# Center not is divisibility-closed

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) does not always satisfy a particular subgroup property (i.e., divisibility-closed subgroup)
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## Statement

It is possible to have a group $G$ such that there exists a natural number $n$ such that:

1. $G$ is $n$-divisible: For every $g \in G$, there exists $x \in G$ satisfying $x^n = g$.
2. The center $Z(G)$ is not $n$-divisible: There exists an element $z \in Z(G)$ such that there is no $x \in Z(G)$ satisfying $x^n = z$.

## Proof

### Example of a Lie group

We will construct an example for the case $n = 2$.

Let $G$ be the group of unit quaternions under multiplication, i.e., it is the group $S^3 \cong SU(2,\mathbb{C})$. Explicitly:

$G = \{ a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} \mid a,b,c,d \in \R, a^2 + b^2 + c^2 + d^2 = 1 \}$

with multiplication given by multiplication of quaternions.

We claim the following:

#### Every element of the group is a square

• Case $a = -1, b = c = d = 0$: In this case, the element $\mathbf{i} = 0 + 1\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}$ is a square root.
• All other cases: The following element works as a square root:

$\sqrt{\frac{1 + a}{2}} + \frac{b}{\sqrt{2(1 +a)}}\mathbf{i} + \frac{c}{\sqrt{2(1 + a)}}\mathbf{j} + \frac{d}{\sqrt{2(1 + a)}}\mathbf{k}$

Note that the intuition is as follows: it turns out that any element with $a = 0, b^2 + c^2 + d^2 = 1$ is a square root of the element $-1 = -1 + 0\mathbf{i} + \mathbf{j} + \mathbf{k}$. Thus, every element of the group can be put inside a copy of $\mathbb{C}$ inside $\mathbb{H}$ where the imaginary part is the normalized imaginary part of the element. Now, taking the square root is like taking square roots inside $\mathbb{C}$.

Note also that with the exception of -1, every element has only two square roots, which are negatives of each other.

#### There is an element of the center that is not the square of any element in the center

The center is the subgroup $\{ -1,1 \}$. The element $-1$ in the center does not have any square root in the center. As noted above, its set of square roots is the set:

$\{ a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} \mid a = 0, b,c,d \in \R, b^2 + c^2 + d^2 = 1 \}$

None of these elements is central.