EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 7, No. 4, 2014, 462-471
ISSN 1307-5543 – www.ejpam.com

Characterization of U1(Z[Cn × K4])

Ismail Gokhan Kelebek∗, Tevfik Bilgin

Department of Mathematics, Fatih University, Istanbul, Turkey

Abstract. Constructing the group of units U(ZG) of the integral group ring ZG, for a finite group G, is
a classical but open problem. In this study, it is shown that

U1(Z[Cn × K4]) = U1(ZCn)× (1+ K x)× (1+ K y)× (1+ K x y).

This structure theorem is applied to give precise characterization of U1(Z[Cn × K4]) for cyclic groups
C5 and C7.

2010 Mathematics Subject Classifications: 16U60, 16S34
Key Words and Phrases: Integral group ring, unit problem, generators of unit group

1. Introduction

Let us denote ZA the integral group ring of a finite abelian group A with the coefficients
from the ring of integers Z. Let U(ZA) be the group of units in ZA. Higman [4] obtained the
following result:

Theorem 1. If A is a finite abelian group then

U(ZA) = ±A× F,

where F is a free abelian group.

Here torsion units are trivial, torsion free units are finite but the rank is not determined.
The rank of torsion free part is determined by Ayoub and Ayoub [2].

Theorem 2. If A is a finite abelian group then U(ZA) = ±A× F with the rank

ρ =
1
2
(|A|+ 1+ n2 − 2l), (1)

where n2 is the number of elements of A of order 2 and l is the number of cyclic subgroups of A.

∗Corresponding author.

Email addresses: gkelebek@fatih.edu.tr (I.G. Kelebek), tbilgin@fatih.edu.tr (T. Bilgin)

http://www.ejpam.com 462 c© 2014 EJPAM All rights reserved.



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 463

The structures of the unit groups for ZC5, ZC8 were given by Karpilovsky [5] as follows:

U(ZC5) =± C5×< −1+ a+ a4 > and

U(ZC8) =± C8×< 2+ (a+ a7)− (a3 + a5)− a4 > .

Aleev and Panina [1] described the structure of U(ZC7) and U(ZC9):

U(ZC7) =± C7×< −1+ a+ a6,−1+ 2a2 − a3 − a4 + 2a5 > and

U(ZC9) =± C9×< −1− (a+ a8)− (a2 + a7) + 2(a4 + a5)>

×< −1− (a+ a8) + (a2 + a7)> .

The unit group U(ZC12) was characterized by Bilgin [3] as,

U(ZC12) = ±C12×< 3+ 2(a+ a11) + (a2 + a10)− (a4 + a8)− 2(a5 + a7)− 2a6 > .

Low [6] gave a generalization of the structure of the the unit group for Cn × C2 using exact
sequences.

Remark 1. Since U(ZG) = ±U1(ZG), we will use U1(ZG) instead of U(ZG).

In this study, we extend group epimorphisms linearly over Z to ring epimorphisms in the
first place. After that, we determine their kernels to construct exact sequences at ring level.
Then, by restricting these exact sequences to unit level, we characterize U1(Z[Cn× K4]) as an
internal direct product of four subgroups. Finally we describe these subgroups explicitly and
give two concrete examples for cyclic groups C5 and C7.

2. Main Structure Theorem

We can construct the following group epimorphisms by using abelian group

Cn × K4 =< a, x , y : an = x2 = y2 = 1, ax = xa, a y = ya, x y = y x >

as follows:
πx : Cn × K4→ Cn ×




y
�

, πy : Cn × K4→ Cn × 〈x〉
a 7→ a a 7→ a
x 7→ 1 x 7→ x
y 7→ y y 7→ 1

.

If we denote the identity map by ι, then we get the following exact sequences:

〈x〉
ι
−→ Cn × K4

πx−→ Cn ×



y
�

,



y
� ι
−→ Cn × K4

πy
−→ Cn × 〈x〉 .

By extending these epimorphisms linearly over Z, the following ring epimorphisms are ob-
tained:



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 464

πx :Z[Cn × K4] −→ Z[Cn ×



y
�

]

P0 + P1 x + P2 y + P3 x y 7→(P0 + P1) + (P2 + P3)y

πy :Z[Cn × K4] −→ Z[Cn × 〈x〉]
P0 + P1 x + P2 y + P3 x y 7→(P0 + P2) + (P1 + P3)x .

Then, we can calculate the kernels of the epimorphisms.

N x=Kerπx

=
�

P = P0 + P1 x + P2 y + P3 x y ∈ ZCn : πx(P) = 0
	

=
�

P = P0 + P1 x + P2 y + P3 x y ∈ ZCn : (P0 + P1) + (P2 + P3)y = 0
	

=
�

P = P0 + P1 x + P2 y + P3 x y ∈ ZCn : P0 = −P1, P2 = −P3

	

=
�

(x − 1)P1 + y(x − 1)P3 : P1, P3 ∈ ZCn

	

=
�

(x − 1)[P1 + yP3] : P1, P3 ∈ ZCn

	

=(x − 1)Z[Cn ×



y
�

].

Similarly N y = (y −1)Z[Cn×〈x〉]. By restricting πx and πy to the kernels N y and N x we get
the images K y = (y − 1)ZCn and K x = (x − 1)ZCn respectively. Of course, K y is the kernel of
πy and K x is also the kernel of πx . On the other hand, the ring

K x y =
�

(x − 1)(y − 1)P3 : P3 ∈ ZCn

	

is the kernel of ring epimorphism :

πy :N x −→ K x

(x − 1)[P1 + yP3] 7→ (x − 1)[P1 + P3],

while K x y is the kernel of another ring epimorphism :

πx :N y −→ K y

(y − 1)[P2 + x P3] 7→ (y − 1)[P2 + P3].

Hence, we have the following commutative diagram at ring level:

K x y ι
−→ N x

πy
−→ K x

ı ↓ ı ↓ ı ↓

N y ι
−→ Z[Cn × K4]

πy
−→ Z[Cn × 〈x〉]

πx ↓ πx ↓ πx ↓

K y ι
−→ Z[Cn ×



y
�

]
πy
−→ ZCn.



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 465

Theorem 3. U1(Z[Cn × K4]) = U1(ZCn)× (1+ K x)× (1+ K y)× (1+ K x y).

Proof. By restricting πx and πy to the the unit group U1(Z[Cn×K4]), we get the following
commutative diagram for groups.

1+ K x y ι
−→ 1+ N x

πy
−→ 1+ K x

ı ↓ ı ↓ ı ↓

1+ N y ι
−→ U1(Z[Cn × K4])

πy
−→ U1(Z[Cn × 〈x〉])

πx ↓ πx ↓ πx ↓

1+ K y ι
−→ U1(Z[Cn ×




y
�

])
πy
−→ U1(ZCn).

In the diagram, each row and column are exact sequences. If we define τ as the identity
function in the reverse directions of πx and πy , we can say that each exact sequence splits.
Thus, from column-wise split-short exact sequences we can write,

1+ N y =(1+ K x y)× (1+ K y),

U1(Z[Cn × K4]) =(1+ N x)× U1(Z[Cn ×



y
�

]),

U1(Z[Cn × 〈x〉]) =(1+ K x)× U1(ZCn).

Equivalently, from row-wise split-short exact sequences we get

1+ N x =(1+ K x y)× (1+ K x),

U1(Z[Cn × K4]) =(1+ N y)× U1(Z[Cn × 〈x〉]),

U1(Z[Cn ×



y
�

]) =(1+ K y)× U1(ZCn).

Finally, the unit group U1(Z[Cn × K4]) can be described as an internal direct product of four
subgroups :

U1(Z[Cn × K4]) =(1+ N x)× U1(Z[Cn ×



y
�

])

= U1(ZCn)× (1+ K x)× (1+ K y)× (1+ K x y).

Lemma 1. In U1(Z[Cn × K4]), the subgroups (1+ K x), (1+ K y), (1+ K x y) satisfy the following
conditions.

(i) 1+ K x = {1+ (x − 1)P : 1− 2P ∈ U1(ZCn)}.

(ii) 1+ K y = {1+ (y − 1)P : 1− 2P ∈ U1(ZCn)}.

(iii) 1+ K x y = {1+ (x − 1)(y − 1)P : 1+ 4P ∈ U1(ZCn)}.



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 466

Proof. u ∈ 1+K x ⇔ u= 1+(x−1)P and there exists v = 1+(x−1)Q for some P,Q ∈ ZCn
such that u.v = 1 then

uv = 1⇔[1+ (x − 1)P][1+ (x − 1)Q] = 1

⇔1+ (x − 1)[P +Q− 2PQ] = 1

⇔P +Q− 2PQ = 0

⇔1− 2P − 2Q+ 4PQ = 1

⇔(1− 2P)(1− 2Q) = 1

⇔1− 2P ∈ U1(ZCn).

Similarly we can see that 1+ K y = {1+ (y − 1)P : 1− 2P ∈ U1(ZCn)} and
1+ K x y = {1+ (x − 1)(y − 1)P : 1+ 4P ∈ U1(ZCn)}.

Now consider the surjective ring homomorphism ρm : ZCn −→ ZmCn, where ρm reduces
the coefficients modulo m. If we denote the kernel of ρm by Mm, we have Mm = (mZ)Cn and
the following exact sequence is obtained at ring level:

Mm = (mZ)Cn
ι
−→ ZCn

ρm−→ ZmCn.

By restricting to the unit group U1(ZCn) we get exact sequence of groups

1+Mm
ι
−→ U1(ZCn)

ρm−→ U(ZmCn).

We can define two group isomorphisms,

σx : Cn × K4→±Cn ×



y
�

, σy : Cn × K4→±Cn × 〈x〉
a 7→ a a 7→ a

x 7→ −1 x 7→ x
y 7→ y y 7→ −1

By extending these isomorphisms linearly over Z, we get the following ring epimorphisms:

σx :Z[Cn × K4] −→ Z[Cn ×



y
�

]

P0 + P1 x + P2 y + P3 x y 7→(P0 − P1) + (P2 − P3)y,

σy :Z[Cn × K4] −→ Z[Cn × 〈x〉]
P0 + P1 x + P2 y + P3 x y 7→(P0 − P2) + (P1 − P3)x .

This leads to the diagrams at ring level. For K x we write

K x ι
−→ Z[Cn×< x >]

π x−→ ZCn
σx ↓ σx ↓ ρ2 ↓
M2

ι
−→ ZCn

ρ2−→ Z2Cn,



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 467

for K y we have
K y ι

−→ Z[Cn×< y >]
π y
−→ ZCn

σy ↓ σy ↓ ρ2 ↓
M2

ι
−→ ZCn

ρ2−→ Z2Cn,

and for K x y we get

K x y ι
−→ Z[Cn × K4]

πxπy
−→ ZCn

σxσy ↓ σxσy ↓ ρ4 ↓
M4

ι
−→ ZCn

ρ4−→ Z4Cn.

Corollary 1. The following maps are group isomorphisms:

(i) σx : 1+ K x −→ 1+M2,

(ii) σy : 1+ K y −→ 1+M2,

(iii) σxσy : 1+ K x y −→ 1+M4.

Proof. Consider the following diagram

1+ K x ι
−→ U1(Z[Cn×< x >])

π x−→ U1(ZCn)
σx ↓ σx ↓ ρ

2 ↓

1+M2
ι
−→ U1(ZCn)

ρ
2−→ U1(Z2Cn).

Here if we restrict the ring homomorphism σx to the unit group 1+ K x , we get the group
homomorphism σx(1+ (x − 1)P) = 1− 2P. By Lemma 1 σx is surjective. For u ∈ 1+ K x ,

u ∈ Kerσx⇔u= 1+ (x − 1)P and σx(u) = 1

⇔1− 2P= 1 and P ∈ ZCn

⇔P = 0

⇔u= 1.

Hence σx is injective. Similarly applying the same method to σy and σxσy one can see that
1+ K y ∼= 1+M2 and 1+ K x y ∼= 1+M4 respectively.

Remark 2.

f :Z[Cn × K4] −→ Z[Cn × K4]

P0 + P1 x + P2 y + P3 x y 7→P0 + P2 x + P1 y + P3 x y

is a ring isomorphism.



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 468

3. Applications

Theorem 4.

U1(Z[C5 × K4]) =C5 × K4×< v > ×< 1+ (x − 1)P > ×< 1+ (y − 1)P >

×< 1+ (x − 1)(y − 1)Q >,

where v = −1+ a+ a4, P = 4− 3(a+ a4) + (a2 + a3) and Q = 32− 26(a+ a4) + 10(a2 + a3).

Proof. Karpilovsky [5] showed if C5 =< a : a5 = 1 > then U1(ZC5) = C5× < v >, where
v = −1+ a+ a4. In order to describe 1+ K x consider the following commutative diagram :

1+ K x ι
−→ U1(Z[C5×< x >])

π x−→ U1(ZC5)
σx ↓ σx ↓ ρ2 ↓

1+M2
ι
−→ U1(ZC5)

ρ2−→ U1(Z2C5)
ı ↑ ı ↑ ı ↑

(1+M2)∩ F
ι
−→ F

ρ
2−→ ρ2(F).

Since F =< v >, ρ2(F) =< 1+ a+ a4 >= {1+ a+ a4, 1+ a2 + a3, 1}. So, we get

(1+M2)∩ F = Kerρ2 = {u ∈ F : ρ2(u) = 1}=< v3 > .

On the other hand, u ∈ 1+ K x ⇒ u = 1+ (x − 1)P, (P ∈ ZC5). Since σx is an isomorphism
and σx(u) = 1− 2P, we conclude that 1− 2P = v3. This leads us to,

P = 4− 3(a+ a4) + (a2 + a3).

Consequently we get the second generator as

u= 1+ (x − 1)[4− 3(a+ a4) + (a2 + a3)].

By Remark 2 we write third generator as

1+ K y =< 1+ (y − 1)[4− 3(a+ a4) + (a2 + a3)]> .

In order to construct 1+ K x y , consider the following commutative diagram:

1+ K x y ι
−→ 1+ N y π x−→ 1+ K y

σxσy ↓ σxσy ↓ σy ↓
1+M4

ι
−→ U1(ZC5)

ρ4−→ U1(Z4C5)
ı ↑ ı ↑ ı ↑

(1+M4)∩ F
ι
−→ F

ρ4−→ ρ4(F).

As F =< v > and ρ4(F) =< −1+ a+ a4 > we write,

(1+M4)∩ F = Kerρ4 = {u ∈ F : ρ4(u) = 1}=< v6 > .



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 469

For u ∈ 1+ K x y ⇒ u = 1+ (x − 1)(y − 1)Q, (Q ∈ ZC5). Since σxσy is an isomorphism and
(σxσy)(u) = 1+ 4Q, we conclude that 1+ 4Q = v6. That is,

Q = 32− 26(a+ a4) + 10(a2 + a3),

then the last generator is

u= 1+ (x − 1)(y − 1)[32− 26(a+ a4) + 10(a2 + a3)].

Theorem 5.

U1(Z[C7 × K4]) =C7 × K4×< v1, v2 > ×< 1+ (x − 1)P1 > ×< 1+ (x − 1)P2 >

×< 1+ (y − 1)P1 > ×< 1+ (y − 1)P2 >

×< 1+ (x − 1)(y − 1)Q1 > ×< 1+ (x − 1)(y − 1)Q2 >,

where

v1 =− 1+ (a+ a6),

v2 =− 1+ (a2 + a6),

P1 =− 4+ (a+ a6) + 4(a2 + a5)− 3(a3 + a4),

P2 =4− (a+ a6)− 3(a2 + a5) + 2(a3 + a4),

Q1 =72− 16(a+ a6)− 65(a2 + a5) + 45(a3 + a4),

Q2 =40− 9(a+ a6)− 36(a2 + a5) + 25(a3 + a4).

Proof. Let C7 =< a : a7 = 1>. Aleev and Panina [1] showed that,

U1(ZC7) = C7×< −1+ a+ a6,−1+ 2a2 − a3 − a4 + 2a5 > .

Since (−1+ a+ a6)2(−1+ a2 + a5) = −1+ 2a2 − a3 − a4 + 2a5, we can write

< −1+ a+ a6,−1+ 2a2 − a3 − a4 + 2a5 >=< −1+ a+ a6,−1+ a2 + a5 > .

Take v1 = −1+ (a+ a6) and v2 = −1+ (a2 + a6).

To describe 1+ K x , consider the following commutative diagram :

1+ K x ι
−→ U1(Z[C7×< x >])

π x−→ U1(ZC7)
σx ↓ σx ↓ ρ2 ↓

1+M2
ι
−→ U1(ZC7)

ρ2−→ U1(Z2C7)
ı ↑ ı ↑ ı ↑

(1+M2)∩ F
ι
−→ F

ρ
2−→ ρ2(F)



I.G. Kelebek, T. Bilgin / Eur. J. Pure Appl. Math, 7 (2014), 462-471 470

F =< v1, v2 > implies ρ2(F) =< ρ2(v1), ρ2(v2)>=< 1+ a+ a6, 1+ a2 + a5 >. Since
(1+ a+ a6)(1+ a2 + a5)3 = 1 and (1+ a+ a6)3(1+ a2 + a5)2 = 1. The kernel of ρ2 is

(1+M2)∩ F =< v1v3
2 , v3

1 v2
2 > .

On the other hand, u ∈ 1+ K x ⇒ u = 1+ (x − 1)P, (P ∈ ZC7). Since σx is an isomorphism
and σx(u) = 1− 2P, we conclude that

1− 2P1 = v1v3
2

1− 2P2 = v3
1 v2

2

�

⇒
P1 = −4+ (a+ a6) + 4(a2 + a5)− 3(a3 + a4)
P2 = 4− (a+ a6)− 3(a2 + a5) + 2(a3 + a4)

�

.

Hence 1+ K x =< 1+ (x − 1)P1, 1+ (x − 1)P2 >. By Remark 2 we have
1+K y =< 1+(y −1)P1, 1+(y −1)P2 >. In order to construct 1+K x y consider the following
commutative diagram :

1+ K x y ι
−→ 1+ N y π x−→ 1+ K y

σxσy ↓ σxσy ↓ σy ↓
1+M4

ι
−→ U1(ZC7)

ρ4−→ U1(Z4C7)
ı ↑ ı ↑ ı ↑

(1+M4)∩ F
ι
−→ F

ρ4−→ ρ4(F)

Since F =< v1, v2 >, ρ4(F) =< −1+ a+ a6,−1+ a2 + a5 >. Then, the kernel of ρ4 is

(−1+ a+ a6)2(−1+ a2 + a5)6 = 1
(−1+ a+ a6)6(−1+ a2 + a5)4 = 1

�

⇒ (1+M4) =< v2
1 v6

2 , v6
1 v4

2 > .

For u ∈ 1+ K x y ⇒ u = 1+ (x − 1)(y − 1)Q, (Q ∈ ZC7). Since σxσy is an isomorphism and
(σxσy)(u) = 1+ 4Q , we conclude that

1+ 4Q1 = v2
1 v6

2
1+ 4Q2 = v6

1 v4
2

�

⇒
Q1 = 72− 16(a+ a6)− 65(a2 + a5) + 45(a3 + a4),
Q2 = 40− 9(a+ a6)− 36(a2 + a5) + 25(a3 + a4).

Thus
1+ K x y =< 1+ (x − 1)(y − 1)Q1, 1+ (x − 1)(y − 1)Q2 > .



REFERENCES 471

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