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

582 S. Sze, D. Kahrobaei, R. Dambreville, M. Dupas
Proof. The identity element 1 must be the product of two elements of X.
Since inverse pairs always commute and distinct elements of X do not commute,
we must have x 2 = 1 for some x ∈ X. Now, x = 1 because 1 ∈ Z(G). Hence, x
must have order 2, so |G| has even order. It follows that |G| is a multiple of 4
since it is a perfect square.
Lemma 5.3. If G has a perfect square root, then |Z(G)| 2 < |G|.
Proof. Let X be a perfect square root of G. Since the elements of X do
not commute, no two elements of X are in the same coset of Z = Z(G): for
if z1, z2 ∈ Z, then (az1)(az2) = (az2)(az1). Let |G| = n 2 so that there are at
least n cosets of Z from which to choose as elements of X. That is, we have
[G : Z] = |G|
|Z| ≥ n, or n ≥ |Z|. Now we can conclude that |Z|2 < |G| since X
contains no elements of Z.
Lemma 5.4. Let G be a finite group with perfect square root X. Then
for each z ∈ Z(G), there is some x ∈ X such that x 2 = z.
Proof. Each z ∈ Z(G) is the product of two elements of X, say x1 and
x2. Since z ∈ Z(G) and the pairs x1x2 and x2x1 are conjugate, we must have
x1x2 = x2x1. This contradicts the fact that elements of X do not commute.
Hence, x1 = x2 and z = x 2 1 .
Definition 5.5. Let G be nilpotent of class 2. That is, G satisfies
[[G,G],G] = {1}. We define Q(G) = {q ∈ G|q 2 ∈ Z(G)}.
q 2 1
2.
We have the following results:
Lemma 5.6. Q = Q(G) is a normal subgroup of G.
Proof. We first show that Q is a subgroup of G: let q1, q2 ∈ Q. Then
(q1q −1
2 )2 = q1q −1
2
= q 2 1q −1
2
q1q −1
2
(q2q −1
−1 −1
= q1(q1q2 q2q1 )q−1 2
1 q−1 2
−1
q1)q2 = q2 1q −2
2 [q−1
q1q −1
2
2 ,q1] ∈ Z(G).
Next, to see that Q is normal in G, let g ∈ G. Then (g −1 q1g) 2 = g −1 q 2 1
∈ Z(G).
Lemma 5.7. Let q ∈ Q(G). Then every commutator [q,g] has order 1 or
Proof. [q,g] 2 = [q 2 ,g] = 1.
g =