On a theorem of Freiman
arXiv:1412.0353v2 [math.CO] 25 Nov 2015
R. Balasubramanian, Prem Prakash Pandey
November 26, 2015
1
Abstract
One of the many theorems Freiman proved, in the second half of
the twentieth century, in the subject which later came to be known as
“structure theory of set addition”, was ‘Freiman’s 3k − 4 theorem’ for
subsets of Z. In this article we introduce concept of a new ‘structure’
on finite subsets of integers. Sets with this structure are quite useful in
additive number theory, in some contexts. Also we give some criterion
for subsets to posses this structure. Then, this is used to establish
Freiman’s 3k − 4 theorem for the groups Z × G, where G is any group.
These ideas are used to give some inverse theorems for non-abelian
groups, for example,an alternate proof of the main theorem of [7] is
provided, up to extremal cases.
1
Introduction
The cardinality of a finite set X, will be denoted by |X|. When X ⊂ Z, we
will use min(X) and max(X) to denote the smallest and largest element of
X respectively. Also we will write gcd(X) to denote the greatest common
divisor of nonzero elements of X. When G is an abelian group, written
additively, for two subsets A and B of G we write A + B = {a + b : a ∈ A, b ∈
B} and A − B = {a − b : a ∈ A, b ∈ B}. Associativity of the group allows
us to define A + B − A, A + A − A + B − B etc. without any ambiguity.
Commutativity of G allows us not to differentiate between A + B and B + A
and so on.
The subject of additive combinatorics explores two directions: first, given
two such sets, what can we say about sumset and second, if we have some
information about sumsets what can we infer about the sets themselves? The
first question lies in the domain of “direct problems” whereas second falls in
the area “inverse problems”. When G = Z, one of the first result under
“direct problems” is the following lower bound on sumsets;
|A + B| ≥ |A| + |B| − 1.
(1)
The equality in equation (1) holds if and only if A and B are arithmetic progressions with same common difference. Another example of direct problem
is the theorem of Cauchy and Davenport [16]. For subsets A and B of Z/pZ,
cyclic group with p elements, where p is a prime number, one has
|A + B| ≥ min{p, |A| + |B| − 1}.
(2)
To study inverse problems, we have following definition
Doubling constant: For a finite subset A of a group G, the doubling constant
of A is defined to be the constant c satisfying |A + A| = c|A|.
The subject Freiman initiated was: if A is a finite subset of a group G, and
has a ‘small’ doubling constant, can we describe structure of A? In this
connection, Freiman established the following [6]:
Theorem A. If A is a subset of integers of cardinality k ≥ 2 and the inequality |A+A| = 2k −1+b ≤ 3k −4 holds then A is a subset of an arithmetic
progression of length k + b.
This theorem is known as Freiman’s 3k − 4 theorem. We remark that the
Lemma 2, in section 3 of the article, is a reformulation of Freiman’s 3k − 4
theorem. Seeking application to Diophantine Frobenius problem, Lev [14]
generalized Freiman’s 3k − 4 theorem for j−fold sum A + . . . + A for any
j ≥ 2. For some more work in these direction we refer to [6, 11].
The question which interests us is, whether Theorem A holds for groups
other than Z. There is considerable amount of literature on this question
[2, 7, 9, 10].
In this article we establish Freiman’s 3k − 4 theorem for the group Z × G, in
a restricted sense, where G is any group. A special case of this theorem finds
an immediate application, as is illustrated in [1]. In fact this application was
the motivating factor for this project. We prove the following theorem
Theorem 1. Let k ≥ 3 be any integer and G be any group. Consider a
subset A = {(ai , xi ) : 1 ≤ i ≤ k} of Z × G such that projection to the first
coordinate restricted to A is injective. If |A + A| ≤ 3k − 4, then A is subset
of an arithmetic progression of length k + b, where |A + A| = 2k − 1 + b, for
A = {ai : 1 ≤ i ≤ k}.
Actually we prove a little more: we prove that the first coordinate is part
of an arithmetic progression of length k + b, which governs the second coordinates too, in the sense that there exist x and y in G such that xi = ai x + y
for all i.
In section 2 we introduce a concept of ‘structured set’ in groups Z and Z ×G.
The main theorem of the article is Theorem 3, from which Theorem 1 is
straightforward deduction. The proofs are worked out in section 3. In Corollary 1 we show that a ‘non-structured set’ tend to have large sumset. Corollary 2 gives a sufficient condition for subset of integers to be ‘structured set’.
In section 4 we prove 3k − 4 Theorem for some class of groups G, which
not necessarily are abelian. For subset A of G we will write A2 = {ab : a ∈
A, b ∈ B}. Using the ideas of section 3 we have
Theorem 2. Let G be an ordered group and S be a finite subset of G. If
|S 2 | ≤ 3|S| − 4, then the subgroup generated by S is an abelian subgroup of
G and is generated by 2 elements.
2
Structured Sets
Consider the abelian group Z, written additively. For a pair (X, A) of subsets
of Z with X ⊂ A, we use the notations X (1) = (X + X − X) ∩ A and for i > 1
we shall write X (i) = (X (i−1) )(1) . We define X (∞) = ∪i≥1 X (i) . Note that the
definition of X (∞) depends on the pair (X, A).
Definition: A subset A of Z is called structured set if there is a two element
subset X = {g1, g2 } ⊂ A such that g2 − g1 = 1 and A = ∪i≥1 X (i) .
Motivation for the definition is the following context:
Let (G, +) be an abelian group. Consider a subset A = {(ai , xi ) : 1 ≤ i ≤ s}
of Z × G, such that
(i) π1 (A) = A is structured and
(ii) the implication ai + aj = ak + al =⇒ xi + xj = xk + xl holds always.
Then there are elements x, y ∈ G such that xi = ai x + y, for all i.
Definition: A subset A ⊂ Z × G is structured if the image π1 (A) of the
first projection is structured subset of Z and there are x, y ∈ G satisfying
xi = ai x + y.
Now we state the main theorem of the article.
Theorem 3. Let k ≥ 3 be any integer and G be an abelian group. Consider
a subset A = {(ai , xi ) : 1 ≤ i ≤ k} of Z × G such that projection to the
first coordinate restricted to A is injective. If |A + A| ≤ 3k − 4, then A is
2 − isomorphic to a structured set.
3
Proof of Theorem 3
We have following elementary lemma.
Lemma 1. Let A = {a1 < a2 < . . . < ak } be a subset of k ≥ 3 integers. If
ak−1 and ak are not successive terms of any arithmetic progression containing
A then, for B = {a1 , a2 , . . . , ak−1}, we have |A + A| ≥ |B + B| + 3.
Proof. Note that ak + ak , ak + ak−1 are in A + A but not in B + B. We need
to produce one more element in A + A which is not in B + B.
We consider the decreasing arithmetic progression c1 = ak , c2 = ak−1 , c3 =
ak − 2(ak − ak−1 ), . . .. Then A is not contained in the arithmetic progression
c1 , c2 , . . . .
On the other hand if ak +ak−2 ∈ B +B then ak−2 = c3 . Similarly, we see that
if ak + ai ∈ B + B, for i ≤ k − 2, then ai = cf (i) for some f (i). Consequently
we see that A is contained in the arithmetic progression c1 , c2 , . . . , which is
not possible.
Since both dilation and translation by a fixed element are 2−isomorphism,
hence for A = π1 (A) we will always assume that min(A) = 0 and gcd(A) = 1.
Let A = {a1 = 0 < a2 < . . . < ak }, we define R = min {ak − k + 3, k}. The
following lemma is an immediate consequence of Theorem A.
Lemma 2. |A + A| ≥ 2k + R − 3.
Proof. Note that R ≤ k. If the lemma does not hold, we have
|A + A| < 2k + R − 3 ≤ 3k − 4.
Let |A + A| = 2k − 1 + b, then by Theorem A we see that A is contained
in an arithmetic progression of length k + b. Hence we have ak ≤ k + b − 1
and consequently R ≤ b + 2. This gives 2k + R − 3 ≤ 2k − 1 + b, which is a
contradiction to our assumption 2k − 1 + b < 2k + R − 3.
Proof. (Theorem 3) We will use induction on k. For k = 3, we have |A+A| ≥
5 and 3k − 4 = 5. Thus
|A + A| = |A + A| = 3k − 4.
(3)
Since min(A) = 0 and gcd(A) = 1, so by equation (3) we have A = {0, 1, 2}.
Also equation (3) fprces xi + xj = xk + xl , whenever ai + aj = ak + al . Let
x, y ∈ G be such that x1 = a1 x + y and x2 = a2 x + y. If x3 6= a3 x + y, then
|A + A| > |A + A| = 3k − 4, which is a contradiction. Thus A is a structured
set.
Now we assume k > 3 and let B = {(ai , xi ) : 1 ≤ i ≤ k − 1}.
Case (1): B is structured set.
If A is not structured, then by definition, (B + B) ∩ ((ak , xk ) + B) = ∅. As
a consequence we have
|A + A| ≥ |B + B| + |B|.
(4)
Using the trivial lower bound on |B + B|, coming from the first coordinate,
we see that |A + A| ≥ 3k − 4. The element (ak , xk ) + (ak , xk ) is not yet
considered in the inequality (4), thus we have |A + A| ≥ 3k − 3, which is a
contradiction.
Case (2): B is not structured.
By induction hypothesis we get |B + B| ≥ 3(k − 1) − 3. If ak−1 6= ak − 1, then
using Lemma 1, with A = π1 (A) and B = π1 (B), one immediately obtains
|A + A| ≥ |B + B| + 3 ≥ 3k − 3,
which is a contradiction.
When ak−1 = ak − 1, one can solve for x, y ∈ G satisfying xk = ak x + y
and xk−1 = ak−1 x + y. We see that there are atleast two elements (ak +
ak , xk + xk ) and (ak + ak−1 , xk + xk−1 ) in A + A which are not in B + B. If
(ak +ak−2 , xk +xk−2 ) is in B+B, then ak +ak−2 = 2ak−1 and xk +xk−2 = 2xk−1 .
As a consequence xk−2 = ak−2 x + y. Continuing this way, either we get
|A + A| ≥ |B + B| + 3 ≥ 3k − 3
or xi = ai x + y holds for all i. The first one leads to a contradiction to the
hypothesis on |A + A| and the second one gives that A is structured.
We now deduce Theorem 1 from Theorem 3. By Lemma 2, π1 (A) is
contained in an arithmetic progression of length k + R − 2, so b ≤ R − 2.
If b < R − 2, then |A + A| = 2k − 1 + b < 2k + R − 3, which contradicts
Lemma 2 and hence b = R − 2. Also from Theorem 3 we have x, y ∈ G such
that xi = ai x + y for all i. Consequently A is contained in an arithmetic
progression of length k + R − 2.
As an immediate corollary to Theorem 3 we have
Corollary 1. Let A be a subset of k ≥ 3 integers with min(A) = 0 and
gcd(A) = 1. If A is not structured then |A + A| > 3|A| − 4.
Proof. Let G be any finite abelian group. Consider the subset A = {(a, 0) :
a ∈ A} of Z × G. If |A + A| ≤ 3|A| − 4, then Theorem 3 will give that A is
a structured set and by definition, so is A.
The following corollary gives a sufficient condition for a set to be structured set.
Corollary 2. Let N ≥ 2, and A ⊂ [0, N − 1]. If |A| ≥ 2N/3 + 1, then A is
a structured set.
Proof. Note that for N ≤ 4, it is easy to see that A is structured. So assume
N ≥ 5. Here |A + A| ⊂ [0, 2N − 2] and hence |A + A| ≤ 2N − 1 ≤ 3|A| − 4.
From corollary 1 it follows that A is a structured set.
The set A = [0, N − 1] \ {a < N : a ≡ 2 (mod 3)} is not a structured
set, though |A| ≥ 2N/3. But in this case, we note that the sumset A + A is
bigger than 3|A| − 4.
4
Non-abelian groups
Let G be a group, not necessarily abelian.
Definition: For a finite subset A of G we say that A is a weakly structured
set if there are two commuting elements x, y ∈ G and a positive integer N
such that A ⊂ {yxt : t = −N, . . . , N}.
Thus, weakly structured sets are those contained in a geometric progression for which the first element and the ratio commute. We remark that,
structured sets of Z × G are weakly structured and Theorem 3 is true for
non-abelian G as well.
Now onwards, we will assume that G is an ordered group, i.e, there is a
total order < on G and for a, b ∈ G with a < b, the relation xay < xby holds
for all x, y ∈ G. Note that inequality (1) is valid for any ordered group G.
Along the line of Theorem 3 we prove
Theorem 4. Let S be a finite subset of G with |S| = k ≥ 3. If |S 2 | ≤ 3|S|−4,
then S is weakly structured and one can chose x, y such that S ⊂ {yxi : 0 ≤
i ≤ N − 1} for N = |S 2 | − |S|.
Proof. We will use induction on k. Write S = {x1 < . . . < xk }. For k = 3,
we have |S 2 | ≤ 5. We have five distinct elements x21 < x1 x2 < x22 < x2 x3 < x23
in S 2 . Also x1 x3 ∈ S 2 , so it has to be one from the five elements mentioned.
Using the order relation, we get x1 x3 = x22 . Also we see that x1 x2 = x2 x1 .
2
Let y = x1 and x = x2 x−1
1 , clearly S = {y, yx, yx }.
Now we assume that k ≥ 4 and the theorem is true for any set T with
|T | ≤ k − 1. Put T = {x1 , . . . , xk−1 }.
Case (1): T is weakly structured. Let T ⊂ {yxj : j = −N, . . . , N} with
commuting elements x, y.
If S is weakly structured then nothing to prove. Assume that S is not weakly
structured.
If there are yxi , yxu , yxv ∈ T such that xk yxi = yxu yxv , then we get xk =
yx(u+v−y) . This is a contradiction to our assumption that S is not weakly
structured. Thus we get T 2 ∩ xk T = ∅. Consequently |S 2 | ≥ |T 2 | + |T | ≥
3k − 3, a contradiction.
Case (2): T is not weakly structured.
By induction hypothesis we have |T 2| ≥ 3|T | − 3 = 3|S| − 6. Using order
of G we find that x2k and xk−1 xk are not in T 2 . Further, if xk−1 xk 6= xk xk−1
then we get |S 2 | ≥ |T 2 | + 3, which is a contradiction. Also if xk−2 xk 6= x2k−1 ,
then we already get |S 2 | ≥ |T 2 | + 3, leading to a contradiction.
So let xk−2 xk = x2k−1 . Similarly it follows that xk xk−2 = x2k−1 , put y =
2
xk , x = xk−1 x−1
k . Then x and y commute and xk = y, xk−1 = yx, xk−2 = yx .
Continuing this way either we get S is a weakly structured set or |S 2 | ≥
|T 2 | + 3. The latter condition yields a contradiction. Thus we get that S is
weakly structured.
Let S = {yxti : 1 ≤ i ≤ k}. Clearly S is 2 − isomorphic to the subset
{ti : 1 ≤ i ≤ k} of Z, from which the last assertion of the theorem follows.
Theorem 2 is a straight forward deduction from Theorem 4.
We remark that, Freiman has strengthened his Theorem A, by studying
extremal sets A ⊂ Z for which one has |A + A| = 3|A| − 3, see [6]. With
this strengthening of Theorem A, authors in [7] obtain a stronger theorem
than Theorem 4. Consequently they obtain a stronger version of Theorem 2,
where one only assumes |S 2 | ≥ 3|S| − 2.
References
[1] R. Balasubramanian, Prem Prakash Pandey,
Deshouillers and Freiman, submitted.
On a theorem of
[2] Y. Bilu, Structure of sets with small sumsets, Asterique 258 (1999), 77108.
[3] G. Freiman, ‘Inverse Problems of the additive theory of numbers. On the
addition of sets of residues with respect to a prime modulus’, Dokl. Akad.
Nauk SSSR 141 (1961) 571-573 (Russian), Soviet Math. Dokl. 2 (1961)
1520-1522 (English).
[4] G. Freiman, Foundations of a structural theory of set addition, Translations of Mathematical Monographs 37 (American Mathematical Society,
Providence, RI, 1973).
[5] G. Freiman, ‘Structure theory of set addition’, Asterisque 258 (1999) 133.
[6] G. Freiman, On the addition of finite sets. I. Izv. Vyss. Ucebn. Zaved.
Matematika 6 (13) (1959), 202-213.
[7] G. Freiman, M. Herzog, P. Longobardi, M. Maj, Small doubling in ordered
groups. J. Aust. Math. Soc. 96(2014), no. 3,316-325.
[8] Jean-Marc Deshouillers, Gregory A. Freiman, A step beyong Kneser’s
theorem for abelian finite groups, Proc. London Math. Soc. (3) 86 (2003),
no. 1, 1-28.
[9] B. Green, I. Ruzsa, Freiman’s theorem in an arbitrary abelian group, J.
Lond. Math. Soc. (2) 75 (1) 2007, 163-175.
[10] Y. Hamidoune, A. Llado, O. Sera, On subsets with small product in
torsion free groups, Combinatorica 18 (4) (1998), 529-540.
[11] Y. Hamidoune, A. Plagne, A multiple set version of 3k − 3 theorem,
Rev. Mat. Iberoamericana 21 (2005), no. 1, 133-161.
[12] L. Husbands, ‘Approximate Groups in Additive Combinatorics: A review
of Methods and Literature’ Masters Thesis, University of Bristol, 2009.
[13] M. Kneser, ’Abschatzung der asymptotischen Dichte von Summenmengen’, Math. Z. 58(1953), 459-484.
[14] V. Lev, Structure theorem for multiple addition and the Frobenius problem. J. Number Theory 58 (1996), 79-88.
[15] J. Steinig . ‘On freiman’s theorems concerning the sum of two finite sets
of integers, Asterisque 258 (1999) 129-140.
[16] T. Tao, Van, Vu, Additive Combinatorics, Cambridge Studies in Mathematics 105, 2006.