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)
View subgroup property satisfactions for subgroup-defining functions

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View subgroup property dissatisfactions for subgroup-defining functions

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 $x^n=z$.

Related facts

Opposite facts

• Center is local powering-invariant
• Center is quotient-powering-invariant

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 $S^3\cong SU(2,\mathbb{C})$ where $\mathbb{C}$ is the field of complex numbers. Explicitly:

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

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:

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

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 $\mathbb{C}$ (where $\mathbb{C}$ is the field of complex numbers) 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:

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

None of these elements is central.