BASIC ADDITION THEOREMS IN GROUPS
SHALOM ELIAHOU
Abstract. Let G be a group written additively. Given finite subsets A, B ⊂ G, how
small can the Minkowski sum A + B = {a + b | a ∈ A, b ∈ B} be in terms of |A|, |B|?
Here we present basic tools and results related to this problem, both in commutative
groups, where quite satisfactory answers are now available, and in non-commutative
groups where much less is known. We also consider the more delicate analogous
problem for the restricted sum A +0 B = {a + b | a ∈ A, b ∈ B, a 6= b}.
Contents
Introduction
1. Lower bounds on sumset sizes
1.1. The Dyson transform
1.2. The theorem of Cauchy-Davenport
1.3. The theorem of Kneser
1.4. The small sumsets property
1.5. The case of abelian groups
1.6. The theorem of Olson
1.7. The isoperimetric method
2. The polynomial method
2.1. Back to Cauchy-Davenport
2.2. The ex-conjecture of Erdős-Heilbronn
3. More about Cauchy-Davenport
3.1. The theorem of Vosper
3.2. The theorem of Pollard
References
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Introduction
Throughout theses notes, we consider an arbitrary group G, usually written additively, and shall focus on lower bounds for the size of finite sumsets A + B ⊂ G.
Definition 0.1. Let A, B ⊂ G be non-empty subsets. The Minkowski sum of A, B is
the subset
A + B = {a + b | a ∈ A, b ∈ B}.
A natural question arises: if A, B are finite of given cardinalities, how small can
|A + B| be? Let r, s be positive integers such that r, s ≤ |G|. We denote by
µG (r, s) = min{|A + B| | A, B ⊂ G, |A| = r, |B| = s}
Date: July 21, 2013.
1991 Mathematics Subject Classification. 11B75, 20D60, 05A05.
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SHALOM ELIAHOU
the minimal cardinality of the sumsets A + B where A, B range over all subsets of G
satisfying |A| = r, |B| = s.
Thus, the above question amounts to determine the function µG for a given group
G. The first result in this direction dates back to 1813 and is due to Cauchy [5], with
the determination of µG for all groups of prime order, i.e. for G = Z/pZ with p prime.
This result, forgotten and later rediscovered in 1935 by Davenport [6], reads
µZ/pZ (r, s) = min(r + s − 1, p)
for all 1 ≤ r, s ≤ p. Only recently has µG been completely determined for all abelian
groups G [9]. The result and its proof will be presented here. In contrast, the nonabelian case still remains very mysterious.
Getting upper bounds for µG (r, s) can be achieved by an explicit construction of
suitably small sumsets A + B with |A| = r, |B| = s. Proving lower bounds for µG (r, s)
is difficult in general and requires tools, such as:
• Dyson’s and Kemperman’s transforms,
• Kneser’s theorem,
• Alon-Tarsi’s polynomial method,
• Hamidoune’s isoperimetric method.
A simple example occurs with the group of integers G = Z. In that instance, one
has
µZ (r, s) = r + s − 1
for all integers r, s ≥ 1. The upper bound ≤ is provided by the particular pair A =
{0, 1, . . . , r − 1}, B = {0, 1, . . . , s − 1}, for which A + B = {0, 1, . . . , r + s − 2} is of
cardinality r + s − 1. The lower bound ≥ easily follows from the ordered nature of
Z. Indeed, if A = {a1 < · · · < ar }, B = {b1 < · · · < bs }, then A + B contains the
(r + s − 1)−subset
{a1 + b1 < a2 + b1 < · · · < ar + b1 < ar + b2 < · · · < ar + bs }.
Determining µG for other groups may be very demanding. Our goal in these notes
is to present some basic tools and results towards it.
1. Lower bounds on sumset sizes
Let (G, +) be a group and A, B ⊂ G. For c ∈ G, denote by rA,B (c) the number of
distinct ways of representing c as the sum of an element in A and an element in B, i.e.
rA,B (c) = {(a, b) ∈ A × B | c = a + b}.
By definition, we have c ∈ A + B if and only rA,B (c) ≥ 1.
Example 1.1. In G = Z, let A = {1, 2, 3, 4}, B = {1, 2, 4, 6, 8}. Then A + B =
{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, and one easily checks that rA,B (2) = 1, rA,B (7) = 2 and
rA,B (5) = 3.
Besides the sum A + B, we shall also consider the difference of A and B, namely
A − B = {a − b | a ∈ A, b ∈ B}.
Of course, A − B is nothing else than the sum of A and −B, where
−B = {−b | b ∈ B}.
ADDITION THEOREMS IN GROUPS
3
When B = {b} is a singleton, we shall write A + b rather than A + {b}. Note that
|A + b| = |A|,
since the translations in a group are bijective. Similarly,
| − A| = |A|,
since taking the inverse in a group is also a bijection.
Here is one first basic lemma, which easily implies that if G is a finite group, then
for all positive integers r, s, we have
r + s ≥ |G| + 1 =⇒ µG (r, s) = |G|.
Lemma 1.2. Let G be a finite group and A, B ⊂ G. If |A| + |B| ≥ |G| + 1, then
A + B = G. More generally, if |A| + |B| = |G| + t for some t ∈ N, then rA,B (g) ≥ t
for all g ∈ G.
Proof. Let g ∈ G. First note the equality
rA,B (g) = |A ∩ (g − B)|.
Indeed, we have g = a + b for some pair (a, b) ∈ A × B if and only the element a = g − b
belongs to A ∩ (g − B). It remains to estimate |A ∩ (g − B)|. For this, one uses the
well-known equality for finite subsets X1 , X2 :
(1.1)
|X1 ∪ X2 | + |X1 ∩ X2 | = |X1 | + |X2 |.
Applying it to the present situation, we have
|A ∩ (g − B)| =
=
≥
=
|A| + |(g − B)| − |A ∪ (g − B)|
|A| + |B| − |A ∪ (g − B)|
|G| + t − |G|
t.
1.1. The Dyson transform.
Definition 1.3. Let G be an abelian group and A1 , A2 nonempty finite subsets of G.
Let e be any element of G. The Dyson transform of (A1 , A2 ) by e is the pair (A01 , A02 )
= (A1 (e), A2 (e)) defined as follows:
A01 = A1 (e) = A1 ∪ (A2 + e),
A02 = A2 (e) = (A1 − e) ∩ A2 .
This transform enjoys two useful properties, namely:
A01 + A02 ⊂ A1 + A2 ,
|A01 | + |A02 | = |A1 | + |A2 |.
The first property requires the commutativity of G and is easy to check. As for the
second one, using (1.1) once again, we have
|A01 | + |A02 | =
=
=
=
|A1 ∪ (A2 + e)| + |(A1 − e) ∩ A2 |
|A1 ∪ (A2 + e)| + |A1 ∩ (A2 + e)|
|A1 | + |(A2 + e)|
|A1 | + |A2 |.
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SHALOM ELIAHOU
1.2. The theorem of Cauchy-Davenport. We now consider the case of cyclic groups.
Cauchy [5] and Davenport [6] independently obtained an optimal lower bound on sumset sizes for G = Z/mZ with m = p a prime number, and then Chowla partially
generalized their result in 1935 for arbitrary m ∈ N. We now state and prove Chowla’s
result. The Cauchy-Davenport will follow as a corollary.
Theorem 1.1. (Cauchy-Davenport-Chowla) Let G = Z/mZ. Let A, B be two nonempty
subsets of G. If 0 ∈ B and gcd(b, m) = 1 for all b ∈ B \ {0}, then
|A + B| ≥ min(|A| + |B| − 1, m).
Proof. If |A| + |B| > m, it follows from Lemma 1.2 that A + B = G. Thus, we have
|A + B| = m ≥ min(|A| + |B| − 1, m).
Assume now |A| + |B| ≤ m. Then
min(|A| + |B| − 1, m) = |A| + |B| − 1,
and we must show |A + B| ≥ |A| + |B| − 1. We proceed by induction on |B|.
The case |B| = 1 is easy, since then B = {0} and |A + B| = |A| ≥ |A| + 1 − 1.
Assume now |B| ≥ 2. Then there exists b ∈ B such that b 6= 0. There are two cases:
• Case 1: A + b ⊂ A.
Then A + tb ⊂ A for all integer t ≥ 0 . But gcd(b, m) = 1 by hypothesis. Thus the
element b is of order m exactly, whence
{tb | t ≥ 0} = Z/mZ.
It follows that A = Z/mZ, which is impossible since |A| ≤ m − |B| ≤ m − 1.
• Case 2: A + b 6⊂ A.
Then there exists g ∈ A such that g + b ∈
/ A. Consider the Dyson transform (A0 , B 0 )
of (A, B) by g. Note that B 0 = (A − g) ∩ B is nonempty since it contains 0. Moreover,
we have |B 0 | < |B| since b ∈ B \ (A − g). By the induction hypothesis, we have
|A0 + B 0 | ≥ |A0 | + |B 0 | − 1, whence
|A + B| ≥ |A0 + B 0 | ≥ |A0 | + |B 0 | − 1 = |A| + |B| − 1.
Corollary 1.4. Let A, B be two nonempty subsets of Z/pZ with p prime. Then
|A + B| ≥ min(|A| + |B| − 1, p).
Proof. Since p is prime, every 0 6= b ∈ Z/pZ satisfies (b, p) = 1. Thus, up to translating
B in order to have 0 ∈ B, the preceding result applies.
Remark 1.5. This lower bound is optimal. Indeed, consider the subsets
A = {0, 1, . . . , k − 1}, B = {0, 1, . . . , l − 1},
of cardinality k and l, respectively. If k +l ≤ p, then clearly A+B = {0, 1, . . . , k +l−2}
and hence |A + B| = k + l − 1. If on the contrary k + l > p, then by Lemma 1.2 we have
A + B = Z/pZ, of cardinality p. In both cases we have |A + B| = min(|A| + |B| − 1, p).
Corollary 1.6. Let G = Z/pZ with p prime. Then
µG (r, s) = min(r + s − 1, p)
for all 1 ≤ r, s ≤ p.
Proof. The inequality µG (r, s) ≥ min(r + s − 1, p) follows from Corollary 1.4 and the
reverse inequality from Remark 1.5.
ADDITION THEOREMS IN GROUPS
5
We may also consider sums of t ≥ 3 subsets of Z/pZ.
Theorem 1.2. Let A1 , A2 , . . . , At be nonempty subsets of Z/pZ with p prime. Then
|A1 + A2 + · · · + At | ≥ min(|A1 | + |A2 | + · · · + |At | − t + 1, p).
Proof. By induction on t, the case t = 2 having been shown above. Assuming the
inequality true for some t, we have
|(A1 + A2 + · · · + At ) + At+1 | ≥ min(|
t
X
Ai | + |At+1 | − 1, p)
i=1
t+1
X
≥ min(
|Ai | − (t + 1) + 1, p).
i=1
1.3. The theorem of Kneser. We now state and prove an important lower bound on
finite sumset sizes in abelian groups due to Kneser [14]. The presentation below follows
that of [15]. Let us first fix a notation. Let G be an abelian group and X a nonempty
subset of G. We shall denote by Per(X) = {g ∈ G : X + g = X} the stabilizer, also
called the period, of X in G. It is easy to see that Per(X) is a subgroup of G.
Theorem 1.3 (Kneser). Let G be an abelian group and A, B ⊂ G two nonempty finite
subsets of G. Then there exists a finite subgroup H ≤ G such that
|A + B| ≥ |A + H| + |B + H| − |H|.
Moreover, this inequality holds with the subgroup H = Per(A + B).
This theorem is an easy consequence of the following proposition.
Proposition 1.7. Let G be an abelian group and A, B ⊂ G two nonempty finite subsets
of G. Let X ⊂ A + B be a nonempty subset of A + B. Then there exists a superset
X ⊂ Y ⊂ A + B such that
|Y | ≥ |A| + |B| − |Per(Y )|.
Proof that Proposition 1.7 implies the theorem of Kneser. Let H = Per(A + B). Then
we have A + B = A + B + H, and hence A + B = (A + H) + (B + H) since G is
abelian. We apply Proposition 1.7 to the pair of subsets A0 = A + H, B 0 = B + H and
to X = A0 + B 0 = A + B. We then obtain, with Y = X necessarily, the inequality
|A0 + B 0 | ≥ |A0 | + |B 0 | − |Per(A0 + B 0 )|,
i.e. |A + B| ≥ |A + H| + |B + H| − |H|, as desired.
It remains to show Proposition 1.7, which we will do by induction on |X|. The case
|X| = 1 uses Dyson’s transform, whereas for the induction step, we shall need the
following lemma.
Lemma 1.8. Let G be an arbitrary group and H1 , H2 two finite normal subgroups of
G. Let E1 be a finite union of H1 -classes and E2 a finite union of H2 -classes. Assume
E1 6⊂ E2 and E2 6⊂ E1 . Then we have
|E1 \ E2 | ≥ |H2 \ H1 | or |E2 \ E1 | ≥ |H1 \ H2 |.
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SHALOM ELIAHOU
Proof. Set k1 = |H1 |, k2 = |H2 |.
(1) Assume first H1 ∩H2 = {1}. We then have |H1 \H2 | = k1 −1 and |H2 \H1 | = k2 −1.
Fix an element e ∈ E2 \ E1 and set R = eH1 H2 . It suffices to show that either
|R ∩ (E1 \ E2 )| ≥ |H2 \ H1 |, or |R ∩ (E2 \ E1 )| ≥ |H1 \ H2 |.
Observe that R is both a finite disjoint union of H1 −classes and of H2 −classes, since
R = eH1 H2 = H1 eH2 . We claim that |R| = k1 k2 . Indeed, if ef1 f2 = eg1 g2 with
fi , gi ∈ Hi for i = 1, 2, then g1 −1 f1 = g2 f2 −1 ∈ H1 ∩ H2 , and since H1 ∩ H2 = {1}, it
follows that f1 = g1 and f2 = g2 . Summarizing, R is a finite disjoint union of k2 classes
mod H1 and a finite disjoint union of k1 classes mod H2 .
Consider the set R ∩ E1 . Then R ∩ E1 is a finite disjoint union of u2 classes mod H1 ,
for a certain 0 ≤ u2 ≤ k2 . We then have |R ∩ E1 | = k1 u2 and |R \ E1 | = k1 (k2 − u2 ).
Similarly, R ∩ E2 is a finite disjoint union of u1 classes mod H2 , for some 0 ≤ u1 ≤ k1 ,
and we have |R ∩ E2 | = u1 k2 and |R \ E2 | = (k1 − u1 )k2 .
We claim that |R ∩ (E1 \ E2 )| = (k1 − u1 )u2 .
This follows from a more general statement: if X1 ⊂ R is a finite union of classes
mod H1 and X2 ⊂ R a finite union of classes mod H2 , then
(1.2)
|X1 ∩ X2 | =
|X1 ||X2 |
.
k1 k2
Indeed, X1 is a finite disjoint union of H1 −classes of the form eH1 f2 with f2 ∈ H2 ,
and X2 is a finite disjoint union of H2 −classes of the form ef1 H2 with f1 ∈ H1 . But the
intersection of two such classes eH1 f2 and ef1 H2 contains exactly one element, namely
ef1 f2 . Indeed, as above, if eg1 f2 = ef1 g2 with g1 ∈ H1 and g2 ∈ H2 , then g1 = f1 and
g2 = f2 . The announced formula for |X1 ∩ X2 | is now proved.
By applying (1.2) to X1 = R ∩ E1 and X2 = R \ E2 , it follows that |R ∩ (E1 \ E2 )| =
u2 (k1 − u1 ), as stated. Similarly, we have |R ∩ (E2 \ E1 )| = u1 (k2 − u2 ).
Now, as e ∈ R ∩ (E2 \ E1 ), it follows that u1 ≥ 1 and k2 − u2 ≥ 1. We shall distinguish two cases: either u1 = k1 , or u1 ≤ k1 − 1.
Case 1 : u1 = k1 . Then |R ∩ (E2 \ E1 )| = k1 (k2 − u2 ) ≥ k1 ≥ |H1 \ H2 |, as desired.
Case 2 : u1 ≤ k1 − 1. Assembling and summing our formulas, and using the two
inequalities k1 ≥ u1 + 1 and k2 ≥ u2 + 1, we obtain
|R ∩ (E1 \ E2 )| + |R ∩ (E2 \ E1 )| − |H1 \ H2 | − |H2 \ H1 |
= u1 (k2 − u2 ) + (k1 − u1 )u2 − (k1 − 1) − (k2 − 1)
= (u1 − 1)k2 + k1 (u2 − 1) − 2u1 u2 + 2
≥ (u1 − 1)(u2 + 1) + (u1 + 1)(u2 − 1) − 2u1 u2 + 2
= 0.
It follows that either |R ∩ (E1 \ E2 )| − |H2 \ H1 | ≥ 0, or |R ∩ (E2 \ E1 )| − |H1 \ H2 | ≥ 0,
as desired.
(2) Assume now |H1 ∩ H2 | ≥ 2. As H1 ∩ H2 is a normal subgroup of G, we may
apply case (1) to the quotient group G = G/(H1 ∩ H2 ) and to its normal subgroups
Hi = Hi /(H1 ∩ H2 ) for i = 1, 2. Considering the quotient map π : G → G, let
Ei = π(Ei ) for i =1, 2. As Ei = π −1 (E i ), we
still have
E1 6⊂ E2 and E2 6⊂ E1 . By
(1), we either have E1 \ E2 ≥ H2 \ H1 , or E2 \ E1 ≥ H1 \ H2 . Multiplying these
inequalities by |H1 ∩ H2 |, we obtain the desired alternative.
We are now in a position to prove Proposition 1.7.
ADDITION THEOREMS IN GROUPS
7
Proof. We proceed by induction on |X|.
• Assume first |X| = 1. Without loss of generality, we may further assume X ⊂ A
and 0 ∈ B. Indeed, by hypothesis X = {g} for a certain g = a + b with a ∈ A and
b ∈ B. Let us set X 0 = X − b = {a} and B 0 = B − b. If Y 0 satisfies the required
properties for the pair A, B 0 , then Y = Y 0 + b satisfies the required properties for the
pair A, B. Therefore, from now on, we shall assume X ⊂ A and 0 ∈ B. Consequently,
we have A ⊂ A+B. We now proceed by induction on |B|, and we distinguish two cases.
Case 1. A = A+B. It follows that B ⊂ Per(A). Then, the set Y = A has the required
properties. Indeed, X ⊂ A = Y , and we have |Y | = |A|+|B|−|B| ≥ |A|+|B|−|Per(Y )|.
Note that this case contains in particular the case B = {0}, i.e. the case |B| = 1 which
starts the induction.
Case 2. A 6= A + B. Then there exist a1 ∈ A and b1 ∈ B such that a1 + b1 ∈
/ A.
We then define A1 = A ∪ (a1 + B) and B1 = (A − a1 ) ∩ B. The pair A1 , B1 is the
Dyson transform of the pair A, B by a1 . We have |B1 | < |B|, since B1 ⊂ B and B \ B1
contains b1 as b1 ∈
/ (A − a1 ). Let us also note that 0 ∈ B1 . By the induction hypothesis
on |B|, there exists a set Y satisfying
X ⊂ Y ⊂ A1 + B1 and |Y | ≥ |A1 | + |B1 | − |Per(Y )|.
Now, it follows from the general properties of the Dyson transform that A1 + B1 ⊂
A + B and |A1 | + |B1 | = |A| + |B|. Thus we also have
X ⊂ Y ⊂ A + B and |Y | ≥ |A| + |B| − |Per(Y )|,
as desired. This finishes the case |X| = 1.
• Assume now X = {g1 , . . . , gr } with r ≥ 2. We shall use the above lemma for
the induction step. Let us set X1 = {g1 , . . . , gr−1 } and X2 = {gr }. By the induction
hypothesis on r, there exist sets Y1 , Y2 satisfying
X1 ⊂ Y1 ⊂ A + B,
X2 ⊂ Y2 ⊂ A + B,
|Y1 | ≥ |A| + |B| − |H1 |,
|Y2 | ≥ |A| + |B| − |H2 |,
where H1 = Per(Y1 ) and H2 = Per(Y2 ). It follows that Yi is a finite union of Hi −classes
for i = 1, 2.
If Y1 ⊂ Y2 , then Y = Y2 satisfies the required constraints: X ⊂ Y ⊂ A + B and
|Y | ≥ |A| + |B| − |Per(Y )|. The case Y2 ⊂ Y1 is treated in the same way.
Assume now Y1 6⊂ Y2 and Y2 6⊂ Y1 . By the above lemma, we either have |Y1 \ Y2 | ≥
|H2 \ H1 |, or |Y2 \ Y1 | ≥ |H1 \ H2 |.
Assume first |Y2 \ Y1 | ≥ |H1 \ H2 |. Let Y = Y1 ∪ Y2 and H = H1 ∩ H2 . Then
H ⊂ Per(Y ). It suffices to show that |Y | ≥ |A| + |B| − |H|. We know from the above
that |Y1 | + |H1 | ≥ |A| + |B|. Now
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SHALOM ELIAHOU
|Y | =
=
≥
=
≥
|Y1 ∪ Y2 |
|Y1 | + |Y2 \ Y1 |
|Y1 | + |H1 \ H2 |
|Y1 | + |H1 | − |H|
|A| + |B| − |H|,
as desired.
The case |Y1 \ Y2 | ≥ |H2 \ H1 | may be treated in the same way. This finishes the
proofs of Proposition 1.7 and of Kneser’s theorem.
1.4. The small sumsets property. In view of generalizing Cauchy-Davenport’s theorem to arbitrary abelian groups, we shall need the following condition on a group F
which guarantess the existence of sufficiently small sumsets [9, 10].
Definition 1.9. Let (F, +) be a group. We say that F has the small sumsets property
(SSP) if for all positive integers r, s < |F |, we have
µF (r, s) ≤ r + s − 1.
Said otherwise, F satisfies SSP if for all 1 ≤ r, s < |F |, there exist subsets A, B ⊂ F
such that |A| = r, |B| = s and |A + B| ≤ |A| + |B| − 1 = r + s − 1.
For example, every cyclic group C satisfies SSP. Indeed, for C = Z or Z/nZ, with
its natural ordering, it suffices to take the initial segments A = {0, 1, . . . , r − 1},
B = {0, 1, . . . , s − 1}. It is then clear that |A + B| ≤ r + s − 1. Note also that every
group containing a copy of Z (i.e. containing an element of infinite order) satisfies SSP.
Let us now show that every abelian group satisfies SSP. This is an easy consequence
of the following lemma.
Lemma 1.10. Let F be a group satisfying the small sumsets property, and let C =
Z/nZ. Then C × F also satisfies the small sumsets property.
Proof. If F is infinite, then of course every group containing F also satisfies SSP. Thus,
we may assume that F is finite, say of order |F | = d. Let r, s < |C × F | = nd be
positive integers. Consider the euclidean division of r, s by d:
r = r1 d + r2
s = s1 d + s2
with 0 ≤ r2 , s2 ≤ d − 1. Choose subsets A, B ⊂ C × F of the form
A = A1 × F ∪ {a} × A2
B = B1 × F ∪ {b} × B2
where A1 ∪ {a}, B1 ∪ {b} ⊂ C; A2 , B2 ⊂ F ; a 6∈ A1 , b 6∈ B1 ; and |Ai | = ri , |Bi | = si for
i = 1, 2. Thus, |A| = r1 d + r2 = r, and similarly |B| = s1 d + s2 = s.
For the natural ordering of C = Z/nZ, choose A and A ∪ {a}, more specifically, as
being the initial segments in C of length r1 and r1 + 1, respectively. Similarly, B and
B ∪ {b} are chosen as the initial segments in C of length s1 and s1 + 1. Now, given
that F satisfies SSP, we may assume that A2 , B2 ⊂ F satisfy
|A2 + B2 | ≤ |A2 | + |B2 | − 1 = r2 + s2 − 1.
We claim that |A + B| ≤ |A| + |B| − 1.
ADDITION THEOREMS IN GROUPS
9
If one among A1 , A2 , B1 , B2 is empty, the proof of this inequality is easy and left to
the reader. Assume now that A1 , A2 , B1 , B2 are all nonempty. We then have
[
[
A + B ⊂ (A1 ∪ {a} + B1 ) × F
(A1 + B1 ∪ {b}) × F
{a + b} × (A2 + B2 ),
as easily verified. A key point to note is that, in C = Z/nZ, a sum of initial segments
is still an initial segment. Thus, we have the alternative
A1 ∪ {a} + B1 ⊂ A1 + B1 ∪ {b}
or
A1 ∪ {a} + B1 ⊃ A1 + B1 ∪ {b}.
Symmetrically, we may assume that A1 ∪ {a} + B1 contains A1 + B1 ∪ {b}. It follows
that
[
A + B ⊂ (A1 ∪ {a} + B1 ) × F
{a + b} × (A2 + B2 ).
Hence, |A + B| ≤ (r1 + s1 )d + (r2 + s2 − 1) = r + s − 1, as desired.
Proposition 1.11. Let G be an abelian group. Then G satisfies the small sumsets
property.
Proof. If G is finite, then it is isomorphic to a product of finite cyclic groups, and hence
it satisfies SSP by repeated application of Lemma 1.10. If G is infinite, then either it
contains a copy of Z, or else it contains finite subgroups of arbitrarily large order. In
both cases, it follows from the above that G satisfies SSP.
Remark 1.12. A slight modification of the proof leads to the slightly more general
following result.
Lemma 1.13. Let 0 → F → G → C → 0 be a short exact sequence of groups. Assume
that F satisfies SSP and that C is cyclic. Then G also satisfies SSP.
It is not known whether the conclusion of the lemma still holds if one replaces the
hypothesis that C is cyclic by the weaker one that C satisfies SSP. Here is an easy
consequence of the lemma.
Proposition 1.14. Let G be a finite solvable group. Then G satisfies SSP.
It is not known which groups satisfy SSP. The smallest open case is that of the
simple group A5 of order 60. We know that µA5 (15, 15) ≤ 30 by construction, and we
conjecture the equality µA5 (15, 15) = 30. If that were true, it would show that A5 does
not satisfy SSP.
1.5. The case of abelian groups. We shall now determine the function µG (r, s) for
an arbitrary abelian group G. This result [9] generalizes earlier partial results for more
specific groups [5, 6, 20, 4, 8, 11].
Notation. Let G be a group. We denote by H(G) the set of orders of finite subgroups
of G, i.e.
H(G) = {d ∈ N | there exists a subgroup H ≤ G such that d = |H|}.
Note that if G is finite of order n, then H(G) is a subset of the set D(n) of divisors of
n, with H(G) = D(n) if G is abelian. Note also that H(G) = {1} if and only if G is
torsion-free.
Theorem 1.4 ([9]). Let G be an abelian group. Then
µG (r, s) = min
r/d + s/d − 1 d.
d∈H(G)
for all integers 1 ≤ r, s < |G|.
10
SHALOM ELIAHOU
It is useful to introduce a notation for the right term above:
κG (r, s) = min
r/d + s/d − 1 d.
d∈H(G)
The proof of the theorem is in two parts. In order to establish the lower bound
µG (r, s) ≥ κG (r, s), we use Kneser’s theorem. On the other hand, the inequality
µG (r, s) ≤ κG (r, s) requires the construction of sumsets reaching the bound κG (r, s).
The key of this construction is the fact that every abelian group G0 satisfies the SSP,
i.e. that µG0 (r, s) ≤ r + s − 1 for all positive integers r, s ≤ |G0 |.
Proof. We distinguish two cases.
• µG (r, s) ≥ κG (r, s) : Let A, B ⊂ G be finite subsets with |A| = r, |B| = s. By
Kneser’s theorem, there exists a finite subgroup H ≤ G such that
|A + B| ≥ |A + H| + |B + H| − |H|.
(Recall that we may take for H the stabilizer of A + B.) Factorizing |H| in the above
expression, we obtain
A + B ≥ |A + H|/|H| + |B + H|/|H| − 1 H .
The key observation is that A + H is a disjoint union of classes mod H. It follows that
the rational number |A + H|/|H| is in fact an integer, of course greater than or equal
to |A|/|H|. Therefore, |A + H|/|H| ≥ d|A|/|H|e = dr/he, where h = |H|. Similarly,
we have |B + H|/|H| ≥ ds/he. Hence
A + B ≥ dr/he + ds/he − 1 h.
Since h ∈ H(G), it follows that
A + B ≥ min
d∈H(G)
r/d + s/d − 1 d = κG (r, s).
• µG (r, s) ≤ κG (r, s) : Let h ∈ H(G), and let H ≤ Gbe asubgroupof G of order h.
Let π : G → G/H be the quotient map. Let us set r = r/h and s = s/h . Since the
group G/H is abelian and hence satisfies SSP, we have µG/H (r, s) ≤ r + s −
1. Thus,
there exist subsets A, B ⊂ G/H, of cardinality r, s respectively, such that A + B ≤
r + s − 1. Let A = π −1 (A), B = π −1 (B). Then
rh,
cardinality
A, B ⊂ G are subsets of
−1
A + B h. Thus,
sh respectively.
Since
A
+
B
=
π
A
+
B
,
we
also
have
|A
+
B|
=
|A + B| = A + B h ≤ (r + s − 1)h, and hence µG (rh, sh) ≤ (r + s − 1)h.
Now, r ≤ rh, s ≤ sh. Since A0 + B0 ⊂ A + B for all subsets A0 ⊂ A, B0 ⊂ B, it
follows that µG (r, s) ≤ µG (rh, sh). In summary, we have
µG (r, s) ≤ µG (rh, sh) ≤ (r + s − 1)h = r/h + s/h − 1 h.
Since h ∈ H(G) is arbitrary, we indeed obtain
µG (r, s) ≤ min r/d + s/d − 1 d.
d∈H(G)
ADDITION THEOREMS IN GROUPS
11
1.6. The theorem of Olson. We now formulate and prove Olson’s theorem [18],
analogous to Kneser’s theorem for non-abelian groups. It also allows one to establish
an inequality of the form
|A + B| ≥ |A| + |B| − |H|
for the sum of two finite subsets A, B of G, with H a suitable finite subgroup of G.
Unfortunately, we can no longer guarantee, as in Kneser’s theorem, that one of these
subgroups H stabilizes A + B.
The essential tool of the proof is Kemperman’s transform [13], a non-commutative
version of Dyson’s transform.
1.6.1. The Kemperman transform. Let G be an arbitrary group and A, B ⊂ G nonempty
finite subsets of G. Let x ∈ G. We define
A0 = A ∪ (A + x)
B 0 = B ∩ (−x + B)
and
A00 = A ∩ (A − x)
B 00 = B ∪ (x + B).
These pairs A0 , B 0 and A00 , B 00 , dependent on x, are the Kemperman transforms of the
pair A, B. Here are some of their elementary properties. First of all,
A0 + B 0 , A00 + B 00 ⊂ A + B
as easily verified. Next, we have the equivalence
B 0 6= ∅ ⇐⇒ x ∈ B − B.
Indeed, for b1 , b2 ∈ B, we have x = b2 − b1 if and only if b1 = −x + b2 belongs to
B 0 = B ∩ (−x + B). Similarly, we have the equivalence
A00 6= ∅ ⇐⇒ x ∈ −A + A.
Finally, it is straightforward to verify that
|A0 | > |A| ⇐⇒ A + x 6= A
and
|B 00 | > |B| ⇐⇒ x + B 6= B.
Lemma 1.15. Let G be a group and A, B ⊂ G two nonempty finite subsets of G. Set
D = (−A + A) ∩ (B − B).
If A + D 6= A or D + B 6= B, then there exists a pair of nonempty finite subsets
A1 , B1 ⊂ G, obtained by Kemperman transform of A, B, such that
(1) A1 + B1 ⊂ A + B
(2) either |A1 | + |B1 | ≥ |A| + |B| and |A1 | > |A|, or |A1 | + |B1 | > |A| + |B|.
Proof. By hypothesis, there exists d ∈ D such that A + d 6= A or d + B 6= B. Let us
set p = |(A + d) \ A|, q = |(d + B) \ B|. We have max(p, q) > 0. We distinguish the
cases p ≥ q and p < q.
• Case p ≥ q. Consider the first Kemperman transform relative to d, i.e. A1 =
A = A ∪ (A + d) and B1 = B 0 = B ∩ (−d + B). We claim that |A1 | = |A| + p and
|B1 | = |B| − q. Indeed, we have
0
|A1 | = |A ∪ (A + d)| = |A| + |(A + d) \ A| = |A| + p,
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SHALOM ELIAHOU
and
|B1 | = |B ∩ (−d + B)| = |B| − |B \ (−d + B)|
= |B| − |(d + B) \ B| = |B| − q.
It follows that |A1 | + |B1 | = |A| + |B| + p − q ≥ |A| + |B|. Moreover, since we have
p = max(p, q) > 0, it follows that A + d 6= A, and hence that |A1 | > |A|.
• Case p < q. Here we consider the second Kemperman transform relative to d, i.e.
A1 = A00 = A ∩ (A − d) and B1 = B 00 = B ∪ (d + B). By analogous computations as
in the first case, we find that
|A1 | = |A| − p, |B1 | = |B| + q.
It follows that |A1 | + |B1 | = |A| + |B| − p + q > |A| + |B|, as desired.
Here is what we obtain by iterating this lemma, i.e. by applying as long as possible
successive Kemperman transforms starting from the pair A, B.
Proposition 1.16. Let G be a group and A, B ⊂ G two nonempty finite subsets. There
exist nonempty subsets E, F ⊂ G such that
(1) E + F ⊂ A + B
(2) |E| + |F | ≥ |A| + |B|
(3) If D = (−E + E) ∩ (F − F ), then E + D = E and D + F = F.
Proof. By successive applications of the preceding lemma, there exists a sequence of
subsets
(A, B) = (A0 , B0 ), (A1 , B1 ), . . . , (Aj , Bj ), . . .
with the following properties: for all j ≥ 1,
(1) Aj + Bj ⊂ Aj−1 + Bj−1 .
(2) In N2 lexicographically ordered,
(|Aj | + |Bj |, |Aj |) > (|Aj−1 | + |Bj−1 |, |Aj−1 |).
Now the integers |Aj |, |Bj | are bounded, since |Aj |, |Bj | ≤ |Aj + Bj | ≤ |A + B| for all
j ≥ 0. It then follows from (2) that the sequence of the (Aj , Bj ) is finite. Hence there
exists an integer n ≥ 0 for which the hypothesis of the lemma no longer applies to the
pair (E, F ) = (An , Bn ). Said otherwise, setting D = (−E + E) ∩ (F − F ), we have
E + D = E and D + F = F . Since Aj + Bj ⊂ A + B and |Aj | + |Bj | ≥ |A| + |B| for
all j ≥ 0, it follows that E + F ⊂ A + B and |E| + |F | ≥ |A| + |B|.
We may now state and prove Olson’s theorem.
Theorem 1.5 ([18]). Let (G, +) be an arbitrary group, not necessarily abelian. Let
A, B ⊂ G be two nonempty finite subsets of G. Then there exists a nonempty subset
S ⊂ A + B and a finite subgroup H of G such that
|S| ≥ |A| + |B| − |H|
and H stabilizes S on the left or on the right, i.e.
H + S = S or S + H = S.
Proof. From the preceding proposition, there exist nonempty finite subsets E, F of
G such that E + F ⊂ A + B and |E| + |F | ≥ |A| + |B|. Moreover, setting D =
(−E + E) ∩ (F − F ), we have E + D = E and D + F = F .
ADDITION THEOREMS IN GROUPS
13
Set S = E + F . It is a nonempty subset of A + B, for the cardinality of which we
would like to give a lower bound in terms of |A|, |B| and |H| for a suitable subgroup
H. We distinguish the cases |E| ≥ |F | and |E| ≤ |F |.
Case |E| ≥ |F |. Here again, we distinguish two cases, according as whether F − F
is contained in −E + E or not.
• Assume first F − F 6⊂ −E + E. Then there exist x1 , x2 ∈ F such that x1 − x2 6∈
−E + E. Thus, we have (E + x1 ) ∩ (E + x2 ) = ∅, and
S = E + F ⊃ (E + x1 ) ∪ (E + x2 ).
This union being disjoint, it follows that |S| ≥ |E + x1 | + |E + x2 | ≥ |E| + |F |. Taking
H = {0} as subgroup of G stabilizing S, we obtain
|S| ≥ |E| + |F | ≥ |A| + |B| ≥ |A| + |B| − |H|,
as required.
• Assume now F − F ⊂ −E + E. It follows that D = F − F . We shall show that
D is then a subgroup of G, that F is a right class of D, and that as the looked-for
subgroup H, a suitable conjugate of D will do.
First of all, 0 ∈ F − F = D, and
D + D = D + F − F = F − F = D.
It easily follows that D is a subgroup of G. Let z ∈ F arbitrary. We have
F = F − z + z ⊂ D + z ⊂ D + F = F.
Hence F = D + z, a right class of D. Set
H = −z + D + z.
Then H, a conjugate of D, is a subgroup of G. Let us show that it stabilizes S on the
right:
S + H = E + F − z + D + z = E + D + z ⊂ E + F = S.
The equality S + H = S being established, it remains to evaluate |S|. We have
|S| = |E + F | ≥ |E| = |E| + |F | − |F | = |E| + |F | − |H| ≥ |A| + |B| − |H|,
as desired.
Case |E| ≤ |F |. The proof is entirely similar to the preceding case. Here, we obtain
a subgroup H stabilizing S on the left.
We are now in a position to prove the following theorem of Kemperman (1956).
Theorem 1.6. Let G be a torsion-free group. Then µG (r, s) = r + s − 1 for all integers
r, s ≥ 1.
Proof. Let A, B ⊂ G such that |A| = r, |B| = s. By Olson’s theorem above, there
exists a nonempty subset S ⊂ A + B and a finite subgroup H ≤ G such that
|S| ≥ |A| + |B| − |H|.
Now H = {1}, since G is torsion-free and hence contains no non-trivial finite subgroup.
It follows that
|A + B| ≥ |S| ≥ r + s − 1.
This shows that µG (r, s) ≥ r + s − 1. Now this lower bound is easy to realize: let x ∈ G
be a nonzero element, and set
A = {ix | 0 ≤ i ≤ r − 1},
B = {ix | 0 ≤ i ≤ s − 1}.
14
SHALOM ELIAHOU
Then |A| = r, |B| = s, because x is of infinite order, and A+B = {ix | 0 ≤ i ≤ r+s−2}
is of cardinality r +s−1 for the same reason. We conclude that µG (r, s) = r +s−1. In the same way, here is another easy consequence of Olson’s theorem.
Second proof of Cauchy-Davenport’s theorem. Let G = Z/pZ and A, B ⊂ G nonempty
subsets of cardinality |A| = r, |B| = s. We have to show |A + B| ≥ min(r + s − 1, p).
By Olson’s theorem, there exists a subgroup H ≤ G such that |A + B| ≥ r + s − |H|,
and H stabilizes a nonempty subset S ⊂ A + B. Here G only has two subgroups,
namely {0} and G.
• If H = {0}, then |A + B| ≥ r + s − 1 and we are done in this case.
• If H = G, then S = A + B = G. This finishes the proof of the theorem.
1.7. The isoperimetric method. The method presented here is due to Hamidoune
[12]. Let G be a group in multiplicative notation. In this section, we fix a finite subset
C ⊂ G with 1 ∈ C. In particular, we have
X ⊂ XC
for all X ⊂ G, and if X is finite, then |XC \ X| = |XC| − |X|.
Definition 1.17. For all X ⊂ G, we denote
∂C X = XC \ X.
We call ∂C X the boundary of X with respect to C. For simplicity, we shall omit the
reference to C and write ∂X instead of ∂C X.
Note that ∂ commute with left translations, i.e.
∂(gX) = g∂(X)
for all g ∈ G and all X ⊂ G. Indeed, we have ∂(gX) = gXC \gX = g(XC \X) = g∂X.
The isoperimetric method consists in considering the sets X ⊂ G minimizing |∂X|.
More precisely, we introduce the following definitions (relative to C).
Definition 1.18. Let k ≥ 1 be an integer. A finite subset F ⊂ G is said to be
k−critical if:
• k ≤ |F | and |F C| ≤ |G| − k,
• |∂F | is minimal for the preceding condition.
A k−atom is a k−critical subset F of minimal cardinality.
We shall see, by restricting ourselves to the case of infinite groups, that the 1-atoms
are translates of subgroups of G. This we yield an alternative proof of Kemperman’s
theorem on torsion-free groups.
Let us start with easy observations.
Lemma 1.19. We have:
• If F1 , F2 are k−critical, then |∂F1 | = |∂F2 |.
• If F is k−critical, then for all g ∈ G, the translate gF is k−critical.
• Similarly, all translates gA of a k−atom A still are k−atoms.
Proof. The first two items are evident, and the third one follows from the first two ones
and the equality |gA| = |A|.
The next formula plays a key role in the sequel.
ADDITION THEOREMS IN GROUPS
15
Proposition 1.20. Let G be a group and X, Y ⊂ G finite subsets. Then
|∂(X ∩ Y )| + |∂(X ∪ Y )| ≤ |∂X| + |∂Y |.
Proof. We have (X ∩ Y )C ⊂ XC ∩ Y C and (X ∪ Y )C = XC ∪ Y C, as easily seen.
Hence
∂(X ∩ Y ) ⊂ ∂X ∩ ∂Y, ∂(X ∪ Y ) = ∂X ∪ ∂Y.
It follows that
|∂(X ∩ Y )| + |∂(X ∪ Y )| ≤ |∂X ∩ ∂Y | + |∂X ∪ ∂Y |
= |∂X| + |∂Y |.
Proposition 1.21. Let G be an infinite group, k ≥ 1, and F1 , F2 ⊂ G two k−critical
subsets. If |F1 ∩ F2 | ≥ k, then F1 ∩ F2 and F1 ∪ F2 are also k−critical.
Proof. Consider the family M of finite subsets X ⊂ G such that |X| ≥ k. Since G
is infinite, the condition |XC| ≤ |G| − k is satisfied for all X ∈ M. Since the Fi are
k−critical, we have F1 , F2 ∈ M and
|∂F1 | = |∂F2 | = min |∂X|.
X∈M
Moreover, by hypothesis on F1 ∩ F2 , we have
|F1 ∩ F2 | ≥ k.
Hence F1 ∩ F2 ∈ M. Of course, we also have F1 ∪ F2 ∈ M. It follows that
(1.3)
|∂(F1 ∩ F2 )| ≥ |∂F1 | and |∂(F1 ∪ F2 )| ≥ |∂F1 |.
Therefore, the inequality of Proposition 1.20 implies
2|∂F1 | ≤ |∂(F1 ∩ F2 )| + |∂(F1 ∪ F2 )| ≤ 2|∂F1 |
and finally, by the estimates in (1.3),
|∂F1 | = |∂(F1 ∩ F2 )| = |∂(F1 ∪ F2 )|.
Hence F1 ∩ F2 and F1 ∪ F2 are indeed k−critical.
Corollary 1.22. Let G be an infinite group and A1 , A2 ⊂ G two distinct k−atoms
where k ≥ 1. Then |A1 ∩ A2 | ≤ k − 1.
Proof. Assume on the contrary that |A1 ∩ A2 | ≥ k. It then follows, by the preceding
proposition, that A1 ∩ A2 is k−critical. Now, as A1 , A2 are k-atoms, their cardinalities
|A1 |, |A2 | are equal, and minimal in the set of cardinalities of the k−critical sets. In
particular, |A1 | = |A2 | ≤ |A1 ∩ A2 |. It follows that A1 = A2 .
We now restrict to the case k = 1, with the aim of reproving Kemperman’s theorem.
Corollary 1.23. Let G be an infinite group and A a 1−atom containing 1. Then A is
a subgroup of G.
Proof. We know that any translate of a 1−atom is also a 1−atom. Let a ∈ A. We have
a ∈ A ∩ aA, since A contains 1, and hence |A ∩ aA| ≥ 1. The preceding corollary then
implies aA = A. It now easily follows from the equality A A = A and the finiteness of
A that A is a subgroup of G.
16
SHALOM ELIAHOU
Kemperman’s theorem establishes the minoration |X C| ≥ |X| + |C| − 1 for all
nonempty finite subsets X, C of a torsion-free group G. Of course it suffices to show
this inequality for C containing 1. Here is a proof of this result with the isoperimetric
method.
Second proof of Kemperman’s theorem. Let C ⊂ G be a finite subset containing 1. Let
A ⊂ G be a 1−atom of G with respect to C, containing 1. By the preceding corollary,
A is a subgroup of G, and hence A = {1} since G is torsion-free. Let X ⊂ G be a
nonempty finite subset of G. We have |∂A| ≤ |∂X| since A is 1−critical. Now
∂A = A C \ A = C \ {1},
and
∂X = X C \ X.
Hence |C| − 1 ≤ |X C| − |X|. It follows that |X C| ≥ |X| + |C| − 1.
2. The polynomial method
The polynomial method was introduced by Noga Alon and Michael Tarsi in [3] and
has had many applications since, see e.g. [1]. It is based on a lemma which generalizes
in n variables the well-known following fact. Let F be a commutative field. If f ∈ F [X]
is a non-zero polynomial of degree d, then f admits at most d roots in F .
Let now R = F [X1 , . . . , Xn ] be the ring of polynomials in n variables X1 , . . . , Xn
with coefficients in F . We view R as being graded in the usual way by deg(Xi ) = 1 for
all i. Thus R decomposes as
M
Rd ,
R=
d≥0
where Rd is the set of all f ∈ R which are homogeneous of degree d, union {0}. That
is, every polynomial f ∈ R uniquely decomposes as a sum f = f0 + f1 + · · · + fm of
homogeneous components fi ∈ Ri for all i.
Notation Let f ∈ R \ {0}. We set top(f ) = the homogeneous component of f of
highest degree. We also set top(0) = 0.
For example, top(X12 + X1 X2 X3 − X13 ) = X1 X2 X3 − X13 .
Proposition 2.1. Let A1 , . . . , An ⊂ F be finite subsets, of cardinality |Ai | = ri for all
i. Let f ∈ R be a polynomial vanishing on A1 ×· · ·×An . Then top(f ) ∈ (X1r1 , . . . , Xnrn ).
Proof. By induction on n. The case n = 1 is true, since if f (X1 ) vanishes on r1 points
of F , then deg(f ) ≥ r1 , and hence top(f ) is divisible by X1r1 .
Assume now n ≥ 2 and the result true for n − 1. If all the monomials in top(f ) are
divisible by Xnrn , then top(f ) ∈ (X1r1 , . . . , Xnrn ) and we are done. Thus, we may assume
that top(f ) does not belong to (Xnrn ).
Let
Y
g=
(Xn − a) ∈ F [Xn ] ⊂ F [X1 , . . . , Xn ].
a∈An
Then degXn (g) = rn , and g(A1 × · · · × An ) = {0}. We may write
g = Xnrn − g 0
with g 0 ∈ F [Xn ] and degXn (g 0 ) < rn . We have Xnrn ≡ g 0 mod (g). In f , we may thus
replace all monomials multiple of Xnrn by g 0 without modifying the class of f modulo
(g). This implies that there exists a polynomial f¯ ∈ F [X1 , . . . , Xn ] such that
ADDITION THEOREMS IN GROUPS
17
(1) f ≡ f¯ mod (g)
(2) degXn (f¯) < rn
(3) top(f ) ∈ (Xnrn , top(f¯)).
rn−1
). Note that f¯ vanishes on A1 ×· · ·×An ,
It suffices to show that top(f¯) ∈ (X1r1 , . . . , Xn−1
since it is the case of f and g. We may write
f¯ = ϕ0 + ϕ1 Xn + · · · + ϕd X d
n
with d < rn and ϕi ∈ F [X1 , . . . , Xn−1 ] for all 0 ≤ i ≤ d.
Let α = (a1 , . . . , an−1 ) ∈ A1 × · · · × An−1 , and let us write
f¯α (Xn ) = ϕ0 (α) + ϕ1 (α)Xn + · · · + ϕd (α)Xnd ∈ F [Xn ].
Then f¯α vanishes on An . Hence f¯α = 0 ∈ F [Xn ], since deg f¯α < rn . It follows that
ϕi (α) = 0 for all α ∈ A1 × · · · × An−1 and all 0 ≤ i ≤ d. By the induction hypothesis,
we have
rn−1
top(ϕi ) ∈ (X1r1 , . . . , Xn−1
)
for all i. But it is clear that
r
top(f¯) ∈ top(ϕ0 ), . . . , top(ϕd ) ⊂ (X r1 , . . . , X n−1 ).
1
Hence top(f ) ∈
(X1r1 , . . . , Xnrn ),
n−1
as announced.
An ideal in R generated by monomials is called a monomial ideal. The question
whether a given polynomial g ∈ R belongs to a given monomial ideal is easy to decide.
Lemma 2.2. Let I = (u1 , . . . , ur ) be a monomial ideal generated by monomials u1 , . . . , ur ∈
R = F [X1 , . . . , Xn ]. Let g ∈ R. Then g ∈ I if and only if every monomial occurring
in g with a non-zero coefficient is divisible by one of the ui .
Proof. Let M = {X1a1 · · · Xnan | ai ∈ N ∀i} be the set of all monomials in R. Then
M is an F −basis of R. If g = 0, the condition is empty whence satisfied. Assume
g ∈ I \ {0}. Then there exist g1 , . . . , gr ∈ R such that
g = u1 g1 + · · · + ur gr .
On the other hand, there exist unique scalars {λu }u∈M and {µv,i }v∈M,i≥1 such that
X
X
g=
λu u and gi =
µv,i v
u∈M
v∈M
for all i = 1, . . . , r. Hence
X
λu u = g =
u∈M
X
µv,i ui v.
v∈M,i≥1
Since M is an F -basis of R, it follows from the above equality that every monomial
u ∈ M such that λu 6= 0 is of the form ui v for suitable i ≥ 1 and v ∈ M .
Therefore, it follows from the above two statements that if f ∈ R is a nonzero
polynomial which vanishes on a finite grid A1 ×· · ·×An ⊂ F n , then for every monomial
u of maximal degree in f , there exists i ∈ {1, . . . , n} such that
|Ai |
Xi
divides u.
18
SHALOM ELIAHOU
2.1. Back to Cauchy-Davenport. As a first application, here is a new proof of
Cauchy-Davenport’s theorem, very different from the preceding ones.
Third proof of Cauchy-Davenport’s theorem. Let A, B be nonempty subsets of Fp =
Z/pZ, of cardinality |A| = r, |B| = s. We know that if r + s > p, then A + B = Fp ,
and hence |A + B| = p = min(r + s − 1, p).
Assume now r + s ≤ p. Setting n = |A + B|, we wish to show n ≥ r + s − 1.
• If n = p we are done.
• If n < p, consider the following polynomial f ∈ Fp [X, Y ]:
Y
f (X, Y ) =
(X + Y − c).
c∈A+B
By construction, we have f (A×B) = {0}. Hence, by Proposition 2.1, we have top(f ) ∈
(X r , Y s ). Now
Y
(X + Y ) = (X + Y )n .
top(f ) =
c∈A+B
Thus, we obtain the following condition on n:
(X + Y )n ∈ (X r , Y s )
in the polynomial ring Fp [X, Y ]. Since the monomials X i Y j constitute a basis of
Fp [X, Y ] as a vector space on the field Fp , this condition is equivalent to require that
all monomials of (X + Y )n belong to the ideal (X r , Y s ). We have
n X
n
n
(X + Y ) =
X i Y n−i ,
i
i=0
the binomial coefficients being considered
as elements of Fp , i.e. reduced modulo p.
But now, since n < p, we have ni 6≡ 0 mod p. It follows that, for all 0 ≤ i ≤ n, the
monomial X i Y n−i belongs to the ideal (X r , Y s ). In particular, for i = r − 1, we must
have
X r−1 Y n−r+1 ∈ (X r , Y s ),
and hence necessarily n − r + 1 ≥ s. It follows that n ≥ r + s − 1, as desired.
2.2. The ex-conjecture of Erdős-Heilbronn. Let A, B be subsets of a group G.
The restricted sum of A, B is defined as
A +0 B = {a + b | a ∈ A, b ∈ B, a 6= b}.
It has been conjectured by Erdős and Heilbronn in 1964 that in G = Z/pZ with p
prime, one should have
|A +0 A| ≥ min(2|A| − 3, p)
for all nonempty subsets A ⊂ G. This conjecture has been confirmed by Dias da Silva
and Hamidoune in 1994 using exterior algebra [7]. It has been confirmed again in 1995
by Alon, Nathanson and Ruzsa using the polynomial method [2]. Here is their result
implying the ex-conjecture of Erdős-Heilbronn.
Theorem 2.3 (Alon-Nathanson-Ruzsa). Let G = Z/pZ with p prime. Let r, s be
integers such that 1 ≤ r, s ≤ p, and let A, B ⊂ G with |A| = r, |B| = s. Then
min(2r − 3, p)
if r = s
0
(2.1)
|A + B| ≥
min(r + s − 2, p) if r 6= s.
ADDITION THEOREMS IN GROUPS
19
Proof. The proof is easy if p = 2, if r = 1 or if s = 1. Assume now p > 2 and r, s ≥ 2.
We distinguish two cases.
Case 1. r + s ≥ p + 2. We then claim A +0 B = A + B. Indeed, let g ∈ G. By
Lemma 1.2, we have rA,B (g) ≥ 2. Now, there is at most one representation of g of
the form g = c + c with c ∈ A ∩ B, for if 2c = 2c0 then c = c0 , as 2 is invertible
mod p. Hence, since rA,B (g) ≥ 2, there is at least one representation g = a + b with
(a, b) ∈ A × B and a 6= b. The stated bound now follows from Cauchy-Davenport.
Case 2. r + s ≤ p + 1. Hence min(r + s − 2, p) = r + s − 2. Set n = |A +0 B|. If
n ≥ r + s − 2, we are done. We may therefore assume
(2.2)
n ≤ r + s − 3,
and then must show that n = r + s − 3 and r = s.
• If n ≥ p, we are done since p − 1 ≥ r + s − 2.
• If n ≤ p − 1, consider the following polynomial g ∈ Fp [X, Y ]:
Y
(X + Y − c).
g(X, Y ) = (X − Y )
c∈A+0 B
By construction, we have g(A×B) = {0}. It follows from Proposition 2.1 that top(g) ∈
(X r , Y s ). Now
Y
top(g) = (X − Y )
(X + Y ) = (X − Y )(X + Y )n .
c∈A+0 B
Therefore
(2.3)
(X − Y )(X + Y )n ∈ (X r , Y s )
in the polynomial ring Fp [X, Y ]. Hence all monomials occurring in (X − Y )(X +
Y )n with non-zero coefficient are divisible by either X r or Y s . A straightforward
computation yields the following development of top(g):
n X
n
n
n
n+1
n+1
(2.4)
(X − Y )(X + Y ) = X
−Y
+
−
X i Y n+1−i ,
i
−
1
i
i=1
the binomial coefficients being considered as reduced mod p. Setting i = r − 1, the
monomial X r−1 Y n+1−(r−1) in (2.4) does not belong to the ideal (X r , Y s ) since n−r+2 <
s by the hypothesis (2.2) on n. By (2.3), it follows that the coefficient of this monomial
in top(g), i.e. in (2.4), must vanish. Therefore, we have
n
n
≡
mod p.
(2.5)
r−2
r−1
Writing these binomial coefficients as usual as quotients of factorials, equation (2.5) is
straightforward to solve and yields
r − 1 ≡ n − (r − 2) mod p,
i.e. n ≡ 2r − 3 mod p. Symmetrically, we also obtain n ≡ 2s − 3 mod p, whence
r ≡ s mod p. Therefore r = s since r, s ≤ p. Now, since we are in Case 2, we have
r + s ≤ p + 1, whence 2r − 3 ≤ p − 2. It now follows from the above congruence
n ≡ 2r − 3 mod p and the hypothesis n < p that n = 2p − 3, as desired.
20
SHALOM ELIAHOU
3. More about Cauchy-Davenport
The Cauchy-Davenport’s theorem may be generalized in several other directions. We
shall present here two such results. The first one is due to Vosper and characterizes the
cases where equality is attained in the Cauchy-Davenport bound. This provides a socalled inverse theorem. The second one is due to Pollard and involves the representation
numbers rA,B (x).
3.1. The theorem of Vosper. We consider the group Z/pZ with p prime. By the
Cauchy-Davenport theorem, if A, B are two nonempty subsets of this group, then
(3.1)
|A + B| ≥ min(|A| + |B| − 1, p).
The aim of this section is to characterize the cases where equality is attained.
• If |A| + |B| ≥ p + 1, i.e. if min(|A| + |B| − 1, p) = p, it is easy. Indeed, in this case,
an earlier lemma implies A + B = Z/pZ. The bound (3.1) then becomes the equality
|A + B| = p = min(|A| + |B| − 1, p).
• If |A| + |B| ≤ p, it is more difficult. In that case the bound (3.1) becomes
(3.2)
|A + B| ≥ |A| + |B| − 1,
and characterizing the case of equality is the content of the theorem of Vosper. This
theorem states that, besides special cases, the equality is only attained in (3.2) if A, B
are arithmetic sequences with the same difference.
Let us start with a general definition.
Definition 3.1. Let G be a group. A pair (A, B) of nonempty finite subsets of G is a
critical pair if
(1) A + B 6= G,
(2) |A + B| ≤ |A| + |B| − 1.
Il is clear that if (A, B) is a critical pair, then (x + A, B + y) is also a critical pair
for all x, y ∈ G. In particular, we may freely assume 0 ∈ A ∩ B.
Moreover, if (A, B) is a critical pair, then |A| + |B| ≤ |G|. Indeed, this follows from
the condition A + B 6= G and Lemma 1.2.
Here is a first example of a critical pair in an arbitrary group G.
Example 3.2. If |A| = 1 and B 6= G, then (A, B) is a critical pair.
Indeed, if A = {a} for example, then |A + B| = |B| = |A| + |B| − 1. The case A 6= G
and |B| = 1 is of course similar.
Here is a second example of a critical pair, valid in any finite group G. For a subset
X of G, we shall denote
X = G \ X,
the set-complement of X in G.
Example 3.3. Let G be a finite group and A 6= G a nonempty subset. Let c ∈ G and
B = −A + c. Then (A, B) is a critical pair.
Indeed, we have |A| + |B| = |−A + c| + |−A + c| = |G|. On the other hand, we have
c∈
/ A + B, since (−A + c) ∩ B = ∅. Il follows that |A + B| ≤ |G| − 1 = |A| + |B| − 1.
ADDITION THEOREMS IN GROUPS
21
Here is a third example of a critical pair in an abelian group G. A finite subset A of
G is said to be an arithmetic progression with difference d if it is of the form
A = {a0 , a0 + d, . . . , a0 + (k − 1)d} = a0 + {0, d, . . . , (k − 1)d}
with a0 , d ∈ G and with d of order at least k. Observe that this set is also an arithmetic progression with difference −d. Note finally that every subset A = {a0 , a1 } of
cardinality 2 is an arithmetic progression, with difference d = −a0 +a1 or d = −a1 +a0 .
Example 3.4. Let G be an abelian group and A, B two arithmetic progressions in G
with the same difference. If |A| + |B| ≤ |G|, then (A, B) is a critical pair.
Indeed, if
A = a0 + {0, d, . . . , (k − 1)d},
B = b0 + {0, d, . . . , (l − 1)d},
with |A| = k, |B| = l, then A + B = a0 + b0 + {0, d, . . . , (k + l − 2)d}. In particular,
A + B is an arithmetic progression with the same difference d, and of length |A + B| ≤
k + l − 1 = |A| + |B| − 1.
Let us go back to G = Z/pZ with p prime. It follows from Cauchy-Davenport’s
theorem that, if A, B are two nonempty subsets of Z/pZ, then
(A, B) is a critical pair if and only if |A + B| = |A| + |B| − 1.
Vosper ’s theorem states that in Z/pZ, there are no other critical pairs besides
examples (1), (2) and (3). Example (2) is easy to recognize. Indeed, it is characterized
by the condition |A + B| = 1.
Proposition 3.5. Let (A, B) be a critical pair in G = Z/pZ. Then |A + B| = 1 ⇔
there exists c ∈ G such that B = c − A.
Proof. We have |A + B| = |A| + |B| − 1 by hypothesis.
• If |A + B| = p − 1, then |A| + |B| = p. Set A + B = {c}. Then c ∈
/ A + B and hence
(c − A) ∩ B = ∅. This implies c − A ⊂ B. But |c − A| = |A| = p − |B| = |B|. Thus,
we have equality c − A = B.
• If B = c − A, then c ∈
/ A + B and |A| + |B| = p. Hence |A + B| = p − 1 and
A + B = {c}.
Here is the statement of Vosper’s theorem.
Theorem 3.1 (Vosper). Let A, B be nonempty finite subsets of Z/pZ such that |A| +
|B| ≤ p. Then (A, B) is a critical pair if and only if the following conditions hold:
(1) |A| = 1 or |B| = 1;
(2) |A + B| = 1 and B = c − A, where {c} = A + B;
(3) A, B are two arithmetic progressions with the same difference.
At this point, we know all critical pairs (A, B) of Z/pZ such that |A| = 1, |B| = 1
or |A + B| = 1. It remains to understand the general case, for which we introduce the
following terminology.
Definition 3.6. Let G be a group. A pair (A, B) of subsets of G is an essential critical
pair if
(1) (A, B) is a critical pair,
(2) |A| ≥ 2, |B| ≥ 2 and |A + B| ≥ 2.
22
SHALOM ELIAHOU
Note that in Z/pZ, an essential critical pair (A, B) is caracterized by the conditions
|A + B| = |A| + |B| − 1, |A| ≥ 2, |B| ≥ 2 and |A + B| ≥ 2.
In order to finish the proof of Vosper’s theorem, it remains to prove the following
statement.
Proposition 3.7. Let (A, B) be an essential critical pair of Z/pZ. Then A, B are two
arithmetic progressions with the same difference.
This will follow from a sequence of three lemmas. The first two show how, starting
from an essential critical pair, one may construct new essential critical pairs.
Lemma 3.8. Let (A, B) be an essential critical pair of Z/pZ. Then the pair (A + B, −A)
is also an essential critical pair.
Proof. Let us set D = A + B. By hypothesis, we have
|A| ≥ 2, |B| ≥ 2, |D| ≥ 2 and |A + B| = |A| + |B| − 1.
Knowing that |−A| = |A|, it remains to show that
(3.3)
|D − A| ≥ 2 and |D − A| = |D| + |A| − 1.
• Step 1. We have
D − A = B.
(3.4)
Indeed, observe first that
(3.5)
|D| = p − |A + B| = p − |A| − |B| + 1.
Then, by Cauchy-Davenport, we have
|D − A| ≥ min(|D| + |A| − 1, p)
= min(p − |B|, p)
= p − |B|
= B.
Finally, since D ∩ (A + B) = ∅, we have D − A ⊂ B. We conclude that D − A = B,
as announced.
• Step 2. We have
|D − A| = |D| − |A| + 1.
Indeed, this follows from the formulas |D − A| = p − |B| and |D| = p − |B| − |A| + 1
established in (3.4) and (3.5).
• Step 3. Since |D − A| = |B| ≥ 2, formulas (3.3) are now established. Knowing that
|D|, |A| ≥ 2, we conclude that (D, −A) is an essential critical pair.
Let us now show how to obtain new essential critical pairs by Dyson’s transforms.
Lemma 3.9. Let (A, B) be an essential critical pair of Z/pZ. Assume 0 ∈ B and
|B| ≥ 3. Then there exists e ∈ A such that the Dyson’s transform (A(e), B(e)) is an
essential critical pair, with 2 ≤ |B(e)| < |B|.
ADDITION THEOREMS IN GROUPS
23
Proof. Recall that for e ∈ G, we denote A(e) = A ∪ (B + e) and B(e) = (A − e) ∩ B.
If e ∈ A, then B(e) 6= ∅ since it contains 0.
• Let us first show that, for all e ∈ A, the transformed pair (A(e), B(e)) is a critical
pair. By the general properties of Dyson’s transforms, we have A(e) + B(e) ⊂ A + B
and |A(e)| + |B(e)| = |A| + |B|. Hence
|A(e) + B(e)| ≤ |A + B| = |A| + |B| − 1 = |A(e)| + |B(e)| − 1.
Il follows that (A(e), B(e)) is a critical pair. Of course, we have |A(e)| ≥ 2 and
|A(e) + B(e)| ≥ 2. Hence, for that critical pair to be essential, it suffices to have
|B(e)| ≥ 2. Let us show that this happens for at least one element e ∈ A.
• Set A = X1 t X2 t X3 , where
X1 = {e ∈ A | |B(e)| = 1},
X2 = {e ∈ A | 2 ≤ |B(e)| < |B|},
X3 = {e ∈ A | |B(e)| = |B|}.
We have to show that |X2 | ≥ 1. For this, we shall show that |X1 | ≤ 1 and that |X3 | is
not too big.
• Since 0 ∈ B(e) for all e ∈ A, we have
X1 = {e ∈ A | (A − e) ∩ B = {0}}.
Let us set B 0 = B \ {0}. Then X1 = {e ∈ A | A ∩ (e + B 0 ) = ∅}. It follows that
X1 + B 0 ⊂ (A + B) \ A,
and hence |X1 + B 0 | ≤ |B| − 1. But since |X1 | + |B 0 | ≤ |A| + |B| < p, the theorem of
Cauchy-Davenport implies
|X1 + B 0 | ≥ |X1 | + |B 0 | − 1 = |X1 | + |B| − 2.
Thus, we have |X1 | + |B| − 2 ≤ |X1 + B 0 | ≤ |B| − 1, which gives
|X1 | ≤ 1.
(3.6)
• For all e ∈ X3 , we have B(e) = B and hence B ⊂ A − e. Il follows that
X3 + B ⊂ A.
Hence, if X3 6= ∅, we have
|A| ≥ |X3 + B| ≥ |X3 | + |B| − 1
by Cauchy-Davenport. Il follows that |X3 | = 0 or |X3 | ≤ |A| − |B| + 1, and hence
(3.7)
|X3 | ≤ max(0, |A| − |B| + 1).
• It is now easy to show that |X2 | ≥ 1. By (3.6) and (3.7), we have
|A| = |X1 | + |X2 | + |X3 | ≤ 1 + |X2 | + max(0, |A| − |B| + 1),
and hence
|X2 | ≥ min(|A| − 1, |B| − 2) ≥ 1.
• Let e ∈ X2 , i.e. such that 2 ≤ |B(e)| < |B|. Then, from the above, the pair
(A(e), B(e)) is an essential critical pair.
Finally, here is a lemma on the interaction between essential critical pairs and arithmetic progressions.
24
SHALOM ELIAHOU
Lemma 3.10. Let (A, B) be an essential critical pair of Z/pZ. If A is an arithmetic
progression with difference d, then B is also an arithmetic progression with difference
d.
Proof. We denote and order the elements of Z/pZ as follows:
0 < 1 < · · · < p − 1.
For all i ≤ j in Z/pZ, we denote [i, j] = {i, i + 1, . . . , j}. This is an arithmetic
progression with difference 1, to be called simply an interval.
Set |A| = k. Up to a multiplying A, B by d−1 , we may assume d = 1. Up to a
translation of A, we may assume
A = [0, k − 1].
We must show that B is also an interval. Up to a translation of B, we may assume
p − 2 ∈ A + B,
(3.8)
p−1∈
/ A + B.
Let us then show that
(3.9)
B ∩ [p − k, p − 1] = ∅.
Indeed, since p−1 ∈
/ A+B, we have B ∩(p−1−A) = ∅. But (p−1−A) = [p−k, p−1].
Let us now show that
(3.10)
max(B) = p − k − 1.
Indeed, max(B) ≤ p − k − 1 by (3.9). Moreover, since p − 2 ∈ A + B, there exists
a ∈ A, i.e. 0 ≤ a ≤ k − 1, such that p − 2 − a ∈ B. Still by (3.9), we necessarily have
a = k − 1, and hence p − k − 1 ∈ B.
We claim that
(3.11)
A + B = B ∪ [p − k, p − 2].
First of all, we have the inclusion B ∪ [p − k, p − 2] ⊂ A + B. Indeed, B ⊂ A + B since
0 ∈ A, and since p − k − 1 ∈ B by (3.10), we also have
[p − k, p − 2] = [1, k − 1] + (p − k − 1) ⊂ A + B.
But these two sets have the same cardinality, since B ∩ [p − k, p − 2] = ∅ by (3.9), and
hence
|A + B| = k + |B| − 1 = |B| + |[p − k, p − 2]| = |B ∪ [p − k, p − 2]|.
This establishes (3.11). It is now easy to show that B is an interval. For that, it suffices
to show
b ∈ B and b < max(B) ⇒ b + 1 ∈ B.
But if b ∈ B, then b + 1 ∈ A + B. Moreover, if b < max(B) = p − k − 1, we have
b + 1 < p − k. By (3.11), it follows that b + 1 ∈ (A + B) \ [p − k, p − 2] = B, and the
proof is complete.
We are now ready to conclude the proof of Proposition 3.7 and hence of Vosper’s
theorem.
Proof of Proposition 3.7. We want to show that every essential critical pair (A, B) is
made of arithmetic progressions with the same difference. We may assume 0 ∈ B and
proceed by induction on |B|.
ADDITION THEOREMS IN GROUPS
25
• If |B| = 2, then B is an arithmetic progression of a certain difference d. It follows
from Lemma 6 that A is also an arithmetic progression with difference d, and we are
done in this case.
• Assume |B| ≥ 3. By Lemma 3.9, there exists e ∈ A such that the Dyson’s transform
(A(e), B(e)) is also an essential critical pair, with |B(e)| < |B|. By the induction hypothesis, we have that A(e), B(e) are two arithmetic progressions with the same difference. It follows that A(e)+B(e) is an arithmetic progression. But A(e)+B(e) = A+B.
It follows that A + B, and hence its complement A + B, are arithmetic progressions.
But by Lemma 3.8, the pair (A + B, −A) is an essential critical pair, one member
of which is an arithmetic progression. By Lemma 3.10, the other member of this
pair, namely −A, is an arithmetic progression. Hence A also is one. Finally, still by
Lemma 3.10, we conclude that B is an arithmetic progression with the same difference
as A.
3.2. The theorem of Pollard. The following result is due to Pollard (1974). For two
nonempty subsets A, B of Z/pZ, we seek here to count those x ∈ Z/pZ having several
representations of the form x = a + b with a ∈ A, b ∈ B. Recall that we denote by
rA,B (x) the number of such representations. For t ≥ 0, we set
Nt = |{x ∈ Z/pZ | rA,B (x) ≥ t}|.
In particular, N0 = p, N1 = |A + B|, and Nt = 0 if t > min(|A|, |B|).
Theorem 3.2. Let p be a prime number and A, B two nonempty subsets of Z/pZ of
cardinality k, l, respectively. For all 1 ≤ t ≤ min(k, l), set S(A, B, t) = N1 + N2 + · · · +
Nt . Then
S(A, B, t) ≥ t min(p, k + l − t).
Proof. First note that S(A, B, t) may be rexpressed as follows:
X
S(A, B, t) =
min(t, rA,B (x)).
x∈Z/pZ
We may assume l ≤ k, and we proceed by double induction on t and l. The case t = 1
is exactly the Cauchy-Davenport theorem. Assume now 2 ≤ t ≤ l ≤ k.
If t = l, then clearly S(A, B, l) = lk = l min(p, k) and we are done.
Assume now t < l. We distinguish the following cases:
• Case 1 : k + l − t > p.
We have 2 ≤ t ≤ p − k + t < l ≤ p. Let us set l0 = p − k + t, and let B 0 ⊂ B of
cardinality |B 0 | = l0 . We have S(A, B, t) ≥ S(A, B 0 , t), and by the induction hypothesis,
S(A, B 0 , t) ≥ t min(p, k + l0 − t) = tp.
• Case 2 : k + l − t ≤ p.
In particular, k < p. We have to show that S(A, B, t) ≥ t(k + l − t). The basic idea
is to replace the pair A, B by the pair A ∪ B, A ∩ B. But first we need to show that
we may assume A ∩ B 6= ∅.
Up to translation of B, we may assume 0 ∈ B. Since |B| = l ≥ 2, there exists
0 6= b ∈ B. It is impossible to have A + b ⊂ A, for otherwise we would have A + jb ⊂ A
for all j ≥ 0 thereby forcing A = Z/pZ, in contradiction with k < p. Let then a ∈ A
be such that a + b 6∈ A. Replacing A by A − a, we have 0 ∈ A and b ∈ B \ A.
Let us set U = A ∪ B, I = A ∩ B, and c = |U |, d = |I|. By the above, we have 0 ∈ I
and 1 ≤ |I| < |B|, i.e. 1 ≤ d < l. Note also that k + l = c + d.
26
SHALOM ELIAHOU
Set A0 = A \ I, B 0 = B \ I. We thus have
A + B = (A0 + B 0 ) ∪ (A0 + I) ∪ (I + B 0 ) ∪ (I + I).
Since U = A0 t B 0 t I, it follows that, for all x ∈ Z/pZ, we have:
rA,B (x) = rA0 ,B 0 (x) + rU,I (x) ≥ rU,I (x).
Let us consider the following two cases:
• Case 2a : t ≤ d.
We then have
X
X
S(A, B, t) =
min(t, rA,B (x)) ≥
min(t, rU,I (x)) = S(U, I, t).
x∈Z/pZ
Since d < l, the induction hypothesis on l implies:
S(U, I, t) ≥ t min(p, d + c − t) = t min(p, k + l − t) = t(k + l − t),
and this case is done.
• Case 2b : t > d.
Let us set t0 = t − d ≥ 1. With the formulas rA,B (x) = rA0 ,B 0 (x) + rU,I (x) and
t = t0 + d, and applying min(m + n, u + v) ≥ min(m, u) + min(n, v), for all x we have
min(t, rA,B (x)) ≥ min(t0 , rA0 ,B 0 (x)) + min(d, rU,I (x)).
Therefore
X
min(t, rA,B (x)) ≥
x∈Z/pZ
X
X
min(t0 , rA0 ,B 0 (x)) +
x∈Z/pZ
min(d, rU,I (x)).
x∈Z/pZ
This implies
S(A, B, t) ≥ S(A0 , B 0 , t0 ) + cd.
Let us set k 0 = |A0 | = k − d, l0 = |B 0 | = l − d. Then
1 ≤ t0 = t − d < l − d = l0 ,
and k 0 + l0 − t0 ≤ p. By the induction hypothesis on l, we have
S(A0 , B 0 , t0 ) ≥ t0 (k 0 + l0 − t0 ).
Therefore
S(A, B, t) ≥
≥
=
=
=
=
=
S(A0 , B 0 , t0 ) + cd
t0 (k 0 + l0 − t0 ) + cd
(t − d)((k − d) + (l − d) − (t − d)) + cd
(t − d)(k + l − d − t) + cd
(t − d)(c − t) + cd
t(c + d − t)
t(k + l − t).
Remark 3.11. The bound in Pollard’s theorem is optimal, attained by the initial segments A = {0, 1, . . . , r − 1}, B = {0, 1, . . . , s − 1} in Z/pZ.
Remark 3.12. Pollard’s theorem does not hold in a general cyclic group. Indeed, consider the following example. In G = Z/9Z, let A = {0, 3, 6}, B = {0, 1, 4, 7}. Then
N1 + N2 = 6 + 3 = 9. However, we have 2 min(9, 4 + 3 − 2) = 10. Note that the
conditions of Chowla’s theorem are satisfied.
ADDITION THEOREMS IN GROUPS
27
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