Center of Quaternion Group

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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.

$\blacksquare$


Sources

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