/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 8, No. 4, 2015, 450-457

ISSN 1307-5543 – www.ejpam.com

On Hadamard Groups with Relatively Large 2-Subgroup

Kristijan Tabak

Rochester Institute of Technology, Zagreb Campus, D.T. Gavrana 15, 10000 Zagreb, Croatia

Abstract. A Hadamard group is any group of order 4u2 that contain a difference set. In this paper

we obtain some new conditions for Hadamard groups with relatively large 2-subgroup. We use norm

invariant polynomials f (ǫ) ∈ Z[ǫ], | f (ǫ t)| = const., where ǫ is root of unity of order 2n. Necessary

condition on a size of normal cyclic 2-subgroup are given. Also, we have covered cases when 2-subgroup

has generators similar to a modular or dihedral 2-group. Additionally, we construct such two infinite

series of groups. Obtained results are natural generalization of a case when entire group is 2-group.

2010 Mathematics Subject Classifications: 05B10, 20E07, 20E28

Key Words and Phrases: Difference set, Norm invariance, Hadamard group, Group representation

1. Introduction and Known Results

We start by introducing a standard definition of a difference set. Let G be a group where

|G|= 4u2. If D ⊆ G, where |D|= 2u2 − u, is such that

{d1d−1
2 | d1, d2 ∈ D}= {(u2 − u) · g | g ∈ G \ {1G}},

then we call D a difference set. In that case group G is called a Hadamard group (see [1, 2]).

In difference set theory is usual to deal with linear combinations of some roots of unity. Good

example is classical paper [6]. We will analyze polynomials like f (ǫ) =
∑w

j=1 k jǫ
r j , where

ki ∈ Z and ǫ is some root of unity.

As it can be seen from some publications (i.e. [5]), algebraic approach to difference set

problems induce a significant progress. Elements of representation theory played important

role in this paper. The main theoretical result which served as technical tool in difference set

theory is

Theorem 1. Let D be a subset of size k of a group G of order v. Let S be a complete set of distinct,

inequivalent, nontrivial, irreducible representations for G. If φ(D)φ(D(−1)) = (k − λ)I for all

φ ∈ S, then D is a (v, k,λ) difference set in G.

Email address: kxtcad@rit.edu

http://www.ejpam.com 450 c© 2015 EJPAM All rights reserved.



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 451

If hypothesized Hadamard group has some cyclic image of order equal to the order of ǫ,

then using representation theory tools it can be shown that there is some polynomial f (ǫ) such

that | f (ǫp)| is constant for every p ∈ Z (assuming f (ǫp) is nontrivial).

In paper [3] Hadamard groups of order 22d+2 were considered. Consequent polynomials

were fully described, after which new necessary conditions have been proven. Pairwise ab-

breviation theorem [3] is also used in this paper. Results that follow will show us again that

’Fourier type’ approach, by comparing coefficients of respective powers, can lead us to some

new conditions.

Main goal of this paper is to offer generalization of results in [3]. That will be provided by

proving result where norm of polynomials is not necessary some power of 2.

We will denote f (ǫ) as nontrivial if it has some nonzero coefficient next to some ǫi 6= 1.

Definition 1. Let ǫ be a root of unity and f (ǫ) ∈ Z[ǫ] is nontrivial. If there is some c, such that

| f (ǫp)|= c, for all p ∈ Z such that f (ǫp) is nontrivial, then we shall say f (ǫ) is norm invariant.

Following result describes pairwise abbreviation.

Theorem 2. Let ǫ be a root of unity of order 2k, k ≥ 1. Suppose that ǫα1 + ǫα2 + · · ·+ ǫαl = 0,

where αi ’s need not to be mutually different. Then l is even and there is a partition of the multiset

{α1,α2, . . . ,αl} in 2-element subsets {αi ,α j} such that ǫαi + ǫα j = 0.

Next result (see [3]) gives a characterization of norm invariant polynomials of norm 2d .

Theorem 3. Let f (ǫ) = ǫr1 + · · ·+ ǫrq be a norm invariant polynomial of norm 2d where

q = 2d(2d+1 − 1) and ǫ is a root of unity of order 22d+2. Let 2n = max{o(ǫri )}. Then for every

k = 0,1,2, . . . , n − 1 there is an r(k) ∈ Z such that f (ǫ2k

) = 2dǫr(k) . We call such polynomials

f (ǫ2k

) maximally abbreviated.

2. Some Algebraic Tools

Let us introduce some notation. For n ∈ N we denote [n] = {1,2, . . . , n} and

[n]0 = {0,1,2, . . . , n}. Next result describes distribution of coefficients standing by ǫ0 in norm

invariant polynomial from Z[ǫ], where ǫ2t

= 1.

Theorem 4. Let ǫ be root of unity of order 2t and F(ǫ) =
∑

kiǫ
i , k ∈ N0. Let sum of coefficients

is 2u2 − u, for some u ∈ N. If every nontrivial F(ǫ2s

) is of norm u, then

∑

i∈A0

k2
i + 2

s
∑

j=1

∑

(a,b)∈A j

kakb − 2
∑

(a,b)∈As+1

kakb = u2

where A0 = [2
t − 1]0 and A j = {(a, b) ∈ A2

0 | a < b, 2 j−1(a− b)≡ 2t−1(mod 2t)}.

Proof. We start by taking s = 0. By assumption we have |F(ǫ)|= u. Then

u2 = |F(ǫ)|2 =
∑

i, j∈A0
kik jβ

i− j , where A0 = [2
t − 1]0. By Theorem 2, after comparing coeffi-

cients next to 1, we get
∑

ǫi− j=1

kik jβ
i− j −

∑

ǫi− j=−1

kik jβ
i− j = u2. (1)



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 452

If ǫi− j = 1, then ki = k j . If ǫi− j = −1, then i − j ≡ 2t−1(mod 2t). It is clear that also

j − i ≡ 2t−1(mod 2t). Therefore, it is natural to introduce

A1 = {(a, b) ∈ A2
0 | a < b, a− b ≡ 2t−1 (mod 2t)}.

Then, from (1) we get
∑

i∈A0

k2
i − 2

∑

(a,b)∈A1

kakb = u2. (2)

Now in next step, let’s take |F(ǫ2)|. After expanding we get |F(ǫ2)|2 =
∑

i, j∈A0
kik jǫ

2(i− j) = u2.

Comparing coefficients next to 1 gives
∑

ǫ2(i− j)=1

kik jǫ
2(i− j) −

∑

ǫ2(i− j)=−1

kik jǫ
2(i− j) = u2. (3)

If ǫ2(i− j) = 1, then ǫi− j = 1 or ǫi− j = −1. Hence
∑

ǫ2(i− j)=1

kik jǫ
2(i− j) =

∑

i∈A0

k2
i + 2

∑

(a,b)∈A1

kakb. (4)

On the other hand, if ǫ2(i− j) = −1, then 2(i− j)≡ 2t−1(mod 2t) and 2( j− i)≡ 2t−1(mod 2t).

Thus
∑

ǫ2(i− j)=−1

kik jǫ
2(i− j) = 2

∑

(a,b)∈A2

kakb

where A2 = {(a, b) ∈ A2
0 | a < b, 2(a− b)≡ 2t−1(mod 2t)}.

If we continue in similar manner, for every nontrivial F(ǫ2s

) we get

∑

i∈A0

k2
i + 2

s
∑

j=1

∑

(a,b)∈A j

kakb − 2
∑

(a,b)∈As+1

kakb = u2. (5)

There is also one simple question that should be covered. Sometimes, we need to know

how many different powers we have such that ǫ2s(i− j) is an involution. Next lemma gives the

answer.

Lemma 1. Let (i, j) ∈ A2
0, where i < j and A0 = [2

t − 1]0. Then there is unique s ∈ {1,2, . . . , t}
such that 2s−1(i − j)≡ 2t−1(mod 2t).

Proof. Take s1 6= s2 from set [t]. Let as assume that 2s1−1(i − j) ≡ 2t−1(mod 2t) and

2s2−1(i − j) ≡ 2t−1(mod 2t). Then, there are K1, K2 ∈ Z such that 2s1−1(i − j) = 2t−1 + K1 · 2
t

and 2s2−1(i − j) = 2t−1 + K2 · 2
t . We may assume that s1 > s2, thus

�

2s1−s2 − 1
�

2s2−1(i − j) = (K1 − K2)2
t .

Now we get (i− j)≡ 0(mod 2t+1−s2), hence 2s2−1(i− j)≡ 0(mod 2t). But, using first assump-

tion we would have 2t−1 ≡ 0(mod 2t+1−s2), which is obvious contradiction. Therefore K1 = K2,

but then we get s1 = s2, again contradiction. Thereby, we have proved the assertion.

Next assertion rise from norm invariance analysis.



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 453

Lemma 2. Let Ω0, . . . ,Ωt ∈ N0 such that where t ∈ N

(i) Ω0 + 2
∑s

i=1Ωi − 2Ωs+1 = u2, s ∈ [t − 1],

(ii) Ω0 + 2
∑t

i=1Ωi = (2u2 − u)2,

If u1 is maximal odd divisor of u, then Ωi ≡ 0(mod u2
1) for every i ∈ [t]0.

Additionally: if u ∈ N is odd, then Ωi ≡ 0(mod u2) for every i ∈ [t]0, and also there is some

integer K such that u= 1+ 2t−1K.

Proof. Solving system of linear equations we get Ωi = 2i−2(Ω0 − u2), i ∈ [t]. Now, using

this and second assumption in our statement, we get (2u2−u)2 = Ω0+(Ω0−u2)(2t−1), hence

2t−2(Ω0 − u2) = u3(u− 1). (6)

Firstly, let assume that u is odd. Therefore u− 1 is divisible by 2t−2. So, there is some integer

K1 such that u= 1+ K1 · 2
t−2. Therefore, Ω0− u2 = K1 · u

3 and Ω0 = u2(1+ K1 · u). This gives

us Ω0 ≡ 0(mod u2). Since Ω1 =
1
2(Ω0−u2) = 1

2 K1u3 ∈ N0 and u odd, we get K1 = 2K for some

integer K . Finally, u = 1+ K · 2t−1 and Ω1 ≡ 0(mod u2). We already proved that Ωi+1 = 2Ωi ,

i ≥ 1. This gives us Ωi ≡ 0(mod u2).

Now, if u= 2Ru1 where u1 is odd, using (6) we getΩ0 ≡ 0(mod u2
1), henceΩ1 ≡ 0(mod u2

1).

Therefore, Ωi ≡ 0(mod u2
1), i ∈ [t].

Now, we are going to need one classical result (for example see [4]).

Lemma 3. Let A be an element of the group ring Z[G], where G is an abelian group. Let χ be a

character of G of order w. Let a prime p be self-conjugated modulo w, i.e. p j ≡ −1(mod w′) for

some j ∈ N where m= paw′, w′ not divisible by p. If χ(A)χ(A)≡ 0(mod p2i), then

χ(A)≡ 0(mod pi).

We present helpful algebraic claim that is most critical for proving main results of this

paper.

Theorem 5. Let ǫ be a root of unity of order 2t . Let F(ǫ) ∈ Z[ǫ] with nonnegative coefficients

whose sum is 2u2 − u, for some u ∈ N. If F(ǫ) is norm invariant of the norm u, then for every

nontrivial F(ǫ2s

) there is some rs ∈ Z such that F(ǫ2s

) = uǫrs .

Proof. Let F(ǫ) =
∑

i∈A0
kiǫ

i , where A0 = [2
t − 1]0. Motivated by previous results, we

introduce notation Ω0 =
∑

i∈A0
k2

i
, Ω j =

∑

(a,b)∈A j
kakb, where

A j = {(a, b) ∈ A2
0 | a < b, 2 j−1(a− b)≡ 2t−1(mod 2t)}.

If we compare coefficients next to 1 in |F(ǫ2s

)|2 and use pairwise abbreviation theorem, we

get Ω0 + 2
∑s

i=1Ωi − 2Ωs+1 = u2. Assumption from theorem claim gives us

|F(ǫ2t

)|2 = Ω0 + 2
∑t

i=1Ωt . Therefore, we are within conditions of Lemma 2.

In the first case we will assume that u is odd. Then Ωi ≡ 0(mod u2). Therefore, we can

define F ′(ǫ) = 1
u F(ǫ), Ω′

i
= 1

u2Ωi . Notice that for every s ∈ [t − 1] we get the system

Ω′0 + 2
∑s

i=1Ω
′
i
− 2Ω′s+1 = 1, or in other words |F ′(ǫ)|= 1.



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 454

Second case deals with even u. In this one we will need more severe tools. Put u = 2Ru1,

where u1 is odd. Then Ωi ≡ 0(mod u2
1). We will use Lemma 3. Let us introduce

A=
∑2t−1

i=0 ki x
i ∈ Z[x], where 〈x〉 ∼= 〈ǫ〉 ∼= C2t . Take character χ : A→ C given by χ(x) = ǫ.

Now we have χ(A)χ(A) = |F(ǫ)|2 = F(ǫ)F(ǫ) = 22Ru2
1 ≡ 0( mod 22R). Using the notation

from Lemma 3, we may write w = 2t , p = 2. Furthermore w = 2t w′, where w′ = 1. It is

obvious that 2 j ≡ −1(mod w′). By Lemma 3, we have F(ǫ) = χ(A) ≡ 0(mod 2R). Therefore,

|F(ǫ)|2 ≡ 0(mod u2). Therefore, we can define F ′ and Ω′
i

as in previous case.

So, both cases lead us to |F(ǫ)′| = 1. It is clear that F ′ has coefficients from Q. Let us

write F(ǫ)′ = (ǫ′)i1+ . . .+(ǫ′)iq , where (ǫ′)i j = 1
uǫ

i j . Let us assume that some two (ǫ′)i j can be

abbreviated in pairs. Additionally, we may assume that we have done all possible abbreviations

in F(ǫ)′. Therefore, for roots that ’survived’ abbreviations we may write

F(ǫ)′ = (ǫ′)s1 + . . .+ (ǫ′)sq1 . Hence,
∑q1

i, j=1
(ǫ′)si−s j = 1. By pairwise abbreviation we get

1= |F(ǫ)′|= |
q1
∑

i=1

(ǫ′)si | ≤
q1
∑

i=1

|(ǫ′)si |=
q1

u
.

Thus q1 ≥ u. Also, we get
∑q1

i 6= j
(ǫ′)si−s j + q1 − 1 = 0. The consequence is that, in previous

sum, we must have at least one term equal to (−1). For example (ǫ′)s1−s2 = −1. It is clear

that we would get (ǫ′)s1 +(ǫ′)s1 = 0, contrary to our assumption about maximal abbreviation.

Therefore, the only option is that for every ji , j2 ∈ [q1] equation (ǫ′)s j1
−s j2 = 1 holds. Now,

it is straight forward q1 = u. Thus F(ǫ)′ =
q1

u ǫ
r0 for some r0 ∈ Z. Finally, this gives us

F(ǫ) = uF(ǫ)′ = uǫr0 .

3. Modular Case

This section deals with groups of order 4u2 with maximal 2-group of modular type. By

that we mean that generators of such 2-group look like those in 2-generated modular 2-group.

We will use o(g) for order of element g.

Theorem 6. Let G = H × L be a group of 4u2 where |L|= u2
1. Let H = 〈x , y1, . . . , ys〉 is of order

22d+2 and o(x) = 2t where x yi ∈ {x , x2t−1+1}. If H ′ ∩ 〈x〉 ≤ 〈x2t−p

〉 and o(x) ≥ 2p+3u, then G

is not a Hadamard group.

Proof. Let us assume that G is a Hadamard group. Then G posses a difference set D with

parameters (4u2, 2u2 − u,u2 − u). Now we will construct a 1-dimensional representations on

H, which is also a representation on G.

Notice that u = 2du1. Condition H ′ ≤ 〈x2t−p

〉 is needed, since for any homomorphism

ϕ : G→ C we also have G′ ≤ Ker(ϕ).

Now, we take difference set and divide it over classes modulo 〈x〉. Then we get

D =

a
∑

i=1

xni +

r
∑

j=1

 

t j
∑

s=1

xm js

!

g j .



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 455

Every g ∈ G can be written in a form g = x i yi1
yi2

. . . yin
, where yi j

∈ 〈y1, . . . , ys〉. Now, for

w = 1,2, . . . , 2t−p we define ϕw : G → C by ϕw(x
i yi1

yi2
. . . yin

) = ǫ2pwi , where ǫ is a root of

unity of order 2t . Also, G′ = H ′ × L′. We notice that ϕw is well defined, since

G′∩
�

〈x〉 × {1L}
�

≤ 〈x2t−p

〉 and 〈ϕw(x
2t−p

)〉= 〈1〉. Now it is clear that G′ ≤ Ker(ϕw) is fulfilled.

By Theorem 1 we have ϕw(D)ϕw(D
−1) = |ϕw(D)|

2 = u2, for every nontrivial ϕw, i.e. for

w = 1,2, . . . , 2t−p − 1. Hence ϕw(D) =
∑a

i=1 ǫ
2pwni +

∑r

j=1

�

∑t j

s=1
ǫ2pwm j

�

is norm invariant

polynomial of norm u. Therefore, by Theorem 5, for every w= 1,2, . . . , 2t−p−1 there is some

rw ∈ Z such that ϕw(D) =
∑a

i=1 ǫ
2pwni +

∑r

j=1

�

∑t j

s=1
ǫ2pwm j

�

= uǫ2p rw . Then, by Theorem 2,

in a sum ϕ1(D) =
∑r

j=0ϕ1(D ∩ 〈x〉g j) we must have u addends ǫ2p r1 distributed over r + 1

classes. This could be interpreted is of u copies (items or objects) of ǫ2p r1 are distributed over

r + 1 boxes. Then we can use Dirichlet’s principle. Number of boxes should be

[G : 〈x〉] = 22d+2−tu2
1. By Dirichlet’s principle, there is some j′ ∈ {0,1, . . . , r} such that ϕ1(D∩

〈x〉g j′) has Ω copies of ǫ2p r1 . We estimate value for Ω as follows:

Ω =

�

u− 1

22d+2−tu2
1

�

+ 1=

�

2du1 − 1

22d+2−tu2
1

�

+ 1≥

�

2du1

22d+2−tu2
1

�

− 1+ 1

=

�

2t−d−2

u1

�

= {since 2t ≥ 2p+3u= 2p+32du1}=

�

2t · 2−d−2

u1

�

≥

�

2p+32du12−d−2

u1

�

= 2p+1.

Therefore, Ω ≥ 2p+1. Before we get a final contradiction, additional observation is necessary.

We have

(ϕ1|〈x〉)
−1(ǫ2p r1) = {x i | ϕ1(x

i) = ǫ2p r1}= {x i | ǫ2p i = ǫ2p r1}

={x i | ǫ2p(i−r1) = 1}= {x i | ϕ1(x
i−r1) = 1}= x r1 Ker(ϕ1).

Hence,

2p+1 ≤|{i | 0≤ i < 2t , ϕ1(x
i g j′) = ǫ

2p r1}|

=|{i | 0≤ i < 2t , ϕ1(x
i) = ǫ2p r1}|

=|(ϕ1|〈x〉)
−1(ǫ2p r1)|

=|Ker(ϕ1|〈x〉)|

=2p,

which is an obvious contradiction.

4. One Infinite Series of Groups for Modular Case

We construct one example of infinite series of groups that can be covered by previous result.

Take G = H × L, where

H = 〈x , y1, y2 | x
22t1
= y23

= y23

2 = 1, x y1 = x y2 = x22t1−1+1, [y1, y2] = x22t1−4

〉.



K.Tabak / Eur. J. Pure Appl. Math, 8 (2015), 450-457 456

L could be any group of order u2
1. Then |H|= 22t1+6 = 22d+2, where t1 = d−2. Put |G|= 4u2,

where u = 2du1. Let’s take p = 4. Then we have [x , yi] = x−1 x yi = x22t1−1

∈ 〈x22t1−4

〉.
Therefore H ′ ≤ 〈x22t1−4

〉 = 〈x22t1−p

〉. Additionally, if we put 2t1 ≥ 29u1, using t1 = d − 2, we

get 22d−4 ≥ 2d · 27u1 = 2p+3u. Therefore, we are within conditions of Theorem 6.

For example, u1 = 36, t1 = 15, p = 4 and d = 17. Then

2t1 = 215 = 32768> 29u1 = 18432.

In that case group order would be |G|= 22d+2u2
1 = 236 · 362.

5. Dihedral Case

This section covers types of groups of order 4u2 which also contain a 2-subgroup of dihedral

type. This means that 2-subgroup has a normal cyclic subgroup, while outer elements either

commute or invert generator of that normal subgroup.

Theorem 7. Let G = H × L be a group of order 4u2 and |L| = u2
1. Let H = 〈x , y1, . . . , ys〉 has

order 22d+2 and o(x) = 2t where x yi ∈ {x , x−1}. If CH(x)
′ ∩ 〈x〉 ≤ 〈x2t−p

〉 and o(x) ≥ 2p+3u,

then G is not a Hadamard group.

Proof. We will use the same approach as in proof of Theorem 6. Let us assume that G

posses a difference set D with parameters (4u2, 2u2 − u,u2 − u). This time, we will have to

develop a 2-dimensional representations. Like before u = 2du1. Since [H : CH(x)] = 2, every

h ∈ H can be written as h = c ym, where c ∈ CH(x) and m ∈ {0,1}. So, every g ∈ G can be

written as g = x ic1 yml, where c1 ∈ CH(x) \ 〈x〉 and l ∈ L.

Then, for every w= 1,2, . . . , 2t−p we define Φw : G→ GL(2,C) by

Φw(x
ic1 yml) =

�

ǫ2pwi 0

0 ǫ−2pwi

��

0 1

1 0

�m

.

Because of CH(x)
′ ∩ 〈x〉 ≤ 〈x2t−p

〉, map Φw is well defined 2-dimensional representation. Dif-

ference set D, after been taken modulo 〈x〉, becomes D =
∑a

i=1 xni +
∑r

j=1

�

∑t j

s=1
xm js

�

g j .

This time, we need to separate indices. Let’s use j1 if x g j1 = x , and j2 when g j2 = x−1. Thus,

from Φw(D)Φw(D
(−1)) = u2 I2, where I2 is a 2× 2 identity matrix, we get

�

�

�

�

�

�

∑

i

ǫ2pwn1 +
∑

j1

ǫ2pwm j1 +
∑

j2

ǫ2pwm j2

�

�

�

�

�

�

= u

where w= 1,2, . . . , 2t−p−1. Therefore, polynomial
∑

i ǫ
2pn1+

∑

j1
ǫ2pm j1+

∑

j2
ǫ2pm j2 is a norm

invariant of norm u. Now, proof continues as for Theorem 6.



REFERENCES 457

6. One Infinite Series of Groups for Dihedral Case

This section offers one construction of infinite series of groups that are within conditions

of Theorem 7. We start by taking a group H = 〈x , y1, y2〉, where o(x) = 22t1 , o(yi) = 2si

and x yi = x−1. It is easy to see that [H : CH(x)] = 2. Firstly, let us show that assumption

[y1, y2] ∈ 〈x〉 forces that 〈y2
1 , y2

2 〉 ≤ Z(H). From assumption we get that there is some α such

that [y1, y2] = xα. Then y1 y2 = y2 y1 xα and y2 y1 = y1 y2 x−α. Using this we get,

y2
2 y1 = y2 y2 y1 = y2 y1 y2 x−α = y1 y2 x−α y2 x−α = y1 y2 y2 xαx−α = y1 y2

2 .

It is clear that [y2
2 , y2] = [y

2
2 , x] = 1, therefore y2

2 ∈ Z(H). Due to symmetry, similar works

for y2
1 . Thus, 〈y2

1 , y2
2 〉 ≤ Z(H).

Now, let us show that CH(x) = 〈x , y2
1 , y2

2 , y1 y2〉 ≤ H is abelian subgroup of index 2. By

previous argument and because of x y1 y2 = x , it is clear that C1 := 〈x , y2
1 , y2

2 , y1 y2〉 ≤ CH(x).

We obtain |H|= 22t1+s1+s2 and |C1|= 22t1+s1+s2−2. Take C2 = 〈C1, y1 y2〉. Since

(y1 y2)
2 = y2

1 y2
2 xα ∈ C1, we conclude that [C2 : C1] = 2, thus [H : C2] = 2. Hence C2 = CH(x).

It is clear that CH(x) is abelian. Then CH(x)
′ ∩ 〈x〉= {1}.

After this, let us take concrete numbers. Put s1 = s2 = 3 and [y1, y2] = x2t1−1 (although,

by previously proven facts, it is sufficient to assume [y1, y2] ∈ 〈x〉). Let L1 be a group of

order u2
1 and G = H × L of order 4u2 = 22t1+6u2

1. Define d = t1 + 2. Then |H| = 22d+2

and, as proved, CH(x) = 〈x , y2
1 , y2

2 , y1 y2〉. As shown, CH(x)
′ ∩ 〈x〉 ≤ 〈x2t1−p〉 for any p. If

o(x) ≥ 2p+3u then we would be within conditions of Theorem 7. Because of u = 2t1+2u1, we

need o(x) = 22t1 ≥ 2p+3 · 2t1+2u1. This gives us a conditions that chosen group parameters

should fulfil: 2t1−p−5 ≥ u1 ≥ 1. So, let us, for example choose u1 = 10, p = 2, t1 = 11. Then

2t1−p−5 = 24 = 16≥ 10= u1. Then o(x) = 22t1 = 222. Thus, the group

G = 〈x , y1, y2 | x
222

= y8
1 = y8

2 = 1, x yi = x−1, [y1, y2] = x221

〉 × L,

where |L1|= u2
1 = 102 is one concrete example. Group order is |G|= 228 · 102.

References

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Wien-Zürich, 1985.

[2] D. Jungnickel, A. Pott and K.W. Smith. Difference Sets, In C. J. Colburn, J.H. Dinitz, editors,

The Handbook of Combinatorial Designs, Second Edition, 419-435. CRC Press, 2007.

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of Algebraic Combinatorics, 38, 785-795. 2013

[4] A. Pott. Finite Geometry and Character Theory, Springer-Verlag, Berlin-Heidelberg, 1995.

[5] B. Schmidt. Characters and Cyclotomic Fields in Finite Geometry, Springer, Berlin-

Heidelberg, 2002.

[6] R. J. Turyn. Character sums and difference sets, Pacific Journal of Mathematics, 15, 319-

346. 1965.