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A NOTE ON FREIMAN MODELS IN HEISENBERG GROUPS
NORBERT HEGYV´ARI AND FRANC ¸OIS HENNECART
Abstract. Green and Ruzsa recently proved that for any s ≥ 2, any small squaring set
A in a (multiplicative) abelian group, i.e. |A · A| < K|A|, has a Freiman s-model: it
means that there exists a group G and a Freiman s-isomorphism from A into G such that
|G| < f(s,K)|A|.
In an unpublished note, Green proved that such a result does not necessarily hold in non
abelian groups if s ≥ 64. The aim of this paper is improve Green’s result by showing that
it remains true under the weaker assumption s ≥ 6.
1. Introduction
We will use the notation |X| for the cardinality of any set or group X. If X and Y are
subsets of a given (multiplicative) group, the product X · Y or simply XY denotes the set
{xy | x ∈ X,y ∈ Y }. For X = Y we write XY = X2. The set X−1is formed by all the
inverse elements x−1, x ∈ X.
Let s ≥ 2 be an integer and A ⊂ H and B ⊂ G be subsets of arbitrary (multiplicative)
groups. A map π : A → B is said to be a Freiman s-homomorphism if for any 2s-tuple
(a1,...,as,b1,...,bs) of elements of A and any signs ?i= ±1, i = 1,...,s, we have
a?1
1...a?s
s= b?1
1...b?s
s=⇒ π(a1)?1...π(as)?s= π(b1)?1...π(bs)?s.
Observe that in the case of abelian groups, we may set, without loss of generality, all the
signs to +1. If moreover π is bijective and π−1is also a Freiman s-homomorphism, then π is
called a Freiman s-isomorphism from A into G. In this case, A and B are said to be Freiman
s-isomorphic.
Green and Ruzsa proved in [2] that a structural result holds for small squaring sets in
an abelian (multiplicative) group. The key argument in their proof is Proposition 1.2 of [2]
asserting that any small squaring finite set A in an abelian group has a good Freiman model,
that is a relatively small finite group G and a Freiman s-isomorphism from A into G. More
precisely, they showed the following effective result:
Date: September 12, 2010.
Research is partially supported by OTKA grants K 67676, K 81658 and Balaton Program Project.
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2 N. HEGYV´ARI AND F. HENNECART
Let s ≥ 2 and K > 1. There exists a constant f(s,K) = (10sK)10K2such that A is a
subset of an abelian group H satisfying the small squaring property |A · A| < K|A|, then
there exists an abelian group G such that |G| < f(s,K)|A| and A is Freiman s-isomorphic
to a subset of G.
It is not difficult to see that this result cannot be literally extended to nonabelian groups
by considering a set A such that |A · A|/|A| is small and |A · A · A|/|A| is large (see [6,
page 94] for such an example). However it is known (by combining [4, section 1.11] and [6,
Proposition 2.40]) that if |A · A|/|A| ≤ K then for any n-tuple of signs ?1,...,?n∈ {−1,1},
we have |X?1·X?2···X?n|/|X| ≤ KO(n)for some large subset X of A satisfying |X| ≥ |A|/2.
Despite this fact, the existenceness of a good Freiman s-model for some large subset of an
arbitrary set A0satisfying the small squaring property |A0· A0| < 2|A0| is not guaranteed.
Indeed in his unpublished note [3], Green gave an example of such a set A0with arbitrarily
large cardinality and the following property: let s ≥ 64 and δ = 1/23; then for any A ⊂ A0
with |A| ≥ |A0|1−δand any finite group G such that there is a Freiman s-isomorphism from
A into G, we have |G| ≥ |A|1+δ. There is no doubt from his proof that the admissible range
for s could be somewhat improved (s ≥ 32 is seemingly the best range that can be read from
his proof).
Our aim is to improve Green’s result by showing:
Theorem 1. Let n be any positive integer and ε be any positive real number. Then there
exists a finite (nonabelian) group H and a subset A0in H with the following properties:
i) |A0| > n and |A0· A0| < 2|A0|;
ii) For any A ⊂ A0with |A| ≥ |A0|43/44and for any finite group G such that there exists
a Freiman 6-isomorphism from A onto G, we have |G| ≥ |A|33/32−ε.
Our proof in Section 4 is partially based on Green’s approach but also includes new mate-
rials. It exploits arguments coming from group theory and Fourier analysis with additional
tools, e.g. a recent incidence theorem due to Vinh [7]. It also needs some additional combi-
natorial arguments.
In Section 3, we include for comparison the proof of a weaker statement that does not use
the new materials, but which optimizes, in some sense, Green’s ideas.
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FREIMAN MODELS IN HEISENBERG GROUPS3
Let p be a prime number and F the fields with p elements. We denote by H the Heisenberg
linear group over F consisting of the upper triangular matrices
0 0 1
[x,y,z] =
1 x z
0 1 y
, x,y,z ∈ F.
We recall the product rule in H:
[x,y,z] · [x?,y?,z?] = [x + x?,y + y?,xy?+ z + z?].
As shown in [3], this group provides an example of a nonabelian group in which there exists
some subset A0with small squaring property, namely |A2
Freiman model. That is there is no relatively big isomorphic image of A0by a Freiman s-
0| < 2|A0|, and not having a good
isomorphism with a given s in any group G. We will also use the Heisenberg group in order
to derive our results.
The proof of Theorem 1 goes in the following manner. We will show that: firstly there
exists a non trivial p-subgroup in the subgroup generated by π(A) in G; secondly any element
in π−1(G) is the product of at most 6 elements from A or A−1. The rest of the proof is based
on some group-theoretical properties which are mainly taken from [3].
As indicated in [3], there is no hope to obtain an optimal result by this approach, namely
a similar result with s0= 2.
2. Some properties of finite nilpotent groups and of the Heisenberg group H
For any group G, we denote by 1Gthe identity element of G. Thus [0,0,0] = 1H.
We will use the following partially classical properties:
1. H is a two-step nilpotent group (or nilpotent of class two). Indeed, the commutator
of a1= [x1,y1,z1] ∈ H and a2= [x2,y2,z2] ∈ H denoted by [a1;a2] is equal to
[a1;a2] = a1a2a−1
1a−1
2
= [0,0,x1y2− x2y1].
For any a3= [x3,y3,z3] ∈ H, we obtain
[[a1;a2];a3] = [0,0,0] = 1H,
for the double commutator. Hence the result.
2. Any finite nilpotent group is the direct product of its Sylow subgroups (see 6.4.14 of
[5]).
3. Any finite p-group of order p or p2is abelian (see 6.3.5 of [5]).
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4 N. HEGYV´ARI AND F. HENNECART
4. Assume that A ⊂ H and π is a Freiman s-homomorphism from A into G with s ≥ 5.
We denote by ?π(A)? the subgroup generated by π(A). Then ?π(A)? is a two-step
nilpotent group. Indeed, for any a,b,c ∈ A, one has
aba−1b−1c = caba−1b−1
since H is a nilpotent group of class two. Hence
π(a)π(b)π(a)−1π(b)−1π(c) = π(c)π(a)π(b)π(a)−1π(b)−1
since π is a Freiman s-homomorphism with s ≥ 5. It thus follows that double com-
mutators satisfy [[a1;b1];c1] = 1Gfor any a1,b1,c1∈ π(A). In [3], the author observed
from a direct argument that it remains true for any a1,b1,c1∈ ?π(A)?: since ?π(A)?
is finite, the result will follow from the next lemma (cf. [3]).
Lemma 2. Let Γ be any group and X a maximal subset of Γ such that
(1)[[a;b];c] = 1Γ, for any a,b,c ∈ X.
Then X in closed under multiplication.
For the the sake of completeness we include the proof which is in the same way as
in [3].
Proof. By (1) and the following identity
(2)[xy;z] = [x;[y;z]] · [y;z] · [x;z], x,y,z ∈ Γ,
we obtain for any a,b,c,d ∈ X, [[ab;c];d] = [[b;c] · [a;c];d]. Applying again (2) with
x = [b;c], y = [a;c] and z = c, yields in view of (1),
(3) [[ab;c];d] = 1Γ,for any a,b,c,d ∈ X.
By a further application of (2) with x = a, y = b and z = [ab;c], we get by (3)
[ab;[ab;c]] = 1Γfor any a,b,c ∈ X. By the maximal property of X, we obtain ab ∈ X
for any a,b ∈ X.
?
3. Approach of the proof with a slightly weaker result
Before proving our main result, we explain the principle of the approach by showing the
following weaker result in which only Freiman s-isomorphisms with s 7 are considered.
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FREIMAN MODELS IN HEISENBERG GROUPS 5
Theorem 3. Let n be a positive integer and θ be a real number such that
11
12≤ θ ≤ 1
and let
ϕθ=12θ − 9
2
.
Then there exists a finite group H and a subset A0in H satisfying the following properties:
i) |A0| > n and |A0· A0| < 2|A0|;
ii) For any A ⊂ A0with |A| ≥ |A0|θand for any finite group G such that there exists a
Freiman 7-isomorphism from A onto G, we have |G| ≥ |A|ϕθ.
For θ = 13/14, it yields the following corollary which can be compared to Theorem 1:
Corollary 4. Let n be any positive integer. Then there exists a finite group H and a subset
A0in H satisfying the following properties:
i) |A0| > n and |A0· A0| < 2|A0|;
ii) For any A ⊂ A0with |A| ≥ |A0|13/14and for any finite group G such that there exists
a Freiman 7-isomorphism from A onto G, we have |G| ≥ |A|15/14.
Let α ∈ (0,1) and A0be the subset of H
(4) A0:= {[x,y,z] | (x,y,z) ∈ [0,pα) × F × F}.
For p large enough, we plainly have
|A0· A0| = 2|A0| − p2,
thus A0is a small squaring subset of H.
Let θ be such that 0 < θ ≤ 1, on which an additional assumption will be given later. Let
A be any subset of A0whose cardinality satisfies
(5)
|A| ≥ |A0|θ.
By an averaging argument, there exists x0,y0,z0,z?
0,u,v ∈ F and X,Y,Z ⊂ F such that
[X,y0,z0] ∪ [x0,Y,z?
|X| ≥|A|
0] ∪ [u,v,Z] ⊂ A
|A|
p1+α,
(6)
p2,
|Y | ≥|Z| ≥
|A|
p1+α. (7)
Observe that |X||Y ||Z|2≥ p3if
(8)A p(8+3α)/4,
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6N. HEGYV´ARI AND F. HENNECART
which holds true if we fix α such that
(9)θ =8 + 3α
8 + 4α,
that is
(10)α =8(1 − θ)
4θ − 3,
assuming that the following condition on θ holds:
θ ≥11
12.
Let a = [x,y0,z0], b = [x0,y,z?
0]. These are elements of A. Moreover the commutator of a
and b is
aba−1b−1= [0,0,xy − x0y0].
Let c = [u,v,z] and d = [u,v,z?] in [u,v,Z] ⊂ A. We thus have
aba−1b−1cd−1= [0,0,xy + z − z?− x0y0].
For any element t in F, let N(t) be the number of representations of t under the form
t = xy + z − z?− x0y0,x ∈ X,y ∈ Y,z,z?∈ Z.
One has
N(t) =1
p
p−1
?
h=0
?
y∈Y
z,z?∈Z
x∈X
e
?h(xy − x0y0+ z − z?− t)
p
?
,
where e(α) is the usual notation for exp(2iπα). We get
N(t) ≥|X||Y ||Z|2
p
−1
p
p−1
?
h=1
|S(h)||T(h)|2,
where
S(h) =
?
(x,y)∈X×Y
e
?hxy
p
?
,T(h) =
?
z∈Z
e
?hz
p
?
.
By Vinogradov’s inequality
|S(h)| ≤
?
p|X||Y |
(if p ? h)
and Parseval’s identity
1
p
p
?
h=1
|T(h)|2= |Z|,
we deduce the lower bound
N(t) >|X||Y ||Z|2
p
−
?
p|X||Y ||Z|.
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FREIMAN MODELS IN HEISENBERG GROUPS7
Hence by (10), N(t) is positive. We thus deduce
[0,0,F] ⊂ B := A2A−2AA−1.
Let G be any finite group and π any Freiman s-isomorphism from A into G. Our goal is
to show that |G| is big compared to |A|. We thus may assume that G = ?π(A)?.
We assume in the sequel that s ≥ 7. We start from the property that is proven just above:
π([0,0,F]) ⊂ π(B).
For any z ∈ F, we let
gz= π([0,0,z]).
If h = π([u,v,w]) ∈ π(A), then for s ≥ 7 we have
(11)π([−u,−v,uv − w + z]) = π([u,v,w]−1[0,0,z]) = h−1gz= gzh−1.
We now show that for some i ?= j,
gλ(i−j)= g(λ−1)(i−j)gi−j,0 < λ ≤ p.
Since [u,v,Z] ⊂ A and |Z| > 1 by (7) and (8), A contains at least two distinct elements
[u,v,i] and [u,v,j]. We denote hk = π([u,v,k]) for k = i,j. Since π is a Freiman s-
isomorphism from A into G and s ≥ 7, we get h−1
in (11)
jhi= gi−jand by a similar calculation as
g(λ+1)(i−j)h−1
i
= gλ(i−j)h−1
j,
hence
g(λ+1)(i−j)= gλ(i−j)+jh−1
jhi= gλ(i−j)gi−j.
We deduce by induction
gλ(i−j)= gλ
i−j, for any λ ≥ 1.
Thus the order of gi−j in G is either 0 or p. Since s ≥ 2, we have hi?= hj hence gi−j =
h−1
jhi?= 1G. This shows that gi−j is of order p in G. We then deduce that p divides the
order of G.
Let Gpbe the Sylow p-subgroup of G. Since s ≥ 5 and H is a two-step nilpotent group, G
is also a two-step nilpotent group by Property 4 of Section 2. Then by Property 2 of Section
2, G can be written as the direct product G = Gp× K. The projection σ of G onto Gpis a
homomorphism thus ˜ π = σ ◦π is a Freiman s-homomorphism. Since for z ?= 0, hzhas order
p in G, σ(hz) has also order p in Gp.
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8 N. HEGYV´ARI AND F. HENNECART
Let a1= [x1,y1,z1] and a2= [x2,y2,z2] be any elements in A. We have a1a2a−1
1a−1
2
=
[0,0,x1y2− x2y1]. If Gpwere abelian we would obtain by using s ≥ 4
1G= ˜ π(a1)˜ π(a2)˜ π(a1)−1˜ π(a2)−1= ˜ π(a1a2a−1
1a−1
2) = ˜ π([0,0,x1y2− x2y1]) = σ(gx1y2−x2y1),
hence x1y2− x2y1= 0. We would conclude that |A| ≤ p2, a contradiction by the fact that
|A| ≥ |A0|θ≥ p(2+α)θ> p2by (9).
Consequently by Property 3 given in Section 2, Gpis not abelian and |Gp| ≥ p3. Finally
|G| ≥ p3= |A0|3/(2+α)≥ |A|(12θ−9)/2.
The proof of Theorem 3 finishes by choosing the prime p large enough in order to have
|A0| > n.
4. Proof of the main result Theorem 1
Again, A0denotes the set
A0= {[x,y,z] : 0 ≤ x < pα, y,z ∈ F},
and A any subset of A0such that |A| ≥ |A0|θ. The parameters α ∈ (0,1) and θ ∈ (0,1) will
be specified below. Again, we have |A0| ≥ p2+αthus
(12)
|A| ≥ p(2+α)θ.
We recall that there exist x0,y0,z0,z?
0,u,v ∈ F and X,Y,Z ⊂ F such that :
[X,y0,z0] ∪ [x0,Y,z?
|X| ≥|A|
0] ∪ [u,v,Z] ⊂ A
|A|
p1+α,
p2,
|Y | ≥|Z| ≥
|A|
p1+α.(13)
For (x,y,z) ∈ X × Y × Z, one has
[x,y0,z0][x0,y,z?
0][x,y0,z0]−1[x0,y,z?
0]−1[u,v,z] = [u,v,xy + z − x0y0].
Our first goal is to show that [u,v,t] is in A2A−2A except for t belonging to a small subset
E of exceptions.
First step: For any t in F, let r(t) be the number of triples (x,y,z) ∈ X ×Y ×Z such that
t = xy + z − x0y0.
One cannot prove that r(t) > 0 for any t. Nevertheless, we will show that except for a small
part of elements t, this property holds. Let C be the set of those elements of t for which
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FREIMAN MODELS IN HEISENBERG GROUPS9
r(t) > 0. Then by the Cauchy-Schwarz inequality
(14)
|C| ≥(|X||Y ||Z|)2
?
tr(t)2
.
Furthermore?
tr(t)2coincides with the number of solutions of
xy + z = x?y?+ z?, x,x?∈ X, y,y?∈ Y, z,z?∈ Z.
If we fix x = x1, x?= x?
1and z?= z?
1, it gives the equation of an hyperplan Dx1,x?
1,z?
1in F3:
x1y − x?
1y?+ z − z?
1= 0.
All these hyperplanes are different and there are |X|2|Z| such hyperplanes. The possible
number of points (y,y?,z) ∈ Y × Y × Z is |Y |2|Z|.
In [7], L.A. Vinh established a Szemeredi-Trotter type result by obtaining an incidence
inequality for points and hyperplanes in Fd. It is connected to the Expander Mixing Lemma
(see Corollary 9.2.5 in [1]). We have:
Lemma 5 (L.A. Vinh [7]). Let d ≥ 2. Let P be a set of points in Fdand H be a set of
hyperplanes in Fd. Then
|{(P,D) ∈ P × H : P ∈ D}| ≤|P||H|
p
+ (1 + o(1))p(d−1)/2(|P||H|)1/2.
By this result with d = 3, we get for any large p
?
which yields by (14)
t
r(t)2≤(|X||Y ||Z|)2
p
+ 2p|X||Y ||Z|,
|C| ≥ p −
2p3
|X||Y ||Z|.
Thus the set E of exceptions t ∈ F with r(t) = 0 has cardinality
(15)
|E| ≤
2p3
|X||Y ||Z|.
Second step: We fix z1any element in Z and let Z1= Z?{z1}. For any z ∈ Z1, we denote
m(z) = max{m ≤ p : z1+ j(z − z1) / ∈ E, 2 ≤ j ≤ m}
if the maximum exists and we let m(z) = p otherwise. Let
(16)T =
?|Z1|
2|E|
?
If we denote by Z?
1the set of the elements z ∈ Z1with m(z) ≤ T, then
?
|Z?
1| =
mT
|{z ∈ Z1 : m(z) = m}| ≤
?
mT
|E| ≤|Z1|
2
,
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10N. HEGYV´ARI AND F. HENNECART
since m = m(z) implies z1+ (m + 1)(z − z1) ∈ E. It follows that m(z) > T for at least one
half of the elements z in Z1. We denote by˜Z1the set of those elements z. We have
(17)
|˜Z1| ≥
|A|
2p1+α.
Lemma 6. Assume that 23/24 < θ ≤ 1 and let γ be a positive real number such that
γ <2(2 + α)θ − (3 + 2α)
(18)
3
.
If |E| < pγ, then there exists an integer t with 1 ≤ t ≤ T and two distinct elements z,z?∈˜Z1
such that
(19)z?− z / ∈ E − E andz?= z1+ t(z − z1)
Proof. For 1 ≤ t ≤ T, we denote by s(t) the number of pairs z,z?of elements of˜Z1with the
required property. It is sufficient to show that
T
?
t=1
s(t) > 0.
This sum can be rewritten as
T
?
t=1
1
p
?
0≤|h|≤p/2
?
z,z?∈−z1+˜Z1
z?−z/ ∈E−E
e
?h(z−1z?− t)
p
?
.
The contribution related to h = 0 is plainly bigger than
T
p(|˜Z1|2− |˜Z1||E − E|),
thus
T
?
t=1
s(t) ≥T
p(|˜Z1|2− |˜Z1||E − E|) −1
p
?
0<|h|<p/2
???
T
?
t=1
e
?−th
p
????
???
?
z,z?∈−z1+˜Z1
z?−z/ ∈E−E
e
?hz−1z?
p
????.
By extending the summation over z and z?, we obtain for any h ?= 0
???
which is less than or equals to
?
z,z?∈−z1+˜Z1
z?−z/ ∈E−E
e
?hz−1z?
p
???? ≤
???
?
z,z?∈−z1+˜Z1
e
?hz−1z?
p
???? + |˜Z1||E − E|,
(√p + |E − E|)|˜Z1|
by using Vinogradov’s inequality for the estimation of the sum over z and z?. Hence by the
bounds
??
T
?
t=1
e
?−ht
p
??? ≤
p
2|h|
for 0 < |h| < p/2,
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FREIMAN MODELS IN HEISENBERG GROUPS11
and
(p−1)/2
?
h=1
1
h≤ lnp,
we get
T
?
t=1
s(t) ≥T
p(|˜Z1|2− |˜Z1||E − E|) − (√p + |E − E|)|˜Z1|lnp.
From the trivial bound |E − E| ≤ |E|2and by (16) and (17), this sum is positive whenever
|E| ≤ pγfor p is large enough, where γ is any positive number such that
?(2 + α)θ − (1 + α)
The second argument in this minimum is less than or equal to the first since θ ≤ 1 and the
third is less than the second since θ > 23/24. Thus condition (20) reduces to (18), and the
(20)γ < min
2
;4(2 + α)θ − (7 + 4α)
2
;2(2 + α)θ − (3 + 2α)
3
?
.
lemma follows.
?
By (13) and (15), we deduce from the lemma that the condition
7 + 2α − 3(2 + α)θ <2(2 + α)θ − (3 + 2α)
3
,
is sufficient in order to ensure that system (19) has at least one solution, assuming p is large
enough. This condition reduces to
θ >
24 + 8α
22 + 11α
or equivalently
(21)α > α0(θ) :=24 − 22θ
11θ − 8.
33. FixingSince α < 1, we must choose θ such that θ >32
(22)α = α0(θ) + ε,
this yields
(23)p3≥ |A|3/(2+α)≥ |A|3(11θ−8)/8−ε,
for any p ≥ p0(?). For θ = 43/44, it will give the desired exponents in Theorem 1.
Third step: We have at our disposal z1,z ∈ Z and t ∈ F such that
(24)z1+ j(z − z1) / ∈ E,j = 2,...,t, andz1+ t(z − z1) ∈ Z.
Let π : A → G, where G is a finite group, be a Freiman 6-isomorphism. As in the proof
of Theorem 3, we will show that p divides |G| and that the p-Sylow subgroup of G cannot
be abelian. It will ensure the boundG p3and the theorem will follow by (23).
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12 N. HEGYV´ARI AND F. HENNECART
Let
(25)h = π([0,0,z − z1]) = π([u,v,z1])−1π([u,v,z]).
Let us show that for any j such that j(z − z1) + z1/ ∈ E, we have π([0,0,j(z − z1)]) = hj.
If 1 ≤ j ≤ t, we proceed by induction: for j = 1, the property is plainly true. Let
2 ≤ j ≤ t. We have
π([u,v,j(z − z1) + z1][u,v,z]−1) = π([u,v,(j − 1)(z − z1) + z1][u,v,z1]−1).
By (24) and by definition of E, both elements [u,v,(j−1)(z−z1)+z1] and [u,v,j(z−z1)+z1]
belong to A2A−2A. Moreover [u,v,z],[u,v,z1] ∈ A hence, by the fact that π is a Freiman
6-homomorphism, we get
π([u,v,j(z − z1) + z1])π([u,v,z])−1= π([u,v,(j − 1)(z − z1) + z1])π([u,v,z1])−1.
Thus, by (25)
π([u,v,j(z − z1) + z1]) = π([u,v,(j − 1)(z − z1) + z1])h.
By multiplying on the left by π([u,v,z1])−1and using again that π is a Freiman 6-homomorphism,
we get
π([0,0,j(z − z1)]) = π([0,0,(j − 1)(z − z1)])h = hj
by the induction hypothesis.
For larger j, we again induct: let j > t be such that j(z −z1)+z1/ ∈ E. Then at least one
of the two elements (j−1)(z−z1)+z1or (j−t)(z−z1)+z1is not in E since z?−z / ∈ E−E.
If (j − 1)(z − z1) + z1/ ∈ E we argue by induction as above. If (j − t)(z − z1) + z1/ ∈ E we
slightly modify the argument: since
π([u,v,j(z − z1) + z1][u,v,t(z − z1) + z1]−1) = π([u,v,(j − t)(z − z1) + z1][u,v,z1]−1)
and π a Freiman 6-isomorphism, we get
π([u,v,j(z − z1) + z1]) = π([u,v,(j − t)(z − z1) + z1])π([u,v,z1])−1π([u,v,t(z − z1) + z1])
= π([u,v,(j − t)(z − z1) + z1])ht,
and finally by induction
π([0,0,j(z − z1)]) = π([u,v,(j − t)(z − z1) + z1])ht= hj−tht= hj.
Since z1 / ∈ E, we obtain hp= 1 in G, thus either h = 1 or h has order p. But z ?= z1hence
[0,0,z − z1] = [u,v,z][u,v,z1]−1?= 1H, hence h ?= 1Gsince π is a Freiman 6-isomorphism.
We deduce that G admits an element of order p, thus the p-Sylow subgroup Gpof G is not
Page 13
FREIMAN MODELS IN HEISENBERG GROUPS13
trivial. By considering the canonical homomorphism σ : G → Gp, ˜ π = σ ◦ π is a Freiman
6-homomorphim of A onto Gp. Hence for any a = [x,y,z] and b = [x?,y?,z?] in A
[˜ π(a); ˜ π(b)] = ˜ π([a;b]) = ˜ π([0,0,xy?− x?y])
which must be equal to 1Gif Gpis assumed to be abelian. It would mean that (x,y) belongs
to a single line for any [x,y,z] ∈ A, giving |A| ≤ p2a contradiction to
ln|A|
lnp
≥ θ(2 + α) > θ(2 + α0(θ)) =
8θ
11θ − 8> 2,
obtained by (12), (21) and (22).
References
[1] Alon, N.; Spencer J.; The probabilistic method, 2nd edition. Wiley Interscience, 2000.
[2] Green, B.; Ruzsa, I. Z.; Freiman’s theorem in an arbitrary abelian group. J. Lond. Math. Soc. (2) 75
(2007), no. 1, 163–175.
[3] Green,
B.;A note onFreiman models(2008).Unpublished note availableon
http://www.dpmms.cam.ac.uk/ bjg23/notes.html
[4] Ruzsa, I. Z.; Sumsets and structure. Combinatorial number theory and additive group theory, 87–210,
Adv. Courses Math. CRM Barcelona, Birkhuser Verlag, Basel, 2009.
[5] Scott, W. R.; Group theory. Second edition. Dover Publications, Inc., New York, 1987. xiv+479 pp.
[6] Tao, T.; Vu V. H.; Additive combinatorics. Cambridge Studies in Advanced Mathematics, 105. Cambridge
University Press, Cambridge, 2006. xviii+512 pp.
[7] Vinh L.A.,Szemer´ edi–Trottertype theoremand sum-product estimatein finitefields,
arXiv:0711.4427v1[CO].
Norbert Hegyv´ ari, ELTE TTK, E¨ otv¨ os University, Institute of Mathematics, H-1117
P´ azm´ any st. 1/c, Budapest, Hungary
E-mail address: hegyvari@elte.hu
Franc ¸ois Hennecart, PRES Universit´ e de Lyon, Universit´ e Jean-Monnet, LAMUSE, 23
rue Michelon, 42023 Saint-´Etienne, France
E-mail address: francois.hennecart@univ-st-etienne.fr
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