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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 15, Issue 1

# Growth functions for some uniformly amenable groups

Janusz Dronka
/ Bronislaw Wajnryb
/ Paweł Witowicz
/ Kamil Orzechowski
Published Online: 2017-04-26 | DOI: https://doi.org/10.1515/math-2017-0049

## Abstract

We present a simple constructive proof of the fact that every abelian discrete group is uniformly amenable. We improve the growth function obtained earlier and find the optimal growth function in a particular case. We also compute a growth function for some non-abelian uniformly amenable group.

MSC 2010: 20F65; 20K21; 22D40

## 1 Introduction

A discrete group G (group G with the discrete topology) is amenable if and only if for every finite subset AG and every t > 0, there exists a non-empty finite subset UG such that |AU| ≤ (1 + t)|U| where |U| denotes the cardinality of U.

A group G is uniformly amenable if the cardinality of the set U in the definition of amenability depends only on t and on the cardinality of A and not on the particular finite set A. More precisely (see [1]).

#### Definition 1.1

A group G is uniformly amenable if there exists a growth function q : ℝ+ × ℕ → ℕ such that for every finite subset AG and every t > 0,

(*) there exists a non-empty finite subset UG, |U| ≤ q(t, |A|) and |AU| ≤ (1 + t)|U|.

The condition in Definition 1.1 is called the uniform Følner condition. There is another approach to uniform amenability due to H. Kesten. It was originally introduced while considering so called symmetric random walks on groups, which are a kind of Markov chains with G as the state space (see [2]). However, Kesten’s approach can be reformulated in a combinatorial manner involving only the group structure without referring to probabilistic issues.

Let A be a finite symmetric subset of G (that means A−1 = A). By m2n(A) we denote the number of all (ordered) 2n-tuples (a1, …, a2n) with aiA, i = 1, …,2n such that the product a1a2n equals e (the identity element of the group). The group G is uniformly amenable if and only if for any natural number k and for any A with |A| = k, $(m2n(A))12n⇉k,$

where the convergence is uniform for all symmetric subsets with the same cardinality k. This condition is called the uniform Kesten condition. The equivalence between the uniform Følner and Kesten conditions was proved by Wysoczański in [10] using ultrapowers of groups.

There is an interesting relationship between the growth function q and values of m2n, namely, according to Kaimanovich, Vershik ([3] and [4]), for a uniformly amenable group G the following inequality holds for all symmetric subsets A with cardinality k $m2n(A)≥k2nsup0

If additionally A does not contain e and elements of order two, the estimation can be refined to $m2n(A)≥k2nsup0

Uniformly amenable groups are obviously amenable. The group S of permutations of ℕ which move only a finite number of elements is an example of a group which is amenable but not uniformly. It is known ([5]) that the class of uniformly amenable groups is closed under extensions and taking subgroups.

More recently some results on uniformly amenable groups were obtained in [6].

Uniformly amenable groups are related to invariant uniform approximation in Banach spaces. We recall briefly this connection in order to motivate our results.

Let G be a compact abelian group, let Γ be the discrete abelian group of the characters of G. Let X = L1(G) be the Banach space of complex functions on G absolutely integrable functions with respect to Haar measure on G. For gX and γ ∈ Γ let $\stackrel{^}{g}\left(\gamma \right)=\int g\left(x\right)\gamma \left(x\right)dx.$ We treat ĝ as a functional on Γ. We say that X has an invariant uniform approximation property (inv. ubap) if there exists a uniform bound β : ℝ × ℕ → ℕ, such that for every ϵ > 0 and every finite subset M ⊂ Γ there exists a function gX with ∥g1 ≤ (1 + ϵ) and |supp(ĝ)| ≤ β(ϵ, |M|).

In [7], Lemma 2.1 it was proven that if Γ has a growth function q(t, k) then X has a uniform bound β(ϵ, k) ≤ q(ϵ, k)2. It was also proven that every Γ has a growth function q(t, k) = (k / t)k and therefore X has a uniform bound β(ϵ, k) ≤ (k / ϵ)2k.

J. Bourgain gave a direct proof in [8] that X has inv. ubap but his proof is difficult to understand and the result is described in terms of entropy numbers.

P. Wojtaszczyk interpreted in [9] (page 209, Theorem 13) the result of Bourgain as a uniform bound β(ϵ, k)≤ (c / ϵ)2k where c is a constant depending on the group G. He also gave a proof of this fact using functional analysis.

In the present paper we give an elementary simple proof that every discrete abelian group has a growth function $q\left(t,k+1\right)=\left(\begin{array}{c}n+k-1\\ k\end{array}\right),$ where t = k / n. By the Stirling formula this is smaller than (2e / t)k. By Lemma 2.1 in [7] it gives a universal bound β(ϵ, k) = (2e / ϵ)2(k − 1) with a small explicit constant c = 2e independant of the group G.

We prove in Lemma 2.5 that our q(t, k + 1) is the minimal growth function if k = 2. We also present an example of a group M of matrices with arbitrary ring coefficients and prove that it is uniformly amenable and we find a growth function for this group. It follows that the group G given in [10] is uniformly amenable because G is a subgroup of M.

## 2 Abelian groups

It is known that every discrete abelian group is uniformly amenable. In this section we shall give a very simple proof of this fact.

We denote by ${C}_{n}^{k}$ the binomial coefficient ${C}_{n}^{k}=\left(\begin{array}{l}n\\ k\end{array}\right).$

#### Lemma 2.1

If a group G satisfies the condition (*) for every finite subset A containing e, then it satisfies (*) for every finite subset A with the same growth function q(t, k).

#### Proof

Let A = {a1, a2, …, ak} ⊂ G and let ${b}_{i}={a}_{1}^{-1}{a}_{i}$ for i = 1, 2, …, k. Let B = {b1, b2, …, bk}. Suppose that U is a finite subset of G such that |BU| ≤ (1 + t)|U|. Then a1BU = AU has the same cardinality as BU so |AU| ≤ (1 + t)|U|.  □

We now pass to the abelian groups. We denote the composition law by + and we denote the neutral element by 0.

#### Lemma 2.2

Let G be an abelian group and let A = {0, a1, …, ak} ⊂ G. Let $U\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left\{\sum _{i=1}^{k}{n}_{i}{a}_{i}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{i}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\in$ {0, 1, …, n − 1}, ∑nin − 1}. Then $|\left(A+U\right)\mathrm{\setminus }U|\le \left(\frac{k}{n}\right)|U|.$

#### Proof

Consider all elements in (A + U) \ U. Every such element can be represented by a sum w = ∑nsas where 0 ≤ nsn and $\sum _{s=1}^{k}{n}_{s}=n\right\}.$ Consider such sums representing all distinct elements of (A + U) \ U. $(A+U)∖U={wi=∑s=1kni,sas:∑s=1kni,s=nfori=1,2,...,d}.$

We fix j ∈ {1, 2, …, k}. We shall prove that the elements {ui,m = wimaj : i = 1, 2, …, d, m = 1, …, ni,j} are distinct elements of U (if ni,j = 0 there are no elements ui,m.)

Since ni,j ≥ 1 and 1 ≤ mni,j the coefficients of ui,m are non-negative and their sum is at most n − 1 so the elements ui,m have the form of the elements in U and they belong to U. Suppose that ui,m = us,r for distinct pairs (i,m) ≠ (s, r). That means wimaj = wsraj. If m = r then wi = ws, which contradicts the assumption that wiws. If mr we may assume 0 < mr < ni,j. Then

ws = wsraj + raj = wimaj + raj = wi − (mr)ajU because 0 < mr < mni,j so wi − (mr)aj has the form of the elements in U. But this contradicts the assumptions that wsU. So we have at least $\sum _{i=1}^{d}{n}_{i,j}$ distinct elements in U.

This is true for j = 1, 2, …, k. On the other hand $\sum _{j=1}^{k}\left(\sum _{i=1}^{d}{n}_{i,j}\right)=\sum _{i=1}^{d}\left(\sum _{j=1}^{k}{n}_{i,j}\right)=dn.$ Therefore there exists j such that $|U|\ge \sum _{i=1}^{d}{n}_{i,j}\ge \frac{dn}{k}.$ Since |(A + U) \ U| = d we have $|\left(A+U\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}U|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\left(\frac{k}{n}\right)|U|.$  □

#### Theorem 2.3

Every abelian group is uniformly amenable with the growth function which satisfies $q\left(\frac{k}{n},k+1\right)={C}_{n+k-1}^{k}$ for natural k and n.

#### Proof

Let k and $t=\frac{k}{n}$ be given. Consider a subset A of G with k + 1 elements. By Lemma 2.1 we may assume that A = {0, a1, …, ak}. Let U be as in Lemma 2.2 for the set A. A sequence of the coefficients (n1, n2, …, nk) of a point in U represents an integer point of a simplex cut off from the non-negative cone of the Euclidean space ℝk (e.g. the non-negative quarter of the plane or the non-negative octant of the space) by the hyperplane $\sum _{i=1}^{k}{n}_{i}=n-1.$ There are ${C}_{n+k-1}^{k}$ integer points in this simplex (an easy induction argument) and therefore there are at most ${C}_{n+k-1}^{k}$ elements in U. By Lemma 2.2 we have $|\left(A+U\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}U|\le \frac{k}{n}|U|\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{hence}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}|A+U|\le \left(1+\frac{k}{n}\right)|U|$ as required.  □

#### Remark 2.4

For the other values of t we can choose the smallest n such that $\frac{k}{n} and let $q\left(t,k+1\right)=q\left(\frac{k}{n},k+1\right).$ We believe that the value of q(t,k + 1) in the theorem is the smallest possible. We have not enough evidence to make it a conjecture. We can prove it for k = 2.

#### Lemma 2.5

For every abelian group G (when G has elements of the infinite order or of an arbitrarily high order) the minimal growth function q(t, k) has values $q\left(\frac{2}{n},3\right)={C}_{n+1}^{2}$ (and for every uniformly amenable group G which has elements of the infinite order or of an arbitrarily high order every growth function satisfies $q\left(\frac{2}{n},3\right)\ge {C}_{n+1}^{2}$).

#### Proof

We know by Theorem 2.3 that if G is abelian, there exists a growth function with these values. Let us fix k = 2 and n. We choose an element gG of a sufficiently high order (at least of order 2n4). We choose a1 = g, a2 = n2g and A = {0, a1, a2}. Suppose U is a finite subset of G such that $|A+U|\le \left(1+\frac{2}{n}\right)|U|.$ Since 0 ∈ A we have U ⊂ (A + U) and therefore $|\left(A+U\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}U|\le \frac{2}{n}|U|.$ We shall prove that |(A + U) \ U| ≥ n + 1 and therefore$|U|\ge {C}_{n+1}^{2}=\frac{n\left(n+1\right)}{2}.$

We partition the group G into cosets with respect to the cyclic subgroup H = < g > generated by the element g (the left cosets H b if G is not abelian) and we partition U into the non-empty pieces Ui contained in different cosets. Then A + Ui is contained in the same coset as Ui. If for every Ui we have $|A+{U}_{i}\right)|>\left(1+\frac{2}{n}\right)|{U}_{i}|,$ then $|\left(A+U\right)|>\left(1+\frac{2}{n}\right)|U|$ which contradicts our assumptions. Therefore for some Ui we have $|\left(A+{U}_{i}\right)|\le \left(1+\frac{2}{n}\right)|{U}_{i}|.$ We shall prove that this implies $|{U}_{i}|\ge {C}_{n+1}^{2}$ and therefore $|U|\ge {C}_{n+1}^{2}$. So we may restrict the argument to one coset and we may assume that G = < g >.

We need to prove that $|U|\ge \frac{n\left(n+1\right)}{2}$ so we may assume that |U|≤n2 − 1. If g is of a finite order, let ord(g) = r ≥ 2n4. Every element of U can be written as u = mg = n1a1 + n2a2 where 0 ≤ n1n2 − 1 and if ord(g) = r then 0 ≤ n2 and m = n1 + n2n2r − 1. So every element of U can be identified with a point (n1, n2) in the plane and different pairs (n1, n2) correspond to different elements of UG. If u = mg then the first coordinate of u is the remainder of m modulo n2.

When we translate U in G we get a different correspondence of the elements of U with the points of the plane and different set of pairs (n1, n2) corresponds to the elements of U.

If u = (n1, n2) ∈ U then a1 + u = (n1 + 1, n2) and a2 + u = (n1, n2 + 1).

Claim. After a suitable translation of U we may assume that U is not a union of two subsets U1 and U2 in the plane in a distance more than $\sqrt{2}$(we say that U is connected).

#### Proof

If subsets U1 and U2 in the plane are in a distance more than$\sqrt{2}$ from each other, then sets A + U1 and A + U2 are disjoint as subsets of the plane but some points in the plane coincide as elements of G. We shall prove that sets A + U1 and A + U2 are disjoint as subsets of G.

Since |U| < n2 not every remainder modulo n2 is equal to the first coordinate of some point in U. If G is infinite we first translate U so that the first coordinate of every point in U is different from n2 − 1. Suppose U = U1U2 as in the claim. Two distinct points in the plane (n1, n2) and (p1, p2) represent the same element of G if p1n1 = n2(n2p2), but this does not happen for the points of the sets A + U1 and A + U2.

If ord(g) = r the same claim requires an additional translation of U. We first translate U so that 0 ∈ U. Since |U| < n2 and r ≥ 2 n4 there must exist a gap of length at least 2n2 between some pair of consecutive values m1, m2 such that m1g, m2gU and m2m1 ≥ 2n2. After another translation we may assume that the gap is at the end so mgU for m > r − 2n2. After another suitable translation by less than n2 − 1 we may assume that the first coordinate of every point in U is different from n2 − 1 and mgU for mrn2. Suppose U = U1U2 as in the claim. Now two distinct points in the plane (n1, n2) and (p1, p2) represent the same element of G if either p1n1 = n2(n2p2) or p1 + n2p2n1n2n2 is a multiple of r, but none of these may happen for the points of the sets A + U1 and A + U2.

Since A + U1 and A + U2 are disjoint and $|\left(A+U\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}U|\le \frac{2}{n}|U|$ we must have $|\left(A+{U}_{i}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}{U}_{i}|\le \frac{2}{n}|{U}_{i}|$ for i = 1 or for i = 2 and we may restrict the discussion to this smaller set Ui and prove that $|{U}_{i}|\ge {C}_{n+1}^{2}.$ So we may assume that U is connected.  □

We assume that U is connected. We make another translation, if necessary, to get min{n1 : (∃n2)(n1, n2) ∈ U} = 0 and min{n2 : (∃ n1)(n1, n2) ∈ U} = 0. This does not change the shape of U, it is still connected.

Let (a, b) be a point in U such that a + b is maximal. Since U is connected every vertical line n1 = c with 0 < c < a meets U and every horizontal line n2 = d with 0 < d < b meets U.

If u1 = (n1, n2) ∈ U is the rightmost element of U in its row then u1 + a1 = (n1 + 1, n2) ∈ (A + U) \ U. We call u1 + a1 a horizontal boundary point of U. If u2 = (m1, m2) ∈ U is the top element of U in its column then u2 + a2 = (m1, m2 + 1) ∈ (A + U) \ U. We call u2+a2 a vertical boundary point of U.

Observe that if u = (n1, n2)∈ U then, by the connectivity of U, there are at least n1 + 1 vertical boundary points in (A + U) \ U and at least n2 + 1 horizontal boundary points in (A + U) \ U. We need only n + 1 points therefore we may assume that n1 < n and n2 < n. If a1 + u or a2 + u is a boundary point, then its coordinates (m1, m2) still satisfy m1 < n2 and m1 + n2m2 is smaller than the order of g. It follows that distinct pairs of coordinates define distinct elements of G also for the boundary points.

We shall prove that each horizontal boundary point (n1 + 1, n2) with n2b is distinct from any vertical boundary point (m1, m2 + 1) with m1a.

Suppose (n1 + 1, n2) = (m1, m2 + 1) is a horizontal and a vertical boundary point. If n2 = b then n1 = a and m1 = n1 + 1 = a + 1. If m1 = a then m2 = b and n2 = m2 + 1 = b + 1. If n2 < b and m1 < a then there are no points of U on the vertical line above this point and on the horizontal line to the right of this point and the top right corner U1 = {(c1, c2) ∈ U : c1 > m1, c2 > n2} splits off, U is not connected.

This means that there are at least a + b + 2 points in (A + U) \ U.

Let d = |(A + U) \ U| ≥ a + b + 2. The set U lies in the triangle {(n1, n2) : 0 ≤ n1, 0 ≤ n2, n1 + n2a + b}, so $\frac{\left(a+b+1\right)\left(a+b+2\right)|}{2}\ge |U|$ and by our assumptions $|U|\ge \frac{n}{2}d\ge \frac{n\left(a+b+2\right)}{2}.$ It follows that a + b + 1 ≥ n and |(A + U) \ U| = da + b + 2 ≥ n + 1.  □

## 3 Some linear uniformly amenable groups

Let R be a commutative ring with a unit 1.

We consider the following group of matrices over R $M={A=1xz01y001:x,y,z∈R}.$

An element of M can be identified with a triple (x, y, z) of elements of R. Then the composition (the product) of the triples is defined by (x1, y1, z1)(x, y, z) = (x1 + x, y1 + y, z1 + z + x1y).

Later we shall also perform the addition of the triples as elements of RRR.

The elements (0,0, z) lie in the center of M and form an abelian normal subgroup C of M. The quotient M/C is abelian so the group M is solvable and it is known that solvable groups are uniformly amenable. We want to find a growth function q(t, k) for this group, which will also give a direct proof of the fact, that M is uniformly amenable.

This example was inspired by a paper [10] by J. Wysoczański. He considered a group G which is a direct sum of groups M over all prime fields Fp = ℤ/ pℤ. If we take for the ring R the direct sum of the fields Fp and we adjoin a unit - the element with each coordinate equal to 1 in the corresponding field we get a ring R1 with a unit. The group G is a subgroup of the group M over R1.

We now consider the group M over an arbitrary commutative ring R with a unit. Let A = {a0, a1, …, ak−1} ⊂ M where ai = (−xi, −yi, −zi) for i = 0, 1, …, k − 1.

The minus sign is for the convenience of the subsequent description where the negative coefficients play a special role.

We want to estimate the number |AU|. By Lemma 2.1 we may assume that a0 = (0, 0, 0) is the unit of the group M. For a fixed n ∈ ℕ we define UnM, $Un={u=(u1,u2,u3):u1=Σitixi,u2=Σitiyi,u3=Σitizi+Σi,jti,jxiyj}$

where i, j ∈ {1, 2, …, k − 1}, ti ∈ {0, 1, …, n − 1}, ti,j ∈ {0, 1, …, n2 − 1}.

An element of Un is determined by (k − 1)2 + (k − 1) integer coefficients. The set J of indices of these coefficients consists of integers and pairs of integers J = {i : 1 ≤ ik − 1, (i, j) : 1 ≤ i, jk − 1}.

We shall denote an element uU by u = (tα)αJ. Then, by the definition of the composition law in M, w = asu has a similar form, w = (fα), where fs = ts − 1, fs,j = ts,jtj and fα = tα for other αJ. In particular fα may be negative if α = s or α = (s, j) and fs ≥ − 1 and fs,j ≥ 1 − n.

#### Lemma 3.1

For any s ∈ {1, 2, …, k − 1} we have $|{a}_{s}{U}_{n}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}{U}_{n}|<\left(\frac{{2}^{k}}{n}\right)|{U}_{n}|.$

#### Proof

We fix an integer s ∈ {1, 2, …, k − 1}. We consider elements of asUn \ Un and for each such element we choose one representative w = (fα). For any non-empty subset of indices I ⊂ {s,(s, 1), (s, 2), …, (s, k − 1)} we consider the set BI of all chosen representatives w = (fα) of the elements in asUn \ Un which have the negative coefficients fα exactly for αI. Suppose that BI = {w1, w2, wd}.

Let eI = (b1, b2, b3) ∈ RRR where b1 = xs, b2 = ys, b3 = zs + n(s, j) ∈ Ixsyj if sI and b1 = 0, b2 = 0, b3 = n(s, j)∈Ixsyj if sI.

We shall prove that the elements wi + meI, i = 1, 2, …, d, m = 1, 2, …,n belong to Un and are distinct. Only the coefficients fα, αI of w change. If sI then fs = −1 and it increases by m when we add meI where 1mn so fs falls into the proper range for the elements in Un. If (s, j) ∈ I then 1nfs, j ≤ −1 and fs, j increases by mn so it also falls into the proper range for the elements in Un.

Suppose that wi + meI = wj + reI for different pairs (i,m) and (j, r). If m = r then wi = wj which contradicts the assumptions. If mr we may assume that m > r. Then wj = wj + reIreI = wi + (mr)eIUn which contradicts our assumptions. Therefore |Un| ≥ nd and $|{B}_{I}|\le \frac{1}{n}|{U}_{n}|.$ There are (2k − 1) non-empty subsets of {s, (s, 1), (s, 2), …, (s, k − 1)} and each element of asUn \ Un lies in one of them so |asUn \ Un| < $\left(\frac{{2}^{k}}{n}\right)|{U}_{n}|.$  □

#### Corollary 3.2

For every commutative ring R with unit the group M is uniformly amenable with a growth function $q(ϵ,k)=(1+(k−1)(2k)ϵ)2(k−1)2+k−1$

#### Proof

For a given k and ϵ choose $n=\left[\frac{\left(k-1\right){2}^{k}}{ϵ}\right]+1.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\left(k-1\right){2}^{k}}{ϵ} Let A = {a0 = (0, 0, 0), a1, …, ak−1} ⊂ M. We choose the set Un as in the previous lemma. Then |Un| ≤ n2(k − 1)2+k − 1$\left(1+\frac{\left(k-1\right){2}^{k}}{ϵ}{\right)}^{2\left(k-1{\right)}^{2}+k-1}=q\left(ϵ,k\right).$ It suffices to prove that |AUn| < (1 + ϵ)|Un|. We have a0Un = Un and by the previous lemma $|A{U}_{n}|\le |{U}_{n}|+\left(k-1\right)\left(\frac{{2}^{k}}{n}\right)|{U}_{n}|<\left(1+ϵ\right)|{U}_{n}|.$  □

#### Remark 3.3

Many of the sets BI in the proof of the previous lemma may have very small contribution to the ″boundary″ AUn \ Un. In particular if different sets of coefficients (tα) represent different elements of Un then for n large |AUn \ Un| is of order $\frac{{k}^{2}}{n}|{U}_{n}|.$ Indeed if a set I in the proof of Lemma 3.1 has only one element then |BI| is equal about n2(k − 1)2+k−2. There are k such sets for each s = 1, 2, …, k − 1. If I has more than one element then |BI| is at least n times smaller therefore for n large |A Un \ Un| is equal about k2n2(k − 1)2+k−2. Moreover |Un| = n2(k − 1)2+k − 1 therefore |A Un \ Un| is equal about $\frac{{k}^{2}}{n}|{U}_{n}|.$ In this case (when n is large and ais areindependent″) it suffices to have $n>\frac{{k}^{2}}{ϵ}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}|{U}_{n}|=\left(\frac{{k}^{2}}{ϵ}{\right)}^{2\left(k-1{\right)}^{2}+k-1}.$ Probably this is the correct size of q(ϵ, k) but we cannot prove it.

## Acknowledgement

We would like to thank Marek Bożejko and Jakub Gismatulin for many helpful remarks.

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Accepted: 2016-12-13

Published Online: 2017-04-26

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 502–507, ISSN (Online) 2391-5455,

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