Open problems about sumsetsin finite abelian groups

Bela Bajnok

January 5, 2016

Bela Bajnok Open problems about sumsets

Three types of sumsets

G : finite abelian group written additively, |G | = nA = {a1, . . . , am} ⊆ G , |A| = mh ∈ N

h-fold sumset of A:

hA = {Σmi=1λiai : (λ1, . . . , λm) ∈ Nm

0 , Σmi=1λi = h}

h-fold restricted sumset of A:

h A = {Σmi=1λiai : (λ1, . . . , λm) ∈ {0, 1}m, Σm

i=1λi = h}

h-fold signed sumset of A:

h±A = {Σmi=1λiai : (λ1, . . . , λm) ∈ Zm, Σm

i=1|λi | = h}

h A ⊆ hA ⊆ h±A Rarely equal

Bela Bajnok Open problems about sumsets

Three functions

ρ(G ,m, h) = min{|hA| : A ⊆ G , |A| = m}

ρ (G ,m, h) = min{|h A| : A ⊆ G , |A| = m}

ρ±(G ,m, h) = min{|h±A| : A ⊆ G , |A| = m}

ρ(G ,m, h) known; ρ (G ,m, h) and ρ±(G ,m, h) not fully known

ρ (G ,m, 1) = ρ(G ,m, 1) = ρ±(G ,m, 1) = m

ρ (G ,m, h) ≤ ρ(G ,m, h) ≤ ρ±(G ,m, h) Often equal

ρ (Z23, 4, 2) = 5, ρ(Z2

3, 4, 2) = 7, ρ±(Z23, 4, 2) = 8

Bela Bajnok Open problems about sumsets

ρ(G , m, h)

ρ(G ,m, h) = min{|hA| : A ⊆ G , |A| = m}

Cauchy, 1813; . . . ; Plagne, 2006⇓

ρ(G ,m, h) is known for all G , m, h

How to make hA small?G = Zn

Put A in a coset:

A ⊆ a + H ⇒ hA ⊆ h · a + H

Put A in an AP (arithmetic progression):

A ⊆ ∪k−1i=0 {a + i · g} ⇒ hA ⊆ ∪hk−h

i=0 {a + i · g}

Bela Bajnok Open problems about sumsets

ρ(G , m, h)

Put A in an AP of cosets:

Choose d ∈ D(n)

H ≤ Zn with |H| = d so H = ∪d−1j=0 {j · n/d}

Use dm/de cosets of H

m = cd + k with 1 ≤ k ≤ d

Ad = Ad(n,m) = ∪c−1i=0 (i + H)

⋃∪k−1

j=0 {c + j · n/d}

⇓hAd = ∪hc−1

i=0 (i + H)⋃∪hk−h

j=0 {hc + j · n/d}

⇓

|hAd | = min{n, hcd + min{d , hk − h + 1}}

Bela Bajnok Open problems about sumsets

ρ(G , m, h)

fd = fd(m, h) = hcd + d = (hdm/de − h + 1) · d

|hAd | = min{n, hcd + min{d , hk − h + 1}}= min{n, fd , hm − h + 1}= min{fn, fd , f1}

u(n,m, h) = min{fd(m, h) : d ∈ D(n)}

Theorem (Plagne)

For all G , m, and h ρ(G ,m, h) = u(n,m, h)

Proof:≤: construction as above but in any G≥: generalized Kneser’s Theorem

u(n,m, h) ∼ Hopf–Stiefel functionBela Bajnok Open problems about sumsets

ρ(G , m, h)—Inverse Problems

|A| = m, |hA| = ρ(G ,m, h) ⇒ A =?

Examples:

ρ(Z15, 6, 2) = 9⇓

A = (a1 + H) ∪ (a2 + H) with |H| = 3

ρ(Z15, 7, 2) = 13⇓

A = (a1 + H) ∪ (a2 + H) ∪ {a3} with |H| = 3 or

A = (a1 + H) ∪ {a2, a3} with |H| = 5 or

A = ∪6i=0{a + i · g}

Bela Bajnok Open problems about sumsets

ρ(G , m, h)—Inverse Problems

Special Case

p = min{p prime : p|n} and m ≤ p⇓

ρ(G ,m, h) = min{p, hm − h + 1}

Theorem (Kemperman)

hm − h + 1 < p and |hA| = ρ(G ,m, h) = hm − h + 1⇓h = 1 (and A arbitrary), or

A = AP

Conjecture

m ≤ p < hm − h + 1 and |hA| = ρ(G ,m, h) = p⇓

A ⊆ a + H with |H| = p

Bela Bajnok Open problems about sumsets

ρ (Zn, m, h)

ρ (G ,m, h) = min{|h A| : A ⊆ G , |A| = m}

How to make h A small? G = Zn

H ≤ G with |H| = d so H = ∪d−1j=0 {j · n/d}

m = cd + k with 1 ≤ k ≤ d

Ad = Ad(n,m) = ∪c−1i=0 (i + H)

⋃∪k−1

j=0 {c + j · n/d}

|hAd | = min{n, fd , hm − h + 1}

fd = fd(m, h) = hcd + d = (hdm/de − h + 1) · d

|h Ad | =

min{n, fd , hm − h2 + 1} if h ≤ min{k, d − 1}

min{n, hm − h2 + 1− δd} otherwise

δd = δd(m, h) is an explicitly computed “correction term”

Bela Bajnok Open problems about sumsets

ρ (Zn, m, h)

u (n,m, h) = min{|h Ad | : d ∈ D(n)}

ρ (Zn,m, h) ≤ u (n,m, h)

|h A1| = |h An| = min{n, hm − h2 + 1}

ρ (Zn,m, h) ≤ min{n, hm − h2 + 1}

Theorem (Dias da Silva, Hamidoune; Alon, Nathanson, Ruzsa)

p prime ⇒ ρ (Zp,m, h) = min{p, hm − h2 + 1}

For n ≤ 40,m, h:

ρ (Zn,m, h) =

u (n,m, h) more than 99.9% of time

u (n,m, h)− 1 otherwise

Bela Bajnok Open problems about sumsets

ρ (Zn, m, h)

m = k1 + (c − 1)d + k2 with 1 ≤ k1, k2 ≤ d , k1 + k2 > d

Bd = ∪k1−1j=0 {j ·n/d}

⋃∪c−1

i=1 (i ·g +H)⋃

∪k2−1j=0 {c ·g +(j0+ j)·n/d}

Bd is still in dm/de cosets of H but with ≤ 2 cosets not fully in Bd

|h Bd | < |h Ad | ⇔ n,m, h are very special

w (n,m, h) = min{|h Bd | : d ∈ D(n)}

ρ (Zn,m, h) ≤ min{u (n,m, h),w (n,m, h)}

Problem

ρ (Zn,m, h) = min{u (n,m, h),w (n,m, h)} ?

True for all n,m, h with n ≤ 40

Bela Bajnok Open problems about sumsets

ρ (Zn, m, 2)

For h = 2 this becomes:

Conjecture

ρ (Zn, m, 2) =

8>><>>:min{ρ(Zn, m, 2), 2m − 4} if 2|n and 2|m, or

(2m − 2)|n and log2(m − 1) 6∈ N;

min{ρ(Zn, m, 2), 2m − 3} otherwise.

Conjecture (Lev)

ρ (G ,m, 2) ≥ min{ρ(G ,m, 2), 2m − 3− |Ord(G , 2)|}

Theorem (Eliahou, Kervaire)

p odd prime ⇒ ρ (Zrp,m, 2) ≥ min{ρ(Zr

p,m, 2), 2m − 3}

Theorem (Plagne)

ρ (G ,m, 2) ≤ min{ρ(G ,m, 2), 2m − 2}

Bela Bajnok Open problems about sumsets

ρ (G , m, h)

Recall:

Special Case

p = min{p prime : p|n} and m ≤ p⇓

ρ(G ,m, h) = min{p, hm − h + 1}

Conjecture

p = min{p prime : p|n} and m ≤ p⇓

ρ (G ,m, h) = min{p, hm − h2 + 1}

Theorem (Karolyi)

Conjecture true for h = 2.

Bela Bajnok Open problems about sumsets

ρ (G , m, h)—Inverse Problems

Theorem (Kemperman)

hm − h + 1 < p and |hA| = ρ(G ,m, h) = hm − h + 1⇓h = 1 (and A arbitrary), or

A = AP

Conjecture

hm − h2 + 1 < p and |hA| = ρ (G ,m, h) = hm − h2 + 1⇓h ∈ {1,m − 1} (and A arbitrary),

h = 2, m = 4, and A = {a, a + g1, a + g2, a + g1 + g2}, or

A = AP

Theorem (Karolyi)

Conjecture true for h = 2.

Bela Bajnok Open problems about sumsets

ρ (G , m, h)—Inverse Problems

Conjecture

m ≤ p < hm − h + 1 and |hA| = ρ(G ,m, h) = p⇓

A ⊆ a + H with |H| = p

Conjecture

m ≤ p < hm − h2 + 1 and |hA| = ρ (G ,m, h) = p⇓

A ⊆ a + H with |H| = p

Bela Bajnok Open problems about sumsets

ρ±(G , m, h)

ρ±(G ,m, h) = min{|h±A| : A ⊆ G , |A| = m}

All work with Matzke, 2014-2015

Surprises:

ρ±(G ,m, h) depends on structure of G not just |G | = n

E.g. ρ±(Z23, 4, 2) = 8, ρ±(Z9, 4, 2) = 7

Usually |h±A| > |hA|, but often ρ±(G ,m, h) = ρ(G ,m, h)

E.g. n ≤ 24 ⇒ ρ±(G ,m, h) = ρ(G ,m, h) except ρ±(Z23, 4, 2)

Symmetric A (i.e. A = −A) is not always best

Sometimesnear-symmetric A (i.e. A \ {a} is symmetric) orasymmetric A (i.e. A ∩ −A = ∅)is best

But one of these three types will always yield ρ±(G ,m, h)

Bela Bajnok Open problems about sumsets

ρ±(G , m, h)

Theorem

For cyclic G ,ρ±(G ,m, h) = ρ(G ,m, h).

Proof: For each d ∈ D(n), find R ⊆ G so that

R is symmetric,

|R| ≥ m,

|h±R| = |hR| ≤ fd(m, h).

(A symmetric set A with |A| = m and |hA| ≤ fd(m, h) may notexist.)

Bela Bajnok Open problems about sumsets

ρ±(G , m, h)

Theorem

For G of type (n1, . . . , nr ),

ρ±(G ,m, h) ≤ u±(G ,m, h),

where

u±(G ,m, h) = min {Πρ±(Zni ,mi , h) : mi ≤ ni ,Πmi ≥ m}= min {Πu(ni ,mi , h) : mi ≤ ni ,Πmi ≥ m}= min{fd(m, h) : d ∈ D(G ,m)}

with

D(G ,m) = {d ∈ D(n) : d = Πdi , di ∈ D(ni ), dnr ≥ drm}

Note: for cyclic G , D(G ,m) = D(n).

Bela Bajnok Open problems about sumsets

ρ±(G , m, h)

Corollary

u(n,m, h) ≤ ρ±(G ,m, h) ≤ u±(G ,m, h),

where

u(n,m, h) = min{fd(m, h) : d ∈ D(n)}u±(G ,m, h) = min{fd(m, h) : d ∈ D(G ,m)}

Corollary

G is a 2-group ⇒ ρ±(G ,m, h) = ρ(G ,m, h)

Corollary

G is such that 6 ∃H ≤ G ,H ∼= Zrp, p > 2, r ≥ 2

⇓ρ±(G ,m, h) = ρ(G ,m, h)

Bela Bajnok Open problems about sumsets

ρ±(G , m, h)

Can we have ρ± (G ,m, h) < u±(G ,m, h)?

Proposition

d ∈ D(n) odd, d ≥ 2m + 1 ⇒ ρ± (G ,m, 2) ≤ d − 1.

Proof:

∃H ≤ G , |H| = d ,

∃A ⊆ H, |A| = m,A ∩ (−A) = ∅0 6∈ 2±A.

Conjecture

Do(n) = {d ∈ D(n) : d odd, d ≥ 2m + 1}.

ρ± (G ,m, h) = u±(G ,m, h) for all h ≥ 3.

ρ± (G ,m, 2) =

{u±(G ,m, 2) if Do(n) = ∅,min{u±(G ,m, 2), dm − 1} if dm = minD0(n)

Bela Bajnok Open problems about sumsets

ρ±(Zrp, m, h)

Elementary abelian groups Zrp with p odd prime

Theorem

p ≤ h ⇒ ρ±(Zrp,m, h) = ρ(Zr

p,m, h)

Theorem

h ≤ p − 1δ = 0 if h|p − 1 and δ = 1 if h 6 |p − 1k max s.t. pk + δ ≤ hm − h + 1c max s.t. (hc + 1) · pk + δ ≤ hm − h + 1

m ≤ (c + 1) · pk ⇒ ρ±(Zrp,m, h) = ρ(Zr

p,m, h)

Conjecture

m > (c + 1) · pk ⇒ ρ±(Zrp,m, h) > ρ(Zr

p,m, h)

Bela Bajnok Open problems about sumsets

ρ±(Z2p, m, 2)

Conjecture holds for r = 2 and h = 2:

Theorem

ρ±(Z2p,m, 2) = ρ(Z2

p,m, 2),m

m ≤ p,

m ≥ p2+12 , or

∃c ≤ p−12 s.t. c · p + p+1

2 ≤ m ≤ (c + 1) · p

Proof: Via results on critical pairs by Vosper, Kemperman, andLev.

Note:∣∣m : ρ±(Z2

p,m, 2) > ρ(Z2p,m, 2)

∣∣ = (p−1)2

4

Conjecture

|m : ρ±(G ,m, h) > ρ(G ,m, h)| < n4 for every G.

Bela Bajnok Open problems about sumsets

ρ±(Z2p, m, 2)

Theorem

m = cp + v with 0 ≤ c ≤ p − 1 and 1 ≤ v ≤ p⇓

c v ρ(Z2p, m, 2) ρ±(Z2

p, m, 2) u±(Z2p, m, 2)

v ≤ p−12

2m − 1 = 2m − 1 = 2m − 10

v ≥ p+12

p = p = p

v ≤ p−12

2m − 1 < (2c + 1)p = (2c + 1)p1 ≤ c ≤ p−3

2 v ≥ p+12

(2c + 1)p = (2c + 1)p = (2c + 1)p

v ≤ p−12

2m − 1 < p2 − 1 < p2

c = p−12 v ≥ p+1

2p2 = p2 = p2

c ≥ p+12

any v p2 = p2 = p2

The two boxed entries were proven by Lee.Bela Bajnok Open problems about sumsets

ρ±(G , m, h)—Inverse Problems

A(G ,m) = Sym(G ,m) ∪Nsym(G ,m) ∪Asym(G ,m) where

Sym(G ,m) = {A ⊆ G : |A| = m,A = −A}Nsym(G ,m) = {A ⊆ G : |A| = m,A \ {a} = −(A \ {a})}Asym(G ,m) = {A ⊆ G : |A| = m,A ∩ (−A) = ∅}

Theorem

ρ±(G ,m, h) = min{|h±A| : A ∈ A(G ,m)}

Note: each type is essential.

Problem

When is

ρ±(G ,m, h) = min{|h±A| : A ∈ Sym(G ,m)}?ρ±(G ,m, h) = min{|h±A| : A ∈ Nsym(G ,m)}?ρ±(G ,m, h) = min{|h±A| : A ∈ Asym(G ,m)}?

Bela Bajnok Open problems about sumsets

The h-critical number

χ(G , h) = min{m : A ⊆ G , |A| = m ⇒ hA = G}

χ±(G , h) = min{m : A ⊆ G , |A| = m ⇒ h±A = G}

χ (G , h) = min{m : A ⊆ G , |A| = m ⇒ h A = G}

χ(G , h) exists ∀G , h

χ±(G , h) exists ∀G , h

χ (G , h) existsm

h ∈ {1, n − 1}, ∀Gh ∈ {2, n − 2},G 6∼= Zr

2

3 ≤ h ≤ n − 3, ∀G

Bela Bajnok Open problems about sumsets

χ(G , h)

χ(G , h) = min{m : A ⊆ G , |A| = m ⇒ hA = G}

Theorem

χ(G , h) = v1(n, h) + 1

where vg (n, h) = max{(⌊

d−1−gcd(d ,g)h

⌋+ 1

)· n

d : d ∈ D(n)}

.

Theorem (Diamanda, Yap)

max{m : ∃A ⊆ Zn, |A| = m,A ∩ 2A = ∅} = v1(n, 3)

Theorem

max{m : ∃A ⊆ Zn, |A| = m,A ∩ 3A = ∅} = v2(n, 4)

Theorem (Hamidoune, Plagne)

k > l , gcd(k − l , n) = 1 ⇒max{m : ∃A ⊆ Zn, |A| = m, kA ∩ lA = ∅} = vk−l(n, k + l)

Bela Bajnok Open problems about sumsets

χ (G , h)

χ (G , h) = min{m : A ⊆ G , |A| = m ⇒ h A = G}

χ (G , 1) = χ (G , n − 1) = n

Proposition

G 6∼= Zr2 ⇒ χ (G , 2) = (n + |Ord(G , 2)|+ 3)/2

Proposition

(n + |Ord(G , 2)| − 1)/2 ≤ h ≤ n − 2 ⇒ χ (G , h) = h + 2

Problem

3 ≤ h ≤ (n + |Ord(G , 2)| − 3)/2 ⇒ χ (G , h) =?

Problem

3 ≤ h ≤ bn/2c − 1 ⇒ χ (Zn, h) =?

Bela Bajnok Open problems about sumsets

χ (G , h)

Theorem (Dias da Silva, Hamidoune; Alon, Nathanson, Ruzsa)

p prime ⇒ ρ (Zp,m, h) = min{p, hm − h2 + 1}

Corollary

p prime ⇒ χ (Zp, h) = b(p − 2)/hc+ h + 1

Theorem (Gallardo, Grekos, Habsieger, Hennecart, Landreau,Plagne)

n ≥ 12, even ⇒ χ (Zn, 3) = n/2 + 1

Theorem

n ≥ 12, even ⇒ χ (Zn, h) =

n/2 + 1 if h = 3, 4, . . . , n/2− 2;

n/2 + 2 if h = n/2− 1.

Bela Bajnok Open problems about sumsets

χ (Zn, 3)

Case 1: {d ∈ D(n) : d ≡ 2 (3)} 6= ∅, p smallestRecall:

Theorem

χ(G , 3) = v1(n, 3) + 1 =(1 + 1

p

)n3 + 1

Theorem

n ≥ 16 ⇒

χ (Zn, 3) ≥

(1 + 1

p

)n3 + 3 if n = p(

1 + 1p

)n3 + 2 if n = 3p(

1 + 1p

)n3 + 1 otherwise

Problem

χ (Zn, 3) = values above?

True for n prime, n even, n ≤ 50Bela Bajnok Open problems about sumsets

χ (Zn, 3)

Case 2: {d ∈ D(n) : d ≡ 2 (3)} = ∅Recall:

Theorem

χ(G , 3) = v1(n, 3) + 1 =⌊

n3

⌋+ 1

Theorem

n ≥ 11 ⇒

χ (Zn, 3) ≥

⌊

n3

⌋+ 4 if 9|n⌊

n3

⌋+ 3 otherwise

Problem

χ (Zn, 3) = values above?

True for n prime, n ≤ 50

Bela Bajnok Open problems about sumsets

The critical number

χ(G , N0) = min{m : A ⊆ G , |A| = m ⇒ ∪∞h=0hA = G}χ±(G , N0) = min{m : A ⊆ G , |A| = m ⇒ ∪∞h=0h±A = G}χ (G , N0) = min{m : A ⊆ G , |A| = m ⇒ ∪∞h=0h A = G}

p = min{d ∈ D(n) : d > 1}

∪∞h=0hA = ∪∞h=0h±A = 〈A〉⇓

χ(G , N0) = χ±(G , N0) = n/p + 1

Theorem (Dias Da Silva, Diderrich, Freeze, Gao, Geroldinger,Griggs, Hamidoune, Mann, Wou)

n ≥ 10 ⇒ χ (G , N0) =b2√

n − 2c+ 1 if G cyclic with n = p or n = pq whereq is prime, 3 ≤ p ≤ q ≤ p + b2

√p − 2c+ 1

n/p + p − 1 otherwise

Bela Bajnok Open problems about sumsets

χ (G , N0)—Inverse problems

A ⊆ Zn ↔ A ⊆ (−n/2, n/2]||A|| = Σa∈A|a|

Proposition

||A|| ≤ n − 2 ⇒ ∀b ∈ Zn,Σ(b · A) 6= Zn

Conjecture

p prime, A ⊆ Zp, |A| = χ (Zp, N0)− 1 = b2√

p − 2c⇓

ΣA 6= Zp ⇔ ∃b ∈ Zp \ {0}, ||b · A|| ≤ p − 2

Theorem (Nguyen, Szemeredi, Vu)

p prime, A ⊆ Zp, |A| ≥ 1.99√

p⇓

ΣA 6= Zp ⇒ ∃b ∈ Zp \ {0}, ||b · A|| ≤ p + O(√

p)

Bela Bajnok Open problems about sumsets

THANK YOU!

DANKE SCHON!

Bela Bajnok Open problems about sumsets