) = | S | , and the result is proved.With greater care, and with extra conditions, the result of Lemma 3.2can be greatly improved; one can find an element c ∈ C with λ ( c ) almost | C | /
2. The extra conditions we need are that C ∩ ( − C ) = φ , that | C | islarge, and that def H ( S ) / | C | is large. We will again use the subadditivity of λ ; in our sketch proof of Lemma 3.2 we used that if λ ( c ) < η then for each r we have λ ( d ) < rη for all d ∈ rC ∗ and so ρ ( d ) > | S | − rη for all d ∈ rC ∗ . Toobtain an the improved result we must use this statement simultaneouslyfor many values of r rather than once for a single value of r . Again it is8mportant to have a lower bound on | rC ∗ | (although in the sequel we take adifferent set C ∗ ), we first prove Lemma 3.3 which gives an improved lowerbound on | rC ∗ | , we will then prove Lemma 3.4 which gives a new bound on λ ( c ). We define C ∗ = C ∪ ( − C ) ∪ { } . Lemma 3.3.
Let C ⊂ H be such that H = h C i , C ∩ ( − C ) = φ , Σ( C ) hastrivial stabiliser and | C | ≥ k +11 then for all positive integers r we havethat either rC ∗ = H or | rC ∗ | ≥ (cid:16) − k +2 (cid:17) r | C | Proof.
Let K be the stabiliser of rC ∗ . If C ⊂ K then K = H and rC ∗ = H and we are done. So we may assume C ∗ meets at least one non-trivial cosetof K , note that C ∗ also meets K as 0 ∈ C ∗ ∩ K . Let Q , ..., Q p be the non-trivial cosets of K meeting C ∗ , it follow that rC ∗ is the set of elements of thecosets in r { K, Q , ..., Q p } . By the definition of K the set r { K, Q , ..., Q p } has trivial stabiliser in H/K so that an application of Kneser’s theoremyields, | r { K, Q , ..., Q p }| ≥ r ( p + 1) − ( r − ≥ rp and so | rC ∗ | ≥ rp | K | . So to complete the proof we must show that, | K | ≥ (cid:16) − k +2 (cid:17) | C | p In the case that | C ∩ K | ≥ | C | / k +2 we use the fact that K contains Σ( C ∩ K )together with Theorem 1.1 (which we may apply to C ∩ K , as Σ( C ∩ K )has trivial stabiliser) to deduce, | K | ≥ | Σ( C ∩ K ) | ≥ | C ∩ K | ≥ | C | k +4 ≥ | C | the final inequality follows from | C | ≥ k +11 , and we are done. We maynow assume that | C ∩ K | ≤ | C | / k +2 , this implies | C \ K | ≥ (1 − − ( k +2) ) | C | and likewise | C ′ \ K | ≥ (1 − − ( k +2) ) | C ′ | , where C ′ denotes C ∪ ( − C ). Alter-natively | C ′ ∩ S pi =1 Q i | ≥ (1 − − ( k +2) ) | C ′ | , so by the pigeon hole principlesome coset Q must have, | C ′ ∩ Q | ≥ (cid:16) − k +2 (cid:17) | C ′ | p = 2 (cid:16) − k +2 (cid:17) | C | p and so noting | K | = | Q | ≥ | C ′ ∩ Q | we are done.9e now deduce the required strengthening of Lemma 3.2, Lemma 3.4.
Let C ⊂ H be such that H = h C i , C ∩ ( − C ) = φ , Σ( C ) hastrivial stabiliser and | C | ≥ k +11 , and let S ⊂ H be such that def H ( S ) ≥ k +1 | C | then there exists c ∈ C with λ ( c ) ≥ (1 − − k ) | C | Proof.
For all d ∈ H we have λ S ( d ) = λ H \ S ( d ), so that exchanging S with H \ S if required, we may assume def H ( S ) = | S | ≤ | H | /
2. As in previousproofs we shall proceed by assuming the result fails and from this deducethat P d ρ ( d ) > | S | , a contradiction. On this occasion we must be muchmore precise in our lower bound on P d ρ ( d ). It will be useful to use thefact that P d ρ ( d ) = R | S | | D t | dt where D t = { d ∈ H : ρ ( d ) ≥ t } . Suppose that every element c ∈ C has λ ( c ) < (1 − − k ) | C | , this implies λ ( c ) < (1 − − k ) | C | for all c ∈ C ∗ , since λ (0) = 0 and λ ( − c ) = λ ( c ).So that by the subadditivity of λ we have for all r and for all d ∈ rC ∗ that λ ( d ) < r (1 − − k ) | C | , and so ρ ( d ) > | S | − r (1 − − k ) | C | . For each r = 1 , ..., ⌊| S | / | C |⌋ , we define, I r = ( | S | − ( r + 1)(1 − − k ) | C | , | S | − r (1 − − k ) | C | ]For t ∈ I r we have t ≤ | S | − r (1 − − k ) | C | and so D t ⊃ rC ∗ , applyingLemma 3.3 we deduce, | D t | ≥ − − ( k +2) ) r | C | .Note also that for t ≤ | S | − ( ⌊| S | / | C |⌋ + 1)(1 − − k ) | C | we have D t ⊃⌊| S | / | C |⌋ C ∗ and so | D t | ≥ − − ( k +2) ) ⌊| S | / | C |⌋| C | . Using the fact that ⌊| S | / | C |⌋ ≥ ( | S | − | C | ) / | C | ≥ (1 − − ( k +1) ) | S | / | C | we have, | D t | ≥ (cid:16) − k +2 (cid:17)j | S || C | k | C | ≥ (cid:16) − k +2 (cid:17)(cid:16) − k +1 (cid:17) | S | for all t ≤ | S | − ( ⌊| S | / | C |⌋ + 1)(1 − − k ) | C | . Note that ⌊| S | / | C |⌋ + 1 ≤| S | / | C | + 1 ≤ (1 + 2 − ( k +1) ) | S | / | C | so that the above bound on | D t | hold forall t satisfying t ≤ | S | − (1 − − k )(1 + 2 − ( k +1) ) | S | and so certainly for all t ≤ | S | / k +1 . We obtain, X d ∈ H ρ ( d ) = Z | S | | D t | dt ≥ ⌊| S | / | C |⌋ X r =1 Z t ∈ I r | D t | dt + Z | S | / k +1 | D t | dt We choose this representation, rather than the sum P | S | t =1 | D t | , as it means we donot have to concern ourselves with integer parts, etc. X d ∈ H ρ ( d ) ≥ ⌊| S | / | C |⌋ X r =1 (cid:16) − k − (cid:17)(cid:16) − k +2 (cid:17) r | C | + 3 | S | k +1 (cid:16) − k +2 (cid:17)(cid:16) − k +1 (cid:17) | S | Since ⌊| S | / | C |⌋ ≥ (1 − − ( k +1) ) | S | / | C | we have that P ⌊| S | / | C |⌋ r =1 r = ( ⌊| S | / | C |⌋ )( ⌊| S | / | C |⌋ +1) / ≥ (1 − − ( k +1) ) | S | / | C | , and so, X d ∈ H ρ ( d ) ≥ (cid:16) − k +1 (cid:17)(cid:16) − k − (cid:17)(cid:16) − k +2 (cid:17) | S | + 32 k (cid:16) − k +2 (cid:17)(cid:16) − k +1 (cid:17) | S | However this quantity is larger than | S | , a contradiction. Let us recall the sequences ( n k ) k ≥ and ( α k ) k ≥ which appear in the state-ment of Theorem 1.6. The sequence ( n k ) k ≥ is defined by n = 1, and for k ≥
10 by n k = 2 k . The only information we shall use about this sequenceis that for k ≥
10 it satisfies, n k > k +15 and n k k ≥ n k − (2)We recall the sequence ( α k ) k ≥ defined by, α = 1 /
64 and for k ≥
10 by, α k = min n α k − , − k − o Having proved all the necessary preliminary results we now proceed to-wards our proof of Theorem 1.6. Our proof is by induction on k , the case k = 9 follows from Theorem 1.1, so we turn to the induction step, we let k ≥
10 and suppose the result holds for all smaller values of k . Let A bea set of size n satisfying the conditions of Theorem 1.6, we may assume n ≥ n k , else there is nothing to prove. As mentioned previously we shallbuild up a set B ⊂ A with Σ( B ) large. We define a function g ( t ) for t ∈ {⌊ n/ k ⌋ , ..., ⌈ n − n/ k ⌉} by g ( ⌊ n/ k ⌋ ) = 0 and then inductively by, g ( t + 1) = g ( t ) + (cid:16) − k − (cid:17) ( n − t )We shall deduce Theorem 1.6 from the following lemma.11 emma 4.1. Let A be as above then for all t ∈ {⌊ n/ k ⌋ , ..., ⌈ n − n/ k ⌉} thereis a subset B ⊂ A of cardinality t with either | Σ( B ) | ≥ g ( t ) or | Σ( B ) | ≥ α k n . Theorem 1.6 now follows by taking B of cardinality ⌈ n − n/ k ⌉ witheither | Σ( B ) | ≥ g ( ⌈ n − n/ k ⌉ ) or | Σ( B ) | ≥ α k n , in the latter case we aredone immediately as | Σ( A ) | ≥ | Σ( B ) | . In the former case we note that, g ( ⌈ n − n/ k ⌉ ) ≥ (cid:16) − k − (cid:17) n − n/ k X t = n/ k n − t = (cid:16) − k − (cid:17) n − n/ k X t = n/ k t and we may bound the sum by, n − n/ k X t = n/ k t ≥ (cid:16) − k (cid:17) n − n k +1 ≥ (cid:16) − k (cid:17) n and so, | Σ( A ) | ≥ (cid:16) − k − (cid:17)(cid:16) − k (cid:17) n ≥ (cid:16) − k − (cid:17) n ≥ α k n and Theorem 1.6 is proved.We must now prove Lemma 4.1. We do this by induction on t . Theresult is trivial for t = ⌊ n/ k ⌋ . If we ever find a set B with | Σ( B ) | ≥ α k n then the induction is trivial from that point on. So for the induction step welet B be a set of size t with | Σ( B ) | ≥ g ( t ), and set S = Σ( B ) and C = A \ B ,we will then show either that | S | ≥ α k n or that, λ ( c ) ≥ (cid:16) − k − (cid:17) | C | = (cid:16) − k − (cid:17) ( n − t ) for some c ∈ C (3)the induction step is then completed by considering B ∪ { c } , as B ∪ { c } isthen a set of size t + 1 with Σ( B ∪ { c } ) = S ∪ ( S + c ) and so, | Σ( B ∪ { c } ) | ≥ | S | + λ ( c ) ≥ g ( t ) + (1 − − k )( n − t ) = g ( t + 1)Let us first note a lower bound on | S | which we shall use during theproof, since | B | = t ≥ ⌊ n/ k ⌋ ≥ n/ k +1 , it is immediate from Theorem 1.1that, | S | ≥ | B | ≥ n k +8 ≥ k +7 n (4)12et H = h C i . As in [2], we must analyse the intersection of S withthe cosets of H . We begin with some lower bounds on | H | that we shalluse during the proof. The set C is contained in A so that C ∩ ( − C ) = φ and 0 C , also this implies that Σ( C ) has trivial stabiliser, note also that | C | = n − t ≥ n/ k . An application of Theorem 1.1 yields, | H | ≥ | Σ( C ) | ≥ | C | ≥ n k +6 ≥ k +9 n while an application of the induction hypothesis of Theorem 1.6 yields, | H | ≥ | Σ( C ) | ≥ α k − | C | ≥ α k n k We say that a coset Q of H is sparse if | S ∩ Q | ≤ k +1 n and Q is very sparse if | S ∩ Q | ≤ | C | /
2, while Q is dense if | Q \ S | ≤ k +1 n , we note that if Q isdense then | S ∩ Q | must be fairly large, in particular, | S ∩ Q | ≥ | H | − k +1 n ≥ α k n / k − k +1 n ≥ α k n / k +1 (5)as α k n/ k +2 ≥ k +1 . Lemma 4.2.
If any of the following conditions (i),(ii) or (iii) hold thenthere is an element c ∈ C with λ ( c ) ≥ (1 − − k ) | C | .(i) There is a coset Q which is neither sparse nor dense.(ii) There are at least k +5 cosets that are sparse but not very sparse(iii) No coset Q is dense.Proof. (i) Suppose Q is neither sparse nor dense, then we have def Q ( S ) ≥ k +1 n ≥ k +1 | C | , an application of Lemma 3.4 to a shift of S ∩ Q then showsthat there exists c ∈ C with λ S ∩ Q ( c ) ≥ (1 − − k ) | C | , and we are done, as λ S ( c ) ≥ λ S ∩ Q ( c ).(ii) Suppose that Q , ..., Q k +5 are sparse but not very sparse, for each i = 1 , ..., k +5 let S i = S ∩ Q i and let ¯ S = S k +5 i =1 S i , then, λ ( c ) = λ S ( c ) ≥ λ ¯ S ( c ) = k +5 X i =1 λ S i ( c )13e write λ i ( c ) for λ S i ( c ) and ¯ λ ( c ) for λ ¯ S ( c ), we shall show that,¯ λ ( c ) = k +5 X i =1 λ i ( c ) ≥ | C | for some c ∈ C We first note that | ¯ S | ≥ k +5 | C | / ≥ k +4 | C | . We work as in the proofs ofresults in Section 3. Suppose for contradiction that ¯ λ ( c ) < | C | for all c ∈ C ,then by the subadditivity of ¯ λ (which follows from the subadditivity of the λ i ), we have that ¯ λ ( d ) < k +3 | C | for all d ∈ k +3 ( C ∪{ } ). For each i we let ρ i ( d ) = | ( S i + d ) ∩ S i | , so that, ρ i ( d ) = | S i | − λ i ( d ) and, P d ∈ G λ i ( d ) = | S i | .We let ¯ ρ ( d ) = P i ρ i ( d ), it follows that, ¯ ρ ( d ) = | ¯ S | − ¯ λ ( d ). So that,¯ ρ ( d ) > | ¯ S | − k +3 | C | ≥ | ¯ S | / d ∈ k +3 ( C ∪ { } )the final inequality follows from the bound | ¯ S | ≥ k +4 | C | obtained earlier.We shall reach a contradiction by putting a lower bound on | k +3 ( C ∪{ } ) | , and using this to deduce that P d ∈ G ρ i ( d ) > | S i | for some i , a con-tradiction. We show that | k +3 ( C ∪ { } ) | ≥ k +2 n . Let K be the stabiliserof 2 k +3 ( C ∪ { } ), if C ⊂ K then K = H and | K | = | H | ≥ k +2 n . Hencewe may assume C ∪ { } meets a non-trivial coset of K , let R , ..., R p bethe non-trivial cosets of K which have non-empty intersection with C , thenthe elements of 2 k +3 ( C ∪ { } ) are exactly the elements of the cosets in2 k +3 { K, R , ..., R p } . The definition of K implies that 2 k +3 { K, R , ..., R p } has trivial stabiliser in H/K , so that an application of Kneser’s theoremyields, | k +3 ( C ∪{ } ) | = | K || k +3 { H, R , ..., R p }| ≥ | K | (cid:0) k +3 ( p +1) − (2 k +3 − (cid:1) ≥ k +3 p | K | An application of the pigeon hole principle shows that, | K | ≥ | C | / ( p + 1) ≥| C | / p , it is now immediate that | k +3 ( C ∪ { } ) | ≥ k +2 | C | ≥ k +2 n . Wenow obtain the required contradiction, as we have, X d ∈ k +3 ( C ∪{ } ) ¯ ρ ( d ) > k +2 n | ¯ S | k +1 n | ¯ S | and so, using that ¯ ρ = P i ρ i and | ¯ S | = P i | S i | , we must have for some i that, X d ∈ k +3 ( C ∪{ } ) ρ i ( d ) > k +1 n | S i | ≥ | S i | k +5 ormore cosets which are sparse but not very sparse, then we are done by (ii).Hence we may assume there are at most 2 k +5 cosets which are sparse butnot very sparse, since | S ∩ Q | ≤ k +1 n for all sparse cosets we may assume atmost 2 k +6 elements of S are in cosets that are sparse but not very sparse.By (4) we have | S | ≥ k +7 n so at least 2 k +6 n ≥ n elements of S belongto very sparse cosets. Applying Lemma 3.1 to each very sparse coset, andaveraging appropriately, we deduce that there exists c ∈ C with λ ( c ) ≥ X Q def Q ( S ) = 12 X Q | S ∩ Q | ≥ n where the summation is taken over very sparse cosets Q , and so we aredone.A very simple addition theorem in finite abelian groups is that if A, B ⊂ G satisfy | A | + | B | > | G | then A + B = G . Let D denote the set of densecosets, D = { Q ∈ G/H : Q is dense } ⊂ G/H
Let ¯ D denote the set of all elements of dense cosets,¯ D = [ Q ∈D Q ⊂ G Claim. S is not contained in ¯ D .Proof. We have from Theorem 1.1 that, | Σ( C ) | ≥ | C | / ≥ n / k +6 > k +1 n . Let Q be a dense coset, then | S ∩ Q | ≥ | H | − k +1 n , and so | Σ( C ) | + | S ∩ Q | > | H | so by considering an appropriate shift of S ∩ Q we obtain (fromthe simple addition theorem stated above) that Q ⊂ Σ( C )+( S ∩ Q ) ⊂ Σ( A ).Suppose the Claim is false, then every coset Q meeting S is dense, then itfollows that Σ( A ) is a union of cosets of H and so has non-trivial stabiliser,a contradiction.If either of the conditions (i) or (iii) of Lemma 4.2 hold then we are done,using this together with the above Claim, we note that we may proceed withthe following assumptions, 15 Every coset is either sparse or denseII There is a dense cosetIII S is not contained in ¯ D From this information we shall deduce that | S | ≥ α k n , completing theproof. To prove this it suffices by (5) to show that there are at least 2 k +1 dense cosets, ie. show |D| ≥ k +1 . We begin by recalling a Claim from [2]. Claim. [2] If Q is dense and b ∈ B then one of the cosets Q + b or Q − b is dense.Proof. If b ∈ H then this is trivial. So suppose b H , let S − be the set ofelements in S ∩ Q which may be represented as a sum of elements of B \ { b } ,and let S + be the set of elements in S ∩ Q which may be represented as b plus a sum of elements of B \ { b } . We note that S ∩ ( Q + b ) ⊃ S − + b so that | S ∩ ( Q + b ) | ≥ | S − | and S ∩ ( Q − b ) ⊃ S + − b so that | S ∩ ( Q − b ) | ≥ | S + | .As | S − | + | S + | ≥ | H | − k +1 n > k +1 n we deduce that one of these sets andhence one of the sets S ∩ ( Q + b ) , S ∩ ( Q − b ) has cardinality greater than2 k +1 n the claim is then proved as we may assume all cosets are either sparseor dense.Fix Q ∈ D and let K be a maximal subgroup of G/H for the propertythat Q + K ⊂ D . Let us also define a subgroup ¯ K of G by,¯ K = [ Q ∈ K Q Lemma 4.3. If | B ∩ ¯ K | ≥ | B | / then | S | ≥ α k n Proof.
We know that B is not contained in ¯ K , for if it were then we wouldhave S ⊂ ¯ K ⊂ ¯ D , contradicting III. Let b ∈ B \ ¯ K , we have from the aboveClaim that for each Q ∈ Q + K that either Q + b or Q − b is dense. Thisimplies that there must be at least | K | / Q + K ,so that the total number of dense cosets is at least 3 | K | /
2. Let us nowestimate the size of ¯ K . Since C ⊂ ¯ K and | B ∩ ¯ K | ≥ | B | /
10 we deducethat | A ∩ ¯ K | ≥ | A | /
10. Since A ′ = A ∩ ¯ K is a subset of A , we have thatΣ( A ′ ) has trivial stabiliser, so that by applying the induction hypothesisof Theorem 1.6 to A ′ we obtain that | Σ( A ′ ) | ≥ α k − n / A ′ ) ⊂ ¯ K it follows that | ¯ K | ≥ α k − n / | ¯ K | = | K || H | , it16ollows that | H | ≥ α k − n / | K | . We may assume that | K | ≤ k +1 since we are done if there are at least 2 k +1 dense cosets. Let Q be a densecoset then, | S ∩ Q | ≥ | H | − k +1 n ≥ α k − n | K | − k +1 n ≥ α k − n | K | where the final inequality follows from α k − n/ | K | ≥ n/ k +15 ≥ k .Since there are at least 3 | K | / Q we have that, | S | ≥ | K | α k − n | K | = 6 α k − n ≥ α k n Hence we may assume that, | B \ ¯ K | ≥ | B | ≥ n k +4 ≥ k +3 (6)Suppose that B \ ¯ K meets at least 2 k +2 distinct cosets of H , then it ispossible to find cosets Q , ..., Q k +1 not in ¯ K which meet B and have theproperty that Q i = − Q j for each i, j . From the Claim we have that for each i = 1 , ..., k +1 that one of the cosets Q + Q i or Q − Q i is dense. Sincewe find a different dense coset for each i = 1 , ..., k +1 we find at least 2 k +1 dense cosets and we are done.Hence we may assume that B meets at most 2 k +2 cosets Q of H with Q not in ¯ K . It follows from this, and (6), that for one such coset Q wehave | B ∩ Q | ≥ k +1 . We write order( Q ) for the order of Q in G/H , thefollowing lemma will allow us to deduce that order( Q ) ≥ k +1 . Lemma 4.4.
Let Q be a coset with | B ∩ Q | ≥ order( Q ) and order( Q ) ≤ k +1 , and let R be a dense coset, then R + Q is denseProof. Let p = order( Q ) ≤ k +1 and let b , ..., b p − be p − B ∩ Q . Since R is dense we have | S ∩ R | ≥ | H | − k +1 n . We partition S ∩ R into, S + = (cid:0) ( b + ... + b p − ) + Σ( B \ { b , ..., b p − } ) (cid:1) ∩ R the elements which can be expressed as a sum of elements of B in such away that all the elements b , ..., b p − are included, and S − the elements of17 ∩ R which can be expressed as a sum of elements of B , in which not all ofthe elements b , ..., b p − are used. We note that − ( b + ... + b p − ) ∈ Q so wehave that, S ∩ ( R + Q ) ⊃ S + − ( b + ... + b p − ) and so | S ∩ ( R + Q ) | ≥ | S + | .We will also be able to relate | S ∩ ( R + Q ) | to | S − | . Consider the bipartitegraph with vertex sets S − and S ∩ ( R + Q ) in which a vertex x ∈ S − isjoined to y ∈ S ∩ ( R + Q ) if y − x = b i for some i = 1 , ..., p −
1; the verticesof S − all have positive degree, while the degree of vertices in S ∩ ( R + Q ) iscertainly at most p −
1, it follows that | S ∩ ( R + Q ) | ≥ | S − | / ( p − | S ∩ ( R + Q ) | ≥ | S + | + | S − | /p ≥ ( | S + | + | S − | ) /p ≥ | S ∩ R | /p , recallthat R is dense so that by (5) we have, | S ∩ R | ≥ α k n / k +1 ≥ n / k +7 ,using this together with the fact p ≤ k +1 we obtain, | S ∩ ( R + Q ) | ≥ | S ∩ R | p ≥ n k +9 > k +1 n the final inequality follows from the bound n ≥ n k ≥ k +10 . This shows usthat R + Q is not sparse, since we are assuming that all cosets are eithersparse or dense it follows that R + Q is dense.So if it were the case that order( Q ) ≤ k +1 then the above lemma wouldshow that Q is in the stabiliser of D , but now the subgroup h K ∪ { Q }i contradicts the maximality of K . Hence we may assume order( Q ) ≥ k +1 ,our proof of the theorem is completed with this final lemma. Lemma 4.5.
Suppose there is a coset Q with | B ∩ Q | ≥ k +1 and order( Q ) ≥ k +1 , then there are at least k +1 dense cosets.Proof. We know there is at least one dense coset R . Consider the cosets R − iQ for positive integers i , if they are all dense then we have foundorder( Q ) ≥ k +1 dense cosets and we are done, hence we may assume thereis some non-negative integer i such that R − iQ is dense, but R − ( i + 1) Q is not. Set Q = R − ( i + 1) Q and Q = R − iQ , and for j = 2 , ..., k +1 let Q j = Q + ( j − Q . We show for each j = 2 , ..., k +1 that Q j is dense.Let b , ..., b k +1 be 2 k +1 elements of B ∩ Q . We partition of S ∩ Q into S − = Σ( B \ { b , ..., b k +1 } ), the set of elements which may be expressed asa sum of elements of B without using any of the elements b , ..., b k +1 , and S + the elements of S ∩ Q which can be expressed as a sum of elementsof B , in such a way that at least one of the elements b , ..., b k +1 is used.We can relate the size of S + to the size of S ∩ Q and this will allow us18o bound the size of S + . Consider the bipartite graph with vertex sets S + and S ∩ Q in which a vertex x ∈ S + is joined to y ∈ S ∩ Q if x − y = b i for some i = 1 , ..., k +1 ; the vertices of S + all have positive degree, whilethe degree of vertices in S ∩ Q is certainly at most 2 k +1 , it follows that | S + | ≤ k +1 | S ∩ Q | . However we have that Q is not dense, since we areassuming all cosets are either sparse or dense we may assume Q is sparse,so that | S ∩ Q | ≤ k +1 n , it follows that | S + | ≤ k +2 n . As Q is dense wehave a lower bound on | S ∩ Q | from (5), noting that S ∩ Q = S + ∪ S − weobtain, | S − | ≥ | S ∩ Q | − | S + | ≥ α k n k +1 − k +2 n ≥ n k +7 − k +2 n > k +1 n the final inequality following from the fact n ≥ n k > k +10 . Now foreach i = 2 , ..., k +1 we have S ∩ Q i ⊃ S − + b + ... + b i − and so for each i = 2 , ..., k +1 we have | S ∩ Q i | > k +1 n . This implies that these cosets arenot sparse, so they are dense. In this section we prove Theorem 1.5. The first half of the section is devotedto proving Lemma 5.2, which is the appropriate variant of Lemma 3.4 forthe new setting of A ⊂ Z ∗ n ⊂ Z n . We then turn in the second half of thesection to applying this, and previous lemmas, to prove the required result.As in Section 3, it is important to find lower bounds on the cardinalities ofthe sets rC ∗ , where C ∗ = C ∪ ( − C ) ∪ { } . Lemma 5.1.
Let C ⊂ H be such that H = h{ c }i for each c ∈ C and C ∩ ( − C ) = φ , then for all positive integers r we have that either rC ∗ = H or, | rC ∗ | ≥ r | C | Proof.
Let K be the stabiliser of rC ∗ . If there is an element c in C ∩ K then K = H and rC ∗ = H and we are done. So we may assume C ∗ ∩ K = { } .Let Q , ..., Q p be the non-trivial cosets of K meeting C ∗ , it follow that rC ∗ is the set of all elements of the cosets in r { K, Q , ..., Q p } . By thedefinition of K the set r { K, Q , ..., Q p } has trivial stabiliser in H/K so that19n application of Kneser’s theorem yields, | r { K, Q , ..., Q p }| ≥ r ( p + 1) − ( r − ≥ rp and so | rC ∗ | ≥ rp | K | . So to complete the proof we must show that, | K | ≥ | C | p However we have that the set C ′ = C ∪ ( − C ) has size 2 | C | and is containedin S pi =1 Q i , so that we may deduce from the pigeon hole principle that forsome coset Q we have, | C ′ ∩ Q | ≥ | C ′ | p = 2 | C | p and so noting | K | = | Q | ≥ | C ′ ∩ Q | we are done.We now deduce the key result we need for finding elements c ∈ C withlarge λ ( c ). Lemma 5.2.
Let C ⊂ H be such that H = h{ c }i for each c ∈ C and C ∩ ( − C ) = φ . Let γ > and let S ⊂ H be such that def H ( S ) ≥ γ | C | thenthere exists c ∈ C with λ ( c ) ≥ (1 − γ − ) | C | Proof.
For all d ∈ H we have λ S ( d ) = λ H \ S ( d ), so that exchanging S with H \ S if required, we may assume def H ( S ) = | S | ≤ | H | /
2. As in the proofsof Section 3 we shall proceed by assuming the result fails and from thisdeduce that P d ρ ( d ) > | S | , a contradiction. It will be useful to use thefact that P d ρ ( d ) = R | S | | D t | dt where D t = { d ∈ H : ρ ( d ) ≥ t } .Suppose that every element c ∈ C has λ ( c ) < (1 − γ − ) | C | , this implies λ ( c ) < (1 − γ − ) | C | for all c ∈ C ∗ , since λ (0) = 0 and λ ( − c ) = λ ( c ). Sothat by the subadditivity of λ we have for all r and for all d ∈ rC ∗ that λ ( d ) < r (1 − γ − ) | C | , and so ρ ( d ) > | S | − r (1 − γ − ) | C | .For each r = 1 , ..., ⌊| S | / | C |⌋ , we define, I r = ( | S | − ( r + 1)(1 − γ − ) | C | , | S | − r (1 − γ − ) | C | ]For t ∈ I r we have t ≤ | S | − r (1 − γ − ) | C | and so D t ⊃ rC ∗ , applyingLemma 5.1 we deduce, | D t | ≥ r | C | .20ote also that for t ≤ | S | − ( ⌊| S | / | C |⌋ + 1)(1 − γ − ) | C | we have D t ⊃⌊| S | / | C |⌋ C ∗ and so | D t | ≥ ⌊| S | / | C |⌋| C | , so using the fact that ⌊| S | / | C |⌋ ≥ ( | S | − | C | ) / | C | ≥ (1 − γ − ) | S | / | C | we have, | D t | ≥ j | S || C | k | C | ≥ − γ − ) | S | for all t ≤ | S | − ( ⌊| S | / | C |⌋ + 1)(1 − γ − ) | C | . Note that ⌊| S | / | C |⌋ + 1 ≤| S | / | C | + 1 ≤ (1 + γ − ) | S | / | C | so that the above bound on | D t | hold forall t satisfying t ≤ | S | − (1 − γ − )(1 + γ − ) | S | and so certainly for all t ≤ γ − | S | . We obtain, X d ∈ H ρ ( d ) = Z | S | | D t | dt ≥ ⌊| S | / | C |⌋ X r =1 Z t ∈ I r | D t | dt + Z γ − | S | | D t | dt and so, X d ∈ H ρ ( d ) ≥ ⌊| S | / | C |⌋ X r =1 − γ − ) r | C | + 6 γ − (1 − γ − ) | S | Since ⌊| S | / | C |⌋ ≥ (1 − γ − ) | S | / | C | we have that P ⌊| S | / | C |⌋ r =1 r = ( ⌊| S | / | C |⌋ )( ⌊| S | / | C |⌋ +1) / ≥ (1 − γ − ) | S | / | C | , and so, X d ∈ H ρ ( d ) ≥ (1 − γ − )(1 − γ − ) | S | + 6 γ − (1 − γ − ) | S | However this quantity is larger than | S | , a contradiction. Proof of Theorem 1.5
Since the required result is asymptotic it suffices to prove it for n ≥ n , forsome n . Let n be chosen such that √ n ≥ n / and log / ( n ) ≤ n / for all n ≥ n . For n ≥ n we define f ( n ) = 1 + 15 n − / and f ( n ) =2( f ( n ) + n − / ) = 2 + 30 n − / + 2 n − / . It is clear that f ( n ) is of the form2 + o (1). To prove Theorem 1.5 we show that a set A ⊂ Z ∗ n ⊂ Z n with | A | ≥ f ( n ) √ n must have Σ( A ) = Z n . Let us note that if | A | ≥ f ( n ) √ n =2 f ( n ) √ n + 2 then it is possible to partition A into subsets A and A , each21ith cardinality at least f ( n ) √ n and satisfying A i ∩ ( − A i ) = φ for i = 1 , | Σ( A i ) | > n/ i = 1 , A ) ⊃ Σ( A ) + Σ( A ) and S + T ⊃ Z n whenever S, T ⊂ Z n are such that | S | + | T | > n .During the proof we will rely essentially on the results we have provedwhich give us an element c ∈ C with large value of λ ( c ) we recall now thethree bounds which we shall need for the sequel. We write λ ( S, C ) for themaximum value of λ S ( c ) over elements c ∈ C . λ ( S, C ) ≥ | S | Z n ( S ) ≤ | C | / λ ( S, C ) ≥ | C | Z n ( S ) ≥ | C | / λ ( S, C ) ≥ (1 − n − / ) | C | for def Z n ( S ) ≥ n / | C | These bounds are taken from Lemmas 3.1, 3.2 and 5.2 respectively. It isvalid to apply these lemmas for sets C ⊂ A i , i = 1 ,
2, because this certainlyimplies C ∩ ( − C ) = φ and that C ⊂ Z ∗ n and so Z n is generated by eachelement c ∈ C .Fix i ∈ { , } , we show that | Σ( A i ) | > n/
2. We show this by buildingup a set B ⊂ A i with Σ( B ) large. We define, g ( t ) = max B ⊂ A i , | B | = t | Σ( B ) | Given B ⊂ A i of cardinality t , and such that | Σ( B ) | = g ( t ), we let S =Σ( B ), and C = A i \ B , let λ ( S, C ) = max c ∈ C λ S ( c ). By considering B ∪ { c } a set of cardinality t + 1 we deduce that, g ( t + 1) ≥ g ( t ) + λ ( S, C )Note that we may assume at all times that | S | ≤ n/ Z n ( S ) = | S | ,for we are immediately done if | S | > n/
2, so we use the three inequalitiesabove concerning λ ( S, C ), with | S | in the place of def Z n ( S ). Lemma 5.3.
Let n ≥ n . Let A ⊂ Z ∗ n ⊂ Z n with A ∩ ( − A ) = φ and | A | ≥ √ n + 15 n / , then | Σ( A ) | > n/ . roof. It suffices to prove this when the size of A is the integer above √ n +15 n / , so we may assume | A | ≤ √ n/
8. We put lower bounds on g ( t )using the bounds on λ ( S, C ) given above. We work in three main stages,corresponding to the three different lower bounds we have on λ ( S, C ). Stage 1:
Let t be the least t such that g ( t ) ≥ ( | A | − t ) /
2. We note that g (1) = 2. Let 2 ≤ t < t , let B be a set of size t with S = Σ( B ) satisfying | S | = g ( t ) and let C = A \ B , we have | S | ≤ ( | A | − t ) / | C | /
2. From ourbounds on λ we have, λ ( S, C ) ≥ | S | / g ( t + 1) ≥ g ( t ) / g ( t ) ≥ / t − ≥ (3 / t . It follows that, t ≤ log / ( | A | ) ≤ log / ( n ) ≤ n / . Stage 2:
Let t be the least t such that g ( t ) ≥ n / /
8. Let t ≤ t ≤ min { t + 9 n / , t } , let B be a set of size t with S = Σ( B ) satisfying | S | = g ( t ), and let C = A \ B , we have that | S | ≥ | C | / | C | = | A | − t ≥ √ n ,so that λ ( S, C ) ≥ | C | / ≥ √ n/ g ( t + 1) ≥ g ( t ) + √ n/
8. So that g ( t ) ≥ g ( t ) + ( t − t ) √ n/ t ≤ t + 9 n / ≤ n / Stage 3:
Let t be the least t such that g ( t ) > n/
2. Let t ≤ t < t and let B be a set of size t with S = Σ( B ) satisfying | S | = g ( t ), and let C = A \ B . We have that | S | = g ( t ) ≥ g ( t ) ≥ n / / ≥ n / | C | (certainlywe have that | C | ≤ | A | ≤ √ n/ λ ( S, C ) ≥ (1 − n − / ) | C | = (1 − n / )( | A | − t )which gives us that, g ( t + 1) ≥ g ( t ) + (1 − n − / )( | A | − t ) for t ≤ t < t ,we now use these to calculate g ( t ) in this range, g ( t ) ≥ g ( t ) + (1 − n − / ) t X t ′ = t ( | A | − t ′ ) = g ( t ) + (1 − n − / ) | A |− t X t ′ = | A |− t t ′ using the bound P st ′ = r t ′ ≥ ( s − r ) / s + r )( s − r ) / ≥ s ( s − r ) / | A | − t ≥ √ n + 5 n / we obtain, g ( t ) ≥ (1 − n − / ) ( | A | − t )( t − t )2 ≥ (1 − n − / ) ( √ n + 5 n / )( t − t )2 > √ n t − t )it is immediate then that t − t ≤ √ n , whence t ≤ √ n + 10 n / . Of coursewe now have, | Σ( A ) | ≥ g ( t ) > n/ n stages (an initial stage as Stage 1 above, followed by log n stages in which23 ( t ) doubles) allows one to deduce Lemma 5.3 with the condition on | A | weakened to | A | ≥ √ n + 10 log n . This then implies that any subset A ⊂ Z ∗ n ⊂ Z n with | A | ≥ √ n + 20 log n + 2 must have Σ( A ) = Z n . References [1] J. A. Dias da Silva and Y. O. Hamidoune, Cyclic spaces for Grassmannderivatives and additive theory,
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