Research Article | Open Access
The Fifth Dimension Subgroup for Metabelian 2 Groups
Given a finite metabelian -group , the object of this paper is to discuss some cases under which . Further, some examples of groups of class , for which but , are discussed.
Let be the integral group ring of group and let be the augmentation ideal of . For , let be th dimension subgroup and let be th term in lower central series of group . The famous dimension subgroup conjecture states that for all and for all groups . This conjecture has been proved for (see, e.g., ). It has been proved in  that if is a finite -group, odd prime, then . Keeping in view this result, Rips  gave an example of -group for which . Further Tahara  gave the structure of and proved that the exponent of is divisible by . It has been shown in  that, for a metabelian group, the exponent of is divisible by , and hence for a metabelian -group odd prime . Gupta  constructed a -group , generated by four elements, such that . Now, the question arises that whether for a metabelian -group , with at most three generators, or not.
In this paper, we prove that if is metabelian group generated by at most two elements, then (Theorem 2). Further we prove that under certain conditions, a metabelian -group generated by elements satisfies the dimension subgroup conjecture for (Theorem 3). Finally, in Section 3, we give some examples of groups of class , for which .
2. Main Results
We first recall the structure of given by Tahara . Let be a finite group of class 4. Consider the lower central series of . Let be basis of , with as the order of : that is, belongs to . Similarly, let be basis of , with as the order of , and let be basis of , with as the order of . Moreover, these basis elements are chosen in such a way that divides , divides , and divides . Thus, we haveThe following theorem gives the structure of .
Theorem 1 (see ). With the above notations, is equal to the subgroup generated by the elements with the following conditions:
Theorem 2. Let be metabelian group with as sum of at most two cyclic groups. Then .
Proof. It is obvious that if metabelian groups are such that is a cyclic group then .
Let be metabelian group and let be sum of two cyclic groups. It is enough to prove the result for a group of class 4. Let , where is a cyclic group generated by with order of and divides . An arbitrary element of is of the form subject to the conditions given in Theorem 1. For and , (9) becomes Thus, Now belongs to and belongs to ; thus we haveas for (14) becomes . Now, (16), for , becomes which implies thatNow,Using (6) and (7), we get thatHence, (24) reduces toNow,Also (7) implies that Thus, (28) becomesas by (7), .
Using (23), (27), and (30) in (21), we get thatThus, , and hence .
Theorem 3. Let be finite metabelian 2 group and , where are cyclic groups of order . Let , . Then .
Proof. It is enough to prove the result for a group of class 4. Any element of is of the form Since for , , (13) reduces to and hence . Now,Also for , (7) implies that , which gives thatNow for , (12) implies that , which gives that Thus, (34) reduces toHence,Now, for , (9) becomes which givesEquation (14) for becomes Thus,Using (16), we get thatNowAlso it can be seen easily from (7) that Now using (38), (40), (43), (44), and (45), we get that Hence, .
3. Groups of Class 3
It has been proved that if is a finite -group of class , odd prime, then . Clearly for such a group . Now the question arises that if is group of class 3 with , then is it possible that ? We will discuss some examples of groups of class 3 in which , but .
3.1. Example 1
Let be a group generated by four elements . Set and , , and . Suppose that satisfies the relations(i), , , and ;(ii), ;(iii);(iv), ;(v);(vi). It has been proved in  that is group of class , and . we will show that .
From the above relations we conclude that(i), , and ;(ii), , , , , and , where denotes the order of an element .ConsiderIt is easy to see that , , , , , , , , , , and . Also with the help of (1) and (2), we get that , , , , , and .
Since is a group of class , so an arbitrary element of can be written asEquation (9) for and becomes which gives thatAlso (17), for , gives thatFor and , (18) becomes Now, (50), (51), and (52) imply thatAgain, for and , (9) becomesSince order of is , combining (53) and (54), we get that .
3.2. Example 2
Let be a group generated by four elements Set , , , and . Suppose that satisfies the following relations:ConsiderIt is easy to see that , , , , , , , , and . Also we have , , , , , , , , , , and . It has been proved in  that is a group of class and . We will prove that .
An arbitrary element of can be written as To show that , it is enough to show that , since order of is .
Now, for and , (9) becomes which implies that
Also (9) for and gives that