105 - J Group Theory 6(2003 223-228 Journal of Group Theory...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: J. Group Theory 6 (2003), 223-228 Journal of Group Theory cg de Gruyter 2003 The unit group of 1 + A(G)A(A) is torsion-free Zbigniew Marciniak and Sudarshan K. Sehgal* (Communicated by I. B. S. Passi) Abstract. Let A be an abelian normal subgroup of a finite group G. Then all units of the inte- gral group ring 7lG which are of the form 1+ J with 0 ""J E d( G)d(A), where d denotes the augmentation ideal, are of infinite order. 1 Introduction Let TLG be the integral group ring of a group G. Graham Higman asked in 1940, in [4], the following question. The IsomorphismProblem. Suppose that the rings TLG and TLH are isomorphic. Does it follow that the groups G andH are isomorphic? Today we know that for some classes of groups the answer is positive; see [9, p. 207] for an account. On the other hand, Hertweck [3] has recently constructed two (large) non-isomorphic finite groups with isomorphic integralgroup rings. However, the situation is still far from being completely understood. Higman himself positively solved this problem for finite abelian groups A, by identifying + A with the set of torsion elements in the group O//(7LA) of units of 7LA. He proved that O//(7LA) = :t:A x all(1 + .1(A)2), where .1(A) is the kernel of the augmentation homomorphism e: TLA-4 7L defined by e(~ ngg) = ~ ng. Higman proved that the subring 1 + .1(A)2 has no torsion units, hence the subgroup A < O//(7LA) has a torsion-free normal complement. Higman's description of units in 7lA was generalized in [1] for groups G with a normal abelian subgroup A such that G / A is abelian of exponent 2, 3, 4 or 6. This time O//(7LG) = +G. all(1 + .1(G).1(A)), *This research was supported by NSERC Canada and Polish KBN Grant No. 2PO3AO0218.-- ~~- 224 Zbigniew Marciniak and Sudarshan K. Sehgal where ~(G)~(A) is the two-sided ideal in 7lG generated by all elements of the form (g- I)(a - I) with g E G, a E A. Again, the subring I + ~(G)~(A) has no torsion units. It is easy to show that in general, if a finite group G has a torsion-free normal complement in the unit group rJ/i(71G) and 7lG ~ 7lH then G ~ H. Hence it was asked by Dennis [2]whether the embedding G --* rJ/i(7LG) always splits and the nor- mal complement arising as the kernel of the splitting homomorphism is torsion- free. Roggenkamp and Scott [7] constructed counter-examples to this even for meta- belian groups. On the other hand, Cliff, Sehgal and Weiss [I] gave an affirmative answer for metabelian groups A <J G--* G / A with I G / A I odd. Again, normal com- plements arise from ideals related to ~(G)~(A). Moreover, they proved that the group rJ/i(1+ il(G)il(A)) is torsion-free for...
View Full Document

This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell.

Page1 / 6

105 - J Group Theory 6(2003 223-228 Journal of Group Theory...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online