# The Fifth Dimension Subgroup for Metabelian 2 Groups

Authors: Shalini Gupta
Journal: Chinese Journal of Mathematics
 (2016)

---

## Abstract

Given a finite metabelian2-group G, the object of this paper is to discuss some cases under which D
5
(
G
)
=
γ
5
(
G
). Further, some examples of 2 groups of class 3, for which D
4
(
G
)
≠
{
1
} but D
5
(
G
)
=
{
1
}, are discussed.


---

## Body

## 1. Introduction

LetZ
(
G
) be the integral group ring of group G and let I
(
G
) be the augmentation ideal of Z
(
G
). For n
≥
1, let D
n
(
G
)
=
G
∩
{
1
+
I
n
(
G
)
} be nth dimension subgroup and let γ
n
(
G
) be nth term in lower central series of group G. The famous dimension subgroup conjecture states that D
n
(
G
)
=
γ
n
(
G
) for all n
≥
1 and for all groups G. This conjecture has been proved for n
≤
3 (see, e.g., [1]). It has been proved in [2] that if G is a finite p-group, p odd prime, then D
4
(
G
)
=
γ
4
(
G
). Keeping in view this result, Rips [3] gave an example of 2-group G for which D
4
(
G
)
≠
γ
4
(
G
). Further Tahara [4] gave the structure of D
5
(
G
) and proved that the exponent of D
5
(
G
)
/
γ
5
(
G
) is divisible by 6. It has been shown in [5] that, for a metabelian group, the exponent of D
5
(
G
)
/
γ
5
(
G
) is divisible by 2, and hence for a metabelian p-group G
,
p odd prime D
5
(
G
)
=
γ
5
(
G
). Gupta [6] constructed a 2-group G, generated by four elements, such that D
5
(
G
)
≠
γ
5
(
G
). Now, the question arises that whether for a metabelian 2-group G, with at most three generators, D
5
(
G
)
=
γ
5
(
G
) or not.In this paper, we prove that ifG is metabelian 2 group generated by at most two elements, then D
5
(
G
)
=
γ
5
(
G
) (Theorem 2). Further we prove that under certain conditions, a metabelian 2-group G generated by n elements satisfies the dimension subgroup conjecture for n
=
5 (Theorem 3). Finally, in Section 3, we give some examples of 2 groups of class 3, for which D
5
(
G
)
=
γ
5
(
G
).


## 2. Main Results

We first recall the structure ofD
5
(
G
) given by Tahara [4]. Let G be a finite group of class 4. Consider the lower central series G
=
γ
1
(
G
)
⊇
γ
2
(
G
)
⊇
γ
3
(
G
)
⊇
γ
4
(
G
)
⊇
γ
5
(
G
)
=
{
1
} of G. Let {
x
1
i
γ
2
(
G
)
}
1
≤
i
≤
s be basis of G
/
γ
2
(
G
), with d
(
i
) as the order of x
1
i
γ
2
(
G
): that is, x
1
i
d
(
i
) belongs to γ
2
(
G
). Similarly, let {
x
2
j
γ
3
(
G
)
}
1
≤
j
≤
t be basis of γ
2
(
G
)
/
γ
3
(
G
), with e
(
j
) as the order of x
2
j
γ
3
(
G
), and let {
x
3
k
γ
4
(
G
)
}
1
≤
k
≤
u be basis of γ
3
(
G
)
/
γ
4
(
G
), with f
(
k
) as the order of x
3
k
γ
4
(
G
). Moreover, these basis elements are chosen in such a way that d
(
i
) divides d
(
i
+
1
), e
(
i
) divides e
(
i
+
1
), and f
(
i
) divides f
(
i
+
1
). Thus, we have(1)
x
1
i
d
i
=
∏
1
≤
j
≤
t
x
2
j
b
i
j
∏
1
≤
k
≤
u
x
3
k
c
i
k
y
4
i
,
y
4
i
∈
γ
4
G
,
1
≤
i
≤
s
;
(2)
x
2
k
e
k
=
∏
1
≤
k
≤
u
x
3
l
d
k
l
y
4
k
′
,
y
4
k
′
∈
γ
4
G
,
1
≤
k
≤
t
;
(3)
x
1
i
d
i
,
x
1
j
=
∏
1
≤
l
≤
u
x
3
l
α
l
i
j
y
4
i
j
′
′
,
y
4
i
j
′
′
∈
γ
4
G
,
1
≤
i
≤
s
.The following theorem gives the structure of D
5
(
G
).Theorem 1 (see [4]).
With the above notations,D
5
(
G
) is equal to the subgroup generated by the elements (4)
∏
1
≤
i
<
j
≤
s
x
1
i
d
i
,
x
1
j
u
i
j
d
j
/
d
i
∏
1
≤
i
≤
s
,
1
≤
k
≤
t
∏
k
<
l
x
2
l
,
x
2
k
b
i
l
v
i
k
∏
1
≤
i
≤
j
≤
k
≤
s
x
1
i
d
i
,
x
1
j
,
x
1
k
w
i
j
k
,with the following conditions:(5)
w
i
i
i
=
0
,
1
≤
i
≤
s
;
(6)
u
i
j
d
j
d
i
d
i
2
+
w
i
i
j
d
i
+
w
i
i
j
′
′
d
j
=
0
,
1
≤
i
<
j
≤
s
;
(7)
-
u
i
j
d
j
2
+
w
i
j
j
d
i
+
w
i
j
j
′
d
j
=
0
,
1
≤
i
<
j
≤
s
;
(8)
w
i
j
k
d
i
+
w
i
j
k
′
d
j
+
w
i
j
k
′
′
d
k
=
0
,
1
≤
i
<
j
<
k
≤
s
;
(9)
∑
i
<
h
u
i
h
b
h
k
-
∑
h
<
i
u
h
i
d
i
d
h
b
h
k
+
v
i
k
d
i
+
v
i
k
′
e
k
=
0
,
1
≤
i
≤
s
,
1
≤
k
≤
t
;
(10)
u
i
j
d
j
d
i
d
i
3
+
w
i
i
j
d
i
2
≡
0
mod
⁡
d
i
,
1
≤
i
<
j
≤
s
;
(11)
w
i
i
j
d
i
2
+
w
i
i
j
′
′
d
j
2
≡
0
mod
⁡
d
i
,
1
≤
i
<
j
≤
s
;
(12)
-
u
i
j
d
j
3
+
w
i
j
j
′
d
j
2
≡
0
mod
⁡
d
i
,
1
≤
i
<
j
≤
s
;
(13)
w
i
j
k
d
i
2
,
w
i
j
k
′
d
j
2
,
w
i
j
k
′
′
d
k
2
≡
0
mod
⁡
d
i
,
1
≤
i
<
j
<
k
≤
s
;
(14)
v
i
k
d
i
2
-
∑
h
≤
i
w
h
i
i
b
h
k
-
∑
i
<
h
w
i
i
h
′
′
b
h
k
≡
0
mod
⁡
d
i
,
e
k
,
1
≤
i
≤
s
,
1
≤
k
≤
t
;
(15)
∑
h
≤
i
w
h
i
j
b
h
k
+
∑
i
<
h
≤
j
w
i
h
j
′
b
h
k
+
∑
j
<
h
w
i
j
k
′
′
b
h
k
≡
0
mod
⁡
d
i
,
f
l
,
1
≤
i
≤
s
,
1
≤
l
≤
u
;
(16)
-
∑
h
<
i
u
h
i
d
i
d
h
α
l
h
i
+
∑
i
<
h
u
i
h
c
h
i
-
∑
h
<
i
u
h
i
d
i
d
h
c
h
i
-
∑
k
v
i
k
′
d
k
l
-
∑
g
≤
i
≤
h
w
g
i
h
α
l
g
h
-
∑
g
≤
h
≤
i
w
g
h
i
α
l
g
h
-
∑
i
<
g
≤
h
w
i
g
h
′
α
l
g
h
≡
0
mod
⁡
d
i
,
f
l
,
1
≤
i
≤
s
,
1
≤
l
≤
u
;
(17)
∑
i
v
i
k
b
i
k
≡
0
mod
⁡
e
k
,
1
⩽
k
⩽
t
;
(18)
∑
i
v
i
k
b
i
l
+
∑
i
v
i
l
b
i
k
≡
0
mod
⁡
e
k
,
1
⩽
k
<
l
⩽
t
.Theorem 2.
LetG be metabelian 2 group with G
/
γ
2
(
G
) as sum of at most two cyclic groups. Then D
5
(
G
)
=
γ
5
(
G
).Proof.
It is obvious that if metabelian2 groups are such that G
/
γ
2
(
G
) is a cyclic group then D
5
(
G
)
=
γ
5
(
G
).
LetG be metabelian 2 group and let G
/
γ
2
(
G
) be sum of two cyclic groups. It is enough to prove the result for a group of class 4. Let G
/
γ
2
(
G
)
=
C
1
⊕
C
2, where C
i is a cyclic group generated by x
1
i
γ
2
(
G
) with order of x
1
i
γ
2
(
G
)
=
d
(
i
) and d
(
1
) divides d
(
2
). An arbitrary element g of D
5
(
G
) is of the form [
x
11
d
(
1
)
,
x
12
]
u
12
(
d
(
2
)
/
d
(
1
)
)
[
x
11
d
(
1
)
,
x
12
,
x
12
]
w
122 subject to the conditions given in Theorem 1. For 1
≤
k
≤
t and i
=
2, (9) becomes (19)
-
u
12
d
2
d
1
b
1
k
+
v
2
k
d
2
+
v
2
k
′
e
k
=
0
.Thus, (20)
1
=
∏
k
=
1
t
x
2
k
,
x
12
-
u
12
d
2
/
d
1
b
1
k
+
v
2
k
d
2
+
v
2
k
′
e
k
=
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
-
u
12
d
2
/
d
1
x
2
k
,
x
12
,
x
2
k
u
12
d
2
/
d
1
b
1
k
2
∏
k
=
1
t
x
2
k
,
x
12
d
2
v
2
k
x
2
k
,
x
12
,
x
12
-
v
2
k
d
2
2
∏
k
=
1
t
x
2
k
e
k
,
x
12
v
2
k
′
x
2
k
,
x
12
,
x
2
k
-
v
2
k
′
e
k
2
.Now [
x
2
k
,
x
12
,
x
2
k
] belongs to γ
5
(
G
) and [
x
2
k
,
x
12
d
(
2
)
] belongs to [
γ
2
(
G
)
,
γ
2
(
G
)
]; thus we have(21)
1
=
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
-
u
12
d
2
/
d
1
∏
k
=
1
t
x
2
k
,
x
12
,
x
12
-
v
2
k
d
2
2
∏
k
=
1
t
x
2
k
e
k
,
x
12
v
2
k
′
=
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
-
u
12
d
2
/
d
1
∏
k
=
1
t
x
2
k
,
x
12
,
x
12
-
v
2
k
d
d
2
2
∏
k
=
1
t
∏
l
=
1
u
x
3
l
d
k
l
,
x
12
v
2
k
′
=
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
-
u
12
d
2
/
d
1
∏
k
=
1
t
x
2
k
,
x
12
,
x
12
-
w
122
b
1
k
∏
l
=
1
u
x
3
l
,
x
12
∑
k
v
2
k
′
d
k
l
,as for i
=
2 (14) becomes v
2
k
d
(
2
)
2
-
w
122
b
1
k
≡
0
(
m
o
d
(
d
(
2
)
,
e
(
k
)
)
). Now, (16), for i
=
2, becomes (22)
-
u
12
d
2
d
1
α
l
12
-
u
12
d
2
d
1
c
1
l
-
∑
k
v
2
k
′
d
k
l
-
2
w
122
α
l
12
-
w
112
α
l
11
≡
0
mod
⁡
d
2
,
f
l
,which implies that(23)
∏
l
=
1
u
x
3
l
,
x
12
∑
k
v
2
k
′
d
k
l
=
∏
l
=
1
u
x
3
l
,
x
12
-
u
12
d
2
/
d
1
α
l
12
-
u
12
d
2
/
d
1
c
1
l
-
2
w
122
α
l
12
-
w
112
α
l
11
=
∏
l
=
1
u
x
3
l
c
1
l
,
x
12
-
u
12
d
2
/
d
1
∏
l
=
1
u
x
3
l
α
l
12
,
x
12
-
u
12
d
2
/
d
1
∏
l
=
1
u
x
3
l
α
l
12
,
x
12
-
2
w
122
∏
l
=
1
u
x
3
l
α
l
11
,
x
12
-
w
112
=
∏
l
=
1
u
x
3
l
c
1
l
,
x
12
-
u
12
d
2
/
d
1
x
11
d
1
,
x
12
,
x
12
-
u
12
d
2
/
d
1
x
11
d
1
,
x
12
,
x
12
-
2
w
122
.Now,(24)
x
11
d
1
,
x
12
,
x
12
-
u
12
d
2
/
d
1
=
x
11
,
x
12
,
x
12
-
u
12
d
2
x
12
,
x
11
,
x
11
,
x
12
u
12
d
2
/
d
1
d
1
2
=
x
11
,
x
12
,
x
12
d
2
-
u
12
x
12
,
x
11
,
x
12
,
x
12
-
u
12
d
2
2
x
12
,
x
11
,
x
11
,
x
12
u
12
d
2
/
d
1
d
1
2
.Using (6) and (7), we get that(25)
u
12
d
2
d
1
d
1
2
≡
0
mod
⁡
d
1
,
(26)
-
u
12
d
2
2
≡
0
mod
⁡
d
1
.Hence, (24) reduces to(27)
x
11
d
1
,
x
12
,
x
12
-
u
12
d
2
/
d
1
=
x
11
,
x
12
,
x
12
d
2
-
u
12
=
1
,
as
x
11
,
x
12
,
x
12
d
2
∈
γ
2
G
,
γ
2
G
.Now,(28)
x
11
d
1
,
x
12
,
x
12
-
2
w
122
=
x
11
,
x
12
,
x
12
-
2
w
122
d
1
x
12
,
x
11
,
x
11
,
x
12
2
w
122
d
1
2
=
x
11
,
x
12
,
x
12
-
2
w
122
d
1
.Also (7) implies that (29)
-
2
w
122
d
1
=
-
2
u
12
d
2
2
+
2
w
122
′
d
2
.Thus, (28) becomes(30)
x
11
d
1
,
x
12
,
x
12
-
2
w
122
=
x
11
,
x
12
,
x
12
2
w
122
′
d
2
x
11
,
x
12
,
x
12
-
u
12
d
2
d
2
-
1
=
x
11
,
x
12
,
x
12
d
2
2
w
122
′
x
12
,
x
11
,
x
12
,
x
12
-
2
w
122
′
d
2
2
x
11
,
x
12
,
x
12
-
u
12
d
2
d
2
-
1
=
x
11
,
x
12
,
x
12
d
2
-
u
12
d
2
-
1
x
12
,
x
11
,
x
12
,
x
12
-
u
12
d
2
-
1
d
2
2
=
1
,as by (7), -
u
12
d
2
2
≡
0
(
mod
⁡
d
(
1
)
).
Using (23), (27), and (30) in (21), we get that(31)
1
=
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
-
u
12
d
2
/
d
1
∏
l
=
1
u
x
3
l
c
1
l
,
x
12
-
u
12
d
2
/
d
1
∏
k
=
1
t
x
2
k
b
1
k
,
x
12
,
x
12
-
w
122
=
x
11
d
1
,
x
12
-
u
12
d
2
/
d
1
x
11
d
1
,
x
12
,
x
12
-
w
122
=
g
-
1
.Thus, g
=
1, and hence D
5
(
G
)
=
{
1
}.Theorem 3.
LetG be finite metabelian 2 group and G
/
γ
2
(
G
)
=
C
1
⊕
C
2
⊕
⋯
⊕
C
n, where C
i are cyclic groups of order d
(
i
). Let d
1
=
d
2
=
⋯
=
d
(
n
-
1
)
=
2, d
(
n
)
≥
4. Then D
5
(
G
)
=
γ
5
(
G
).Proof.
It is enough to prove the result for a groupG of class 4. Any element g of D
5
(
G
) is of the form (32)
g
=
∏
1
≤
i
<
j
≤
n
x
1
i
d
i
,
x
1
j
u
i
j
d
j
/
d
i
.
∏
1
≤
i
<
j
<
k
≤
n
x
1
i
d
i
,
x
1
j
,
x
1
k
w
i
j
k
.
∏
1
≤
i
<
j
≤
n
x
1
i
d
i
,
x
1
j
,
x
1
j
w
i
j
j
=
A
·
B
·
C
say
.Since for 1
≤
i
<
j
<
k
≤
n, d
(
i
)
=
d
(
j
)
=
2, (13) reduces to (33)
w
i
j
k
≡
0
mod
⁡
d
jand hence B
=
1. Now,(34)
A
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
u
i
j
d
j
/
d
i
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
u
i
n
d
n
/
d
i
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
u
i
j
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
u
i
n
d
n
/
d
i
as  for
1
≤
i
<
j
≤
n
-
1
,
d
i
=
d
j
.Also for 1
≤
i
<
j
≤
n, (7) implies that -
u
i
j
+
2
w
i
j
j
+
2
w
i
j
j
′
=
0, which gives that(35)
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
u
i
j
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
2
w
i
j
j
+
2
w
i
j
j
′
,
as
d
j
=
2
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
d
j
w
i
j
j
+
w
i
j
j
′
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
d
j
w
i
j
j
+
w
i
j
j
′
x
1
i
d
i
,
x
1
j
,
x
1
j
-
w
i
j
j
+
w
i
j
j
′
d
j
2
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
,
x
1
j
-
w
i
j
j
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
,
x
1
j
-
w
i
j
j
′
as
x
1
i
d
i
,
x
1
j
d
j
∈
γ
2
G
,
γ
2
G
,
d
j
=
2
.Now for 1
≤
i
<
j
≤
n, (12) implies that w
i
j
j
′
≡
0
(
mod
⁡
d
(
j
)
), which gives that (36)
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
u
i
j
=
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
,
x
1
j
-
w
i
j
j
.Thus, (34) reduces to(37)
A
=
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
u
i
n
d
n
/
d
i
∏
1
≤
i
<
j
≤
n
-
1
x
1
i
d
i
,
x
1
j
,
x
1
j
-
w
i
j
j
.
Hence,(38)
g
=
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
u
i
n
d
n
/
d
i
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
w
i
n
n
.Now, for i
=
n, (9) becomes (39)
S
=
-
∑
j
=
1
n
-
1
u
j
n
d
n
d
j
b
j
k
+
v
n
k
d
n
+
v
n
k
′
e
k
=
0
,which gives(40)
1
=
∏
k
=
1
t
x
2
k
,
x
1
n
S
=
∏
k
=
1
t
x
2
k
,
x
1
n
-
∑
j
=
1
n
-
1
u
j
n
d
n
/
d
j
b
j
k
∏
k
=
1
t
x
2
k
,
x
1
n
v
n
k
d
n
∏
k
=
1
t
x
2
k
,
x
1
n
v
n
k
′
e
k
=
∏
k
=
1
t
∏
j
=
1
n
-
1
x
2
k
b
j
k
,
x
1
n
-
u
j
n
d
n
/
d
j
∏
k
=
1
t
x
2
k
,
x
1
n
d
n
v
n
k
x
2
k
,
x
1
n
,
x
1
n
-
v
n
k
d
n
2
∏
k
=
1
t
x
2
k
e
k
,
x
1
n
v
n
k
′
=
∏
k
=
1
t
∏
j
=
1
n
-
1
x
2
k
b
j
k
,
x
1
n
-
u
j
n
d
n
/
d
j
∏
k
=
1
t
x
2
k
,
x
1
n
,
x
1
n
-
v
n
k
d
n
2
∏
k
=
1
t
∏
l
=
1
u
x
3
l
d
k
l
,
x
1
n
v
n
k
′
=
∏
k
=
1
t
∏
j
=
1
n
-
1
x
2
k
b
j
k
,
x
1
n
-
u
j
n
d
n
/
d
j
∏
k
=
1
t
x
2
k
,
x
1
n
,
x
1
n
-
v
n
k
d
n
2
∏
l
=
1
u
x
3
l
,
x
1
n
∑
k
v
n
k
′
d
k
l
.
Equation (14) for i
=
n becomes (41)
v
n
k
d
n
2
≡
∑
j
=
1
n
-
1
w
j
n
n
b
j
k
m
o
d
d
n
,
e
k
.Thus,(42)
∏
k
=
1
t
x
2
k
,
x
1
n
,
x
1
n
-
v
n
k
d
n
2
=
∏
i
=
1
n
-
1
∏
k
x
2
k
b
j
k
,
x
1
n
,
x
1
n
-
w
i
n
n
=
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
w
i
n
n
.Using (16), we get that(43)
∏
l
=
1
u
x
3
l
,
x
1
n
∑
k
v
n
k
′
d
k
l
=
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
u
i
n
d
n
/
d
i
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
2
w
i
n
n
∏
i
=
1
n
-
1
∏
l
x
3
l
c
i
l
,
x
1
n
-
u
i
n
d
n
/
d
i
.Now(44)
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
u
i
n
d
n
/
d
i
=
∏
i
=
1
n
-
1
x
1
i
,
x
1
n
,
x
1
n
d
n
-
u
i
n
x
1
n
,
x
1
i
,
x
1
i
,
x
1
n
u
i
n
d
n
/
d
i
d
i
2
x
1
n
,
x
1
i
,
x
1
n
,
x
1
n
u
i
n
d
n
2
=
1
,
as
2
=
d
i
d
i
v
i
d
e
s
d
n
d
i
.
Also it can be seen easily from (7) that (45)
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
2
w
i
n
n
=
1
.Now using (38), (40), (43), (44), and (45), we get that (46)
1
=
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
u
i
n
d
n
/
d
i
∏
i
=
1
n
-
1
x
1
i
d
i
,
x
1
n
,
x
1
n
-
w
i
n
n
=
g
-
1
.Hence, g
=
1.


## 3. Groups of Class 3

It has been proved that ifG is a finite p-group of class 3, p odd prime, then D
4
(
G
)
=
{
1
}. Clearly for such a group G
,
D
5
(
G
)
=
{
1
}. Now the question arises that if G is 2 group of class 3 with D
4
(
G
)
≠
{
1
}, then is it possible that D
5
(
G
)
=
{
1
}? We will discuss some examples of 2 groups of class 3 in which D
4
(
G
)
≠
{
1
}, but D
5
(
G
)
=
{
1
}.


### 3.1. Example  1

LetG be a group generated by four elements r
,
a
,
b
,
c. Set x
0
=
y
0
=
z
0
=
r and x
i
=
[
x
i
-
1
,
a
], y
i
=
[
y
i
-
1
,
b
], and z
i
=
[
z
i
-
1
,
c
]
,
i
=
1,2
,
3. Suppose that G satisfies the relations(i)
r
2
7
=
1, a
2
6
=
y
1
4
z
1
2, b
2
4
=
x
1
-
4
z
1, and c
2
2
=
x
1
-
2
y
1
-
1;
(ii)
z
2
=
y
2
4, y
2
=
x
2
4;
(iii)
x
3
=
y
3
=
z
3
=
1;
(iv)
[
a
,
b
,
g
]
=
1, [
b
,
c
,
g
]
=
[
a
,
c
,
g
]
=
1
∀
g
∈
G;
(v)
[
x
i
,
b
]
=
[
x
i
,
c
]
=
[
y
i
,
a
]
=
[
y
i
,
c
]
=
[
z
i
,
a
]
=
[
z
i
,
b
]
=
1
∀
i
≥
1;
(vi)
[
x
i
,
x
j
]
=
[
x
i
,
y
j
]
=
[
x
i
,
z
j
]
=
[
y
i
,
y
j
]
=
[
y
i
,
z
j
]
=
[
z
i
,
z
j
]
=
1
∀
i
,
j
≥
0. It has been proved in [6] that G is 2 group of class 3, and D
4
(
G
)
≠
{
1
}. we will show that D
5
(
G
)
=
{
1
}.From the above relations we conclude that(i)
[
a
,
b
]
2
4
=
x
2
4, b
,
c
2
2
=
y
2
=
x
2
4, and a
,
c
2
2
=
x
2
2;
(ii)
o
(
x
1
)
=
2
7, o
(
x
2
)
=
2
6, o
(
y
1
)
=
2
5, o
(
y
2
)
=
2
4, o
(
z
1
)
=
2
3, and o
(
z
2
)
=
2
2, where o
(
g
) denotes the order of an element g.Consider(47)
x
11
=
c
,
x
12
=
b
,
x
13
=
a
,
x
14
=
r
;
x
21
=
a
,
c
,
x
22
=
b
,
c
,
x
23
=
r
,
c
,
x
24
=
a
,
b
,
x
25
=
r
,
b
,
x
26
=
r
,
a
;
x
31
=
r
,
a
,
a
=
x
2
.It is easy to see that d
(
1
)
=
2
2, d
(
2
)
=
2
4, d
(
3
)
=
2
6, d
(
4
)
=
2
7, e
(
1
)
=
2
2, e
(
2
)
=
2
2, e
(
3
)
=
2
2, e
(
4
)
=
2
4, e
(
5
)
=
2
4, e
(
6
)
=
2
6, and f
(
1
)
=
2
6. Also with the help of (1) and (2), we get that b
15
=
-
1, b
16
=
-
2, b
23
=
1, b
26
=
-
4, b
33
=
2, and b
35
=
4.SinceG is a group of class 3, so an arbitrary element of D
5
(
G
) can be written as(48)
g
=
∏
1
≤
i
<
j
≤
4
x
1
i
,
x
1
j
u
i
j
d
j
=
c
,
b
16
u
12
c
,
a
64
u
13
c
,
r
128
u
14
b
,
a
64
u
23
b
,
r
128
u
24
a
,
r
128
u
34
=
x
2
-
16
u
12
-
32
u
13
-
16
u
23
.Equation (9) for i
=
1 and k
=
6 becomes (49)
0
=
u
12
b
26
+
u
13
b
36
+
u
14
b
46
+
v
16
d
1
+
v
16
′
e
6
=
-
4
u
12
+
4
v
16
+
64
v
16
′
,which gives that(50)
-
u
12
+
v
16
≡
0
mod
⁡
4
.Also (17), for k
=
6, gives that(51)
v
16
+
2
v
26
≡
0
mod
⁡
16
.For k
=
3 and l
=
6, (18) becomes (52)
v
26
≡
0
mod
⁡
2
.Now, (50), (51), and (52) imply that(53)
u
12
≡
0
mod
⁡
4
.Again, for i
=
3 and k
=
6, (9) becomes(54)
32
u
13
+
16
u
23
≡
0
mod
⁡
64
.Since order of x
2 is 64, combining (53) and (54), we get that g
=
1.


### 3.2. Example  2

LetG be a group generated by four elements x
1
,
x
2
,
x
3
,
x
4
. Set y
1
=
[
x
2
,
x
3
], y
2
=
[
x
1
,
x
2
], y
3
=
[
x
1
,
x
3
], and y
4
=
[
x
1
,
x
4
]. Suppose that G satisfies the following relations:(55)
x
2
,
y
2
=
y
1
4
,
x
3
,
y
3
=
y
1
-
16
,
x
4
,
y
4
=
y
1
64
;
x
1
,
y
q
=
1
,
1
≤
q
≤
4
;
x
2
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
2
;
x
3
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
3
;
x
4
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
4
;
x
1
2
k
+
l
+
4
=
y
1
-
2
k
+
l
+
4
y
2
2
k
+
l
+
3
;
x
2
2
k
+
4
=
y
3
2
k
y
4
-
2
k
-
1
y
1
3
.
2
k
+
3
;
x
3
2
k
+
2
=
y
1
-
5
.
2
k
+
1
y
2
2
k
y
4
2
k
-
2
;
x
4
2
k
=
y
1
2
k
y
2
2
k
-
1
y
3
2
k
-
2
.Consider(56)
x
11
=
x
4
,
x
12
=
x
3
,
x
13
=
x
2
,
x
14
=
x
1
;
x
21
=
y
1
,
x
22
=
y
4
,
x
23
=
y
3
,
x
24
=
y
12
;
x
31
=
x
2
,
y
2
=
y
1
4
.It is easy to see that d
(
1
)
=
2
k, d
(
2
)
=
2
k
+
2, d
(
3
)
=
2
k
+
4, d
(
4
)
=
2
k
+
l
+
4, e
(
1
)
=
4, e
(
2
)
=
2
k, e
(
3
)
=
2
k
+
2, e
(
4
)
=
2
k
+
4, and f
(
1
)
=
2
k
+
2. Also we have b
11
=
2
k, b
13
=
2
k
-
2, b
14
=
2
k
-
1, b
21
=
-
5
.
2
k
+
1, b
22
=
2
k
-
2, b
24
=
2
k, b
31
=
3
.
2
k
+
3, b
32
=
-
2
k
-
1, b
33
=
2
k, b
41
=
-
2
k
+
l
+
4, and b
44
=
2
k
+
l
+
3. It has been proved in [7] that G is a group of class 3 and D
4
(
G
)
≠
1. We will prove that D
5
(
G
)
=
{
1
}.An arbitrary elementg of D
5
(
G
) can be written as (57)
g
=
∏
1
≤
i
<
j
≤
4
x
1
i
,
x
1
j
u
i
j
d
j
=
x
4
,
x
3
u
12
2
k
+
2
x
4
,
x
2
u
13
2
k
+
4
x
4
,
x
1
u
14
2
k
+
l
+
4
x
3
,
x
2
u
23
2
k
+
4
x
3
,
x
1
u
24
2
k
+
l
+
4
x
2
,
x
1
u
34
2
k
+
l
+
4
=
y
1
2
k
+
4
u
12
-
2
u
13
-
u
23
+
2
l
+
1
u
34
.To show that g
=
1, it is enough to show that u
12
-
2
u
13
-
u
23
+
2
l
+
1
u
34
≡
0
(
mod
⁡
4
), since order of y
1 is 2
k
+
6.Now, fori
=
3 and k
=
4, (9) becomes (58)
0
=
u
34
b
44
-
u
13
d
3
d
1
b
14
-
u
23
d
3
d
2
b
24
+
v
34
d
3
+
v
34
′
e
4
=
2
k
+
2
2
l
+
1
u
34
-
2
u
13
-
u
23
+
4
v
34
+
4
v
34
′
,which implies that(59)
2
l
+
1
u
34
-
2
u
13
-
u
23
≡
0
mod
⁡
4
.Also (9) for i
=
1 and k
=
2 gives that (60)
0
=
u
12
b
22
+
u
13
b
32
+
u
14
b
42
+
v
12
d
1
+
v
12
′
e
2
=
2
k
-
2
u
12
-
2
k
-
1
u
13
+
2
k
v
12
+
2
k
v
12
′
.That is, we have (61)
u
12
=
2
u
13
-
4
v
12
-
4
v
12
′
.Again (9), for i
=
1, k
=
3, gives that(62)
u
13
≡
-
v
13
mod
⁡
4
.Now (17) for k
=
3 gives that(63)
v
13
≡
0
mod
⁡
4
.Combining (61), (62), and (63), we get that(64)
u
13
≡
0
mod
⁡
4
,which together with (59) gives that g
=
1.


## 3.1. Example  1

LetG be a group generated by four elements r
,
a
,
b
,
c. Set x
0
=
y
0
=
z
0
=
r and x
i
=
[
x
i
-
1
,
a
], y
i
=
[
y
i
-
1
,
b
], and z
i
=
[
z
i
-
1
,
c
]
,
i
=
1,2
,
3. Suppose that G satisfies the relations(i)
r
2
7
=
1, a
2
6
=
y
1
4
z
1
2, b
2
4
=
x
1
-
4
z
1, and c
2
2
=
x
1
-
2
y
1
-
1;
(ii)
z
2
=
y
2
4, y
2
=
x
2
4;
(iii)
x
3
=
y
3
=
z
3
=
1;
(iv)
[
a
,
b
,
g
]
=
1, [
b
,
c
,
g
]
=
[
a
,
c
,
g
]
=
1
∀
g
∈
G;
(v)
[
x
i
,
b
]
=
[
x
i
,
c
]
=
[
y
i
,
a
]
=
[
y
i
,
c
]
=
[
z
i
,
a
]
=
[
z
i
,
b
]
=
1
∀
i
≥
1;
(vi)
[
x
i
,
x
j
]
=
[
x
i
,
y
j
]
=
[
x
i
,
z
j
]
=
[
y
i
,
y
j
]
=
[
y
i
,
z
j
]
=
[
z
i
,
z
j
]
=
1
∀
i
,
j
≥
0. It has been proved in [6] that G is 2 group of class 3, and D
4
(
G
)
≠
{
1
}. we will show that D
5
(
G
)
=
{
1
}.From the above relations we conclude that(i)
[
a
,
b
]
2
4
=
x
2
4, b
,
c
2
2
=
y
2
=
x
2
4, and a
,
c
2
2
=
x
2
2;
(ii)
o
(
x
1
)
=
2
7, o
(
x
2
)
=
2
6, o
(
y
1
)
=
2
5, o
(
y
2
)
=
2
4, o
(
z
1
)
=
2
3, and o
(
z
2
)
=
2
2, where o
(
g
) denotes the order of an element g.Consider(47)
x
11
=
c
,
x
12
=
b
,
x
13
=
a
,
x
14
=
r
;
x
21
=
a
,
c
,
x
22
=
b
,
c
,
x
23
=
r
,
c
,
x
24
=
a
,
b
,
x
25
=
r
,
b
,
x
26
=
r
,
a
;
x
31
=
r
,
a
,
a
=
x
2
.It is easy to see that d
(
1
)
=
2
2, d
(
2
)
=
2
4, d
(
3
)
=
2
6, d
(
4
)
=
2
7, e
(
1
)
=
2
2, e
(
2
)
=
2
2, e
(
3
)
=
2
2, e
(
4
)
=
2
4, e
(
5
)
=
2
4, e
(
6
)
=
2
6, and f
(
1
)
=
2
6. Also with the help of (1) and (2), we get that b
15
=
-
1, b
16
=
-
2, b
23
=
1, b
26
=
-
4, b
33
=
2, and b
35
=
4.SinceG is a group of class 3, so an arbitrary element of D
5
(
G
) can be written as(48)
g
=
∏
1
≤
i
<
j
≤
4
x
1
i
,
x
1
j
u
i
j
d
j
=
c
,
b
16
u
12
c
,
a
64
u
13
c
,
r
128
u
14
b
,
a
64
u
23
b
,
r
128
u
24
a
,
r
128
u
34
=
x
2
-
16
u
12
-
32
u
13
-
16
u
23
.Equation (9) for i
=
1 and k
=
6 becomes (49)
0
=
u
12
b
26
+
u
13
b
36
+
u
14
b
46
+
v
16
d
1
+
v
16
′
e
6
=
-
4
u
12
+
4
v
16
+
64
v
16
′
,which gives that(50)
-
u
12
+
v
16
≡
0
mod
⁡
4
.Also (17), for k
=
6, gives that(51)
v
16
+
2
v
26
≡
0
mod
⁡
16
.For k
=
3 and l
=
6, (18) becomes (52)
v
26
≡
0
mod
⁡
2
.Now, (50), (51), and (52) imply that(53)
u
12
≡
0
mod
⁡
4
.Again, for i
=
3 and k
=
6, (9) becomes(54)
32
u
13
+
16
u
23
≡
0
mod
⁡
64
.Since order of x
2 is 64, combining (53) and (54), we get that g
=
1.


## 3.2. Example  2

LetG be a group generated by four elements x
1
,
x
2
,
x
3
,
x
4
. Set y
1
=
[
x
2
,
x
3
], y
2
=
[
x
1
,
x
2
], y
3
=
[
x
1
,
x
3
], and y
4
=
[
x
1
,
x
4
]. Suppose that G satisfies the following relations:(55)
x
2
,
y
2
=
y
1
4
,
x
3
,
y
3
=
y
1
-
16
,
x
4
,
y
4
=
y
1
64
;
x
1
,
y
q
=
1
,
1
≤
q
≤
4
;
x
2
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
2
;
x
3
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
3
;
x
4
,
y
q
=
1
,
1
≤
q
≤
4
,
q
≠
4
;
x
1
2
k
+
l
+
4
=
y
1
-
2
k
+
l
+
4
y
2
2
k
+
l
+
3
;
x
2
2
k
+
4
=
y
3
2
k
y
4
-
2
k
-
1
y
1
3
.
2
k
+
3
;
x
3
2
k
+
2
=
y
1
-
5
.
2
k
+
1
y
2
2
k
y
4
2
k
-
2
;
x
4
2
k
=
y
1
2
k
y
2
2
k
-
1
y
3
2
k
-
2
.Consider(56)
x
11
=
x
4
,
x
12
=
x
3
,
x
13
=
x
2
,
x
14
=
x
1
;
x
21
=
y
1
,
x
22
=
y
4
,
x
23
=
y
3
,
x
24
=
y
12
;
x
31
=
x
2
,
y
2
=
y
1
4
.It is easy to see that d
(
1
)
=
2
k, d
(
2
)
=
2
k
+
2, d
(
3
)
=
2
k
+
4, d
(
4
)
=
2
k
+
l
+
4, e
(
1
)
=
4, e
(
2
)
=
2
k, e
(
3
)
=
2
k
+
2, e
(
4
)
=
2
k
+
4, and f
(
1
)
=
2
k
+
2. Also we have b
11
=
2
k, b
13
=
2
k
-
2, b
14
=
2
k
-
1, b
21
=
-
5
.
2
k
+
1, b
22
=
2
k
-
2, b
24
=
2
k, b
31
=
3
.
2
k
+
3, b
32
=
-
2
k
-
1, b
33
=
2
k, b
41
=
-
2
k
+
l
+
4, and b
44
=
2
k
+
l
+
3. It has been proved in [7] that G is a group of class 3 and D
4
(
G
)
≠
1. We will prove that D
5
(
G
)
=
{
1
}.An arbitrary elementg of D
5
(
G
) can be written as (57)
g
=
∏
1
≤
i
<
j
≤
4
x
1
i
,
x
1
j
u
i
j
d
j
=
x
4
,
x
3
u
12
2
k
+
2
x
4
,
x
2
u
13
2
k
+
4
x
4
,
x
1
u
14
2
k
+
l
+
4
x
3
,
x
2
u
23
2
k
+
4
x
3
,
x
1
u
24
2
k
+
l
+
4
x
2
,
x
1
u
34
2
k
+
l
+
4
=
y
1
2
k
+
4
u
12
-
2
u
13
-
u
23
+
2
l
+
1
u
34
.To show that g
=
1, it is enough to show that u
12
-
2
u
13
-
u
23
+
2
l
+
1
u
34
≡
0
(
mod
⁡
4
), since order of y
1 is 2
k
+
6.Now, fori
=
3 and k
=
4, (9) becomes (58)
0
=
u
34
b
44
-
u
13
d
3
d
1
b
14
-
u
23
d
3
d
2
b
24
+
v
34
d
3
+
v
34
′
e
4
=
2
k
+
2
2
l
+
1
u
34
-
2
u
13
-
u
23
+
4
v
34
+
4
v
34
′
,which implies that(59)
2
l
+
1
u
34
-
2
u
13
-
u
23
≡
0
mod
⁡
4
.Also (9) for i
=
1 and k
=
2 gives that (60)
0
=
u
12
b
22
+
u
13
b
32
+
u
14
b
42
+
v
12
d
1
+
v
12
′
e
2
=
2
k
-
2
u
12
-
2
k
-
1
u
13
+
2
k
v
12
+
2
k
v
12
′
.That is, we have (61)
u
12
=
2
u
13
-
4
v
12
-
4
v
12
′
.Again (9), for i
=
1, k
=
3, gives that(62)
u
13
≡
-
v
13
mod
⁡
4
.Now (17) for k
=
3 gives that(63)
v
13
≡
0
mod
⁡
4
.Combining (61), (62), and (63), we get that(64)
u
13
≡
0
mod
⁡
4
,which together with (59) gives that g
=
1.



---

*Source: 5342926-2016-07-25.xml*