# 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:

 $\ds x a$ $=$ $\ds a x$ $\ds \leadsto \ \$ $\ds a^i b^j a$ $=$ $\ds a^{i + 1} b^j$ $\ds \leadsto \ \$ $\ds b^j a$ $=$ $\ds a b^j$

For $j = 1$ this means:

 $\ds a b$ $=$ $\ds b a$ $\ds$ $=$ $\ds a^{-1} \beta$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds a^{-1}$ $\ds \leadsto \ \$ $\ds a^2$ $=$ $\ds e$

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:

 $\ds x b$ $=$ $\ds b x$ $\ds \leadsto \ \$ $\ds a^i b$ $=$ $\ds b a^i$ $\ds$ $=$ $\ds a^{-i} b$ Product of Generating Elements of Quaternion Group $\ds \leadsto \ \$ $\ds a^i$ $=$ $\ds a^{-i}$ $\ds \leadsto \ \$ $\ds a^{2 i}$ $=$ $\ds e$ $\ds$ $=$ $\ds a^4$ Definition of Order of Group Element

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.

$\blacksquare$