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Discrete Mathematics, Algorithms and Applications
Vol. 4, No. 1 (2012) 1250007 (12 pages)
c World Scientific Publishing Company
DOI: 10.1142/S1793830912500073
A SIMPLE WAY TO ESTABLISH
NON-EXISTENCE OF DIFFERENCE SETS
ADEGOKE S. OSIFODUNRIN
Department of Mathematics
Division of Mathematics and Sciences
Livingstone College, Salisbury
North Carolina, 28144, USA
asa [email protected]
[email protected]
Accepted 26 October 2011
Published 15 March 2012
We illustrate that the combination of the variance technique and factorization in the
cyclotomic rings is sufficient to establish non-existence of difference sets.
Keywords: Difference sets; factorization of ideals; representations and groups.
Mathematics Subject Classification 2010: 05B10
1. Introduction
Let G be a multiplicative group of order v. A k-subset D of G is a (v, k, λ) difference
set if every non-identity element of G can be reproduced exactly λ times by the
multi-set {d1 d−1
2 : d1 , d2 ∈ D, d1 = d2 }. The natural number n := k − λ is called
the order of the difference set. Usually, we say that D is abelian (respectively nonabelian or cyclic) difference set if the underlying group G is abelian (respectively
non-abelian or cyclic). Difference sets are closely associated with many fields of
study and its strength lies in the combination of various techniques ranging from
algebraic number theory to geometry, algebra and combinatorics [14].
The study of difference sets was initiated by Paley [13] and Singer [16] used
them while working on symmetric designs obtained from points and hyperplanes of
a finite projective geometry. Nowadays, the most popular method of investigating
difference sets is the use of characters which were developed by Yamamoto [20]
and Turyn [19]. The relationship between symmetric designs and difference sets is
that a symmetric design admitting a group, G, as a regular automorphism group is
isomorphic to the development of the difference set (Theorem 4.2, [7]). This means
that difference sets are useful in the construction of symmetric designs. However,
the converse is not necessarily true (See [4]). It is known that abelian (220, 73, 14)
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difference sets do not exist. This paper illustrates the use of simple methods to rule
out the existence of (220, 73, 14) difference sets in all groups order order 220. Thus,
Theorem 1.1. Suppose that G is a group of order 220. If N is a normal subgroup
of G such that G/N is isomorphic to any group of order 44, C20 , C10 × C2 or D10 ,
then G does not admit (220, 73, 14) difference sets.
This article is organized as follows. Section 2 gives background information
required for this work and we prove the main theorem in Secs. 3 and 4. Ionin
and Shrikhande [6], Lander [7] and Pott [14] give good background information on
difference sets.
2. Preliminary Work
We discuss the basic information required for this work.
2.1. Difference sets
Let Z and C represent the ring of integers and field of complex numbers respectively.
Suppose that G is a group and D is a (v, k, λ) difference set in a group G. We
sometimes view the elements of D as members of the group ring Z[G], which is a
subring of the group algebra C[G]. As a result of this, D represents both subset of
G and element g∈D g of Z[G]. The sum of inverses of elements of D is D(−1) =
−1
. Consequently, D is a difference set if and only if
g∈D g
DD(−1) = n + λG
and DG = kG
(2.1)
Suppose that D is a (v, k, λ) difference set in a group G of order v and ϕ : G → H
is a homomorphism with kernel N . The difference set image in H (also called the
contraction of D with respect to the kernel N ) is the multi-set D/N = ϕ(D) = {dN :
d ∈ D}. If T ∗ = {1 = t0 , t1 , t2 , . . . , th } is a left transversal of N in G, then D/N =
ti ∈T di ti N and the natural number di = |D ∩ ti N | is known as the intersection
number of D with respect to N . In this work, we shall always use the notation D̂ for
ϕ(D) and denote the number of times di equals i by mi ≥ 0. The notation ΩG/N
represents the set of inequivalent difference set images in the factor group G/N .
The following result is very useful in the study of difference sets.
Lemma 2.1. Suppose that D is a (v, k, λ) difference set in a group G and N is a
normal subgroup of G. Suppose that ϕ : G → G/N is a natural epimorphism. Then
(i) D̂D̂(−1) = n · 1G/N + λ|N |(G/N ),
2
(ii)
i di = n + λ|N |.
Lemma 2.1 enables us to search for difference set images in factor groups of group
G by solving equation (i). A counting argument can be used to prove the following
lemma.
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Lemma 2.2. (Variance technique) Suppose that G is a group of order v and H
is a homomorphic image of G with kernel N . Suppose that D̂ is the difference set
image in H and T ∗ is a left transversal of N in G such that {di } is a sequence of
intersection numbers and {mi }, the number of times di equals i. Then
|N |
mi = |H|,
(2.2)
i=0
|N |
imi = k,
(2.3)
i=0
|N |
i(i − 1)mi = λ(|N | − 1).
(2.4)
i=0
2.2. A little about designs
We briefly explore the connection between difference sets and a special type of
design, called symmetric design. An incidence structure X = (P, B, I) is a system
consisting of a set of points P and a set of blocks B with a binary relation I, that
gives the incidence relation. In particular, if the size of P is v, |B| = b, where distinct
points of P are arranged such that each block is incident with k points, each point
is incident with r (known as replication number) distinct blocks and every pair of
points is incident with λ blocks, then we obtain a (v, b, r, k, λ)-design where v > 1
and b are positive integers; r, k and λ are non-negative integers. If v > 1, then the
order of the (v, b, r, k, λ)-design is n = r − λ ≥ 0. The basic equations of designs
are vr = bk and λ(v − 1) = r(k − 1). One example of incidence structure is the
symmetric design. This is a (v, b, r, k, λ)-design in which b = v or r = k. Difference
sets are known to be related to symmetric design in that the development of a
difference set yields a symmetric design [7].
2.3. Representation and algebraic number theories
Suppose that K is the field of real or complex numbers. Then a representation of
a group G is a homomorphism, χ : G → GL(d, K), where GL(d, K) is the group of
invertible d× d matrices over K and d, a positive integer, is the degree of χ. A linear
representation (called character) is a representation of degree one and we denote
the group of characters of G by G∗ . Every group possesses a principal (trivial)
representation χ0 . That is, χ0 (x) = In for all x ∈ G. In this case, the kernel is the
whole group. A faithful character is the character whose kernel is the identity. The
least positive integer m is the exponent of G if g m = 1 for all g ∈ G. The following
orthogonality relations ([14], Lemma 1.2.1 or [9], Chapter 2) summarizes a basic
property of characters.
Lemma 2.3. (Orthogonality relations) Let G be an abelian group of order v and
exponent v ∗ . If the field K contains a primitive v ∗ -th root of unity and characteristic
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of K and v are relatively prime, then for all characters of G
|G|
|G| if χ = χ0
and
χ(g) =
χ(G) =
0
if χ = χ0
0
∗
χ∈G
if g = 1
.
if g =
1
By this lemma, it easy to show that if χ is any non-trivial character of G and χ(D) ∈
Z[ξ], where ξ is a primitive root of unity, then χ(D) χ(D) = n · In + λχ(G) = n · In .
We extend this fact to D̂ as follows:
Lemma 2.4. Suppose that ϕ is an epimorphism of a finite group G with kernel
N and D is a difference set in this group. Then χ(D̂)χ(D̂) = n · 1d , where χ is a
non-trivial representation of G/N of order d which does not have χ0 as a constituent.
Let m be the exponent of an abelian group, G and ξm be the primitive m -th
root of unity. Suppose that K is a field containing ξm . Then K := Q(ξm ) is the
splitting field for G. Without loss of generality, we may replace K with C, the field
of complex numbers. Thus, the central primitive idempotents in C[G] is
χi (1) 1 χi (g)g (−1) =
χi (g)g,
(2.5)
eχi =
|G|
|G|
g∈G
g∈G
where, χi is an irreducible character of G and {eχi : χi ∈ G∗ } is a basis for C[G].
Every element A ∈ C[G] can be expressed uniquely by its image under the char
acter χi ∈ G∗ . That is, if A ∈ C[G], then A = A χi ∈G∗ eχi = χi ∈G∗ Aeχi =
χi ∈G∗ χ(A)eχi and consequently, χi (A)eχi = Aeχi .
Suppose that χ is a representation of G and σ is a Galois automorphism of Km
fixing G. For any g ∈ G, σ acts on the entries of the matrix χ(g) in the natural
way and the function σ(χ) is also a group representation. In this case, χ and σ(χ)
are algebraic conjugates. It is easy to see that algebraic conjugacy is an equivalence
relation. This brings us to an instrument, called an alias that is an interface between
the values of group rings and combinatorial analysis. Aliases are members of group
ring. They enable us to transfer information from C[G] to group algebra Q[G] and
then to Z[G]. Let G be an abelian group and Ω = {χ0 , χ1 , χ2 , . . . , χh }, be the set of
characters of G. The element β ∈ C[G] is known as Ω-alias if for A ∈ C[G] and all
χi ∈ Ω, χi (A) = χi (β). Since A = χi ∈G∗ χi (A)eχi , we can replace the occurrence
of χi (A) which is a complex number by Ω-alias, β an element of Z[G]. Furthermore,
two characters of G are algebraic conjugate if and only if they have the same kernel
and we denote the set of equivalence classes of G∗ by G∗ /∼. Primitive idempotents
give rise to rational idempotents as follows: If K is the Galois field over Q, the
field of rational numbers, then central rational idempotents in Q[G] are obtained by
summing over the equivalence classes Xi on the eχi ’s under the action of the Galois
group of K over Q. That is, [eχi ] = χj ∈Xi eχj , i = 1, . . . , s ≤ h. This yields the
general formula employed in the search of difference set [10].
Theorem 2.5. Let G be an abelian group and K be a field of characteristic 0. Suppose that G∗ /∼ is the set of equivalence classes of characters with
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{χ0 , χ1 , χ2 , . . . , χs } a system of distinct representatives for the equivalence classes.
If A ∈ K[G], then
A=
s
αi [eχi ],
(2.6)
i=0
where αi is any χi -alias for A.
Equation (2.6) is known as the rational idempotent decomposition of A. For
m
example, if G is a cyclic group of the form Cpm = x : xp = 1 (p is prime) whose
characters are of the form χi (x) = ξpi m , i = 0, 1, . . . , m − 1, then the rational
idempotents are
1
x, and 0 ≤ j ≤ m − 1,
p m
1 m −j pm −j−1 [eχpj ] = j+1 p xp
− x
.
p
[eχ0 ] =
(2.7)
(2.8)
Thus, the difference set image in Cpm is D̂ = αχ0 [eχ0 ] + αχpj [eχpj ], where αχ0 and
αχpj are aliases. The following lemma gives information about the character value
of χ(D).
Lemma 2.6. Suppose that G is group of order v with normal subgroup N such that
G/N is abelian. If D̂ ∈ Z[G/N ] and χ ∈ (G/N )∗ , then
k
if χ is a principal character of G/N,
|χ(D̂)| = √
k − λ otherwise.
Dillon [2] proved the following results which will be used to obtain difference set
images in dihedral group of a certain order if the difference images in the cyclic
group of same order are known.
Theorem 2.7. (Dillon Dihedral Technique) Let H be an abelian group and G
be the generalized dihedral extension of H. That is, G = q, H : q 2 = 1, qhq =
h−1 , f or all h ∈ H. If G contains a difference set, then so does every abelian group
which contains H as a subgroup of index 2.
Corollary 2.8. If the cyclic group C2m does not contain a (non-trivial) difference
set, then neither does the dihedral group of order 2m.
A nice way to study difference sets is to use representation theoretic method
made popular by Leibler [10]. Some authors like Iiams [5] and Smith [17] have
used this method in search of difference sets. Our approach is to first obtain the
comprehensive list ΩG/N , of difference set image distribution in least order factor
group of G. We then garner information about D as we gradually increase the size of
the factor group and compute ΩG/N . If at a point the distribution list ΩG/N is empty,
then the group G with factor group G/N does not admit (v, k, λ) difference sets. In
order to obtain ΩG/N , we use Lemmas 2.4, 2.6 and the difference set equatoin (2.6).
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To successfully obtain the difference set images, we need the aliases. Suppose
that G/N is an abelian factor group of exponent m and D̂ is a difference set
image in G/N . If χ is not a principal character of G/N , then by Lemma 2.6,
χ(D̂)χ(D̂) = n. This is an algebraic equation in Z[ξm ] and χ(D̂) is an algebraic
√
number of length n. Also, the image of Z[G/N ] is Z[ξm ]. The determination of the
alias involves knowing how to factor the ideal generated by χ(D̂) in the cyclotomic
ring Z[ξm ], where ξm is the m -th root of unity. If δ := χ(D̂) and based on (2.6),
we seek α ∈ Z[G/N ] such that χ(α) = δ. The task of solving the algebraic equation
δ δ̄ = n is sometimes made easier if we consider the factorization of principal ideals
¯ = (n). To achieve this,
(δ)(δ)
• we must look for all principal ideals π ∈ Z[ξm ] such that ππ̄ = (n),
• for each such ideals, we find a representative element, say δ with δ δ̄ = n and
• for each δ, we find an alias α ∈ Z[G/N ] such that χ(α) = δ.
Using algebraic number theory, we can easily construct the ideal π. However, the
daunting task
is to find an appropriate element δ ∈ π. Suppose we are able to find
ϕ(m )−1
i
di ξm
δ = i=0
∈ Z[ξm ] such that δ δ̄ = n, where ϕ is the Euler ϕ-function. A
theorem due to Kronecker [15] states that any algebraic integer all whose conjugates
have absolute value 1 must be a root of unity. This theorem enables us to describe
all solutions to the algebraic equation. Thus, if there is any other solution to the
j
algebraic equation, then it must be of the form δ = δu [8], where u = ξm
is a
unit. To construct alias from this information, we choose a group element g that is
ϕ(m )−1
di g i such that χ(α) = δ. Hence, the set of
mapped to ξm and set δ = i=0
j
complete aliases is {±αg : j = 0, . . . , m − 1}. We can determine the number of
factors of an ideal in a ring as follows: Suppose p is any prime and m is an integer
such that gcd(p, m ) = 1. Suppose that d is the order of p in the multiplicative group
∗
Zm
of the modular number ring Zm . Then the number of prime ideal factors of the
)
∗
principal ideal p in the cyclotomic integer ring Z[ξm ] is ϕ(m
i.e., ϕ(m ) = |Zm
|
d
[8]. For instance, the ideal generated by 7 has two factors in Z[ξ20 ] while the ideal
generated by 7 is prime in Z[ξm ], m = 3, 23. On the other hand, since 2s is a
power of 2. So the ideal generated by 2 is said to completely ramifies as power of
(1 − ξ2s ) = (1 − ξ2s ) in Z[ξ2s ].
According to Turyn [19], an integer n is said to be semi-primitive modulo m
if for every prime factor p of n, there is an integer i such that pi ≡ −1 mod m .
In this case, −1 belongs to the multiplicative group generated by p. Furthermore,
n is self conjugate modulo m if every prime divisor of n is semi-primitive modulo
m p , m p is the largest divisor of m relatively prime to p. This means that every
prime ideals over n in Z[ξm ] are fixed by complex conjugation. For instance, 75 ≡
−1 mod m , m = 11, 22, 44. Also, 72 ≡ −1 mod m , where m = 5, 10. Thus, 7 is
fixed by conjugation in Z[ξm ]. In this paper, we shall use the phrase m factors
trivially in Z[ξm ] if the ideal generated by m is prime (or ramifies) or m is self
conjugate modulo m . In this case, if χ(D̂) is the difference set image of order n = m2
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in H, where H is a group with exponent m and χ is a non-trivial representation of
i
H, then χ(D̂) = ±mξm
, m is the m -th root of unity.
The ideal generated by 7 has two factors in Z[ξ20 ] and we now find algebraic number δ ∈ Z[ξ20 ] such that δ δ̄ = 49. Consider the automorphism σ ∈ Gal(Q(ξ20 )/Q).
7
. This automorphism split the basis
We define this automorphism as σ(ξ20 ) = ξ20
7
9
3
ξ20 + ξ20
+ ξ20
+ ξ20
and θ̄. Then
elements of Gal(Q(ξ20 )/Q) into two orbits as θ := √
2
implies
that
θ
=
i
5,
where
i
is
the
fourth
root of
θ̄ = −θ and θθ̄ = −θ = 5. This
√
√
unity. Consequently, θ ∈ Z[i 5]. The basis√elements of Z[i 5] are 1 and θ. [Ref. 18]
Thus, we now require δ = a + bθ ∈ Z[i 5] such that δ δ̄ = 49, a, b ∈ Z. This
condition generates the equation a2 − 5b2 = 49. The solutions to the this equation
are (a, b) = (±7, 0), (±2, ±3). Hence δ = ±7, ±(2 + 3θ) or ±(2 + 3θ̄).
In summary, if D̂ is a (220, 73, 14) difference set in Cm , where m =
5, 10, 11, 22, 44, then the possible alias α in the rational idempotent decomposition
of D̂ is
• α = ±7xr , x is a generator of Cm and r = 0, . . . , m − 1.
Furthermore, if D̂ is a (220, 73, 14) difference set in C20 , then the possible alias α
in the rational idempotent decomposition of D̂ is one of the two forms:
• α = ±7xr
• α = 2 + 3(x + x3 + x7 + x9 )g or −2 + 3(x + x3 + x7 + x9 )g, x is a generator of
C20 and g ∈ C20 and r = 0, . . . , 19. [Refs. 11 and 12]
Finally, we look at subgroup properties of a group that can aid the construction of
difference set image. For the convenience of the reader, we reproduce the modified
idea of Gjoneski et al. [4]. Suppose that H is a group of order 2h with a central
involution z. We take T = {ti : i = 1, . . . , h} to be the transversal of z in H
so that every element in H is viewed as ti z j , 0 ≤ i ≤ h, j = 0, 1. Denote the
h
set of all integral combinations, i=1 ai ti of elements of T , ai ∈ Z by Z[T ]. The
subgroup z has two irreducible representations: z → 1 and z → −1. Let ψ0
be the representation induced on H by the trivial representation z → 1 and ψ1 be
the representation induced on H by the non-trivial representation z → −1. Using
the Frobenius reciprocity theorem [9], every irreducible representation of H is a
constituent of ψ0 or ψ1 . Thus, we may write any element X of the group ring Z[H]
in the form
X =X
(1 − z)
(1 + z)
+X
.
2
2
(2.9)
Suppose that A is the group ring element created by replacing every occurrence
of z in X by 1. Also, let B be the group ring element created by replacing every
occurrence of z in X by −1. Then
X=A
(1 − z)
(1 + z)
+B
,
2
2
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h
h
where A = i=1 ai ti and B = j=1 bj tj , ai bj ∈ Z. As X ∈ Z[H], A, B are both
in Z[T ] and A ≡ B mod 2. We may equate A with the homomorphic image of X
in G/z. Consequently, if X is a difference set, then the coefficients of ti in the
expression for A will be intersection number of X in the coset z. In particular, if
K is a subgroup of H such that
H∼
= K × z,
(2.11)
(−1)
= (k −
then we may assume that A and B are in the group ring Z[K] and BB
λ) · 1. The search for the homomorphic image A in K gives considerable information
about the element B as we now illustrate: If the structure of a group H is like (2.11),
then the characters of the group are induced by those of K and z. Let ψ00 be the
characters of H induced by both trivial characters of K and z; ψs1 , induced by
non-trivial characters of K and z; ψ01 , induced by trivial character of K and nontrivial character of z while ψs0 , is the character induced by non-trivial characters
of K and trivial character of z. Suppose that A is a difference set image in K.
Then by Lemma 2.6,
√
√
√
(2.12)
ψ00 (A) = k, |ψs0 (A) = n|, |ψ01 (B) = n|, |ψs1 (B) = n|.
The identity element of Z[K] is K and since A is a rational idempotent, it is of the
√
√
Y
form K
, Y ∈ Z[K]. We subtract k + n or k − n multiples of K from both sides
of ψ00 (A) = k to get
√ √ √
√
k+ n
k− n
ψ00 A −
K = n or ψ00 A −
K = n.
|K|
|K|
√
√
n
k− n
Set α = k+
difference set. The
|K| or α = |K| and B = A − αK, k is the size of
√
√
entries of A are non-negative integers and if |K| divides k + n or k − n, then
BB (−1) = (k − λ) · 1 and
(1 + z)
(1 − z)
+ gB
, g∈H
(2.13)
2
2
Equation (2.13) can be used to determine the existence or otherwise of difference set
√
image in H. However, this approach fails to yield a definite result if |K| (k + n)
√
and |K| (k − n). To buttress the point being made here, consider the parameter
set (70, 24, 8) in the group C70 ∼
= C35 × C2 . Take K = C35 . Then |K| = 35 but
35 does not divide (24 + 4) and (24 − 4). It is known that the group C70 does not
admit this difference set. On the other hand, consider H = (C2 )6 × C5 and take
K = (C2 )5 × C5 . Then |K| = 160 but 160 does not divide (88 + 8) or (88 − 8). Davis
and Jedwab [1] constructed (320, 88, 24) difference set in H.
D̂ = A
3. Difference Set Images in Some Groups of Order 20
In this section, we show that abelian and dihedral groups of order 20 do not admit
(220, 73, 14) difference sets. Our strategy is to first find the difference set images
in C5 and C10 .
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3.1. The C5 image
4
Consider G/N ∼
= C5 = x : x5 = 1. Let D̂ = j=0 dj xj be the difference set image
1
in G/N . Using (2.6)–(2.8), the difference set equation is D̂ = i=0 αχi [eχi ], where
αχ0 = 73 and αχ1 = ±11xr . Consequently, up to translation, difference set image
is A = −7 + 16x.
3.2. The C10 image
Suppose that G/N ∼
= C10 = x, y : x5 = y 2 = [x, y] = 1 and D̂ =
6 1
j k
j=0
k=0 djk x y is the difference set image in this group. Notice that this group
is of the form (2.11) and take K = C5 , α = 16 and A is the unique difference set
image in K. Thus, (2.13) becomes
(1 + y)
(1 − y)
+ gB
, g ∈ G/N,
2
2
where B = A − 16K, A is the difference set images in C5 . Up to translation, the
elements in ΩC10 is A1 = −7 + 8xy. We can use Dillon trick to show that A1 =
−7+8xy is the only difference set image in D5 = x, y : x5 = y 2 = 1, yxy = x−1 .
D̂ = A
3.3. There are no difference set images in C10 × C2 and D10
Suppose that G/N ∼
= C10 × C2 or D10 . Each of these groups is of the form K × C2 ,
where K = C10 or D5 . Using (2.13), we can write the difference set equation in
K × C2 as
(1 + z)
(1 − z)
+ gB1
, g ∈ K × C2 ,
(3.1)
2
2
with B1 = A1 − 8K, z is the generator of C2 , α = 8 and A1 is the unique difference
set image in K. Up to equivalence, the unique solution to (3.1) is −7 + 4xyz,
which has a negative entry. Thus, if G/N ∼
= K × C2 , then G does not admit (220,
73, 14) difference sets.
D̂ = A1
3.4. There are no difference set images in C20
Suppose that G/N ∼
= C20 = x, y : x5 = y 4 = [x, y] = 1 and suppose that the
4 3
difference set in G/N exists and D̂ = i=0 , j=0 dij xi y j . We view this group ring
element as a 4 × 5 matrix with the columns indexed by the powers of x and rows
indexed by powers of y. This group has 6 rational idempotents out of which four
have y 2 in their kernel. The linear combination of these four rational idempotent is
A1 y 2 , where A1 is the difference set image in C10 and
j=0,1
k=0,2 αχij [eχij ] =
2
αχij is an alias. The remaining two rational idempotents are: [eχ01 ] =
[eχ11 ] =
(5−x)(1−y 2 )
.
10
x(1−y 2 )
10
and
Thus, the difference set image in C20 is
D̂ =
A1 y 2 + αχ01 [eχ01 ] + αχ11 [eχ11 ],
2
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(3.2)
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3
7
9
with αχ01 = ±7g3, αχ11 ∈ {7g, 2 + 3(xy + (xy) + (xy) + (xy) )g1 or −2 + 3(xy +
3
7
9
(xy) + (xy) + (xy) )g2 } and g, g1 , g2 , g3 ∈ C20 .
7
((5 − x) − (5 − x)y 2 ), B2 = (2 + 3(xy + (xy)3 + (xy)7 +
Set B1 = 7[eχ11 ] = 10
1
9
(xy) ))[eχ11 ] = 10 ((10 − 2x) + 15(x − x2 − x3 + x4 ) − (10 − 2x) − 15(x − x2 − x3 +
1
(−(10 − 2x) + 15(x −
x4 ))y 3 , B3 = (−2 + 3(xy + (xy)3 + (xy)7 + (xy)9 ))[eχ11 ] = 10
2
3
4
2
2
3
4 3
2
x −x +x )y +(10−2x)y −15(x−x −x +x )y and C = 7[eχ01 ] = 7x
10 (1−y ).
(3.2) becomes
D̂ =
A1 y 2 ± xr y s Bi ± y t C,
2
r = 0, . . . , 4; s, t = 0, . . . , 3; i = 1, 2, 3
(3.3)
2
Observe that two entries of A1 y
are congruent to 10 mod (20) while the remaining
2
entries are congruent to 0 mod (20). Thus, a solution exist if and only if
A1 y 2 (3.4)
+ xr y s (B1 + C),
2
Up to equivalence, the unique solution is −7 + 4xy. The negative entry makes
this solution invalid as difference set image.
D̂ =
4. Difference Set Images in Factor Groups of Order 44
In this section, we compute the difference set images in G/N ∼
= C11 and C22 .
Thereafter, we show that none of the four groups of order 44 admit (220, 73, 14)
difference sets.
4.1. The C11 image
∼ C11 := x : x11 = 1 and D̂ = 10 dj xj is the difference set
Suppose that G/N =
j=0
image in G/N . Using (2.6)–(2.8), the difference set equation is D̂ = 1i=0 αχi [eχi ],
where αχ0 = 73 and αχ1 = ±11xr . Up to translation, difference set image is E =
7 + 6x.
4.2. The C22 and D11 images
Notice that G/N ∼
= C22 = x, y : x11 = y 2 = [x, y] = 1 is of the form (2.11). Thus,
using (2.13) with K = C11 , α = 6, B = E − 6K and E is the unique difference set
image in K, the unique difference set image is E1 = 7 + 3xy.
Suppose that D11 := a, b : a11 = b2 = 1, bab = a−1 and the difference set
10 1
image is i=0, j=0 dij ai bj . This group has one representation of degree two and
four characters. The degree two representation is:
ζ11 0
0 1
, y →
,
χ : x →
−1
1 0
0
ζ11
ζ11 is the eleventh root of unity.
To use Dillon trick, we rewrite the presentation of C22 as C22 = θ : θ22 = 1
and set a := θ2 and b := θ. The difference set image in C22 is now E2 = 7 + 3θ.
However, the above transformation enable us to view E2 as E3 = 7 + 3ab.
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A Simple Way to Establish Non-Existence of Difference Sets
To verify whether or not E3 is a difference set image in D11 , we apply χ and
check if χ(E3 )χ(E3 ) = 49I 2 , where I2 is a 2 × 2 identity matrix. Observe that
χ(E3 ) = (70 07) and the condition χ(E3 )χ(E3 ) = 49I 2 is satisfied. Hence, up to
equivalence, E3 = 7 + 3ab is the unique difference set image in D11 .
4.3. There are no difference set images in any factor group
of order 44
There are four groups of order 44 up to isomorphism. These groups according to
GAP library [3] are:
• The generalized quaternion group, G/N ∼
= Q44 := x, y : x22 = y 4 , x11 =
2
−1
−1
11
= x . If we set z := x , the unique element of order 2, then
y , yxy
(G/N )/z ∼
= D11 .
• The cyclic group, G/N ∼
= C44 := x, y : x11 = y 4 = [x, y] = 1. If we set z := y 2 ,
the unique element of order 2, then (G/N )/z ∼
= C22 .
• The dihedral group, G/N ∼
= D22 := x, y : x22 = y 2 = 1, yxy −1 = x−1 . If we set
z := x11 , the unique element of order 2, then (G/N )/z ∼
= D11 .
11
2
×
C
:=
x,
y
:
x
=
y
=
z 2 = [x, y] = [x, z] =
• The abelian group, G/N ∼
C
= 22
2
[y, z] = 1. Then (G/N )/y ∼
= (G/N )/y ∼
= (G/N )/z ∼
= C22 .
Notice that each of these groups has a factor group of order 22. Our strategy is to
combine the difference set image in the factor group of order 22 with the variance
technique to establish our result.
The distribution of the difference set image in C22 or D11 is 101 321 , which means
that the intersection 10 occurs once while the intersection number 3 occurs 21 times.
If the difference set images in groups of order 44 exist, it must have distributions.
We use variance technique to find these distributions, if they exist. The size of the
kernel of the homomorphism between factor group of order 22 and factor group of
order 44 is 2. This means that each intersection number in groups of order 22 will
split into 2 in the groups of order 44. Furthermore, the coset bound of difference
set image in groups of order 44 is 5. This implies that the intersection numbers of
difference set images in these groups are integers in the interval [0, 5]. Consequently,
the intersection number 10 in groups of order 22 must split as (5, 5) in groups of
order 44. For the remaining intersection numbers, let a be the number of intersection
number 3 that split as (3, 0) and b be the number of intersection number 3 that split
as (2, 1). Using the variance technique notations, m0 = a, m1 = b, m2 = b, m3 = a
and m5 = 2. Consequently, (2.2) and (2.4) yield
2a + 2b + 2 = 44,
(4.1)
6a + 2b + 2(20) = 14(4).
(4.2)
Notice that (2.3) is redundant. Thus, the simplified form of the system of
equations is
a + b = 21,
3a + b = 8.
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(4.4)
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However, this system has no positive integer solutions and there is no difference set
image in any group of order 44.
5. Conclusion
The explorations by GAP show that every group of order 220 is isomorphic to at
least one of C20 , C10 ×C2 , D10 , C44 , D22 , Q44 or C22 ×C2 . Consequently, (220, 73, 14)
difference sets do not exist.
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