tiling groups with di erence sets -...

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Tiling groups with difference sets

Vedran KrcadinacUniversity of Zagreb, Croatia Joint work with:[email protected]

Ante Custic Yue ZhouSimon Fraser University Otto-von-Guericke University

Surrey, Canada Magdeburg, Germany

[email protected] [email protected]

Dedicated to the memory of Axel Kohnert.

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Let G be an additively written group of order v. A (v, k, λ) differenceset in G is a k-subset D ⊆ G such that every nonzero element of Gcan be expressed as a difference x− y with x, y ∈ D in exactly λ ways.

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Tiling = partition of G into disjoint (v, k, λ) difference sets.

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Theorem 1. It is not possible to tile G by difference sets.

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Proof. λ(v − 1) = k(k − 1) ⇒ v = k ·(k−1λ

+ 1k

).

However, k−1λ

+ 1k

can never be an integer.

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Proof. λ(v − 1) = k(k − 1) ⇒ v = k ·(k−1λ

+ 1k

).

However, k−1λ

+ 1k

can never be an integer.

Let’s partition G \ {0} instead!

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Definition. Let G be a finite group of order v with neutral element 0.A (v, k, λ) tiling of G is a collection {D1, . . . , Dt} of mutually disjoint(v, k, λ) difference sets such that D1 ∪ · · · ∪Dt = G \ {0}.

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Example 1. A (7, 3, 1) tiling of Z7:

D1 = {1, 2, 4}, D2 = {3, 5, 6}.

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D1 = {1, 2, 4}, D2 = {3, 5, 6}.


D1 = {1, 5, 11, 24, 25, 27},D2 = {2, 10, 17, 19, 22, 23},D3 = {3, 4, 7, 13, 15, 20},D4 = {6, 8, 9, 14, 26, 30},D5 = {12, 16, 18, 21, 28, 29}.

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An application:

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In the terms of the previous presentation, this matrix is a 6-mosaic

2-(31, 6, 1)⊕ · · · ⊕ 2-(31, 6, 1)︸︷︷︸5 times

⊕ 2-(31, 1, 0).

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2-(31, 6, 1)⊕ · · · ⊕ 2-(31, 6, 1)︸︷︷︸5 times

⊕ 2-(31, 1, 0).

Generally, the development of a (v, k, λ) tiling of G by t difference setsis a (t + 1)-mosaic of symmetric designs

2-(v, k, λ)⊕ · · · ⊕ 2-(v, k, λ)︸︷︷︸t times

⊕ 2-(v, 1, 0).

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2-(31, 6, 1)⊕ · · · ⊕ 2-(31, 6, 1)︸︷︷︸5 times

⊕ 2-(31, 1, 0).

Generally, the development of a (v, k, λ) tiling of G by t difference setsis a (t + 1)-mosaic of symmetric designs

2-(v, k, λ)⊕ · · · ⊕ 2-(v, k, λ)︸︷︷︸t times

⊕ 2-(v, 1, 0).

Another application of (v, k, λ) tilings: hopping sequences for multi-channel wireless networks.

• F. Hou, L.X. Cai, X. Shen, and J. Huang, Asynchronous multichannel MAC design withdifference-set-based hopping sequences, IEEE Transactions on Vehicular Technology, 60 (2011),no. 4, 1728–1739.

• K. Wu, F. Han, F. Han, and D. Kong, Rendezvous sequence construction in cognitive radio ad-hoc networks based on difference sets, 2013 IEEE 24th International Symposium on PersonalIndoor and Mobile Radio Communications, IEEE 2013, p. 1840–1845.

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Necessary existence conditions

The number of difference sets in a (v, k, λ) tiling:

λ(v − 1) = k(k − 1) ⇒ t =v − 1

k=k − 1

λ

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λ(v − 1) = k(k − 1) ⇒ t =v − 1

k=k − 1

λ

Lemma 1. If a (v, k, λ) tiling exists, then k divides v − 1.

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λ(v − 1) = k(k − 1) ⇒ t =v − 1

k=k − 1

λ


The parameters (16, 6, 2) are ruled out by this criterion.

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λ(v − 1) = k(k − 1) ⇒ t =v − 1

k=k − 1

λ



Only parameters (v, k, λ) satisfying Lemma 1 and groups G in whichthere is at least one (v, k, λ) difference set are considered admissible fortilings.

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λ(v − 1) = k(k − 1) ⇒ t =v − 1

k=k − 1

λ



Only parameters (v, k, λ) satisfying Lemma 1 and groups G in whichthere is at least one (v, k, λ) difference set are considered admissible fortilings.

The parameters (31, 10, 3) are not admissible because there is no (31, 10, 3)difference set in Z31 (a nontrivial result!).

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Translates of a (v, k, λ) difference set D ⊆ G are the sets D + x,x ∈ G. They form a symmetric (v, k, λ) block design.

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Lemma 2. If {D1, . . . , Dt} is a (v, k, λ) tiling of a group G, then thedifference sets D1, . . . , Dt are not translates of each other.

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Proposition 1. A (21, 5, 1) tiling of Z21 does not exist.

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Example 3. A (57, 8, 1) tiling of the non-abelian group of order 57,

G = 〈a, b | a3 = b19 = 1, ab7 = ba〉 :

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D1 = {a, b, a2, b2, ab4, ab10, b13, b18},D2 = {ab, ab5, a2b6, a2b13, b15, a2b14, ab15, ab18},D3 = {a2b, a2b7, a2b8, ab9, ab12, b14, ab14, a2b16},D4 = {ab2, b4, a2b3, b9, a2b9, b11, b12, a2b18},D5 = {b3, a2b2, b5, b8, a2b10, a2b11, ab17, a2b17},D6 = {ab3, b6, ab6, ab8, b10, b16, a2b15, b17},D7 = {a2b4, a2b5, b7, ab7, ab11, a2b12, ab13, ab16}.

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Example 4. A (27, 13, 6) tiling of the non-abelian group

G = 〈a, b | a3 = b9 = 1, ab7 = ba〉 :

D1 = {a, b2, ab2, a2b2, b3, ab3, b4, a2b4, ab5, a2b5, ab6, ab7, b8},D2 = {a2, b, ab, a2b, a2b3, ab4, b5, b6, a2b6, b7, a2b7, ab8, a2b8}.

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Example 4. A (27, 13, 6) tiling of the non-abelian group

G = 〈a, b | a3 = b9 = 1, ab7 = ba〉 :

D1 = {a, b2, ab2, a2b2, b3, ab3, b4, a2b4, ab5, a2b5, ab6, ab7, b8},D2 = {a2, b, ab, a2b, a2b3, ab4, b5, b6, a2b6, b7, a2b7, ab8, a2b8}.

Using a computer, we examined all admissible parameters and grups oforder v ≤ 50.

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(v, k, λ) Group Tiling

(7, 3, 1) Z7 Paley-Hadamard(11, 5, 2) Z11 Paley-Hadamard(13, 4, 1) Z13 No(15, 7, 3) Z15 No(19, 9, 4) Z19 Paley-Hadamard(21, 5, 1) Z21 No(21, 5, 1) 〈a, b | a7 = b3 = 1, a2b = ba〉 No(23, 11, 5) Z23 Paley-Hadamard(27, 13, 6) Z3 × Z3 × Z3 Paley-Hadamard(27, 13, 6) 〈a, b | a3 = b9 = 1, ab7 = ba〉 Example 4(31, 6, 1) Z31 Example 2(31, 15, 7) Z31 Paley-Hadamard(35, 17, 8) Z35 No

(37, 9, 2) Z37 F∗37/F(4)37

(40, 13, 4) Z40 No(40, 13, 4) 〈a, b | a5 = b8 = 1, a4b = ba〉 No(43, 21, 10) Z43 Paley-Hadamard(47, 23, 11) Z47 Paley-Hadamard

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If a group G can be tiled by two difference sets {D1, D2}, then theparameters (v, k, λ) are of the form (4n − 1, 2n − 1, n − 1) for somen ≥ 2, i.e. D1 and D2 are Hadamard difference sets.

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A well know construction are the Paley-Hadamard difference sets:

D1 = F(2)q = {x2 | x ∈ F∗q}, q ≡ 3 (mod 4).

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D1 = F(2)q = {x2 | x ∈ F∗q}, q ≡ 3 (mod 4).

The non-squares D2 = F∗q \F(2)q = −D1 are also a difference set, giving

a (q, q−12 ,

q−34 ) tiling of the elementary abelian group (Fq,+).

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D1 = F(2)q = {x2 | x ∈ F∗q}, q ≡ 3 (mod 4).


a (q, q−12 ,


More general: a difference set D is skew Hadamard or antisymmetricprovided D ∪ (−D) = G \ {0} holds. Since −D = {−x | x ∈ D} isalso a difference set, {D,−D} is a tiling of G by two difference sets.

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D1 = F(2)q = {x2 | x ∈ F∗q}, q ≡ 3 (mod 4).


a (q, q−12 ,


More general: a difference set D is skew Hadamard or antisymmetricprovided D ∪ (−D) = G \ {0} holds. Since −D = {−x | x ∈ D} isalso a difference set, {D,−D} is a tiling of G by two difference sets.

Question: does any tiling of G by two difference sets {D1, D2} comefrom a skew Hadamard difference set, i.e. must D2 = −D1 hold?

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Theorem. If {D1, D2} is a tiling of a group G by two (v, k, λ) differ-ence sets, then D2 = −D1 holds, and the two difference sets are skewHadamard.

Proof. Calculation in the group ring Z[G].

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Hence, all results about skew Hadamard difference sets also apply totilings of groups by two difference sets. For example:

Lemma (G. Weng and L. Hu, 2009). A skew Hadamard difference setin an elementary abelian group is necessarily normalized, i.e. the sum ofits elements is 0.

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A few days ago Yue Zhou improved this result by proving it for abeliangroups (without the assumption “elementary abelian”)!

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A few days ago Yue Zhou improved this result by proving it for abeliangroups (without the assumption “elementary abelian”)!

Conjecture. The difference sets in any tiling of an abelian group arenormalized.

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Another generalization of Paley-Hadamard difference sets:

Theorem. Let Fq be a finite field of order q. If there exists a differenceset D in (Fq,+) such that D is a subgroup of (F∗q, ·), then the quotientgroup F∗q/D is a tiling of (Fq,+).

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The theorem also applies to the fourth and eighth power difference sets:

• F(4)q = {x4 | x ∈ F∗q}, q = 4t2 + 1, t odd;

• F(8)q = {x8 | x ∈ F∗q}, q = 8t2 + 1 = 64u2 + 9, t, u odd.

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The theorem also applies to the fourth and eighth power difference sets:

• F(4)q = {x4 | x ∈ F∗q}, q = 4t2 + 1, t odd;

• F(8)q = {x8 | x ∈ F∗q}, q = 8t2 + 1 = 64u2 + 9, t, u odd.

This gives examples of (q, (q−1)/4, (q−5)/16) tilings by four differencesets and (q, (q − 1)/8, (q − 9)/64) tilings by eight difference sets. Thefirst few examples:

• (37, 9, 2), (101, 25, 6), (197, 49, 12), (677, 169, 12) . . .

• (73, 9, 1), (104411704393, 13051463049, 1631432881) . . .

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Example 2 is not covered by this construction because D1 is not asubgroup of F∗31. However, the difference sets in this (31, 6, 1) tiling areunions of two cosets of the subgrup 〈5〉 = {1, 5, 25}:

D1 = {1, 5, 11, 24, 25, 27} = ω0〈5〉 ∪ ω3〈5〉,D2 = {2, 10, 17, 19, 22, 23} = ω4D1,

D3 = {3, 4, 7, 13, 15, 20} = ω8D1,

D4 = {6, 8, 9, 14, 26, 30} = ω2D1,

D5 = {12, 16, 18, 21, 28, 29} = ω6D1.

Here ω = 3 is a primitive element of F31.

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Theorem. Let ω be a primitive element and 〈m〉 a multiplicativesubgroup of order r of the finite field Fq. Suppose we have a (q, k, λ)difference set in (Fq,+) which is a union of cosets

D = ωc1〈m〉 ∪ ωc2〈m〉 ∪ · · · ∪ ωck/r〈m〉,

for c1, c2, . . . , ck/r ∈ {0, 1, . . . , n − 1}, n = (q − 1)/r. Then thereexists a (q, k, λ) tiling of (Fq,+) by multiples of D if and only if thereexist integers b1, b2, . . . , b(k−1)/λ ∈ {0, 1, . . . , n− 1} such that

bi − bj 6≡ cu − cv (mod n)

for all i, j ∈ {1, 2, . . . , (k − 1)/λ}, i 6= j, and u, v ∈ {1, 2, . . . , k/r},u 6= v.

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Theorem. Let ω be a primitive element and 〈m〉 a multiplicativesubgroup of order r of the finite field Fq. Suppose we have a (q, k, λ)difference set in (Fq,+) which is a union of cosets

D = ωc1〈m〉 ∪ ωc2〈m〉 ∪ · · · ∪ ωck/r〈m〉,

for c1, c2, . . . , ck/r ∈ {0, 1, . . . , n − 1}, n = (q − 1)/r. Then thereexists a (q, k, λ) tiling of (Fq,+) by multiples of D if and only if thereexist integers b1, b2, . . . , b(k−1)/λ ∈ {0, 1, . . . , n− 1} such that

bi − bj 6≡ cu − cv (mod n)

for all i, j ∈ {1, 2, . . . , (k − 1)/λ}, i 6= j, and u, v ∈ {1, 2, . . . , k/r},u 6= v.

Singer difference sets: cyclic (qn−1q−1 ,

qn−1−1q−1 , q

n−2−1q−1 ) difference sets.

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The classical examples of Singer difference sets come from PG(n−1, q).Let ω be a primitive element of Fqn, Tr : Fqn → Fq the trace mapping,α ∈ F∗qn and r an integer coprime to qn − 1. Then the following set ofintegers forms a Singer difference set:

{i : 0 ≤ i <qn − 1

q − 1,Tr(αωri) = 0}.

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{i : 0 ≤ i <qn − 1

q − 1,Tr(αωri) = 0}.

Tilings of cyclic groups by Singer difference sets:

• the (7, 3, 1) Paley-Hadamard tiling,

• the (31, 6, 1) tiling of Example 2,

• the (73, 9, 1) eighth powers tiling.

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{i : 0 ≤ i <qn − 1

q − 1,Tr(αωri) = 0}.





Theorem. When qn−1q−1 > q ·

(n+q−2n−1

)+ 1, there is no classical Singer

tiling of the cyclic group of order qn−1q−1 .

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{i : 0 ≤ i <qn − 1

q − 1,Tr(αωri) = 0}.





Theorem. When qn−1q−1 > q ·

(n+q−2n−1

)+ 1, there is no classical Singer

tiling of the cyclic group of order qn−1q−1 .

Thank you!

tiling groups with di erence sets -...

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