On the units of integral group ring of Cn×C₆
There are many kind of open problems with varying difficulty on units in a given integral group ring. In this note, we characterize the unit group of the integral group ring of Cn×C₆ where Cn=⟨a:aⁿ=1⟩ and C₆=⟨x:x⁶=1⟩. We show that U₁(Z[Cn×C₆]) can be expressed in terms of its 4 subgroups. Furthermor...
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Küsmüş, Ö. 2019-06-15T19:58:39Z 2019-06-15T19:58:39Z 2015 On the units of integral group ring of Cn×C₆ / Ö. Küsmüş // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 142-151. — Бібліогр.: 11 назв. — англ. 1726-3255 2010 MSC:16U60, 16S34. https://nasplib.isofts.kiev.ua/handle/123456789/154754 There are many kind of open problems with varying difficulty on units in a given integral group ring. In this note, we characterize the unit group of the integral group ring of Cn×C₆ where Cn=⟨a:aⁿ=1⟩ and C₆=⟨x:x⁶=1⟩. We show that U₁(Z[Cn×C₆]) can be expressed in terms of its 4 subgroups. Furthermore, forms of units in these subgroups are described by the unit group U₁(ZCn). The authors would like to thank to all the members of the journal Algebra and Discrete Mathematics. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On the units of integral group ring of Cn×C₆ Article published earlier |
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On the units of integral group ring of Cn×C₆ |
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There are many kind of open problems with varying difficulty on units in a given integral group ring. In this note, we characterize the unit group of the integral group ring of Cn×C₆ where Cn=⟨a:aⁿ=1⟩ and C₆=⟨x:x⁶=1⟩. We show that U₁(Z[Cn×C₆]) can be expressed in terms of its 4 subgroups. Furthermore, forms of units in these subgroups are described by the unit group U₁(ZCn).
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On the units of integral group ring of Cn×C₆ / Ö. Küsmüş // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 142-151. — Бібліогр.: 11 назв. — англ. |
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2025-11-26T10:17:37Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 20 (2015). Number 1, pp. 142–151
© Journal “Algebra and Discrete Mathematics”
On the units of integral group ring of Cn × C6
Ömer Küsmüş
Communicated by I. Ya. Subbotin
Abstract. There are many kind of open problems with
varying difficulty on units in a given integral group ring. In this
note, we characterize the unit group of the integral group ring of
Cn × C6 where Cn = 〈a : an = 1〉 and C6 = 〈x : x6 = 1〉. We show
that U1(Z[Cn × C6]) can be expressed in terms of its 4 subgroups.
Furthermore, forms of units in these subgroups are described by the
unit group U1(ZCn). Notations mostly follow [11].
1. Introduction
Let G given as a finite group. Its integral group ring is denoted by ZG.
Invertible elements in ZG is called by units and the set of units forms
a group according to the multiplication and is shown by U(ZG). The
group of units with augmentation 1 is displayed by U1(ZG). If one pay
attention to the corresponding literature, that can easily see that the
obtained results mostly arises from finite groups especially finite abelian
groups. Fundamentals of the unit problem have come from the thesis of
G. Higman in 1940. Higman stated and proved the following [4]:
Lemma 1. If U(ZG) = ±G, then U(Z[G × C2]) = ±[G × C2].
Also, the following useful lemma was shown in [4] and [3].
2010 MSC: 16U60, 16S34.
Key words and phrases: group ring, integral group ring, unit group, unit prob-
lem.
Ö. Küsmüş 143
Lemma 2. U(ZG) has a torsion-free complement of finite rank ρ =
1
2(|G| + n2 + 1 − 2l) where n2 shows the number of elements of order 2 in
G and l is the number of all distinct cyclic subgroups of G.
On the other hand, in [7], Li considered the question: If U(ZG) has a
normal complement generated by bicyclic units, does U(Z[G × C2]) has
also a normal complement generated by bicyclic units? Jespers showed
that the answer for this question is yes while G = D6 or D8 [8–10]. Li gave
a counterexample for showing this is not true in general by considering
the group D8 ×C2 ×C2 [7]. However, Li proved that if U(ZG) is generated
by unitary units, then U(Z[G × C2]) is also generated by unitary units [7].
Another description of U(Z[G × C2]) was given by Low in [6] by linearly
extending some group epimorphisms to the group ring homomorphisms.
He also tried to generalize the problem for U(Z[G × Cp]) where p is a
prime integer. In [6], He showed that
U(Z[G × Cp]) = K ⋊ U(ZG) ∼= M ⋊ U(ZG)
where K is the kernel of the natural group homomorphism:
π : U(Z[G × Cp]) −→ U(ZG) and M is a subgroup of finite index in
U(Z[ζ]G) such that ζ is a primitive pth root of unity. Low also explicitly
proved the following 4 lemmas [6]:
Lemma 3. Let G∗ = G × 〈x : x2 = 1〉. Then, U(ZG∗) is obtained as
{u = 1 + (x − 1)α : α ∈ ZG, u ∈ U(ZG∗)} ⋊ U(ZG).
Further, 1 + (x − 1)α ∈ U(ZG∗) ⇔ 1 − 2α ∈ U(ZG).
Lemma 4. Let P = 〈a, b : a4 = b4 = 1, [b, a] = a2〉 be the indecomposable
group of order 16. Then,
U(Z[P × C2]) = ±[F65 ⋊ F9] ⋊ (P × C2)
where Fi denotes a free group of rank i.
Lemma 5. Let C∗
5 = 〈c : c5 = 1〉 × 〈x : x2 = 1〉. Then, the unit group
U(ZC∗
5 ) = 〈1 + (x − 1)P 〉 × 〈v〉 × C∗
5
where P = −3 − c + 3c2 + 3c3 − c4 and v = (c + 1)2 − ĉ.
Lemma 6. Let C∗
8 = 〈c : c8 = 1〉 × 〈x : x2 = 1〉. Then, the unit group
U(ZC∗
8 ) = 〈1 + (x − 1)P 〉 × 〈v〉 × C∗
8
where P = −4−3c+3c3+4c4+3c5−3c7 and v = 3− ĉ+2(c+c7)+(c2+c6).
144 On the units of integral group ring of Cn × C6
Kelebek and Bilgin considered the finite abelian group Cn × K4 where
K4 is the Klein 4-group and characterized the unit group of its integral
group ring in terms of 4 components as follows [1]:
Theorem 1. U1(Z[Cn ×K4]) = U1(ZCn)×(1+Kx)×(1+Ky)×(1+Kxy)
where
1 + Kx = {1 + (x − 1)P : 1 − 2P ∈ U1(ZCn)}
1 + Ky = {1 + (y − 1)P : 1 − 2P ∈ U1(ZCn)}
1 + Kxy = {1 + (x − 1)(y − 1)P : 1 + 4P ∈ U1(ZCn)}
2. Motivation for construction of U1(Z[Cn × C6])
Now, let us begin with some remarks.
Remark 1. The following maps are group epimorphisms:
πx2 : Cn × C6 −→ Cn × 〈x2〉
a 7→ a
x 7→ x2
πx3 : Cn × C6 −→ Cn × 〈x3〉
a 7→ a
x 7→ x3
Remark 2. Ker(πx2) = 〈x3〉 and Ker(πx3) = 〈x2〉 .
Since Cn × 〈x2〉 →֒ Cn × C6, Cn × 〈x3〉 →֒ Cn × C6 and i denotes the
inclusion map, we get the following short exact sequences at group level:
0 −→ 〈x3〉
i
→ Cn × C6
π
x2
−−→ Cn × 〈x2〉 −→ 0
0 −→ 〈x2〉
i
→ Cn × C6
π
x3
−−→ Cn × 〈x3〉 −→ 0
0 −→ 〈x〉
i
→ Cn × C6
π
x2 π
x3
−−−−→ Cn −→ 0
If we linearly extend πx2 and πx3 to integral group rings over Z, we obtain
the following ring homomorphisms:
πx2 : Z[Cn × C6] −→ Z[Cn × 〈x2〉]
5∑
j=0
Pjxj 7→ (P0 + P3) + (P1 + P4)x2 + (P2 + P5)x4
Ö. Küsmüş 145
and
πx3 : Z[Cn × C6] −→ Z[Cn × 〈x3〉]
5∑
j=0
Pjxj 7→ (P0 + P2 + P4) + (P1 + P3 + P5)x3
Lemma 7. Kx2
:= Ker(πx2) = (x3 − 1)Z[Cn × 〈x2〉]
Proof.
Ker(πx2) =
{ 5∑
i=0
Pix
i : πx2(
5∑
i=0
Pix
i) = 0, Pi ∈ ZCn
}
=
{ 5∑
i=0
Pix
i : P0 + P3 = P1 + P4 = P2 + P5 = 0
}
=
{ 5∑
i=0
Pix
i : P0 = −P3, P1 = −P4, P2 = −P5
}
= {−P3 − P4x − P5x2 + P3x3 + P4x4 + P5x5}
= {(x3 − 1)P3 + (x4 + x)P4 + (x5 − x2)P5}
= (x3 − 1)[ZCn ⊕ x2
ZCn ⊕ x4
ZCn]
= (x3 − 1)Z[Cn × 〈x2〉].
Lemma 8. Kx3
:= Ker(πx3) = (x2 − 1)[ZCn ⊕ xZCn ⊕ x2
ZCn ⊕ x3
ZCn]
Proof.
Ker(πx3) =
{ 5∑
i=0
Pix
i : πx3(
5∑
i=0
Pix
i) = 0, Pi ∈ ZCn
}
=
{ 5∑
i=0
Pix
i : P0 + P2 + P4 = P1 + P3 + P5 = 0
}
=
{ 5∑
i=0
Pix
i : P0 = −(P2 + P4), P1 = −(P3 + P5)
}
= {(x2 − 1)[P2 + xP3 + (x2 + 1)P4 + (x2 + 1)xP5]}
= (x2 − 1)[ZCn ⊕ xZCn ⊕ x2
ZCn ⊕ x3
ZCn]
146 On the units of integral group ring of Cn × C6
Similarly, we can write the following ring homomorphism:
πx2πx3 : Z[Cn × C6] −→ ZCn
5∑
j=0
Pjxj 7→
5∑
j=0
Pj .
Lemma 9. Kx2x3
:= Ker(πx2πx3) = ⊕5
j=1(xj − 1)ZCn
Proof.
Ker(πx2πx3) =
{ 5∑
i=0
Pix
i : πx2πx3(
5∑
i=0
Pix
i) = 0, Pi ∈ ZCn
}
=
{ 5∑
i=0
Pix
i :
5∑
i=0
Pi = 0, Pi ∈ ZCn
}
=
{ 5∑
i=0
Pix
i : P0 = −
5∑
i=1
Pi, Pi ∈ ZCn
}
=
{
−
5∑
i=1
Pi +
5∑
i=1
Pix
i : Pi ∈ ZCn
}
=
{ 5∑
j=1
(xj − 1)Pj : Pj ∈ ZCn
}
= ⊕5
j=1(xj − 1)ZCn.
By Remarks 1 and 2, we get the following short exact sequences at group
ring level:
0 −→ Kx2 i
→ Z[Cn × C6]
π
x2
→ Z[Cn × 〈x2〉] −→ 0
0 −→ Kx3 i
→ Z[Cn × C6]
π
x3
→ Z[Cn × 〈x3〉] −→ 0
0 −→ Kx2x3 i
→ Z[Cn × C6]
π
x2 π
x3
→ ZCn −→ 0
If we restrict πx2 and πx3 to the unit level, we conclude that the
followings are also short exact sequences:
1 −→ U1(1 + Kx2
)
i
→ U1(Z[Cn × C6])
π
x2
→ U1(Z[Cn × 〈x2〉]) −→ 1
1 −→ U1(1 + Kx3
)
i
→ U1(Z[Cn × C6])
π
x3
→ U1(Z[Cn × 〈x3〉]) −→ 1
1 −→ U1(1 + Kx2x3
)
i
→ U1(Z[Cn × C6])
π
x2 π
x3
→ U1(ZCn) −→ 1
Ö. Küsmüş 147
Since we can consider embeddings U1(Z[Cn × 〈x2〉]) →֒ U1(Z[Cn × C6])
and U1(Z[Cn × 〈x3〉]) →֒ U1(Z[Cn × C6]), the following split extensions
hold:
U1(Z[Cn × C6]) = U1(1 + Kx2
) × U1(Z[Cn × 〈x2〉])
= U1(1 + Kx3
) × U1(Z[Cn × 〈x3〉])
= U1(1 + Kx2x3
) × U1(ZCn).
Remark 3. In U1(Z[Cn × C6]) the normal subgroups U1(1 + Kx2
),
U1(1 + Kx3
) and U1(1 + Kx2x3
) are determined as in the following forms
respectively:
(i) {u = 1 + (x3 − 1)[P0 + P2x2 + P4x4] : u is a unit};
(ii) {u = 1 + (x2 − 1)[P0 + P1x + P2x2 + P3x3] : u is a unit};
(iii) {u = 1 +
∑5
j=1(xj − 1)Pj : u is a unit}.
3. An explicit characterization of U1(Z[Cn × C6])
In this section, an explicit characterization of U1(Z[Cn × C6]) is given
with the help of the results in the previous section. First, we should
give some restrictions of the maps πx2 , πx3 and πx2πx3 . Let πx3 |
U1(1+Kx2 )
denote the restriction of πx3 on U1(1 + Kx2
).
Lemma 10. W1 := Im(πx3 |
U1(1+Kx2
)
) = 1 + (x3 − 1)ZCn.
Proof. Let us take an element from U1(1 + Kx2
) as γ = 1 + (x3 − 1)[P0 +
P2x2 + P4x4] where Pi ∈ ZCn.Then,
πx3 : γ 7→ 1 + (x3 − 1)[P0 + P2 + P4].
Say P0 + P2 + P4 = P . Thus, Im(πx3 |
U1(1+Kx2
)
) consists of elements of
the form 1 + (x3 − 1)P .
Lemma 11. W2 :=Ker(πx3 |U1(Z[Cn×〈x2〉]))=1+(x2−1)ZCn ⊕ (x4−1)ZCn.
Proof. Let us take an element from U1(Z[Cn × 〈x2〉]) as σ = P0 + P2x2 +
P4x4. Here, we can manipulate the parameter P0 = 1 + P
′
0. Then, we get
πx3 : σ 7→ 1 + P
′
0 + P2 + P4 = 1 ⇐⇒ P
′
0 = −P2 − P4.
This means that the kernel consists of elements of the form
1 + (−P2 − P4) + P2x2 + P4x4 = 1 + (x2 − 1)P2 + (x4 − 1)P4.
Hence the required is obtained.
148 On the units of integral group ring of Cn × C6
Lemma 12.
W3 := Ker(πx3 |
U1(1+Kx2
)
) = 1+(x3−1)(x2−1)ZCn⊕(x3−1)(x4−1)ZCn.
Proof. Again, let us consider an element from U1(1 + Kx2
) as η = 1 +
(x3 − 1)[P0 + P2x2 + P4x4]. Then,
πx3 : η 7→ 1 + (x3 − 1)[P0 + P2 + P4] = 1 ⇐⇒ P0 = −P2 − P4
Thus, Ker(πx3 |
U1(1+Kx2 )
) consists of
1 + (x3 − 1)[P0 + P2x2 + P4x4] = 1 + (x3 − 1)[−P2 − P4 + P2x2 + P4x4]
= 1 + (x3 − 1)[(x2 − 1)P2 + (x4 − 1)P4].
Therefore, by Lemma 10, Lemma 11 and Lemma 12, we can construct
the following commutative diagram:
W3
i
//
i
��
U1(1 + Kx3
)
π
x2
//
i
��
W2
i
��
U1(1 + Kx2
)
i
//
π
x3
��
U1(Z[Cn × C6])
π
x2
//
π
x3
��
U1(Z[Cn × 〈x2〉])
π
x3
��
W1
i
// U1(Z[Cn × 〈x3〉])
π
x2
// U1(ZCn)
Since we can take embeddings as the inverses of πx2 and πx3 , this diagram
splits as follows:
U1(Z[Cn × C6]) = W1 × W2 × W3 × U1(ZCn).
Now, let us characterize explicitly W1, W2 and W3.
Proposition 1. u = 1+(x3−1)P ∈ W1 is a unit ⇐⇒ 1−2P ∈ U1(ZCn)
Proof.
u = 1 + (x3 − 1)P is unit ⇐⇒ ∃v = 1 + (x3 − 1)Q : uv = 1
⇐⇒ 1 + (x3 − 1)[P + Q − 2PQ] = 1
⇐⇒ P + Q − 2PQ = 0
⇐⇒ 1 − 2P − 2Q + 4PQ = 1
⇐⇒ (1 − 2P )(1 − 2Q) = 1
⇐⇒ 1 − 2P ∈ U1(ZCn).
Ö. Küsmüş 149
Proposition 2. u = 1 + (x2 − 1)P + (x4 − 1)Q ∈ W2 is a unit ⇐⇒
P 2 + Q2 − PQ − P − Q = 0
Proof. First, we need to define a closed operation. If we define α = x2 − 1
and β = x4 − 1, we get the following straightforward computations:
α2 = (x2 − 1)2 = −2(x2 − 1) + (x4 − 1) = −2α + β
αβ = (x2 − 1)(x4 − 1) = −(x2 − 1) − (x4 − 1) = −α − β
β2 = (x4 − 1)2 = −2(x4 − 1) + (x2 − 1) = α − 2β
Let us state this operation in a table as follows:
• α β
α −2α + β −α − β
β −α − β α − 2β
Now, we can give a necessary and sufficient condition to be a unit for the
element u. u = 1 + (x2 − 1)P + (x4 − 1)Q ∈ W2 is a unit if and only if
∃v = 1 + (x2 − 1)P
′
+ (x4 − 1)Q
′
such that uv = 1. Hence,
1 + αP + βQ + αP
′
+ βQ
′
+ α2PP
′
+ β2QQ
′
+ αβ(PQ
′
+ P
′
Q) = 1.
By the above operation, we can arrange this equation as
1 + α(P + P
′
) + β(Q + Q
′
) + (−2α + β)PP
′
+ (α − 2β)QQ
′
+ (−α − β)(PQ
′
+ P
′
Q) = 1
That is,
1 + α(P + P
′
− 2PP
′
+ QQ
′
− PQ
′
− P
′
Q)
+ β(Q + Q
′
+ PP
′
− 2QQ
′
− PQ
′
− P
′
Q) = 1.
This equation holds if and only if the following system of matrix has a
unique solution:
[
1 − 2P − Q Q − P
P − Q 1 − 2Q − P
] [
P
′
Q
′
]
=
[
−P
−Q
]
Therefore,
[
1 − 2P − Q Q − P
P − Q 1 − 2Q − P
]
∈ SL2(ZCn)
A straightforward calculation shows that P 2 + Q2 − PQ − P − Q = 0.
150 On the units of integral group ring of Cn × C6
Proposition 3. u = 1 + (x3 − 1)(x2 − 1)P + (x3 − 1)(x4 − 1)Q ∈ W3 is
a unit if and only if the following equation holds:
2P 2 + 2Q2 − 2PQ − P − Q = 0
Proof. First, let λ = (x3 − 1)(x2 − 1) and µ = (x3 − 1)(x4 − 1). One can
easily compute the followings:
λ2 = (x3 − 1)2(x2 − 1)2 = 4λ − 2µ,
λµ = (x3 − 1)2(x2 − 1)(x4 − 1) = 2λ + 2µ,
µ2 = (x3 − 1)2(x4 − 1)2 = −2λ + 4µ.
In a better expression, we write
• λ µ
λ 4λ − 2µ 2λ + 2µ
µ 2λ + 2µ −2λ + 4µ
Now, let us determine the necessary and sufficient condition to be a
unit for an element u. u = 1 + λP + µQ ∈ W3 is a unit if and only if
∃v = 1 + λP
′
+ µQ
′
: uv = 1. Thus, a straight forward computation shows
us that
1 + λ(P + P
′
+ 4PP
′
+ 2P
′
Q + 2PQ
′
− 2QQ
′
)
+ µ(Q + Q
′
− 2PP
′
+ 2P
′
Q + 2PQ
′
+ 4QQ
′
) = 1.
This equation holds if and only if the following system of matrix has a
unique solution:
[
1 + 4P + 2Q 2P − 2Q
2Q − 2P 1 + 2P + 4Q
] [
P
′
Q
′
]
=
[
−P
−Q
]
Then, the required result comes from the following:
[
1 + 4P + 2Q 2P − 2Q
2Q − 2P 1 + 2P + 4Q
]
∈ SL2(ZCn).
Consequently, we can summarize all the obtained results as follows:
Corollary 1. U1(Z[Cn × C6]) = U1(ZCn) × U × V × W where
U = {1 + (x3 − 1)P : 1 − 2P ∈ U1(ZCn)}
V = {1 + αP + βQ : P 2 + Q2 − PQ − P − Q = 0}
W = {1 + λP + µQ : 2P 2 + 2Q2 − 2PQ − P − Q = 0}
such that
α = x2 − 1, β = x4 − 1, λ = (x3 − 1)(x2 − 1), µ = (x3 − 1)(x4 − 1).
Ö. Küsmüş 151
Acknowledgements
The authors would like to thank to all the members of the journal
Algebra and Discrete Mathematics.
References
[1] Kelebek I. G. and T. Bilgin, Characterization of U1(Z[Cn × K4]), Eur. J. Pure
and Appl. Math., 7(4), 462-471, 2014.
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Contact information
Ö. Küsmüş Department of Mathematics, Faculty of Science,
Yuzuncu Yil University, 65080, Van, TURKEY
E-Mail(s): omerkusmus@yyu.edu.tr
Received by the editors: 21.02.2015
and in final form 05.03.2015.
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