GROUP RINGS WITH FC-NILPOTENT UNIT GROUPS

GROUP RINGS WITH FC-NILPOTENT UNIT GROUPS

Let R be a commutative ring with identity, G be a group.U(RG) the group of units of the group ring RG and A(G) the FC-subgroup of G. Define,

Ak+i(G)/Ak(G) = A(G/Ak(G)) for k > 1 and AI(G) = A(G).
A group G is said to be FC-nilpotent if An (G) = G for some n.
In the present note, we determine groups G such that U(RG) is FC-nilpotent, where R is either the ring of integers Z or a field K of characteristic zero.
The main results are Theorem 1.If U(ZG) is FC-nilpotent, then G is FC-nilpotent and T(G) the set of torsion elements of G, is an abelian or a Hamiltonian ,2-group with every subgroup of T(G) normal in G.
Conversely if G is FC-nilpotent with T(G) satisfying the abone conditions and G/T(G) is right ordered, then U(ZG) is FC-nilpotent .Theorem 2. Let G be a finitely generated group and K a field of characteristic zero.If U(KG) is FC-nilpotent, then G is FC-nilpotent and T(G) is an abelian subgroup of G with every idempotent of KT(G) central in KG.
Conversely if G satisfies the abone conditions and G/T(G) is right ordered, then U(KG) is FC-nilpotent .
Proof of Theorem 1: Let U(ZG) be FC-nilpotent, t E T(G) and x E G. Then II = (x, t) is finitely generated FC-nilpotent and so H is nilpotent by finite [1].Also U(ZH) cannot have free noncyclic subgroups because it is FC-nilpotent .Thus, by (3), (t) is a normal subgroup of H. Hence T(G) is either an abelian or a Hamiltonian group with every subgroup of T(G) normal in G.
If T(G) is nonabelian and has an element of odd order, then again by [3] Also, if g E S(G) and y = 3x Proof of Theorem 2: If U(KG) is FC-nilpotent, then U(ZG) is also FCnilpotent .Thus by Theorem 1, G is FC-nilpotent and T(G) is a subgroup which is either abelian or a Hamiltonian 2-group with every subgroup of T(G) normal in G .
If T(G) is non abelian, then Ks C T(G) and thus

QKs=Q®Q®Q®Q®S,
where S is a Quaternion algebra over rationals.By [2], U(S) has a free non cyclic subgroup and so U(QK8) is not FC-nilpotent .Thus T(G) is abelian with every subgroup of T(G) normal in G. Now as G is finitely generated FC-nilpotent, it satisfies maximal condition on subgroups and thus T(G) is finite abelian.By [5,VI.3.12]every idempotent of KT(G) is central in KG as GL,(K), n > 1 has a free non cyclic subgroup .
Conversely, if G is finitely generated FC-nilpotent, then T(G) is finite.Since T(G) is finite abelian therefore KT(G) = ® j :á I Fi, a direct sum of fields.Hence to prove that U(Fi * G/T(G)) is FC-nilpotent, it is sufficient to prove that U(Fi) C 0((U(Fi * G/T(G))) .
Let a E U(Fi) and Qx E U(F, * G/T(G)), where fl E U(Fi) and x is an element of a transversal of T(G) in G. Then (&) -l a(Qx) = x -1 0 -l aQx = x-l ax .Now as a E U(Fi) C_ U(KT(G)) and every subgroup of T(G) is normal in G.So a has finitely many conjugates in U(Fi * GIT(G)) .Thus U(Fi) C_ A(U(Fi * G/T(G))) .This further gives that U(KG) = Dr ,U(Fi * G/T(G)) is FC-nilpotent .
Remark .If G = (x,yIx -1 y 2 x = y -2 , y -1 x2 y = x-2), then by [4, p. 606] G is torsion free but not right ordered .However, G is FC-nilpotent because it is abelian by finite .Hence G is torsion free FC-nilpotent group which is not right ordered .