On the structure of subsets of an orderable group with some small doubling properties
G. A. Freiman, M. Herzog, P. Longobardi, M. Maj, A. Plagne, D. J. S. Robinson, Y.V. Stanchescu
aa r X i v : . [ m a t h . G R ] M a y ON THE STRUCTURE OF SUBSETSOF AN ORDERABLE GROUPWITH SOME SMALL DOUBLING PROPERTIES
G. A. FREIMAN, M. HERZOG, P. LONGOBARDI, M. MAJ,A. PLAGNE, D. J. S. ROBINSON AND Y.V. STANCHESCU Introduction
Let G denote an arbitrary group (multiplicatively written). If S is a subsetof G , we define its square S by the formula S = { x x | x , x ∈ S } . In the abelian context, G will usually be additively written and we shall ratherspeak of sumsets and specifically of the double of S , namely2 S = { x + x | x , x ∈ S } . Here, we are concerned with the following general problem: for two realnumbers α ≥ β , determine the structure of S if S is a finite subset of agroup G satisfying an inequality on cardinalities of the form | S | ≤ α | S | + β when α is small and | S | is typically large.Problems of this kind are called inverse problems of small doubling type inadditive number theory. The coefficient α (or more precisely the ratio | S | / | S | )is called the doubling coefficient of S . This type of problems became the mostcentral issue in additive combinatorics. Inverse problems of small doublingtype have been first investigated by G.A. Freiman very precisely in the additivegroup of the integers (see [4], [5], [6], [7]) and by many other authors in generalabelian groups, starting with M. Kneser [16] (see, for example, [15], [17], [1],[23], [14]). More recently, small doubling problems in non-necessarily abeliangroups have been also studied, see [13], [24] and [3] for recent surveys on theseproblems and [19] and [26] for two important books on the subject. Keywords : Inverse problems; small doubling; nilpotent groups; ordered groups.
Mathematics Subject Classification 2010 : Primary 11P70; Secondary 20F05, 20F99,11B13, 05E15. t is easy to prove that if S is a finite subset of Z , then2 | S | − ≤ | S | ≤ | S | ( | S | + 1)2 . Moreover | S | = 2 | S | − S is a (finite) arithmetic progression,that is, a set of the form { a, a + q, a + 2 q, . . . , a + ( t − q } where a , q and t are three integers, t ≥ q ≥
0. The parameter t is calledthe size of the arithmetic progression and q its difference (we shall use ratio inthe multiplicative notation). In the articles [4] and [5], G.A. Freiman provedthe following more general results. The first result is referred to as the 3 k − Theorem A.
Let S be a finite set of integers with at least three elements.If | S | ≤ | S | − , then S is contained in an arithmetic progression of size | S | − | S | + 1 ≤ | S | − . . The second result goes one step further. It is called the 3 k − Theorem B.
Let S be a finite set of integers with at least elements. If | S | = 3 | S | − , then one of the following holds: (i) S is a subset of an arithmetic progression of size at most | S | − , (ii) S is the union of two arithmetic progressions with the same difference, (iii) | S | = 6 and S is Freiman-isomorphic to the set K , where K = { (0 , , (1 , , (2 , , (0 , , (0 , , (1 , } . Recall that two sets are Freiman isomorphic if they behave with respect toaddition in the same way (there is a one-to-one correspondence between thetwo sets as well as between their two sumsets). Notice that translations anddilations are transformations which send a set on another set which is Freimanisomorphic to it. For instance, in case (iii) of the above-stated theorem, sucha set S of integers is, up to translation and dilation, of the form { , , , u, u + 1 , u } for an integer u ≥ S ofthe additive group Z if | S | = 3 | S | − k −
2) and more generallyof small doubling sumsets of Z d (see for instance [25]).Since then, far-reaching generalizations have been obtained, see [3]. Al-though very powerful and general, these last results are not very precise.In papers [8], [9], [10], [11] and [12], we started the precise investigation ofsmall doubling problems for subsets of an ordered group. We recall that if G s a group and ≤ is a total order relation defined on the set G , then ( G, ≤ )is an ordered group if for all a, b, x, y ∈ G the inequality a ≤ b implies that xay ≤ xby , and a group G is orderable if there exists an order ≤ on the set G such that ( G, ≤ ) is an ordered group. Obviously the group of integers with theusual ordering is an ordered group. More generally, it is possible to prove thata nilpotent group is orderable if and only if it is torsion-free (see, for example,[18] or [20]).Extending Freiman’s results, we proved in [8] (Corollary 1.4) the followingtheorem, which is analogous to the 3 k − Theorem C.
Let G be an orderable group and let S be a finite subset of G with at least three elements. If | S | ≤ | S | − , then h S i is abelian. Moreover,there exists elements x and g in G , such that gx = x g and S is a subset of { x , x g, x g , . . . , x g u } where u = | S | − | S | . In other words, S is contained in an abelian geometricprogression of size | S | − | S | + 1 . Furthermore, in the case when | S | = 3 | S | −
3, we obtained the followinginitial result (this is our Theorem 1.3 in [8])
Theorem D.
Let G be an orderable group and let S be a finite subset of G ofsize at least . If | S | ≤ | S | − , then h S i is abelian. Using results of Freiman and Stanchescu, we proved in [12] (Theorem 1) thefollowing theorem concerning the more precise structure of such small doublingsets S . Theorem E.
Let G be an orderable group and let S be a finite subset of G of size at least satisfying | S | ≤ | S | − . Then h S i is abelian and at most -generated.Moreover, if | S | ≥ , then one of the following two possibilities occurs: (i) S is a subset of a geometric progression of length at most | S | − , (ii) S is a union of two geometric progressions with the same ratio. We also proved the following result concerning the structure of h S i in thecase when | S | = 3 | S | − a, b ] = a − b − ab and a b = b − ab. Theorem F.
Let G be an orderable group and let S be a finite subset of G ofsize at least .If | S | = 3 | S | − , then one of the following holds: (i) h S i is abelian and at most 4-generated, ii) h S i = h a, b i , with [ a, b ] = 1 and [[ a, b ] , a ] = [[ a, b ] , b ] = 1 . In particular, h S i is nilpotent of class 2, (iii) h S i = h a, b i , with a b = aa b and [ a, a b ] = 1 , (iv) h S i = h a, b i , with a b = a . In particular, h S i is a quotient of theBaumslag-Solitar group B(1,2), (v) h S i = h a i × h c, b i , with c b = c , | S | = 4 and S = { a, ac, ac , y } , where y is a suitable element of h c, b i . If h S i is an abelian group which is at most 4-generated and | S | = 3 | S | −
2, then the structure of S can be deduced from previous results of Freimanand Stanchescu (see [12], Theorem 2). The aim of this paper is to go onestep further in the non-abelian case. Namely, we shall present a completedescription of the structure of S if S is a finite subset of an orderable group G with | S | = 3 | S | − h S i is non-abelian. The following result will be ourmain theorem. Theorem 1.
Let ( G, ≤ ) be an ordered group and let S be a finite subset of G of size at least . We assume that | S | = 3 | S | − and that h S i is non-abelian.Then one of the following holds: (i) S = { a, ac, . . . , ac i , b, bc, . . . , bc j } , where [ a, c ] = [ b, c ] = 1 , c > andeither ab = bac or ba = abc , (ii) S = { x, xc, xc , . . . , xc k − } , where c > and either c x = c or ( c ) x = c , (iii) | S | = 4 and the structure of S is of one of the following types: (a) either S = { x, xc, xc x , xc x } or S = { x − , x − c, x − c x , x − c x } ,with c > and c x = cc x = c x c , (b) S = { , c, c , x } , where either c x = c or ( c ) x = c , (c) S = { x, xc, xc , y } , where [ c, x ] = 1 and either [ x, y ] = c = ( c ) y or [ y, x ] = c = c y , (d) S = { x, xc, xc , y } , where [ c, x ] = [ x, y ] = 1 and either c y = c or ( c ) y = c .Moreover, if either (i) or (ii) or case (b) of (iii) holds, then | S | = 3 | S | − . This paper is organized as follows. In Section 2, we record some usefulresults from [12] and [11]. In Section 3, we first study the group h a, b | a b = aa b , [ a, a b ] = 1 i . Then we investigate the structure of S if h S i = h a, b | a b = aa b , [ a, a b ] = 1 i and | S | = 3 | S | −
2. We prove in Theorem 2 that theseassumptions imply that | S | ≤ S if | S | = 4 and | S | = 10 . In Section 4 we record some basic results concerningthe Baumslag-Solitar group B (1 , S of B (1 ,
2) satisfying h S i = G nd | S | = 3 | S | − . Notice that small doubling problems in the Baumslag-Solitar groups have been also studied in [9]. Finally, in Section 5 we proveTheorem 1. We refer to books [2], [6] [21] and [22] for notation and definitions.2.
Some useful results
Here we collect the following useful results from [12] and [11].
Proposition 1 (see [12], Lemma 4) . Let G be an orderable group and let S bea finite subset of G of size at least . If T denotes S \ { max S } , theneither h S i is a 2-generated abelian group, or | T | ≤ | S | − . In the next proposition, we use the notation ˙ ∪ for a disjoint union. Proposition 2 (see [12], Proposition 3) . Let G be an orderable group and let S be a finite subset of G of size at least satisfying | S | = 3 | S | − . Supposethat S = T ˙ ∪{ y } , where h T i is abelian. Then either h S i is abelian, or S = { x, xc, . . . , xc k − , y } and one of the following holds: (i) [ c, x ] = [ c, y ] = 1 and either [ x, y ] = c or [ y, x ] = c , (ii) | S | = 4 , [ c, x ] = 1 and either [ x, y ] = c = ( c ) y or [ y, x ] = c = c y , (iii) | S | = 4 , [ c, x ] = [ x, y ] = 1 and either c y = c or ( c ) y = c . The following two propositions deal with the case when | S | = 3. Proposition 3 (see [12], Proposition 1) . Let ( G, ≤ ) be an ordered group and let x , x , x be three elements of G such that x < x < x . Let S = { x , x , x } .Suppose that h S i is non-abelian and either x x = x x or x x = x x . Then | S | = 7 if and only if one of the following holds: (i) S ∩ Z ( h S i ) = ∅ , (ii) S is of the form { a, a b , b } , where aa b = a b a . Proposition 4 (see [12], Proposition 2) . Let ( G, ≤ ) be an ordered group, andlet x , x , x be three elements of G such that x < x < x . Let S = { x , x , x } and assume that x x = x x and x x = x x . If | S | = 7 , then one of thefollowing statements holds: (i) either S = { x, xc, xc x } or S = { x − , x − c, x − c x } , with c > , c ∈ G ′ and c x = cc x = c x c , (ii) either S = { x, xc, xcc x } or S = { x − , x − c, x − cc x } , with c > , c ∈ G ′ and c x = cc x = c x c , (iii) S = { x, xc, xc } , with either c x = c or ( c ) x = c . Next proposition considers the case when | S | = 4. roposition 5 (see [12], Lemma 6) . Let ( G, ≤ ) be an ordered group andlet x , x , x , x be four elements of G such that x < x < x < x . Let S = { x , x , x , x } and suppose that | S | = 10 . If x x = x x , then either h x , x , x i or h x , x , x i is abelian. The final proposition which we shall need deals with the case of nilpotentgroups of class 2.
Proposition 6 (see [11], Theorem 3.2) . Let G be an orderable nilpotent groupof class and let S be a finite subset of G of size at least , such that h S i is non-abelian. Then | S | = 3 | S | − if and only if S = { a, ac, . . . , ac i , b, bc, . . . , bc j } ,with c > and either ab = bac or ba = abc . Subsets of the group G = h a, b | a b = aa b , aa b = a b a i In this section we shall prove the following theorem.
Theorem 2.
Let G = h a, b | a b = aa b , aa b = a b a i . (i) The group G is orderable.Let ≤ be a total order on G . (ii) Let S be a finite subset of G satisfying h S i = G and | S | = 3 | S | − . Then | S | ≤ . (iii) Moreover, in the case when | S | = 4 , then either S = { x, xc, xc x , xc x } or S = { x − , x − c, x − c x , x − c x } , with c > and c x = cc x = c x c . Throughout this section, we shall denote by G the following group: G = h a, b | a b = aa b , aa b = a b a i . We begin with some remarks concerning G .Define H = ( h u i × h v i ) ⋊ h t i , where t, u, v have infinite order and u t = v , v t = uv . Then the defining relations of G hold in H , namely [ u, u t ] = 1 and u t = uu t and by von Dyck’s theorem (Theorem 2.2.1 in [22]) there is anepimorphism θ : G → H with a θ = u and b θ = t . Since ker θ = 1, it followsthat G = ( h a i × h a b i ) ⋊ h b i , with a b = aa b . Thus G ′ = h a i × h a b i and G is a polycyclic metabelian group.We have a b = aa b , a b = a ( a b ) and it is easy to see, by induction on n ,that(1) a b n = a f n − ( a b ) f n for any n ∈ N , where ( f n ) n ∈ N is the Fibonacci sequence defined in the standard way by in-duction by f = 0 , f = 1 and f n +1 = f n + f n − for n > . In particular, f = 1and a b = a b +1 . urthermore, recall that a group is called an R ⋆ -group if g x . . . g x n = e implies g = e for all n ∈ N and all g, x , . . . , x n ∈ G. Since metabelian R ⋆ -groups are orderable (see [2], Theorem 4.2.2), in order toprove that G is orderable it suffices to show that it is an R ⋆ -group. This iswhat we do now. Proposition 7. G is an R ⋆ -group and hence it is orderable.Proof. It suffices to show that, if n ∈ N , k, u, v ∈ Z and g i ∈ G , then( b k a u ( a b ) v ) g · · · ( b k a u ( a b ) v ) g r = 1implies that b k a u ( a b ) v = 1 . First we notice that by working mod G ′ it follows that k = 0. Hence wemay assume that(2) ( a u ( a b ) v ) b n ( a u ( a b ) v ) b n · · · ( a u ( a b ) v ) b nr = 1 , where n i ≥
2, applying conjugation by a power of b , if necessary. Since by (1) a b n + b n + ··· + b nr = a c + db where c, d are positive integers, relation (2) now becomes( a u ) c + db ( a v ) cb + db = 1 . Equivalently a uc + udb a vcb + vd ( b +1) = 1 , or a ( uc + vd )+( ud + vc + vd ) b = 1 . Hence we have uc + vd = 0 and ud + vc + vd = 0which implies that ud + ( − uc ) c + d ( − uc ) = 0 . If u = 0, then vd = 0 and v = 0, so we are done. Hence we may assume that u = 0. It follows that d − dc − c = 0 . Therefore d/c , which is rational, must satisfy dc = 1 + √ , a contradiction. (cid:3) Corollary 1.
Part (i) of Theorem 2 holds. oreover, the following property holds in G . Proposition 8.
We have C G ′ ( g ) = { } for any g ∈ G \ G ′ . In particular Z ( G ) = { } .Proof. Let g ∈ G \ G ′ . Then g = db n , where d ∈ G ′ and n is an integer differentfrom 0. Since d centralizes G ′ , it suffices to show that C G ′ ( b n ) = { } , where n is a positive integer different from 0.Denote the element a u ( a b ) v of G ′ by ( u, v ), where u, v are integers. Then b acts on ( u, v ) by conjugation via the following function:( u, v ) b = (0 · u + 1 · v, · u + 1 · v ) = ( u, v ) B where B = (cid:20) (cid:21) . Thus, by induction, b n acts by conjugation on ( u, v ) via multiplication by: B n = (cid:20) f n − f n f n f n +1 (cid:21) , where ( f m ) m ∈ N is the Fibonacci sequence.If b n centralizes a non-trivial element of G ′ , then 1 would be an eigenvalue of B n . But the characteristic polynomial of B n is x − ( f n − + f n +1 ) x +( f n − f n +1 − f n ), which has roots ( f n − + f n +1 ± √ f n ) /
2, which are irrational for n > C G ′ ( b n ) = { } , as required. (cid:3) Now, let ≤ be a total order in G such that ( G, ≤ ) is an ordered group. Let S = { x , x , . . . , x k } be a subset of G of size k and suppose that x < x < · · · < x k . We wish to study the structure of S if | S | = 3 | S | − . We beginwith the case k = 3. Proposition 9.
Let S ⊆ G , with h S i = G . Suppose that | S | = 3 . Then | S | = 7 if and only if one of the following holds: (i) S = { , x, y } , with xy = yx , (ii) S = { c, c t , t } , with cc t = c t c , (iii) either S = { t, tc, tc t } or S = { t, tc, tc t } , where c ∈ G ′ , c > and c t = cc t = c t c , (iv) either S = { t − , t − c, t − c t } or S = { t − , t − c, t − c t } , where c ∈ G ′ , c > and c t = cc t = c t c .Furthermore, in any case, either S ∩ G ′ ⊆ { } or | S ∩ G ′ | = 2 .Proof. Suppose that S = { x , x , x } and | S | = 7. If either x x = x x or x x = x x holds, then either (i) or (ii) holds by Proposition 3, since Z ( G ) = { } . f x x = x x and x x = x x , then either (iii) or (iv) holds by Proposition4, since the relation c x = c is impossible in a torsion-free polycyclic group(otherwise there would be an infinite chain of subgroups of G : h c i ⊂ h c x − i ⊂h c x − i ⊂ · · · ⊂ h c x − n i ⊂ · · · ).In particular, if (i) holds, then either S ∩ G ′ = { } or S ∩ G ′ = { , z } with z ∈ { x, y } , since G ′ is abelian and G is non-abelian. In case (ii), c t = cd with d ∈ G ′ \ { } and cc t = c t c implies that cd = dc . Hence, by Proposition 8, c ∈ S ∩ G ′ and since c t = c , it follows that t / ∈ S ∩ G ′ . Thus S ∩ G ′ = { c, c t } .In cases (iii) and (iv) c, c t , c t ∈ G ′ , so t / ∈ G ′ since G is non-abelian. Hence S ∩ G ′ = ∅ in these cases. Thus in all cases either S ∩ G ′ ⊆ { } or | S ∩ G ′ | = 2holds.A direct calculation proves the converse of the main statement. (cid:3) Now we study the structure of S if | S | = 4 and | S | = 10. We begin withthe following four lemmas. Lemma 1.
Let S = { t, tc, tc t , x } ⊆ G , with c t = c t c = cc t and t < tc < tc t Let S = { t, tc, tc t , x } ⊆ G , with c t = c t c = cc t and t < tc Let S = { t − , t − c, t − c t , x } ⊆ G , with c t = c t c = cc t and t − < t − c < t − c t < x . Then | S | > .Proof. We can argue as in Lemma 2, using the result of Lemma 3. (cid:3) Now we can prove part (iii) of Theorem 2. Proof of Theorem 2 (iii). Suppose that | S | = 10. Write, as usual, S = { x , x , x , x } , x < x < x < x , T = { x , x , x } and V = { x , x , x } .If S = A ˙ ∪{ y } with h A i abelian, then by Proposition 2, either (i) of thatproposition holds, in contradiction to Proposition 8, or one of (ii) and (iii)holds, in which case either c y = c or ( c ) y = c , which as shown in the proofof Proposition 9, is impossible in a polycyclic torsion-free group. Therefore wemay assume that each triple of elements in S generates a non-abelian group.In particular, h T i and h V i are non-abelian. By Proposition 1 | T | ≤ h T i is non-abelian, Theorem D implies that | T | = 7.Therefore | S | = | T | + 3 and h T i is non-abelian. The elements x x , x arenot in T because of the ordering, so one of the elements x x , x x belongs to T . Hence x ∈ h T i and G = h T i . Arguing similarly, it follows that | V | = 7, h V i is non-abelian and G = h V i . Therefore T and V satisfy the hypotheses ofProposition 9.If 1 ∈ T , then 1 = x by Proposition 5, and V satisfies either (ii) or (iii) or(iv) of Proposition 9. If V satisfies (ii), then | V ∩ G ′ | = 2 and | S ∩ G ′ | = 3, acontradiction, since G ′ is abelian. If V satisfies either (iii) or (iv) of Proposition9, then S = V ˙ ∪ V ˙ ∪{ } and | S | = 11, a contradiction. Therefore we mayassume that 1 / ∈ T and T satisfies either (ii) or (iii) or (iv) of Proposition 9.If T satisfies (ii) of Proposition 9, then | T ∩ G ′ | = 2 and hence | V ∩ G ′ | ≥ | V ∩ G ′ | = 2. If | T ∩ G ′ | = { x , x } ,then x x = x x and Proposition 5 implies that either h T i or h V i is abelian,a contradiction. Hence x ∈ T ∩ G ′ and | S ∩ G ′ | = 3, yielding again a contra-diction. Consequently T satisfies either (iii) or (iv) of Proposition 9 and thusit is equal to one of the following sets: { t, tc, tc t } , { t − , t − c, t − c t } , { t, tc, tc t } and { t − , t − c, t − c t } , where c ∈ G ′ , c > c t = cc t = c t c . It follows byLemmas 1, 2, 3 and 4 that T must be equal to one of the first two sets and S is as required.A direct calculation proves the converse. (cid:3) Finally, we study the case S ⊆ G , | S | = 5 in the next proposition. roposition 10. Let S ⊆ G , with h S i = G and suppose that | S | = 5 . Then | S | > 13 = 3 | S | − Proof. If | S | = 5, then by Theorem D, | S | ≥ 13. So it suffices to assume that | S | = 13 and to reach a contradiction.Write S = { x , x , x , x , x } , x < x < x < x < x , T = { x , x , x , x } and suppose that | S | = 13. Arguing as in the previous proposition, we mayconclude that | T | = 10, | S | = | T | + 3 and h T i = G . Hence T satisfies thehypotheses of part (iii) of Theorem 2.Suppose first that T = { t, tc, tc t , tc t } , with c > c t = cc t = c t c. Then x = t , x = tc , x = tc t , x = tc t and as shown in the proof of Lemma 1, c, c t ∈ G ′ and t / ∈ G ′ . Moreover, T = { t , t c, t c t , t cc t , t c t c t , t cc t c t , t c c t , t c c t c t , t c ( c t ) , t c ( c t ) } , and in particular(3) if t c α ( c t ) β ∈ T , then α ∈ { , , } , β ∈ { , , , } . Since x x / ∈ T , at least one of the elements x x , x x , x x belongs to T ,implying that x = tc l ( c t ) m for some integers l and m . We have : x x = t c l +1 ( c t ) m , x x = t c m ( c t ) l + m +1 , x x = t c l ( c t ) m ,x x = t c m +1 ( c t ) l + m , x x = t c l ( c t ) m and x x = t c m ( c t ) l + m . If each of x x , x x , x x , x x , x x , x x belongs to T , then l, m ∈ { , } ,since these elements involve { c l , c m , c l +1 , c m +1 } and each element of T involvesonly one of { c , c, c } . Hence either x = t , or x = tc , or x = tc t , or x = tcc t = tc t , a contradiction. Therefore there exists i ∈ { , , } such thateither x i x / ∈ T or x x i / ∈ T .Now x x = t c m +1 ( c t ) l + m +1 and x x = t c l +1 ( c t ) m +2 . Thus x x = x x isimpossible, since it implies that l = m = 1 and x = tcc t = tc t , a contradic-tion. Hence S = T ˙ ∪{ x x , x x , x } . If x i x / ∈ T , then the only possibility is x i x = x x and if x x i / ∈ T , thenthe only possibility is x x i = x x .If x x = x x , then l = m = 0, a contradiction. If x x = x x , then l = 0 , m = − 1, so x = t ( c t ) − < t = x , also impossible. If x x = x x , then l = − , m = − 2, so x = tc − ( c t ) − < t = x , a contradiction. If x x = x x ,then l = 2 , m = 3, so x = tc ( c t ) and x x = t c ( c t ) / ∈ T , leading usto a previous case. If x x = x x , then l = 2 , m = 2, so x = tc ( c t ) nd x x = t c ( c t ) / ∈ T , leading us again to a previous case. Finally, if x x = x x , then l = 1 , m = 2, so x = tc ( c t ) and x x = t c ( c t ) / ∈ T , andagain we are in a previous case.We have reached a contradiction in all cases. Hence if T = { t, tc, tc t , tc t } ,with c > c t = cc t = c t c , then | S | > T = { t − , t − c, t − c t , t − c t } with c > c t = cc t = c t c. Arguing as before we may write x = t − c l ( c t ) m , for some integers l, m .Write ¯ S = { t, tc, tc t , tc t , tc l ( c t ) m } . Then | ¯ S | = | t { , c, c t , c t , c l ( c t ) m } t { , c, c t , c t , c l ( c t ) m }| = | t { , c, c t , c t , c l ( c t ) m } t { , c, c t , c t , c l ( c t ) m }| = |{ , c, c t , c t , c l ( c t ) m }{ , c, c t , c t , c l ( c t ) m } t − | = |{ , c, c t , c t , c l ( c t ) m } t − { , c, c t , c t , c l ( c t ) m }| = | t − { , c, c t , c t , c l ( c t ) m } t − { , c, c t , c t , c l ( c t ) m }| = | S | . But | ¯ S | > 13 by the previous paragraph, so | S | > 13, as required. (cid:3) Now we can conclude the proof of Theorem 2. Proof of Theorem 2, part (ii). Suppose that | S | = k ≥ S = { x , x , . . . , x k − , x k } and T = { x , x , . . . , x k − } . Then, by Proposition 1, | T | ≤ | S | − | T | − 2. If | T | ≤ | T | − 3, then,by Theorem D, h T i is abelian and Proposition 2 yields a contradiction, sincecase (i) of Proposition 2 is impossible as Z ( G ) = { } by Proposition 8 and asshown in the proof of Proposition 9, cases (ii) and (iii) of Proposition 2 areimpossible in a polycyclic torsion-free group.Thus | T | = 3 | T | − | S | − k − ≥ 4, at least one of the k − x x k , x x k , . . . , x k − x k belongs to T . Hence x k ∈ h T i and h T i = h S i = G . Thus, by induction, we may assume that | S | = 5, in whichcase we get the contradiction | S | > 13 by Proposition 10. Therefore | S | ≤ S if | S | = 4 follows from part (iii) of Theorem 2. (cid:3) Subsets of the Baumslag-Solitar group BS (1 , Theorem 3. Let G = h a, b | a b = a i and let S be a generating subset of G .Then the following statements hold: i) If | S | > , then | S | = 3 | S | − if and only if S = { x, xc, . . . , xc k − } ,where c > and either c x = c or ( c ) x = c . (ii) If | S | = 4 , then | S | = 3 | S | − if and only if either S = { , c, c , x } ,where either c x = c or ( c ) x = c , or S = { x, xc, xc , xc } , where c > and either c x = c or ( c ) x = c. Throughout this section we shall denote by G the Baumslag-Solitar group G = BS (1 , 2) = h a, b | a b = a i . We begin with some basic well-known results concerning G . Proposition 11. The following statements hold: (i) G ′ = h a b n | n ∈ Z i is abelian, (ii) G = G ′ ⋊ h b i , (iii) G is orderable, (iv) If c ∈ G ′ , then c b = c , (v) C G ′ ( g ) = { } for any g ∈ G \ G ′ . In particular Z ( G ) = { } .Proof. Clearly G ′ = a G = h a b n | n ∈ Z i . For claims (i)-(iv), see Theorem 10 in[10] and its proof.Concerning (v), let g ∈ G \ G ′ , c ∈ G ′ and suppose that c g = c . By (i) g = db s for some s ∈ Z \ { } and c g = c implies that c b s = c . We may assumethat s ≥ c s = c . Since 2 s > 1, it follows that c = 1,as required. (cid:3) Now, let ≤ be a total order in G such that ( G, ≤ ) is an ordered group. Let S = { x , x , . . . , x k } be a subset of G of order k and suppose that x < x < · · · < x k . We study the structure of S if | S | = 3 | S | − . We begin with thecase k = 3. Proposition 12. Let S ⊆ G , with h S i = G and | S | = 3 . Then | S | = 7 if andonly if one of the following holds: (i) S = { , x, y } , with xy = yx , (ii) S = { c, c t , t } , with cc t = c t c and c, c t ∈ G ′ , (iii) S = { x, xc, xc } , where c > and either c x = c or ( c ) x = c. Furthermore, in any case, either S ∩ G ′ ⊆ { } or | S ∩ G ′ | = 2 .Proof. Suppose that S = { x , x , x } and | S | = 7. If either x x = x x or x x = x x , then either (i) or (ii) holds by Proposition 3, since Z ( G ) = { } .If x x = x x and x x = x x , then (iii) holds by Proposition 4. In fact,cases (i) and (ii) of Proposition 4 cannot occur, since the relation c x = cc x with c > BS (1 , c x = cc x , then c x andhence c belong to G ′ and by Proposition 11 (iv) c b = c . Clearly x / ∈ G ′ , so e may assume that x = b s for some s ∈ Z \ { } . If x = b s with s ≥ c x = c b s = c s , c x = c s and since c x = c x c , we get c s = c s +1 . But c > 1, so 4 s = 2 s + 1, a contradiction. Similarly, if x = b − s with s > c x = cc x implies c = c x − c x − = c s c s and since c > 1, we get thecontradiction 1 = 4 s + 2 s . So it follows that one of (i), (ii) or (iii) must hold.In particular, if (i) holds, then either S ∩ G ′ = { } or S ∩ G ′ = { , z } with z ∈ { x, y } , since G ′ is abelian, and | S ∩ G ′ | = 2, as required.In case (ii), c t = cc t c − = cd with d ∈ G ′ \ { } and cc t = c t c implies that cd = dc . Hence, by Proposition 11 (v), c ∈ S ∩ G ′ and since c t = c , it followsthat t / ∈ S ∩ G ′ . Thus S ∩ G ′ = { c, c t } and | S ∩ G ′ | = 2, as required.Finally, if case (iii) holds, then c ∈ G ′ and x / ∈ G ′ since [ c, x ] = 1. Hence S ∩ G ′ = ∅ , as required.A direct calculation proves the converse of the main statement. (cid:3) Now we study the structure of S if k ≥ 4. We begin with the case S = A ˙ ∪{ y } , where h A i is abelian. Proposition 13. Let S ⊆ G , with h S i = G . Suppose that | S | = k ≥ and S = A ˙ ∪{ y } with h A i abelian. Then | S | = 3 | S | − if and only if k = 4 and S = { , c, c , y } , with either c y = c or ( c ) y = c .Proof. Suppose that | S | = 3 | S | − 2. Since Z ( G ) = { } , h S i = G is not anilpotent group of class at most 2 and by Proposition 2 we have | S | = 4, S = { x, xc, xc , y } and either (ii) or (iii) of Proposition 2 holds. If (iii) holds,then x ∈ Z ( G ) = { } and x = 1. Thus S has the required structure in thiscase.Now suppose that (ii) of Proposition 2 holds. Then c ∈ G ′ and by Proposi-tion 11 (v) x ∈ C G ( c ) ⊆ G ′ . Moreover, y / ∈ G ′ since G ′ is abelian. Hence, byProposition 11 (ii), y = eb s for some s ∈ Z \ { } and e ∈ G ′ .If c y = c , then c b s = c and since c b = c by Proposition 11 (iv), we musthave s = 1. Thus x y = x b = x by Proposition 11 (iv) and c = [ y, x ] =( x − ) y x = x − x = x − . Write d = c − . Then { x, xc, xc , y } = { d , d, , y } with d y = d , as required.If ( c ) y = c , then it follows that ( c ) b s = c and the only possibility for s is s = − 1. Now x b = x , so ( x ) y = ( x ) b − = x and ( c ) y = c = [ x, y ] = x − x y .Thus x y = xc and x = ( x ) y = xcxc = x c , yielding x = c − . Write d = c − .Then { x, xc, xc , y } = { d , d, , y } with ( d ) y = ( d ) b − = d , as required.A direct calculation proves the converse. (cid:3) We can now describe the structure of S if k = 4. Proposition 14. Let S ⊆ G , with h S i = G and suppose that | S | = 4 . Then | S | = 10 if and only if one of the following holds: i) S = { , c, c , y } , with either c y = c or ( c ) y = c , (ii) S = { x, xc, xc , xc } , with c > and either c x = c or ( c ) x = c. Proof. Suppose that S = { x , x , x , x } with x < x < x < x and | S | = 10.Let T = { x , x , x } and V = { x , x , x } . If S = A ˙ ∪{ y } with h A i abelian,then (i) holds by Proposition 13. Therefore we may assume that each triple ofelements in S generates a non-abelian group. In particular, h T i and h V i arenon-abelian.By Proposition 1 | T | ≤ h T i is non-abelian, Theorem D impliesthat | T | = 7. The elements x x , x are not in T because of the ordering,so one of the elements x x , x x belongs to T , since | S | = | T | + 3. Hence x ∈ h T i and G = h T i . Arguing similarly, it follows that | V | = 7 and h V i = G . Therefore T and V satisfy the hypotheses of Proposition 12.If 1 ∈ T , then 1 = x by Proposition 5 and V satisfies either (ii) or (iii)of Proposition 12. But then S = V ˙ ∪ V ˙ ∪{ } and | S | = 11, a contradiction.Therefore we may assume that 1 / ∈ T .Thus, by Proposition 12, T satisfies either (ii) or (iii) of that proposition. If | T ∩ G ′ | = 2, then also | V ∩ G ′ | = 2 and V ∩ G ′ = { x , x } by Proposition 5.Hence | S ∩ G ′ | = 3, which is impossible, since h S ∩ G ′ i is abelian. Therefore T ∩ G ′ is an empty set and T = { x, xc, xc } , with c > c x = c or( c ) x = c .Suppose, first, that c x = c . Then T = { x , x c, x c , x c , x c , x c , x c } .Obviously x / ∈ T , because of the ordering. Since | S | = | T | +3, it follows thatone of the elements xx , xcx , xc x belongs to T . Therefore x = xc s for someinteger s . Similarly one of the elements xc s x = x c s , xc s xc = x c s +1 , xc s xc = x c s +2 belongs to T . Since x = xc s > xc and c > 1, we must have s ≥ x c s ≤ x c , the only possible case is that s = 3. Thus (ii) holds.Now suppose that ( c ) x = c. In this case T = { x , x c, x c , x c x , x c x c, x c x c , x c } and, as before, one of the elements x x, x xc, x xc belongs to T . Thus x xc l = xc i xc j for some integers i, j, l and x = xc i xc j − l x − = xc i +2( j − l ) = xc s for some integer s ≥ 3, since x > xc .Similarly one of the elements xx , xcx , xc x belongs to T . If xc xc s ∈ T ,then x c s +1 ∈ T and s ≤ 2, a contradiction, and if xcxc s ∈ T , then x c x c s ∈ T again implying that s ≤ 2, a contradiction. Therefore the only possibilecase is that x c s ∈ T , forcing s = 3 and yielding (ii) again. Thus either (i) or(ii) holds in all cases.A direct calculation proves the converse. (cid:3) Now we can prove Theorem 3. roof of Theorem 3. Suppose that | S | = k ≥ | S | = 3 k − 2. If k = 4,then S has the required structure by Proposition 14. So suppose that k > . Write S = { x , x , . . . , x k } , with x < x < · · · < x k , T = { x , . . . , x k − } and V = { x , . . . , x k } . Since k ≥ 5, it follows by Proposition 13 that eachsubset of S with k − h T i is non-abelian and by Theorem D | T | ≥ | T | − 2. But by Proposition 1 | T | ≤ | S | − | T | − 2, so it follows that | T | = 3 | T | − | S | − 3. Since k − ≥ 4, one of the k − x x k , x x k , . . . , x k − x k must belong to T .Hence x k ∈ h T i and G = h T i . Similar arguments yield | V | = 3 | V | − h V i = G. It follows by induction that either T = { x, xc, . . . , xc k − } with c > c x = c or ( c ) x = c , or | T | = 4 and T = { , c, c , y } with either c y = c or ( c ) y = c .First suppose that the latter case holds. Then c ∈ G ′ , | T ∩ G ′ | = 3 and | V ∩ G ′ | ≥ 2. But induction applied to V implies that | V ∩ G ′ | ∈ { , } , so | V ∩ G ′ | = 3. If T ∩ G ′ = V ∩ G ′ , then | S ∩ G ′ | = 4 = k − G ′ is abelian. Hence T ∩ G ′ = V ∩ G ′ = { x , x , x } . Now, ifone of the elements x x , x x , x x belongs to { x , x , x } , then h T i is abelianand if one of the elements x x , x x , x x belongs to { x , x , x } , then h V i is abelian, a contradiction in both cases. So neither of the above six elementsbelongs to { x , x , x } and since x x < x x < x x < x x < x x < x x ,it follows that | S | ≥ |{ x , x , x } | ≥ T = { x, xc, . . . , xc k − } , with c > c x = c or ( c ) x = c . Thus x i = xc i − for i = 1 , , . . . , k − c x = c . Since | T | = | S | − k − ≥ 4, there existsan integer i , 1 ≤ i ≤ k − 1, such that xc i x k ∈ T . Hence xc i x k = xc j xc l forsome integers i, j, l and x k = c j − i xc l = xc j − i )+ l = xc s for an integer s . Noweither x k x k − ∈ T or x k x k − / ∈ T .If x k x k − ∈ T , then xc s xc k − = xc r xc k − for some integer r and x c s + k − = x c r + k − . Thus 2 s + k − r + k − 2, which is impossible, since these numbersare of different parity.So suppose that x k x k − / ∈ T . Then S \ T = { x k x k − , x k x k − , x k } and since x k − x k ∈ S \ T , it follows that either x k − x k = x k x k − or x k − x k = x k x k − .If x k − x k = x k x k − , then x c k − s = x c s + k − and 2( k − 2) + s = 2 s + k − s = k − x k ∈ T , a contradiction. Hence x k x k − = x k − x k ,yielding 2 s + k − k − 2) + s . Thus s = k − 1, as required.Now, suppose that ( c ) x = c . Then c x − = c and arguing as in the previouscase, there exists i such that 1 ≤ i ≤ k − x k xc i ∈ T . Hence x k xc i = xc j xc l for some integers i, j, l and x k = xc j xc l − i x − = xc j ( c l − i ) x − = xc s forsome integer s . Now either xc k − xc s ∈ T or xc k − xc s / ∈ T . f xc k − xc s ∈ T , then xc k − xc s = xc k − xc r for some integer r and xc s = cxc r . Thus c s x = c r x , which is impossible.Thus xc k − xc s / ∈ T . Then S \ T = { x k − x k , x k − x k , x k } and since x k x k − ∈ S \ T , it follows that either x k x k − = x k − x k or x k x k − = x k − x k . If x k x k − = x k − x k , then c s xc k − = c k − xc s and s + 2( k − 2) = k − s . Thus s = k − x k ∈ T , a contradiction. Hence x k x k − = x k − x k , yielding s + 2( k − 2) = k − s . Thus s = k − 1, as required. The proof in one direction is complete.Conversely, suppose that | S | = k . If k = 4 and S satisfies the corre-sponding conditions, then | S | = 10 by Proposition 14. If k > S = { x, xc, . . . , xc k − } with c x = c , then S = { xc i xc j = x c i + j | ≤ i, j ≤ k − } . We claim that | S | = 3 k − 2. Since 0 ≤ i + j ≤ k − 3, it suffices toshow that each integer n with 0 ≤ n ≤ k − n = 2 i + j for some 0 ≤ i, j ≤ k − 1. If n = 0, then n = 2 · n > n = 3 a + b , where 0 ≤ a ≤ k − ≤ b ≤ 3. In the latter case, therequired representations are: n = 2 a + ( a + 1) if b = 1, n = 2( a + 1) + a if b = 2 and n = 2( a + 1) + ( a + 1) if b = 3.Hence | S | = 3 | S | − 2, as required. If S = { x, xc, . . . , xc k − } with ( c ) x = c ,then c x − = c and x − S x = { c i xc j x = c i +2 j x | ≤ i, j ≤ k − } . As shownabove | S | = | x − S x | = 3 | S | − 2, as required. (cid:3) Proof of Theorem 1 Now we can prove Theorem 1. Proof of Theorem 1. Let S ⊆ G , | S | = k ≥ | S | = 3 k − h S i is non-abelian. Then h S i satisfies either (ii), or (iii), or (iv), or (v) ofTheorem F. From now on, (ii), (iii), (iv) and (v) denote items of Theorem Fand (1i), (1ii) and (1iii) denote items of Theorem 1.If G is a nilpotent ordered group of class 2 and S is a finite subset of G suchthat | S | = k ≥ h S i is non-abelian, then by Proposition 6 | S | = 3 k − S satisfies (1i). This takes care of h S i satisfying (ii), since in cases(iii), (iv) and (v), h S i is non-nilpotent. If h S i satisfies (iii), then Theorem 2implies that k = 4 and case (a) of (1iii) holds.Suppose, now, that h S i satisfies (iv). Then S satisfies the assumptions ofTheorem 3. Hence, if k > 4, then | S | = 3 k − S satisfies (1ii)and if k = 4, then | S | = 3 k − S satisfies either (1ii) or case(b) of (1iii). Notice that if k = 4, then the second case in item (ii) of Theorem3 is of type (1ii). inally, if h S i satisfies (v), then | S | = 4 and h S i is not a nilpotent group ofclass at most 2. Hence, by Proposition 2, S satisfies one of the cases (c) or (d)of (1iii).Conversely, it follows from our proof that if S satisfies one of the conditions(1i), (1ii) or case (b) of (1iii), then | S | = 3 k − (cid:3) Acknowledgements This work was supported by the ”National Group for Algebraic and Geo-metric Structures, and their Applications” (GNSAGA - INDAM), Italy.A.P. is supported by the ANR grant Caesar number 12 - BS01 - 0011. Hethanks his colleagues from the University of Salerno for their hospitality duringthe preparation of this paper. References [1] Y. Bilu, Structure of sets with small sumset, Ast´erisque (1999), 77–108.[2] R. Botto Mura, A. 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Freiman, Marcel HerzogSchool of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978,Israel E-mail address : [email protected] E-mail address : [email protected] Patrizia Longobardi, Mercede MajDipartimento di Matematica, Universita’ di Salerno, Via Giovanni Paolo II,132, 84084 Fisciano (Salerno), Italy E-mail address : [email protected] E-mail address : [email protected] Alain PlagneCentre de Math´ematiques Laurent Schwartz, ´Ecole polytechnique, 91128Palaiseau Cedex,France E-mail address : [email protected] Derek J. S. RobinsonDepartment of Mathematics, University of Illinois at Urbana-Champaign,Urbana, IL 61801, USA E-mail address : [email protected] Yonutz V. StanchescuAfeka Academic College, Tel Aviv 69107, IsraelandThe Open University of Israel, Raanana 43107, Israel E-mail address : [email protected] and [email protected]@openu.ac.il