Application: The sum of two sets of integers

IA+BI+IHI

4.3 Application: The sum of two sets of integers

Let A and B be nonempty, finite sets of integers. Then IA + BI > IA; + IBI - 1.

By Theorem 1.3, I A + B I - J A I + IBI - I if and only A and B are arithmetic progressions with the same common difference. Our goal in this section is to show that if 1 A+BI is "small," then A and B are "large" subsets of arithmetic progressions with the same common difference. This inverse theorem for the sumset A + B is a generalization of Theorem 1.16.

118 4. Kneser's theorem for groups

Theorem 4.6 Let k, 8 > 2, and let A - {ao, al, ... , ak_I } and B - {bo, b1, ... , bt_1 } be nonempty, finite sets of integers such that

0-ao <a, <... <ak-1,

0 - bo < bl < ... < bt-1, bt-1 < ak-1, and

(a1,

a2,...,ak-1)-l.

Let

0 if bt_1 < ak-1

Then

d- I

if

bt_1 - ak_1.

IA+BI > min(ak_I+l,k+2t-8-2}.

Proof. If I A + B I > k + 2f - 8 - 2, we are done. Therefore, we can assume that

We shall prove that Let G - Z/ak _ 1 Z, and let

IA+BI <k+2t-S-3.

IA+BI > ak-1 +t.

r:Z -* G

be the canonical homomorphism onto the cyclic group G. Then n(A + B) - tr(A) + 7r (B),

Itr(A)I -k - 1

since n (ao) - tr (ak _ 1) - 0, and

(4.15)

I,r(B)I-f-8

since n(bt_ 1) - 0 if and only if bt_1 - ak_1. We can rewrite (4.15) in the form

IA+BI < I,r(A)I+I,r(B)I+t-2.

(4.16) We shall show that there are at least e integers in A + B that lie in the same congruence classes modulo ak-1 as other integers in A + B. If bt_1 < ak_1, then

tr(ao + b,) - a(ak_I + bi)

fori -0, 1,...,1 - 1, and

ao + bo <ao+b1 <... <ao+bc_1

< ak-1 +bo < ak-1 +bi < - < ak-1 +bt_I.

If bt_, - ak-1, then

while

and

ao+bo<ao+bi < <ao+bt_2 <ao+bt_I-ak_I+bo

< ak_1 + bi < .. < ak_1 + be-2 < ak_t + bt_i,

n (ao + bo) - ,r (ao + bt_ t) - a(ak_ t + bt_, )

7r(ao+bi) - n(ak_i +bi)

for i - 1, ... , l - 2. Therefore, by inequality (4.16), we have

I,r(A)+rr(B)I < IA+BI -8 < I,r(A)1+In(B)l -2.

(4.17) We can apply Kneser's theorem to the sunset 7r (A) + rr(B) in the group G. Let H - H(n(A)+rr(B) be the stabilizer of n(A)+n(B)). By Theorem 4.2, we have

I,r(A)+s(B)I - I,r(A)+HI+I,r(B)+HI - IHI.

^(4.18)

Since every subgroup of a cyclic group is cyclic, there is a divisor d of ak_ 1 such that H - d G - d Z/ak_ i Z. We shall prove that d - 1.

Let a : G --> G/H be the canonical homomorphism from G onto the quotient group G/H. We partition A + B as follows:

A+B-C,UC2,

where

Ci

-

{c E A + B : an(C) E a7r(B)}

-

{c E A + B : 7r(c) E 7r(B) + H}

and

C2 - ICE A + B : aYr(C) E a2r(A + B) \ alr(B)}

{c E A + B : ,r(c) §t ,r(B)+H}.

Then C1 f1 C2 -0, and

IA+BI-IC1I+IC21

We shall estimate the cardinalities of the sets C1 and C2. Since

,r(B)+H c 7r(A)+7r(B)+H -rr(A+B)

and

Jr(ao+bi)-n(ak_i +bi)-n(b,) E,r(B) c,r(B)+H

120 4. Kneser's theorem for groups

for i - 0, 1, ... , t- I, it follows (by the same argument used to derive (4.17)) that ICiI

- I{cEA+B:lr(c)EJr(B)+H}I

> £+I{7r(c)E7r(A+B):7r(c)E7r(B)+H}I

=

Z+I,r(B)+HI t+Ia,r(B)IIHI

Next we estimate ICz1. Let

r - Ia r(A+ B) \ a7r(B)I.

It follows from (4.18) that

a,r(A+B)I - Ia7r(A)I+Ia7r(B)I - I

and so

r - Ia r(A)I - 1.

Choose ci , ... , Cr E C2 such that

air(A + B) \ a,r(B) - {a7r(cl),... , a,r(c,)),

and choose ai E A and bi E B such that

ai + bi - ci

fori - 1,...,r. Foreachi - 1....,r, we have

I{c E A + B : a7r(c) - a7r(ci)}I

> I(a E A : a,r(a) - an(al)) + (b E B : a7r(b) - a7r(bi)}I I{a E A a7r(a) - a7r(ai)}I + I{b E B : a,r(b) - a r(bi)}I - 1.

Since

I{a E A : a,r(a) - a7r(ai)}I

I(7r(ai)+H)n,r(A)I

17r(ai)+HI+17r(A)I - I(7r(ai)+H)Uir(A)I

IHI+17r(A)I-I,r(A)+HI

and

I{bE B: a,r(b)-a,r(bi)}I > IHI+I7r(B)I -17r(B)+HI,

it follows from (4.18) that

I{c E A + B ax(c) - a7r(ci)}I

21HI + I7r(A)I + 17r(B)I - Iir(A)+ HI - I7r(B)+ HI - I

-

IHI + I,r(A)I + I,r(B)I - I,r(A + B)I - 1,

and so

IC21

- E I{c E

^{A + B}: a7r(c) - o7r(c,)}I i-I

r(IHI +17r(A)I+17r(B)I -17r(A+B)l - 1).

Using our estimates for IC, I and IC21, we obtain

IA+BI

ICII+IC21

e+lan(B)IIHI+r(IHI +17r(A)I+17f(B)I - In(A+B)I - 1) e+lan(B)IIHI +(lo7r(A)l - 1)IHI

+r (17f(A)I + I7r(B)I - 17r(A + B)I - 1)

l + lo,r(A + B)IIHI +r (17r(A)l + In(B)I -17r(A + B)l - 1) E+ 17r(A+ B)I + r (In(A)l + l,r(B)I - 17r(A+ B)I - 1).

On the other hand, from (4.16) we have

IA+BI -< 17f(A)l+I7r(B)I+t-2.

Combining these upper and lower bounds for IA + B1, we obtain

e + 17r(A + B)I + r (17r(A)I + I7r(B)I - 17r(A + B)I - 1)

< IA+BI

< I71(A)I + 171(B)I + e - 2, and so

(r - 1)(17C(A)I + I7r(B)I - 17r(A + B)I) -< r - 2.

By (4.17), we have

and so Therefore,

17r(A + B)I <- 17r(A)I + 17r(B)I - 2,

2(r-1)<r-2.

r - Io7r(A)I - I - 0.

Then o7r(A) - H in Gill since 0 E A, and so 7r(A) C H, that is,

a, =0 (modd)

for every a, E A. Since (a1....,ak_I) - 1, we must have d - 1, hence H -

Z/ak -1 Z - G and

7r(A+B)-7r(A+B)+H-Z/ak_IZ.

122 4. Kneser's theorem for groups

Suppose that S = 0. Then the congruence classes tr(ao + b,) are pairwise distinct

fori = 0, 1, ..., e - 1. Since tr (ao + bi) - tr (ak _, + bi) fori = 0, 1, ..., e - 1, it

follows that there are at least two distinct integers in A + B that belong to each of the a congruence classes n (ao + bi ), and there is at least one integer in A + B in each of the remaining ak_, - e congruence classes in Z/ak_IZ. Therefore,

I A + B I ? 2e+(ak_, - E)=ak_I +e.

Similarly, if S = 1, then the e - I congruence classes tr(ao+bi) are pairwise distinct

fori-0,1,...,e-2.Since it(ao+bi)_it(ak_,+bi)fori-l,...,e-2, and

n(ao + bo) = n (ak_, + bo) - n (ak_, + bt_,) = n(0),

it follows that there are at least two distinct integers in A + B that belong to each of the e - 2 congruence classes n (ao + bi) f o r i - 1, ... , e - 2, that there are atleast three distinct integers in A + B in the congruence class and there is at least one integer in A + B in each of the remaining ak_I - e + I congruence classes.

Therefore,

A+BI ? 2(e-2)+3+(ak_1 -e+l)-ak_,+

This completes the proof.

Theorem 4.7 Let k, e > 2, and let A - {ao, a, , ... , ak_ I } and B - {bo, b, ,

... ,

bt_, } be nonempty, finite sets of integers such that

0 - ao < a, < < ak_, 0 = bo < b, < ... <

be-, 5 ak-1, and

(a,,...,ak-I,bi,...,bt_,)= 1.

^(4.19)

Let

and let

Then

S =

( 0 if

be-, < ak_,

t

¹

if

be-, = ak_,

m = min(k, e - S).

IA+BI > min(ak_,+e,k+e+m-2).

^(4.20)

Proof. If (a, , ... , ak _,) - 1, then inequality (4.20) follows immediately form Theorem 4.6.

Let

d = (a,,...,ak-,) > 2.

Fori -0, 1,...,d - 1, let

B,-{beB:b-i (modd)}

and let

ei-IB,I-I[0,ak_1-1]nB;I+S;,

where Si - 0 for i ¢ 0 and So - S. Then Bo f 0 since 0 E B. Let s denote the number of nonempty sets B,, or, equivalently, the number of congruence classes modulo d that contain at least one element of B. Then (4.19) implies that Bi 0

for some i f 0, and so 2 < s < d. If C E A + Bi, then c - i

(mod d), and so the sumsets A + B; are pairwise disjoint. Moreover,

d-1

A+B - U(A+B;).

e, R

It follows that

d-1

IA+BI

- IA+Bil

e, R d-1

> (k + ti - 1) e, R

s(k-1)+e

> 2k+e-2

> k+e+m-2

> min(ak _ 1 + e, k + e + m - 2).

This completes the proof of inequality (4.20).

Recall that the diameter of a set A is

diam(A) - sup(la - a'I : a, a' E A).

If A is finite and A - {ao,a,,...,ak_I), where ao < a1 < ,ak_1, then

diam(A) - ak_I - ao.

Theorem 4.8 Let A and B be nonempty, finite sets of integers such that diam(B) < diam(A).

Let

0

if diam(B) < diam(A)

I

if diam(B) - diam(A)

Let I Al -k, I B I - e, and m - min(k, e - S). If

IA+BI-k+e-I+b<k+e+m-3,

then A and B are subsets of arithmetic progressions of length at most k + b with the same common difference.

124 4. Kneser's theorem for groups

Proof. Let A - {ao, a, , ... , ak_) } and B - {bo, b1, ... , bi_, }, where

ao < a, < ... < ak-1,

bo < b, < < bt_,,

and let

d - (at - ao, a2 - ao,...,ak_, - ao,b, - bo,...,bk_, - bo).

Let

and

Let

fori-0,1,...,k-I

'

forj-0,1,...,t-l.

A(N) - f a(iN): i

-Q, I,...,k- I}

and

B

lb(N): j -0, 1,...,e- I}.

Then

min(A(N)) - min(B(N)) - 0

and

(a(N),

...

a,,-,, b,N). .. I t-1)

- 1.

Since diam(B) < diam(A), it follows that

b(N) < a(N)

l-1 - k-1

The sets A(N) and B(N) are constructed from A and B, respectively, by affine transformations, and

IA(N)+B(N)I - JA+BI < k+f+m-3.

It follows from Theorem 4.7 that

SA(N)+B(N)I > a,((_ +B, or, equivalently.

br_ <a,((Ni <JA(N)+B(N)j

-e-k-1+b.

Since a; - ao ±a;N)d for i - 0, 1, ... , k - 1, it follows that

Ac {ao+xd:x-0,...,a,Ni} c {ao+xd:x-0,...,k-I+b}.

Similarly,

Bc{bo+yd:y-0,...,b(N)}c{bo+yd:y-0,...,k-I+b}.

This completes the proof.

Theorem 4.9 Let k, t > 2, and let A - {ao, a, , ... , ak_, } and B = {bo, b, , ...,

b_1 } be nonempty, finite sets of integers such that 0 - ao < a, < <

and

Let

If

then

ak_,,

0-bo<b,

be-, ak-1,

d-(a,,...,ak_,)> 1,

(a,,...,ak-1,b,,...,be_,) - 1.

8 - ( 0 if

be_, < ak_,

if

bf_,

-ak_,

I

ak_, <k+Z-S-2,

.

IA+BI>ak_,+P.

Proof. Since d divides a; f o r all i - 1, ... , k - 1, we have

d(k - 1):!j ak_,.

(4.21)

(4.22) The interval [0, ak -I - I] contains exactly ak -I Id integers in each congruence class modulo d. Let s denote the number of congruence classes modulo d that contain at least one element of B. Since

BC[0,ak-,-1+31,

it follows that

sak_I

C -IBI <

d +S.

Inequalities (4.21) and (4.22) imply that

and so It follows that

d

d(k - 1)(d - s) < ak_,(d - s) < d(k - 2).

s -d,

that is, B intersects every congruence class modulo d. Let ak_,

< k+e-8-2

< k+ sak-1 - 2,

B;-{bEB:b-i (modd)}

126 4. Kneser's theorem for groups and let

ti -IB11-I[O,ak_I - l]fl BI +S,

whereSi - O for i 710 and So- S. By (4.21),

1[0,ak_) - 11\ BI - ak_I - e+S < k-2,

and so

ei

- I[O,ak_) - l]nBiI+Si

ak-1

1[0,ak-1-1]\BI+Si

d

Therefore,

min(adl,k+ei-1,

(4.23)

fori-O,1,...,d-1.

Letbi,o-min(Bi)fori -0,...,d- 1. Let A(N)-(a aEA}

ld

and let

B'(N)- (b-b1.0

:bE B}.

d ^JJJJJJ

Since the elements of A(N) are relatively prime, and since min (A(N) U B(N) ) - 0

and

max (A(N) UB,N))

-

^ak-1

d ' it follows from Theorem 4.6 and (4.23) that

IA + Bil

-

IA(N)+ Bi(N)I

>

min(adl,k+ei -Si -2)+ei

ak-1

d + L.

Since the sets A + Bi are pairwise disjoint for i - 0, 1, ... , d - 1, and A U B - Ud-0 (A + Bi ), we have

d-I

IA+BI - EIA+BiI

i-0

d-1

> E 1 a I +ei

-

^{ak_I +e.}1.0

This completes the proof.