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

FINDING N-TH ROOTS IN NILPOTENT GROUPS... 581
⎛ ⎞
9 1 0
Example 4.3. Let T = J3,9 = ⎝0
9 1⎠.
Find the square root of T.
0 0 9
We have c = 9 and compute N = J3,0
9 =
⎛ ⎞ 1 0 9 0
⎝ 1 0 0 ⎠, 9 N
0 0 0
2 ⎛ ⎞ 1 0 0 81
= ⎝0
0 0 ⎠
0 0 0
and N3 = 0. Then √ ⎛⎛
⎞ T =
1 0 0
3⎝⎝0
1 0⎠
+
0 0 0
1
⎛ ⎞ 1 0 9 0
⎝ 1
2 0 0 ⎠
9 −
0 0 0
1
⎛ ⎞⎞
⎛ ⎞
1
1 1
0 0 81 3 6 −216 ⎝
8 0 0 0 ⎠⎠
= ⎝ 1 0 3 ⎠. 6
0 0 0 0 0 3
The result is similar for n-th roots; if we take
A = n√
c I + 1
1 1 − n N2 N +
n n n 2!
then A n = T = c(I + N).
+ 1
n
1 − n 1 − 2n N3 n n 3!
5. Square Roots of Finite Groups
In this section, we follow the work of Abhyankar and Grossman [1].
Definition 5.1. Let G be a finite group and for X ⊆ G, let
X 2 = {x1x2 | x1,x2 ∈ X}.
Then X ⊆ G is said to be a perfect square root of G if
1. X 2 = G, and
2. |X| 2 = |G|.
+ · · · ,
Note that since X 2 = G and |X| 2 = |G|, every element appears once in the
multiplication table for X. We can then conclude that distinct elements of a
perfect square root do not commute. From this observation, is is clear that if
G has a perfect square root, G is non-Abelian and X does not contain elements
of Z(G). We have the following lemmas that provide necessary conditions for
a group to have perfect square roots:
Lemma 5.2. Let G be a group with perfect square root. Then 4 divides
the order of G.