Research ArticleThe Fifth Dimension Subgroup forMetabelian 2 Groups

Shalini Gupta

Department of Mathematics Punjabi University Patiala 147002 India

Correspondence should be addressed to Shalini Gupta gupta mathyahoocom

Received 18 April 2016 Accepted 14 June 2016

Academic Editor Burkhard Kulshammer

Copyright copy 2016 Shalini GuptaThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Given a finite metabelian 2-group 119866 the object of this paper is to discuss some cases under which 1198635(119866) = 120574

5(119866) Further some

examples of 2 groups of class 3 for which 1198634(119866) = 1 but 119863

5(119866) = 1 are discussed

1 Introduction

Let Z(119866) be the integral group ring of group 119866 and let 119868(119866)

be the augmentation ideal of Z(119866) For 119899 ge 1 let 119863119899(119866) =

119866 cap 1 + 119868119899(119866) be 119899th dimension subgroup and let 120574

119899(119866)

be 119899th term in lower central series of group 119866 The famousdimension subgroup conjecture states that 119863

119899(119866) = 120574

119899(119866)

for all 119899 ge 1 and for all groups 119866 This conjecture has beenproved for 119899 le 3 (see eg [1]) It has been proved in [2]that if 119866 is a finite 119901-group 119901 odd prime then 119863

4(119866) =

1205744(119866) Keeping in view this result Rips [3] gave an example

of 2-group 119866 for which 1198634(119866) = 120574

4(119866) Further Tahara [4]

gave the structure of 1198635(119866) and proved that the exponent of

1198635(119866)1205745(119866) is divisible by 6 It has been shown in [5] that for

a metabelian group the exponent of1198635(119866)1205745(119866) is divisible

by 2 and hence for a metabelian 119901-group 119866 119901 odd prime1198635(119866) = 120574

5(119866) Gupta [6] constructed a 2-group119866 generated

by four elements such that1198635(119866) = 120574

5(119866) Now the question

arises that whether for a metabelian 2-group 119866 with at mostthree generators 119863

5(119866) = 120574

5(119866) or not

In this paper we prove that if 119866 is metabelian 2 groupgenerated by at most two elements then 119863

5(119866) = 120574

5(119866)

(Theorem 2) Further we prove that under certain conditionsa metabelian 2-group 119866 generated by 119899 elements satisfiesthe dimension subgroup conjecture for 119899 = 5 (Theorem 3)Finally in Section 3 we give some examples of 2 groups ofclass 3 for which 119863

5(119866) = 120574

5(119866)

2 Main Results

We first recall the structure of 1198635(119866) given by Tahara [4]

Let 119866 be a finite group of class 4 Consider the lower centralseries 119866 = 120574

1(119866) supe 120574

2(119866) supe 120574

3(119866) supe 120574

4(119866) supe 120574

5(119866) = 1

of 119866 Let 11990911198941205742(119866)1le119894le119904

be basis of 1198661205742(119866) with 119889(119894) as the

order of 11990911198941205742(119866) that is 119909

1119894

119889(119894) belongs to 1205742(119866) Similarly let

11990921198951205743(119866)1le119895le119905

be basis of 1205742(119866)1205743(119866) with 119890(119895) as the order

of 11990921198951205743(119866) and let 119909

31198961205744(119866)1le119896le119906

be basis of 1205743(119866)1205744(119866)

with 119891(119896) as the order of 11990931198961205744(119866) Moreover these basis

elements are chosen in such a way that 119889(119894) divides 119889(119894 + 1)119890(119894) divides 119890(119894 + 1) and 119891(119894) divides 119891(119894 + 1) Thus we have

1199091119894

119889(119894)= prod

1le119895le119905

1199092119895

119887119894119895prod

1le119896le119906

1199093119896

1198881198941198961199104119894

1199104119894

isin 1205744(119866) 1 le 119894 le 119904

(1)

1199092119896

119890(119896)= prod

1le119896le119906

1199093119897

1198891198961198971199101015840

4119896

1199101015840

4119896isin 1205744(119866) 1 le 119896 le 119905

(2)

[1199091119894

119889(119894) 1199091119895] = prod

1le119897le119906

1199093119897

120572(119894119895)

119897 11991010158401015840

4119894119895

11991010158401015840

4119894119895isin 1205744(119866) 1 le 119894 le 119904

(3)

The following theorem gives the structure of 1198635(119866)

Hindawi Publishing CorporationChinese Journal of MathematicsVolume 2016 Article ID 5342926 7 pageshttpdxdoiorg10115520165342926

2 Chinese Journal of Mathematics

Theorem 1 (see [4]) With the above notations1198635(119866) is equal

to the subgroup generated by the elements

prod

1le119894lt119895le119904

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894le1199041le119896le119905

prod

119896lt119897

[1199092119897 1199092119896]119887119894119897V119894119896

sdot prod

1le119894le119895le119896le119904

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

(4)

with the following conditions

119908119894119894119894

= 0 1 le 119894 le 119904 (5)

119906119894119895

119889 (119895)

119889 (119894)(

119889 (119894)

2) + 119908

119894119894119895119889 (119894) + 119908

10158401015840

119894119894119895119889 (119895) = 0

1 le 119894 lt 119895 le 119904

(6)

minus 119906119894119895(

119889 (119895)

2

) + 119908119894119895119895

119889 (119894) + 1199081015840

119894119895119895119889 (119895) = 0

1 le 119894 lt 119895 le 119904

(7)

119908119894119895119896

119889 (119894) + 1199081015840

119894119895119896119889 (119895) + 119908

10158401015840

119894119895119896119889 (119896) = 0

1 le 119894 lt 119895 lt 119896 le 119904

(8)

sum

119894ltℎ

119906119894ℎ119887ℎ119896

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)119887ℎ119896

+ V119894119896119889 (119894) + V1015840

119894119896119890 (119896) = 0

1 le 119894 le 119904 1 le 119896 le 119905

(9)

119906119894119895

119889 (119895)

119889 (119894)(

119889 (119894)

3) + 119908

119894119894119895(

119889 (119894)

2) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(10)

119908119894119894119895

(

119889 (119894)

2

) + 11990810158401015840

119894119894119895(

119889 (119895)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(11)

minus 119906119894119895(

119889 (119895)

3

) + 1199081015840

119894119895119895(

119889 (119895)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(12)

119908119894119895119896

(

119889 (119894)

2

) 1199081015840

119894119895119896(

119889 (119895)

2

) 11990810158401015840

119894119895119896(

119889 (119896)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 lt 119896 le 119904

(13)

V119894119896(

119889 (119894)

2

) minus sum

ℎle119894

119908ℎ119894119894

119887ℎ119896

minus sum

119894ltℎ

11990810158401015840

119894119894ℎ119887ℎ119896

equiv 0

(mod (119889 (119894) 119890 (119896))) 1 le 119894 le 119904 1 le 119896 le 119905

(14)

sum

ℎle119894

119908ℎ119894119895

119887ℎ119896

+ sum

119894ltℎle119895

1199081015840

119894ℎ119895119887ℎ119896

+ sum

119895ltℎ

11990810158401015840

119894119895119896119887ℎ119896

equiv 0

(mod (119889 (119894) 119891 (119897))) 1 le 119894 le 119904 1 le 119897 le 119906

(15)

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)120572119897

(ℎ119894)+ sum

119894ltℎ

119906119894ℎ119888ℎ119894

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)119888ℎ119894

minus sum

119896

V1015840119894119896119889119896119897

minus sum

119892le119894leℎ

119908119892119894ℎ

120572119897

(119892ℎ)minus sum

119892leℎle119894

119908119892ℎ119894

120572119897

(119892ℎ)

minus sum

119894lt119892leℎ

1199081015840

119894119892ℎ120572119897

(119892ℎ)equiv 0

(mod (119889 (119894) 119891 (119897))) 1 le 119894 le 119904 1 le 119897 le 119906

(16)

sum

119894

V119894119896119887119894119896

equiv 0 (mod 119890 (119896)) 1 ⩽ 119896 ⩽ 119905 (17)

sum

119894

V119894119896119887119894119897+ sum

119894

V119894119897119887119894119896

equiv 0 (mod 119890 (119896)) 1 ⩽ 119896 lt 119897 ⩽ 119905 (18)

Theorem 2 Let119866 bemetabelian 2 group with1198661205742(119866) as sum

of at most two cyclic groups Then 1198635(119866) = 120574

5(119866)

Proof It is obvious that if metabelian 2 groups are such that1198661205742(119866) is a cyclic group then 119863

5(119866) = 120574

5(119866)

Let 119866 be metabelian 2 group and let 1198661205742(119866) be sum

of two cyclic groups It is enough to prove the result for agroup of class 4 Let 119866120574

2(119866) = 119862

1oplus 1198622 where 119862

119894is a cyclic

group generated by 11990911198941205742(119866) with order of 119909

11198941205742(119866) = 119889(119894)

and 119889(1) divides 119889(2) An arbitrary element 119892 of 1198635(119866) is of

the form [11990911

119889(1) 11990912]11990612(119889(2)119889(1))[119909

11

119889(1) 11990912 11990912]119908122 subject

to the conditions given inTheorem 1 For 1 le 119896 le 119905 and 119894 = 2(9) becomes

minus11990612

119889 (2)

119889 (1)1198871119896

+ V2119896119889 (2) + V1015840

2119896119890 (119896) = 0 (19)

Thus

1 =

119905

prod

119896=1

[1199092119896 11990912]minus11990612(119889(2)119889(1))1198871119896+V2119896119889(2)+V

1015840

2119896119890(119896)

=

119905

prod

119896=1

[1199092119896

1198871119896 11990912]minus11990612(119889(2)119889(1))

sdot [1199092119896 11990912 1199092119896]11990612(119889(2)119889(1))(

1198871119896

2)

sdot

119905

prod

119896=1

[1199092119896 11990912

119889(2)]V2119896

[1199092119896 11990912 11990912]minusV2119896( 119889(2)2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 11990912]V10158402119896

[1199092119896 11990912 1199092119896]minusV10158402119896(119890(119896)

2)

(20)

Now [1199092119896 11990912 1199092119896] belongs to 120574

5(119866) and [119909

2119896 11990912

119889(2)] belongs

to [1205742(119866) 1205742(119866)] thus we have

1 =

119905

prod

119896=1

[1199092119896

1198871119896 11990912]minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minusV2119896( 119889(2)2)

119905

prod

119896=1

[1199092119896

119890(119896) 11990912]V10158402119896

Chinese Journal of Mathematics 3

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minusV2119896119889( 119889(2)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 11990912]V10158402119896

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minus1199081221198871119896

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

(21)

as for 119894 = 2 (14) becomes V2119896

(119889(2)

2) minus 119908122

1198871119896

equiv 0 (mod(119889(2)119890(119896))) Now (16) for 119894 = 2 becomes

minus 11990612

119889 (2)

119889 (1)120572(12)

119897minus 11990612

119889 (2)

119889 (1)1198881119897

minus sum

119896

V10158402119896119889119896119897

minus 2119908122

120572(12)

119897

minus 119908112

120572(11)

119897equiv 0 (mod (119889 (2) 119891 (119897)))

(22)

which implies that

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

=

119906

prod

119897=1

[1199093119897

11990912]minus11990612(119889(2)119889(1))120572

(12)

119897minus11990612(119889(2)119889(1))1198881119897minus2119908122120572

(12)

119897minus119908112120572

(11)

119897

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897

11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897 11990912]

minus2119908122

[

119906

prod

119897=1

1199093119897

120572(11)

119897

11990912]

minus119908112

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[11990911

119889(1)

11990912 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus2119908122

(23)

Now

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912]minus11990612119889(2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

= [11990911 11990912 11990912

119889(2)]minus11990612

sdot [11990912 11990911 11990912 11990912]minus11990612(119889(2)

2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

(24)

Using (6) and (7) we get that

11990612

119889 (2)

119889 (1)(

119889 (1)

2

) equiv 0 (mod 119889 (1)) (25)

minus11990612

(

119889 (2)

2

) equiv 0 (mod 119889 (1)) (26)

Hence (24) reduces to

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912

119889(2)]minus11990612

= 1 as [11990911 11990912 11990912

119889(2)] isin [120574

2(119866) 120574

2(119866)]

(27)

Now

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]minus2119908122119889(1)

sdot [11990912 11990911 11990911 11990912]2119908122(119889(1)

2)

= [11990911 11990912 11990912]minus2119908122119889(1)

(28)

Also (7) implies that

minus2119908122

119889 (1) = minus211990612

(

119889 (2)

2

) + 21199081015840

122119889 (2) (29)

Thus (28) becomes

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]21199081015840

122119889(2)

sdot [11990911 11990912 11990912]minus11990612119889(2)(119889(2)minus1)

= [11990911 11990912 11990912

119889(2)]21199081015840

122

sdot [11990912 11990911 11990912 11990912]minus21199081015840

122(119889(2)

2)

sdot [11990911 11990912 11990912]minus11990612(119889(2)(119889(2)minus1))

= [11990911 11990912 11990912

119889(2)]minus11990612(119889(2)minus1)

sdot [11990912 11990911 11990912 11990912]minus11990612((119889(2)minus1)(

119889(2)

2))

= 1

(30)

as by (7) minus11990612

(119889(2)

2) equiv 0 (mod 119889(1))

Using (23) (27) and (30) in (21) we get that

1 = [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

sdot [

119905

prod

119896=1

1199092119896

1198871119896 11990912 11990912]

minus119908122

= [11990911

119889(1) 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus119908122

= 119892minus1

(31)

Thus 119892 = 1 and hence 1198635(119866) = 1

4 Chinese Journal of Mathematics

Theorem 3 Let119866 be finite metabelian 2 group and1198661205742(119866) =

1198621oplus1198622oplus sdot sdot sdot oplus 119862

119899 where 119862

119894are cyclic groups of order 119889(119894) Let

119889(1) = 119889(2) = sdot sdot sdot = 119889(119899 minus 1) = 2 119889(119899) ge 4 Then 1198635(119866) =

1205745(119866)

Proof It is enough to prove the result for a group119866 of class 4Any element 119892 of 119863

5(119866) is of the form

119892 = prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894lt119895lt119896le119899

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895 1199091119895]119908119894119895119895

= 119860 sdot 119861 sdot 119862 say

(32)

Since for 1 le 119894 lt 119895 lt 119896 le 119899 119889(119894) = 119889(119895) = 2 (13) reduces to

119908119894119895119896

equiv 0 (mod 119889 (119895)) (33)

and hence 119861 = 1 Now

119860 = prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

as for 1 le 119894 lt 119895 le 119899 minus 1 119889 (119894) = 119889 (119895)

(34)

Also for 1 le 119894 lt 119895 le 119899 (7) implies that minus119906119894119895+2119908119894119895119895

+21199081015840

119894119895119895= 0

which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]2119908119894119895119895+2119908

1015840

119894119895119895

as 119889 (119895) = 2

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119889(119895)(119908119894119895119895+119908

1015840

119894119895119895)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895

119889(119895)](119908119894119895119895+119908

1015840

119894119895119895)

sdot [1199091119894

119889(119894) 1199091119895 1199091119895]minus(119908119894119895119895+119908

1015840

119894119895119895)(119889(119895)

2)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus1199081015840

119894119895119895

as [1199091119894

119889(119894) 1199091119895

119889(119895)] isin [120574

2(119866) 120574

2(119866)] 119889 (119895) = 2

(35)

Now for 1 le 119894 lt 119895 le 119899 (12) implies that 1199081015840119894119895119895

equiv 0 (mod 119889(119895))which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(36)

Thus (34) reduces to

119860 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(37)

Hence

119892 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]119908119894119899119899

(38)

Now for 119894 = 119899 (9) becomes

119878 = minus

119899minus1

sum

119895=1

119906119895119899

119889 (119899)

119889 (119895)119887119895119896

+ V119899119896119889 (119899) + V1015840

119899119896119890 (119896) = 0 (39)

which gives

1 =

119905

prod

119896=1

[1199092119896 1199091119899]119878

=

119905

prod

119896=1

[1199092119896 1199091119899]minussum119899minus1

119895=1119906119895119899(119889(119899)119889(119895))119887119895119896

sdot

119905

prod

119896=1

[1199092119896 1199091119899]V119899119896119889(119899)

119905

prod

119896=1

[1199092119896 1199091119899]V1015840119899119896119890(119896)

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899

119889(119899)]V119899119896

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 1199091119899]V1015840119899119896

Chinese Journal of Mathematics 5

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 1199091119899]V1015840119899119896

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

(40)

Equation (14) for 119894 = 119899 becomes

V119899119896

(

119889 (119899)

2

) equiv

119899minus1

sum

119895=1

119908119895119899119899

119887119895119896

(mod (119889 (119899) 119890 (119896))) (41)

Thus

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

=

119899minus1

prod

119894=1

[prod

119896

1199092119896

119887119895119896 1199091119899 1199091119899]

minus119908119894119899119899

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

(42)

Using (16) we get that

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

sdot

119899minus1

prod

119894=1

[prod

119897

1199093119897

119888119894119897 1199091119899]

minus119906119894119899(119889(119899)119889(119894))

(43)

Now

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

=

119899minus1

prod

119894=1

[1199091119894 1199091119899 1199091119899

119889(119899)]minus119906119894119899

sdot [1199091119899 1199091119894 1199091119894 1199091119899]119906119894119899(119889(119899)119889(119894))(

119889(119894)

2)

sdot [1199091119899 1199091119894 1199091119899 1199091119899]119906119894119899(119889(119899)

2)

= 1

as 2 = 119889 (119894) divides 119889 (119899)

119889 (119894)

(44)

Also it can be seen easily from (7) that

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

= 1 (45)

Now using (38) (40) (43) (44) and (45) we get that

1 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

= 119892minus1

(46)

Hence 119892 = 1

3 Groups of Class 3

It has been proved that if119866 is a finite 119901-group of class 3 119901 oddprime then119863

4(119866) = 1 Clearly for such a group119866119863

5(119866) =

1 Now the question arises that if119866 is 2 group of class 3 with1198634(119866) = 1 then is it possible that 119863

5(119866) = 1 We will

discuss some examples of 2 groups of class 3 inwhich1198634(119866) =

1 but 1198635(119866) = 1

31 Example 1 Let 119866 be a group generated by four elements119903 119886 119887 119888 Set 119909

0= 1199100= 1199110= 119903 and 119909

119894= [119909119894minus1

119886] 119910119894= [119910119894minus1

119887]and 119911

119894= [119911119894minus1

119888] 119894 = 1 2 3 Suppose that 119866 satisfies therelations

(i) 11990327

= 1 11988626

= 1199101

41199111

2 11988724

= 1199091

minus41199111 and 119888

22

=

1199091

minus21199101

minus1(ii) 1199112= 1199102

4 1199102= 1199092

4(iii) 119909

3= 1199103= 1199113= 1

(iv) [119886 119887 119892] = 1 [119887 119888 119892] = [119886 119888 119892] = 1 forall119892 isin 119866(v) [119909

119894 119887] = [119909

119894 119888] = [119910

119894 119886] = [119910

119894 119888] = [119911

119894 119886] = [119911

119894 119887] =

1 forall119894 ge 1(vi) [119909

119894 119909119895] = [119909

119894 119910119895] = [119909

119894 119911119895] = [119910

119894 119910119895] = [119910

119894 119911119895] =

[119911119894 119911119895] = 1 forall119894 119895 ge 0 It has been proved in [6] that 119866

is 2 group of class 3 and 1198634(119866) = 1 we will show

that 1198635(119866) = 1

From the above relations we conclude that

(i) [119886 119887]24

= 1199092

4 [119887 119888]22

= 1199102= 1199092

4 and [119886 119888]22

= 1199092

2(ii) 119900(119909

1) = 27 119900(1199092) = 26 119900(1199101) = 25 119900(1199102) = 24 119900(1199111) =

23 and 119900(119911

2) = 22 where 119900(119892) denotes the order of an

element 119892

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Chinese Journal of Mathematics

Theorem 1 (see [4]) With the above notations1198635(119866) is equal

to the subgroup generated by the elements

prod

1le119894lt119895le119904

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894le1199041le119896le119905

prod

119896lt119897

[1199092119897 1199092119896]119887119894119897V119894119896

sdot prod

1le119894le119895le119896le119904

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

(4)

with the following conditions

119908119894119894119894

= 0 1 le 119894 le 119904 (5)

119906119894119895

119889 (119895)

119889 (119894)(

119889 (119894)

2) + 119908

119894119894119895119889 (119894) + 119908

10158401015840

119894119894119895119889 (119895) = 0

1 le 119894 lt 119895 le 119904

(6)

minus 119906119894119895(

119889 (119895)

2

) + 119908119894119895119895

119889 (119894) + 1199081015840

119894119895119895119889 (119895) = 0

1 le 119894 lt 119895 le 119904

(7)

119908119894119895119896

119889 (119894) + 1199081015840

119894119895119896119889 (119895) + 119908

10158401015840

119894119895119896119889 (119896) = 0

1 le 119894 lt 119895 lt 119896 le 119904

(8)

sum

119894ltℎ

119906119894ℎ119887ℎ119896

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)119887ℎ119896

+ V119894119896119889 (119894) + V1015840

119894119896119890 (119896) = 0

1 le 119894 le 119904 1 le 119896 le 119905

(9)

119906119894119895

119889 (119895)

119889 (119894)(

119889 (119894)

3) + 119908

119894119894119895(

119889 (119894)

2) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(10)

119908119894119894119895

(

119889 (119894)

2

) + 11990810158401015840

119894119894119895(

119889 (119895)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(11)

minus 119906119894119895(

119889 (119895)

3

) + 1199081015840

119894119895119895(

119889 (119895)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 le 119904

(12)

119908119894119895119896

(

119889 (119894)

2

) 1199081015840

119894119895119896(

119889 (119895)

2

) 11990810158401015840

119894119895119896(

119889 (119896)

2

) equiv 0

(mod 119889 (119894)) 1 le 119894 lt 119895 lt 119896 le 119904

(13)

V119894119896(

119889 (119894)

2

) minus sum

ℎle119894

119908ℎ119894119894

119887ℎ119896

minus sum

119894ltℎ

11990810158401015840

119894119894ℎ119887ℎ119896

equiv 0

(mod (119889 (119894) 119890 (119896))) 1 le 119894 le 119904 1 le 119896 le 119905

(14)

sum

ℎle119894

119908ℎ119894119895

119887ℎ119896

+ sum

119894ltℎle119895

1199081015840

119894ℎ119895119887ℎ119896

+ sum

119895ltℎ

11990810158401015840

119894119895119896119887ℎ119896

equiv 0

(mod (119889 (119894) 119891 (119897))) 1 le 119894 le 119904 1 le 119897 le 119906

(15)

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)120572119897

(ℎ119894)+ sum

119894ltℎ

119906119894ℎ119888ℎ119894

minus sum

ℎlt119894

119906ℎ119894

119889 (119894)

119889 (ℎ)119888ℎ119894

minus sum

119896

V1015840119894119896119889119896119897

minus sum

119892le119894leℎ

119908119892119894ℎ

120572119897

(119892ℎ)minus sum

119892leℎle119894

119908119892ℎ119894

120572119897

(119892ℎ)

minus sum

119894lt119892leℎ

1199081015840

119894119892ℎ120572119897

(119892ℎ)equiv 0

(mod (119889 (119894) 119891 (119897))) 1 le 119894 le 119904 1 le 119897 le 119906

(16)

sum

119894

V119894119896119887119894119896

equiv 0 (mod 119890 (119896)) 1 ⩽ 119896 ⩽ 119905 (17)

sum

119894

V119894119896119887119894119897+ sum

119894

V119894119897119887119894119896

equiv 0 (mod 119890 (119896)) 1 ⩽ 119896 lt 119897 ⩽ 119905 (18)

Theorem 2 Let119866 bemetabelian 2 group with1198661205742(119866) as sum

of at most two cyclic groups Then 1198635(119866) = 120574

5(119866)

Proof It is obvious that if metabelian 2 groups are such that1198661205742(119866) is a cyclic group then 119863

5(119866) = 120574

5(119866)

Let 119866 be metabelian 2 group and let 1198661205742(119866) be sum

of two cyclic groups It is enough to prove the result for agroup of class 4 Let 119866120574

2(119866) = 119862

1oplus 1198622 where 119862

119894is a cyclic

group generated by 11990911198941205742(119866) with order of 119909

11198941205742(119866) = 119889(119894)

and 119889(1) divides 119889(2) An arbitrary element 119892 of 1198635(119866) is of

the form [11990911

119889(1) 11990912]11990612(119889(2)119889(1))[119909

11

119889(1) 11990912 11990912]119908122 subject

to the conditions given inTheorem 1 For 1 le 119896 le 119905 and 119894 = 2(9) becomes

minus11990612

119889 (2)

119889 (1)1198871119896

+ V2119896119889 (2) + V1015840

2119896119890 (119896) = 0 (19)

Thus

1 =

119905

prod

119896=1

[1199092119896 11990912]minus11990612(119889(2)119889(1))1198871119896+V2119896119889(2)+V

1015840

2119896119890(119896)

=

119905

prod

119896=1

[1199092119896

1198871119896 11990912]minus11990612(119889(2)119889(1))

sdot [1199092119896 11990912 1199092119896]11990612(119889(2)119889(1))(

1198871119896

2)

sdot

119905

prod

119896=1

[1199092119896 11990912

119889(2)]V2119896

[1199092119896 11990912 11990912]minusV2119896( 119889(2)2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 11990912]V10158402119896

[1199092119896 11990912 1199092119896]minusV10158402119896(119890(119896)

2)

(20)

Now [1199092119896 11990912 1199092119896] belongs to 120574

5(119866) and [119909

2119896 11990912

119889(2)] belongs

to [1205742(119866) 1205742(119866)] thus we have

1 =

119905

prod

119896=1

[1199092119896

1198871119896 11990912]minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minusV2119896( 119889(2)2)

119905

prod

119896=1

[1199092119896

119890(119896) 11990912]V10158402119896

Chinese Journal of Mathematics 3

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minusV2119896119889( 119889(2)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 11990912]V10158402119896

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minus1199081221198871119896

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

(21)

as for 119894 = 2 (14) becomes V2119896

(119889(2)

2) minus 119908122

1198871119896

equiv 0 (mod(119889(2)119890(119896))) Now (16) for 119894 = 2 becomes

minus 11990612

119889 (2)

119889 (1)120572(12)

119897minus 11990612

119889 (2)

119889 (1)1198881119897

minus sum

119896

V10158402119896119889119896119897

minus 2119908122

120572(12)

119897

minus 119908112

120572(11)

119897equiv 0 (mod (119889 (2) 119891 (119897)))

(22)

which implies that

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

=

119906

prod

119897=1

[1199093119897

11990912]minus11990612(119889(2)119889(1))120572

(12)

119897minus11990612(119889(2)119889(1))1198881119897minus2119908122120572

(12)

119897minus119908112120572

(11)

119897

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897

11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897 11990912]

minus2119908122

[

119906

prod

119897=1

1199093119897

120572(11)

119897

11990912]

minus119908112

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[11990911

119889(1)

11990912 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus2119908122

(23)

Now

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912]minus11990612119889(2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

= [11990911 11990912 11990912

119889(2)]minus11990612

sdot [11990912 11990911 11990912 11990912]minus11990612(119889(2)

2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

(24)

Using (6) and (7) we get that

11990612

119889 (2)

119889 (1)(

119889 (1)

2

) equiv 0 (mod 119889 (1)) (25)

minus11990612

(

119889 (2)

2

) equiv 0 (mod 119889 (1)) (26)

Hence (24) reduces to

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912

119889(2)]minus11990612

= 1 as [11990911 11990912 11990912

119889(2)] isin [120574

2(119866) 120574

2(119866)]

(27)

Now

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]minus2119908122119889(1)

sdot [11990912 11990911 11990911 11990912]2119908122(119889(1)

2)

= [11990911 11990912 11990912]minus2119908122119889(1)

(28)

Also (7) implies that

minus2119908122

119889 (1) = minus211990612

(

119889 (2)

2

) + 21199081015840

122119889 (2) (29)

Thus (28) becomes

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]21199081015840

122119889(2)

sdot [11990911 11990912 11990912]minus11990612119889(2)(119889(2)minus1)

= [11990911 11990912 11990912

119889(2)]21199081015840

122

sdot [11990912 11990911 11990912 11990912]minus21199081015840

122(119889(2)

2)

sdot [11990911 11990912 11990912]minus11990612(119889(2)(119889(2)minus1))

= [11990911 11990912 11990912

119889(2)]minus11990612(119889(2)minus1)

sdot [11990912 11990911 11990912 11990912]minus11990612((119889(2)minus1)(

119889(2)

2))

= 1

(30)

as by (7) minus11990612

(119889(2)

2) equiv 0 (mod 119889(1))

Using (23) (27) and (30) in (21) we get that

1 = [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

sdot [

119905

prod

119896=1

1199092119896

1198871119896 11990912 11990912]

minus119908122

= [11990911

119889(1) 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus119908122

= 119892minus1

(31)

Thus 119892 = 1 and hence 1198635(119866) = 1

4 Chinese Journal of Mathematics

Theorem 3 Let119866 be finite metabelian 2 group and1198661205742(119866) =

1198621oplus1198622oplus sdot sdot sdot oplus 119862

119899 where 119862

119894are cyclic groups of order 119889(119894) Let

119889(1) = 119889(2) = sdot sdot sdot = 119889(119899 minus 1) = 2 119889(119899) ge 4 Then 1198635(119866) =

1205745(119866)

Proof It is enough to prove the result for a group119866 of class 4Any element 119892 of 119863

5(119866) is of the form

119892 = prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894lt119895lt119896le119899

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895 1199091119895]119908119894119895119895

= 119860 sdot 119861 sdot 119862 say

(32)

Since for 1 le 119894 lt 119895 lt 119896 le 119899 119889(119894) = 119889(119895) = 2 (13) reduces to

119908119894119895119896

equiv 0 (mod 119889 (119895)) (33)

and hence 119861 = 1 Now

119860 = prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

as for 1 le 119894 lt 119895 le 119899 minus 1 119889 (119894) = 119889 (119895)

(34)

Also for 1 le 119894 lt 119895 le 119899 (7) implies that minus119906119894119895+2119908119894119895119895

+21199081015840

119894119895119895= 0

which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]2119908119894119895119895+2119908

1015840

119894119895119895

as 119889 (119895) = 2

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119889(119895)(119908119894119895119895+119908

1015840

119894119895119895)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895

119889(119895)](119908119894119895119895+119908

1015840

119894119895119895)

sdot [1199091119894

119889(119894) 1199091119895 1199091119895]minus(119908119894119895119895+119908

1015840

119894119895119895)(119889(119895)

2)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus1199081015840

119894119895119895

as [1199091119894

119889(119894) 1199091119895

119889(119895)] isin [120574

2(119866) 120574

2(119866)] 119889 (119895) = 2

(35)

Now for 1 le 119894 lt 119895 le 119899 (12) implies that 1199081015840119894119895119895

equiv 0 (mod 119889(119895))which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(36)

Thus (34) reduces to

119860 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(37)

Hence

119892 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]119908119894119899119899

(38)

Now for 119894 = 119899 (9) becomes

119878 = minus

119899minus1

sum

119895=1

119906119895119899

119889 (119899)

119889 (119895)119887119895119896

+ V119899119896119889 (119899) + V1015840

119899119896119890 (119896) = 0 (39)

which gives

1 =

119905

prod

119896=1

[1199092119896 1199091119899]119878

=

119905

prod

119896=1

[1199092119896 1199091119899]minussum119899minus1

119895=1119906119895119899(119889(119899)119889(119895))119887119895119896

sdot

119905

prod

119896=1

[1199092119896 1199091119899]V119899119896119889(119899)

119905

prod

119896=1

[1199092119896 1199091119899]V1015840119899119896119890(119896)

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899

119889(119899)]V119899119896

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 1199091119899]V1015840119899119896

Chinese Journal of Mathematics 5

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 1199091119899]V1015840119899119896

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

(40)

Equation (14) for 119894 = 119899 becomes

V119899119896

(

119889 (119899)

2

) equiv

119899minus1

sum

119895=1

119908119895119899119899

119887119895119896

(mod (119889 (119899) 119890 (119896))) (41)

Thus

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

=

119899minus1

prod

119894=1

[prod

119896

1199092119896

119887119895119896 1199091119899 1199091119899]

minus119908119894119899119899

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

(42)

Using (16) we get that

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

sdot

119899minus1

prod

119894=1

[prod

119897

1199093119897

119888119894119897 1199091119899]

minus119906119894119899(119889(119899)119889(119894))

(43)

Now

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

=

119899minus1

prod

119894=1

[1199091119894 1199091119899 1199091119899

119889(119899)]minus119906119894119899

sdot [1199091119899 1199091119894 1199091119894 1199091119899]119906119894119899(119889(119899)119889(119894))(

119889(119894)

2)

sdot [1199091119899 1199091119894 1199091119899 1199091119899]119906119894119899(119889(119899)

2)

= 1

as 2 = 119889 (119894) divides 119889 (119899)

119889 (119894)

(44)

Also it can be seen easily from (7) that

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

= 1 (45)

Now using (38) (40) (43) (44) and (45) we get that

1 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

= 119892minus1

(46)

Hence 119892 = 1

3 Groups of Class 3

It has been proved that if119866 is a finite 119901-group of class 3 119901 oddprime then119863

4(119866) = 1 Clearly for such a group119866119863

5(119866) =

1 Now the question arises that if119866 is 2 group of class 3 with1198634(119866) = 1 then is it possible that 119863

5(119866) = 1 We will

discuss some examples of 2 groups of class 3 inwhich1198634(119866) =

1 but 1198635(119866) = 1

31 Example 1 Let 119866 be a group generated by four elements119903 119886 119887 119888 Set 119909

0= 1199100= 1199110= 119903 and 119909

119894= [119909119894minus1

119886] 119910119894= [119910119894minus1

119887]and 119911

119894= [119911119894minus1

119888] 119894 = 1 2 3 Suppose that 119866 satisfies therelations

(i) 11990327

= 1 11988626

= 1199101

41199111

2 11988724

= 1199091

minus41199111 and 119888

22

=

1199091

minus21199101

minus1(ii) 1199112= 1199102

4 1199102= 1199092

4(iii) 119909

3= 1199103= 1199113= 1

(iv) [119886 119887 119892] = 1 [119887 119888 119892] = [119886 119888 119892] = 1 forall119892 isin 119866(v) [119909

119894 119887] = [119909

119894 119888] = [119910

119894 119886] = [119910

119894 119888] = [119911

119894 119886] = [119911

119894 119887] =

1 forall119894 ge 1(vi) [119909

119894 119909119895] = [119909

119894 119910119895] = [119909

119894 119911119895] = [119910

119894 119910119895] = [119910

119894 119911119895] =

[119911119894 119911119895] = 1 forall119894 119895 ge 0 It has been proved in [6] that 119866

is 2 group of class 3 and 1198634(119866) = 1 we will show

that 1198635(119866) = 1

From the above relations we conclude that

(i) [119886 119887]24

= 1199092

4 [119887 119888]22

= 1199102= 1199092

4 and [119886 119888]22

= 1199092

2(ii) 119900(119909

1) = 27 119900(1199092) = 26 119900(1199101) = 25 119900(1199102) = 24 119900(1199111) =

23 and 119900(119911

2) = 22 where 119900(119892) denotes the order of an

element 119892

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

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Stochastic AnalysisInternational Journal of

Chinese Journal of Mathematics 3

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minusV2119896119889( 119889(2)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 11990912]V10158402119896

= [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot

119905

prod

119896=1

[1199092119896 11990912 11990912]minus1199081221198871119896

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

(21)

as for 119894 = 2 (14) becomes V2119896

(119889(2)

2) minus 119908122

1198871119896

equiv 0 (mod(119889(2)119890(119896))) Now (16) for 119894 = 2 becomes

minus 11990612

119889 (2)

119889 (1)120572(12)

119897minus 11990612

119889 (2)

119889 (1)1198881119897

minus sum

119896

V10158402119896119889119896119897

minus 2119908122

120572(12)

119897

minus 119908112

120572(11)

119897equiv 0 (mod (119889 (2) 119891 (119897)))

(22)

which implies that

119906

prod

119897=1

[1199093119897 11990912]sum119896V10158402119896119889119896119897

=

119906

prod

119897=1

[1199093119897

11990912]minus11990612(119889(2)119889(1))120572

(12)

119897minus11990612(119889(2)119889(1))1198881119897minus2119908122120572

(12)

119897minus119908112120572

(11)

119897

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897

11990912]

minus11990612(119889(2)119889(1))

[

119906

prod

119897=1

1199093119897

120572(12)

119897 11990912]

minus2119908122

[

119906

prod

119897=1

1199093119897

120572(11)

119897

11990912]

minus119908112

= [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

[11990911

119889(1)

11990912 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus2119908122

(23)

Now

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912]minus11990612119889(2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

= [11990911 11990912 11990912

119889(2)]minus11990612

sdot [11990912 11990911 11990912 11990912]minus11990612(119889(2)

2)

sdot [11990912 11990911 11990911 11990912]11990612(119889(2)119889(1))(

119889(1)

2)

(24)

Using (6) and (7) we get that

11990612

119889 (2)

119889 (1)(

119889 (1)

2

) equiv 0 (mod 119889 (1)) (25)

minus11990612

(

119889 (2)

2

) equiv 0 (mod 119889 (1)) (26)

Hence (24) reduces to

[11990911

119889(1) 11990912 11990912]minus11990612(119889(2)119889(1))

= [11990911 11990912 11990912

119889(2)]minus11990612

= 1 as [11990911 11990912 11990912

119889(2)] isin [120574

2(119866) 120574

2(119866)]

(27)

Now

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]minus2119908122119889(1)

sdot [11990912 11990911 11990911 11990912]2119908122(119889(1)

2)

= [11990911 11990912 11990912]minus2119908122119889(1)

(28)

Also (7) implies that

minus2119908122

119889 (1) = minus211990612

(

119889 (2)

2

) + 21199081015840

122119889 (2) (29)

Thus (28) becomes

[11990911

119889(1) 11990912 11990912]minus2119908122

= [11990911 11990912 11990912]21199081015840

122119889(2)

sdot [11990911 11990912 11990912]minus11990612119889(2)(119889(2)minus1)

= [11990911 11990912 11990912

119889(2)]21199081015840

122

sdot [11990912 11990911 11990912 11990912]minus21199081015840

122(119889(2)

2)

sdot [11990911 11990912 11990912]minus11990612(119889(2)(119889(2)minus1))

= [11990911 11990912 11990912

119889(2)]minus11990612(119889(2)minus1)

sdot [11990912 11990911 11990912 11990912]minus11990612((119889(2)minus1)(

119889(2)

2))

= 1

(30)

as by (7) minus11990612

(119889(2)

2) equiv 0 (mod 119889(1))

Using (23) (27) and (30) in (21) we get that

1 = [

119905

prod

119896=1

1199092119896

1198871119896 11990912]

minus11990612(119889(2)119889(1))

sdot [

119906

prod

119897=1

1199093119897

1198881119897 11990912]

minus11990612(119889(2)119889(1))

sdot [

119905

prod

119896=1

1199092119896

1198871119896 11990912 11990912]

minus119908122

= [11990911

119889(1) 11990912]minus11990612(119889(2)119889(1))

[11990911

119889(1) 11990912 11990912]minus119908122

= 119892minus1

(31)

Thus 119892 = 1 and hence 1198635(119866) = 1

4 Chinese Journal of Mathematics

Theorem 3 Let119866 be finite metabelian 2 group and1198661205742(119866) =

1198621oplus1198622oplus sdot sdot sdot oplus 119862

119899 where 119862

119894are cyclic groups of order 119889(119894) Let

119889(1) = 119889(2) = sdot sdot sdot = 119889(119899 minus 1) = 2 119889(119899) ge 4 Then 1198635(119866) =

1205745(119866)

Proof It is enough to prove the result for a group119866 of class 4Any element 119892 of 119863

5(119866) is of the form

119892 = prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894lt119895lt119896le119899

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895 1199091119895]119908119894119895119895

= 119860 sdot 119861 sdot 119862 say

(32)

Since for 1 le 119894 lt 119895 lt 119896 le 119899 119889(119894) = 119889(119895) = 2 (13) reduces to

119908119894119895119896

equiv 0 (mod 119889 (119895)) (33)

and hence 119861 = 1 Now

119860 = prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

as for 1 le 119894 lt 119895 le 119899 minus 1 119889 (119894) = 119889 (119895)

(34)

Also for 1 le 119894 lt 119895 le 119899 (7) implies that minus119906119894119895+2119908119894119895119895

+21199081015840

119894119895119895= 0

which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]2119908119894119895119895+2119908

1015840

119894119895119895

as 119889 (119895) = 2

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119889(119895)(119908119894119895119895+119908

1015840

119894119895119895)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895

119889(119895)](119908119894119895119895+119908

1015840

119894119895119895)

sdot [1199091119894

119889(119894) 1199091119895 1199091119895]minus(119908119894119895119895+119908

1015840

119894119895119895)(119889(119895)

2)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus1199081015840

119894119895119895

as [1199091119894

119889(119894) 1199091119895

119889(119895)] isin [120574

2(119866) 120574

2(119866)] 119889 (119895) = 2

(35)

Now for 1 le 119894 lt 119895 le 119899 (12) implies that 1199081015840119894119895119895

equiv 0 (mod 119889(119895))which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(36)

Thus (34) reduces to

119860 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(37)

Hence

119892 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]119908119894119899119899

(38)

Now for 119894 = 119899 (9) becomes

119878 = minus

119899minus1

sum

119895=1

119906119895119899

119889 (119899)

119889 (119895)119887119895119896

+ V119899119896119889 (119899) + V1015840

119899119896119890 (119896) = 0 (39)

which gives

1 =

119905

prod

119896=1

[1199092119896 1199091119899]119878

=

119905

prod

119896=1

[1199092119896 1199091119899]minussum119899minus1

119895=1119906119895119899(119889(119899)119889(119895))119887119895119896

sdot

119905

prod

119896=1

[1199092119896 1199091119899]V119899119896119889(119899)

119905

prod

119896=1

[1199092119896 1199091119899]V1015840119899119896119890(119896)

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899

119889(119899)]V119899119896

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 1199091119899]V1015840119899119896

Chinese Journal of Mathematics 5

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 1199091119899]V1015840119899119896

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

(40)

Equation (14) for 119894 = 119899 becomes

V119899119896

(

119889 (119899)

2

) equiv

119899minus1

sum

119895=1

119908119895119899119899

119887119895119896

(mod (119889 (119899) 119890 (119896))) (41)

Thus

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

=

119899minus1

prod

119894=1

[prod

119896

1199092119896

119887119895119896 1199091119899 1199091119899]

minus119908119894119899119899

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

(42)

Using (16) we get that

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

sdot

119899minus1

prod

119894=1

[prod

119897

1199093119897

119888119894119897 1199091119899]

minus119906119894119899(119889(119899)119889(119894))

(43)

Now

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

=

119899minus1

prod

119894=1

[1199091119894 1199091119899 1199091119899

119889(119899)]minus119906119894119899

sdot [1199091119899 1199091119894 1199091119894 1199091119899]119906119894119899(119889(119899)119889(119894))(

119889(119894)

2)

sdot [1199091119899 1199091119894 1199091119899 1199091119899]119906119894119899(119889(119899)

2)

= 1

as 2 = 119889 (119894) divides 119889 (119899)

119889 (119894)

(44)

Also it can be seen easily from (7) that

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

= 1 (45)

Now using (38) (40) (43) (44) and (45) we get that

1 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

= 119892minus1

(46)

Hence 119892 = 1

3 Groups of Class 3

It has been proved that if119866 is a finite 119901-group of class 3 119901 oddprime then119863

4(119866) = 1 Clearly for such a group119866119863

5(119866) =

1 Now the question arises that if119866 is 2 group of class 3 with1198634(119866) = 1 then is it possible that 119863

5(119866) = 1 We will

discuss some examples of 2 groups of class 3 inwhich1198634(119866) =

1 but 1198635(119866) = 1

31 Example 1 Let 119866 be a group generated by four elements119903 119886 119887 119888 Set 119909

0= 1199100= 1199110= 119903 and 119909

119894= [119909119894minus1

119886] 119910119894= [119910119894minus1

119887]and 119911

119894= [119911119894minus1

119888] 119894 = 1 2 3 Suppose that 119866 satisfies therelations

(i) 11990327

= 1 11988626

= 1199101

41199111

2 11988724

= 1199091

minus41199111 and 119888

22

=

1199091

minus21199101

minus1(ii) 1199112= 1199102

4 1199102= 1199092

4(iii) 119909

3= 1199103= 1199113= 1

(iv) [119886 119887 119892] = 1 [119887 119888 119892] = [119886 119888 119892] = 1 forall119892 isin 119866(v) [119909

119894 119887] = [119909

119894 119888] = [119910

119894 119886] = [119910

119894 119888] = [119911

119894 119886] = [119911

119894 119887] =

1 forall119894 ge 1(vi) [119909

119894 119909119895] = [119909

119894 119910119895] = [119909

119894 119911119895] = [119910

119894 119910119895] = [119910

119894 119911119895] =

[119911119894 119911119895] = 1 forall119894 119895 ge 0 It has been proved in [6] that 119866

is 2 group of class 3 and 1198634(119866) = 1 we will show

that 1198635(119866) = 1

From the above relations we conclude that

(i) [119886 119887]24

= 1199092

4 [119887 119888]22

= 1199102= 1199092

4 and [119886 119888]22

= 1199092

2(ii) 119900(119909

1) = 27 119900(1199092) = 26 119900(1199101) = 25 119900(1199102) = 24 119900(1199111) =

23 and 119900(119911

2) = 22 where 119900(119892) denotes the order of an

element 119892

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Chinese Journal of Mathematics

Theorem 3 Let119866 be finite metabelian 2 group and1198661205742(119866) =

1198621oplus1198622oplus sdot sdot sdot oplus 119862

119899 where 119862

119894are cyclic groups of order 119889(119894) Let

119889(1) = 119889(2) = sdot sdot sdot = 119889(119899 minus 1) = 2 119889(119899) ge 4 Then 1198635(119866) =

1205745(119866)

Proof It is enough to prove the result for a group119866 of class 4Any element 119892 of 119863

5(119866) is of the form

119892 = prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

prod

1le119894lt119895lt119896le119899

[1199091119894

119889(119894) 1199091119895 1199091119896]119908119894119895119896

prod

1le119894lt119895le119899

[1199091119894

119889(119894) 1199091119895 1199091119895]119908119894119895119895

= 119860 sdot 119861 sdot 119862 say

(32)

Since for 1 le 119894 lt 119895 lt 119896 le 119899 119889(119894) = 119889(119895) = 2 (13) reduces to

119908119894119895119896

equiv 0 (mod 119889 (119895)) (33)

and hence 119861 = 1 Now

119860 = prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895(119889(119895)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

as for 1 le 119894 lt 119895 le 119899 minus 1 119889 (119894) = 119889 (119895)

(34)

Also for 1 le 119894 lt 119895 le 119899 (7) implies that minus119906119894119895+2119908119894119895119895

+21199081015840

119894119895119895= 0

which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]2119908119894119895119895+2119908

1015840

119894119895119895

as 119889 (119895) = 2

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119889(119895)(119908119894119895119895+119908

1015840

119894119895119895)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895

119889(119895)](119908119894119895119895+119908

1015840

119894119895119895)

sdot [1199091119894

119889(119894) 1199091119895 1199091119895]minus(119908119894119895119895+119908

1015840

119894119895119895)(119889(119895)

2)

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus1199081015840

119894119895119895

as [1199091119894

119889(119894) 1199091119895

119889(119895)] isin [120574

2(119866) 120574

2(119866)] 119889 (119895) = 2

(35)

Now for 1 le 119894 lt 119895 le 119899 (12) implies that 1199081015840119894119895119895

equiv 0 (mod 119889(119895))which gives that

prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895]119906119894119895

= prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(36)

Thus (34) reduces to

119860 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot prod

1le119894lt119895le119899minus1

[1199091119894

119889(119894) 1199091119895 1199091119895]minus119908119894119895119895

(37)

Hence

119892 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899]119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]119908119894119899119899

(38)

Now for 119894 = 119899 (9) becomes

119878 = minus

119899minus1

sum

119895=1

119906119895119899

119889 (119899)

119889 (119895)119887119895119896

+ V119899119896119889 (119899) + V1015840

119899119896119890 (119896) = 0 (39)

which gives

1 =

119905

prod

119896=1

[1199092119896 1199091119899]119878

=

119905

prod

119896=1

[1199092119896 1199091119899]minussum119899minus1

119895=1119906119895119899(119889(119899)119889(119895))119887119895119896

sdot

119905

prod

119896=1

[1199092119896 1199091119899]V119899119896119889(119899)

119905

prod

119896=1

[1199092119896 1199091119899]V1015840119899119896119890(119896)

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899

119889(119899)]V119899119896

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

sdot

119905

prod

119896=1

[1199092119896

119890(119896) 1199091119899]V1015840119899119896

Chinese Journal of Mathematics 5

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 1199091119899]V1015840119899119896

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

(40)

Equation (14) for 119894 = 119899 becomes

V119899119896

(

119889 (119899)

2

) equiv

119899minus1

sum

119895=1

119908119895119899119899

119887119895119896

(mod (119889 (119899) 119890 (119896))) (41)

Thus

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

=

119899minus1

prod

119894=1

[prod

119896

1199092119896

119887119895119896 1199091119899 1199091119899]

minus119908119894119899119899

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

(42)

Using (16) we get that

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

sdot

119899minus1

prod

119894=1

[prod

119897

1199093119897

119888119894119897 1199091119899]

minus119906119894119899(119889(119899)119889(119894))

(43)

Now

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

=

119899minus1

prod

119894=1

[1199091119894 1199091119899 1199091119899

119889(119899)]minus119906119894119899

sdot [1199091119899 1199091119894 1199091119894 1199091119899]119906119894119899(119889(119899)119889(119894))(

119889(119894)

2)

sdot [1199091119899 1199091119894 1199091119899 1199091119899]119906119894119899(119889(119899)

2)

= 1

as 2 = 119889 (119894) divides 119889 (119899)

119889 (119894)

(44)

Also it can be seen easily from (7) that

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

= 1 (45)

Now using (38) (40) (43) (44) and (45) we get that

1 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

= 119892minus1

(46)

Hence 119892 = 1

3 Groups of Class 3

It has been proved that if119866 is a finite 119901-group of class 3 119901 oddprime then119863

4(119866) = 1 Clearly for such a group119866119863

5(119866) =

1 Now the question arises that if119866 is 2 group of class 3 with1198634(119866) = 1 then is it possible that 119863

5(119866) = 1 We will

discuss some examples of 2 groups of class 3 inwhich1198634(119866) =

1 but 1198635(119866) = 1

31 Example 1 Let 119866 be a group generated by four elements119903 119886 119887 119888 Set 119909

0= 1199100= 1199110= 119903 and 119909

119894= [119909119894minus1

119886] 119910119894= [119910119894minus1

119887]and 119911

119894= [119911119894minus1

119888] 119894 = 1 2 3 Suppose that 119866 satisfies therelations

(i) 11990327

= 1 11988626

= 1199101

41199111

2 11988724

= 1199091

minus41199111 and 119888

22

=

1199091

minus21199101

minus1(ii) 1199112= 1199102

4 1199102= 1199092

4(iii) 119909

3= 1199103= 1199113= 1

(iv) [119886 119887 119892] = 1 [119887 119888 119892] = [119886 119888 119892] = 1 forall119892 isin 119866(v) [119909

119894 119887] = [119909

119894 119888] = [119910

119894 119886] = [119910

119894 119888] = [119911

119894 119886] = [119911

119894 119887] =

1 forall119894 ge 1(vi) [119909

119894 119909119895] = [119909

119894 119910119895] = [119909

119894 119911119895] = [119910

119894 119910119895] = [119910

119894 119911119895] =

[119911119894 119911119895] = 1 forall119894 119895 ge 0 It has been proved in [6] that 119866

is 2 group of class 3 and 1198634(119866) = 1 we will show

that 1198635(119866) = 1

From the above relations we conclude that

(i) [119886 119887]24

= 1199092

4 [119887 119888]22

= 1199102= 1199092

4 and [119886 119888]22

= 1199092

2(ii) 119900(119909

1) = 27 119900(1199092) = 26 119900(1199101) = 25 119900(1199102) = 24 119900(1199111) =

23 and 119900(119911

2) = 22 where 119900(119892) denotes the order of an

element 119892

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Chinese Journal of Mathematics 5

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119905

prod

119896=1

119906

prod

119897=1

[1199093119897

119889119896119897 1199091119899]V1015840119899119896

=

119905

prod

119896=1

[

[

119899minus1

prod

119895=1

1199092119896

119887119895119896 1199091119899]

]

minus119906119895119899(119889(119899)119889(119895))

sdot

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

(40)

Equation (14) for 119894 = 119899 becomes

V119899119896

(

119889 (119899)

2

) equiv

119899minus1

sum

119895=1

119908119895119899119899

119887119895119896

(mod (119889 (119899) 119890 (119896))) (41)

Thus

119905

prod

119896=1

[1199092119896 1199091119899 1199091119899]minusV119899119896( 119889(119899)

2)

=

119899minus1

prod

119894=1

[prod

119896

1199092119896

119887119895119896 1199091119899 1199091119899]

minus119908119894119899119899

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

(42)

Using (16) we get that

119906

prod

119897=1

[1199093119897 1199091119899]sum119896V1015840119899119896119889119896119897

=

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

sdot

119899minus1

prod

119894=1

[prod

119897

1199093119897

119888119894119897 1199091119899]

minus119906119894119899(119889(119899)119889(119894))

(43)

Now

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

=

119899minus1

prod

119894=1

[1199091119894 1199091119899 1199091119899

119889(119899)]minus119906119894119899

sdot [1199091119899 1199091119894 1199091119894 1199091119899]119906119894119899(119889(119899)119889(119894))(

119889(119894)

2)

sdot [1199091119899 1199091119894 1199091119899 1199091119899]119906119894119899(119889(119899)

2)

= 1

as 2 = 119889 (119894) divides 119889 (119899)

119889 (119894)

(44)

Also it can be seen easily from (7) that

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus2119908119894119899119899

= 1 (45)

Now using (38) (40) (43) (44) and (45) we get that

1 =

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119906119894119899(119889(119899)119889(119894))

sdot

119899minus1

prod

119894=1

[1199091119894

119889(119894) 1199091119899 1199091119899]minus119908119894119899119899

= 119892minus1

(46)

Hence 119892 = 1

3 Groups of Class 3

It has been proved that if119866 is a finite 119901-group of class 3 119901 oddprime then119863

4(119866) = 1 Clearly for such a group119866119863

5(119866) =

1 Now the question arises that if119866 is 2 group of class 3 with1198634(119866) = 1 then is it possible that 119863

5(119866) = 1 We will

discuss some examples of 2 groups of class 3 inwhich1198634(119866) =

1 but 1198635(119866) = 1

31 Example 1 Let 119866 be a group generated by four elements119903 119886 119887 119888 Set 119909

0= 1199100= 1199110= 119903 and 119909

119894= [119909119894minus1

119886] 119910119894= [119910119894minus1

119887]and 119911

119894= [119911119894minus1

119888] 119894 = 1 2 3 Suppose that 119866 satisfies therelations

(i) 11990327

= 1 11988626

= 1199101

41199111

2 11988724

= 1199091

minus41199111 and 119888

22

=

1199091

minus21199101

minus1(ii) 1199112= 1199102

4 1199102= 1199092

4(iii) 119909

3= 1199103= 1199113= 1

(iv) [119886 119887 119892] = 1 [119887 119888 119892] = [119886 119888 119892] = 1 forall119892 isin 119866(v) [119909

119894 119887] = [119909

119894 119888] = [119910

119894 119886] = [119910

119894 119888] = [119911

119894 119886] = [119911

119894 119887] =

1 forall119894 ge 1(vi) [119909

119894 119909119895] = [119909

119894 119910119895] = [119909

119894 119911119895] = [119910

119894 119910119895] = [119910

119894 119911119895] =

[119911119894 119911119895] = 1 forall119894 119895 ge 0 It has been proved in [6] that 119866

is 2 group of class 3 and 1198634(119866) = 1 we will show

that 1198635(119866) = 1

From the above relations we conclude that

(i) [119886 119887]24

= 1199092

4 [119887 119888]22

= 1199102= 1199092

4 and [119886 119888]22

= 1199092

2(ii) 119900(119909

1) = 27 119900(1199092) = 26 119900(1199101) = 25 119900(1199102) = 24 119900(1199111) =

23 and 119900(119911

2) = 22 where 119900(119892) denotes the order of an

element 119892

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Chinese Journal of Mathematics

Consider

11990911

= 119888

11990912

= 119887

11990913

= 119886

11990914

= 119903

11990921

= [119886 119888]

11990922

= [119887 119888]

11990923

= [119903 119888]

11990924

= [119886 119887]

11990925

= [119903 119887]

11990926

= [119903 119886]

11990931

= [119903 119886 119886] = 1199092

(47)

It is easy to see that 119889(1) = 22 119889(2) = 2

4 119889(3) = 26 119889(4) = 2

7119890(1) = 2

2 119890(2) = 22 119890(3) = 2

2 119890(4) = 24 119890(5) = 2

4 119890(6) = 26

and 119891(1) = 26 Also with the help of (1) and (2) we get that

11988715

= minus1 11988716

= minus2 11988723

= 1 11988726

= minus4 11988733

= 2 and 11988735

= 4Since 119866 is a group of class 3 so an arbitrary element of

1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [119888 119887]1611990612

[119888 119886]6411990613

sdot [119888 119903]12811990614

[119887 119886]6411990623

[119887 119903]12811990624

[119886 119903]12811990634

= 1199092

minus1611990612minus3211990613minus1611990623

(48)

Equation (9) for 119894 = 1 and 119896 = 6 becomes

0 = 1199061211988726

+ 1199061311988736

+ 1199061411988746

+ V16119889 (1) + V1015840

16119890 (6)

= minus411990612

+ 4V16

+ 64V101584016

(49)

which gives that

minus11990612

+ V16

equiv 0 (mod 4) (50)

Also (17) for 119896 = 6 gives that

V16

+ 2V26

equiv 0 (mod 16) (51)

For 119896 = 3 and 119897 = 6 (18) becomes

V26

equiv 0 (mod 2) (52)

Now (50) (51) and (52) imply that

11990612

equiv 0 (mod 4) (53)

Again for 119894 = 3 and 119896 = 6 (9) becomes

3211990613

+ 1611990623

equiv 0 (mod 64) (54)

Since order of 1199092is 64 combining (53) and (54) we get that

119892 = 1

32 Example 2 Let 119866 be a group generated by four elements1199091 1199092 1199093 1199094 Set 119910

1= [1199092 1199093] 1199102

= [1199091 1199092] 1199103

= [1199091 1199093]

and 1199104

= [1199091 1199094] Suppose that 119866 satisfies the following

relations[1199092 1199102] = 1199101

4

[1199093 1199103] = 1199101

minus16

[1199094 1199104] = 1199101

64

[1199091 119910119902] = 1 1 le 119902 le 4

[1199092 119910119902] = 1 1 le 119902 le 4 119902 = 2

[1199093 119910119902] = 1 1 le 119902 le 4 119902 = 3

[1199094 119910119902] = 1 1 le 119902 le 4 119902 = 4

1199091

2119896+119897+4

= 1199101

minus2119896+119897+4

1199102

2119896+119897+3

1199092

2119896+4

= 1199103

2119896

1199104

minus2119896minus1

1199101

32119896+3

1199093

2119896+2

= 1199101

minus52119896+1

1199102

2119896

1199104

2119896minus2

1199094

2119896

= 1199101

2119896

1199102

2119896minus1

1199103

2119896minus2

(55)

Consider11990911

= 1199094

11990912

= 1199093

11990913

= 1199092

11990914

= 1199091

11990921

= 1199101

11990922

= 1199104

11990923

= 1199103

11990924

= 11991012

11990931

= [1199092 1199102] = 1199101

4

(56)

It is easy to see that 119889(1) = 2119896 119889(2) = 2

119896+2 119889(3) = 2119896+4

119889(4) = 2119896+119897+4 119890(1) = 4 119890(2) = 2

119896 119890(3) = 2119896+2 119890(4) = 2

119896+4and119891(1) = 2

119896+2 Also we have 11988711

= 2119896 11988713

= 2119896minus2 11988714

= 2119896minus1

11988721

= minus52119896+1 11988722

= 2119896minus2 11988724

= 2119896 11988731

= 32119896+3 11988732

= minus2119896minus1

11988733

= 2119896 11988741

= minus2119896+119897+4 and 119887

44= 2119896+119897+3 It has been proved in

[7] that 119866 is a group of class 3 and 1198634(119866) = 1 We will prove

that 1198635(119866) = 1

An arbitrary element 119892 of 1198635(119866) can be written as

119892 = prod

1le119894lt119895le4

[1199091119894 1199091119895]119906119894119895119889(119895)

= [1199094 1199093]119906122119896+2

sdot [1199094 1199092]119906132119896+4

[1199094 1199091]119906142119896+119897+4

[1199093 1199092]119906232119896+4

sdot [1199093 1199091]119906242119896+119897+4

[1199092 1199091]119906342119896+119897+4

= 1199101

2119896+4(11990612minus211990613minus11990623+2

119897+111990634)

(57)

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Chinese Journal of Mathematics 7

To show that 119892 = 1 it is enough to show that 11990612

minus211990613

minus11990623

+

2119897+1

11990634

equiv 0 (mod 4) since order of 1199101is 2119896+6

Now for 119894 = 3 and 119896 = 4 (9) becomes

0 = 1199063411988744

minus 11990613

119889 (3)

119889 (1)11988714

minus 11990623

119889 (3)

119889 (2)11988724

+ V34119889 (3)

+ V101584034119890 (4)

= 2119896+2

(2119897+1

11990634

minus 211990613

minus 11990623

+ 4V34

+ 4V101584034)

(58)

which implies that

2119897+1

11990634

minus 211990613

minus 11990623

equiv 0 (mod 4) (59)

Also (9) for 119894 = 1 and 119896 = 2 gives that

0 = 1199061211988722

+ 1199061311988732

+ 1199061411988742

+ V12119889 (1) + V1015840

12119890 (2)

= 2119896minus2

11990612

minus 2119896minus1

11990613

+ 2119896V12

+ 2119896V101584012

(60)

That is we have

11990612

= 211990613

minus 4V12

minus 4V101584012 (61)

Again (9) for 119894 = 1 119896 = 3 gives that

11990613

equiv minusV13

(mod 4) (62)

Now (17) for 119896 = 3 gives that

V13

equiv 0 (mod 4) (63)

Combining (61) (62) and (63) we get that

11990613

equiv 0 (mod 4) (64)

which together with (59) gives that 119892 = 1

Competing Interests

The author declares no competing interests

References

[1] I B S Passi Group Rings and Their Augmentation Ideals vol715 of Lecture Notes in Mathematics Springer Berlin Germany1979

[2] I B Passi ldquoDimension subgroupsrdquo Journal of Algebra vol 9 pp152ndash182 1968

[3] E Rips ldquoOn the fourth integer dimension subgrouprdquo Israel Jour-nal of Mathematics vol 12 pp 342ndash346 1972

[4] K-I Tahara ldquoThe augmentation quotients of group rings andthe fifth dimension subgroupsrdquo Journal of Algebra vol 71 no 1pp 141ndash173 1981

[5] I B Passi andRMikhailov Lower Central andDimension Seriesof Groups vol 1952 of Lecture Notes in Mathematics SpringerBerlin Germany 2009

[6] N Gupta ldquoThe dimension subgroup conjecturerdquo The Bulletinof the London Mathematical Society vol 22 no 5 pp 453ndash4561990

[7] K-I Tahara ldquoThe fourth dimension subgroups and polynomialmaps IIrdquo Nagoya Mathematical Journal vol 69 pp 1ndash7 1978

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of