FINDING N-TH ROOTS IN NILPOTENT GROUPS AND ...

578 S. Sze, D. Kahrobaei, R. Dambreville, M. Dupas
Proof. Let G be a torsion-free group, x ∈ G, x = 1. Then x generates an
infinite cyclic subgroup which is an abelian torsion-free group of rank 1. Since
the union of an ascending sequence of abelian torsion-free groups of rank 1 is
a group of the same kind, x is contained in at least one subgroup A that is
a maximal abelian torsion-free subgroup of G of rank 1. If G is an R−group,
then A is contained in the isolator I(x) of x: for an arbitrary a ∈ A, the cyclic
subgroups {x} and {a} have intersection other than {1}, but {x} ⊂ I(x).
Suppose x is contained in two distinct maximal abelian torsion-free subgroups
of rank 1, A and B; the intersection of these two subgroups will not
be {1}. Let a ∈ A, b ∈ B be arbitrary elements, other than 1. Since there
exists i,j ∈ Z such that a i = b j , this implies that a and b are permutable.
Therefore, {A,B} is an abelian torsion-free group. However, in such a group
distinct maximal subgroups of rank 1 cannot have intersection other than {1}.
Thus the theorem follows.
Proposition 3.10. Every nilpotent torsion-free group is an R−group.
That is, unique extraction of roots in nilpotent torsion-free groups is possible.
Proof. For abelian torsion-free groups, the statement is clear. We will prove
by induction on the class of the nilpotent group. We assume that the statement
is true for all nilpotent groups of class less than or equal to k − 1, k > 1. Let
G be a torsion-free group of class k with upper central series
{1} = Z0 Z1 · · · Zk−1 Zk = G
and let x, y ∈ G such that x n = y n , n > 0. Consider the subgroup H =
{Zk−1,x}. H is normal in G since G/Zk−1 is abelian. In addition, H is nilpotent
of class less than or equal to k − 1: in any noncommutative group the factor
group of the center cannot be cyclic. Hence, by the inductive hypothesis, H
is an R−group. Therefore, x n = y n implies that y −1 x n = y n−1 = x n y −1 , or
x n = y −1 x n y = (y −1 xy) n . Since x and y −1 x n y belong to the R−group H, we
have x = y −1 xy, or xy = yx. Therefore, x n = y n may be rewritten (y −1 x) n = 1.
Since G is torsion-free, this means that x = y.
This proposition only states that if G is a torsion-free nilpotent group, then
xn = a has a unique solution in G; it does not show us how to find x.
A natural question that arises is if there is a relation between Černikov
complete groups and R−groups. In Section 2, we showed that the Heisenberg
group is not Černikov complete. Since a complete group is Černikov complete,
this implies that the Heisenberg group is not complete. Hence, we may not take
arbitrary n-th roots in the Heisenberg group. However, since the Heisenberg