Center of Quaternion Group

Theorem
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group.

Let $\map Z {\Dic 2}$ denote the center of $\Dic 2$.

Then:
 * $\map Z {\Dic 2} = \set {e, a^2}$

Proof
By definition, the center of $\Dic 2$ is:
 * $\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$

We are given that:


 * $\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$

We have that $\Dic 2$ is generated by $\alpha$ and $\beta$.

Thus:
 * $x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$

Let $x \in \map Z {\Dic 2}$.

We have that $x$ can be expressed in the form:
 * $x = a^i b^j$

for $i \in \set {0, 1, 2, 3}$ and $j \in \set {0, 1}$.

As $x \in \map Z {\Dic 2}$, we have:

For $j = 1$ this means:

But the order of $a$ is $4$, not $2$, and hence:


 * $a^2 \ne e$

So if $x \in \map Z {\Dic 2}$ it follows that $x$ has to be in the form:
 * $x = a^i$

for some $i \in \Z_{\ge 0}$.

Again, as $x \in \map Z {\Dic 2}$, we have:

So either $i = 0$ or $2 i = 4$, as $0 \le i \le 4$.

If $i = 0$ then $x = a^0 = e$.

If $2 i = 4$ then:
 * $x = a^{4 / 2} = a^2$

Hence the result.