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

FINDING N-TH ROOTS IN NILPOTENT GROUPS... 583
Definition 5.8. Let G be nilpotent of class 2. Define M(G) = {g ∈
G|g = g1g2 · · · gn, where each factor gi ∈ G occurs an even number of times,
with occurrences of g −1
i counted together with occurrences of gi}.
Clearly, M = M(G) is a normal subgroup of G.
Lemma 5.9. G/M is an elementary Abelian two-group.
Proof. If gM ∈ G/M, then (gM) 2 = g 2 M = M.
Lemma 5.10. Let CG(Q) = {g ∈ G|gq = qg,q ∈ Q} (that is, CG(Q) is
the centralizer in G of Q). Then M ≤ CG(Q).
Proof. Let q ∈ Q and m = g1g2 · · · gn ∈ M. Then
[q,m] = [q,g1g2 · · · gn] = [q,g1][q,g2] · · · [q,gn],
where distinct g ′ is occur an even number of times. Since G is nilpotent of class
two, each [q,gi] ∈ Z(G). By Lemma 5.7, [q,gi] 2 = 1. Hence, [q,m] = 1.
This last lemma implies that no element of the perfect square root X of G
is contained in M. For if it is, an element of X would commute with Q ∩ X
(which is nonempty by Lemma 5.4), contradicting the fact that no two elements
of X commutes.
Theorem 5.11. If G is nilpotent of class 2, then G has no perfect square
root.
Proof. We will prove by contradiction. Suppose that G is nilpotent of class
2 and that X is a perfect square root of G. Let M be as defined above. We will
denote the image of X in G/M by X. Further, we will consider X as a multiset
instead of a set, since distinct elements of X may be mapped to same cosets
of M. We will look at the occurrences of elements of M in the multiplication
table of X by finding a bound for occurrences of 1 ∈ G/M in the multiplication
table for X.
Let the distinct elements of X be a0, a1, ..., ak−1, so that there are k
distinct cosets of M in X. Let |X| = |X| = n = kq + r, where 0 ≤ r < k. List
the elements of the multiset X as b0,b1,... ,bn−1 as follows:
• Let b0 = a0.
• Suppose bi = aj. Then let
bi+1 =
aj+1 if j < k − 1
a0 if j = k − 1