Order of Elements in Quaternion Group

Theorem
Let $Q = \Dic 2$ be the quaternion group, whose group presentation is given by:


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

Then $\Dic 2$ has:
 * $1$ element of order $2$

and:
 * $6$ elements of order $4$.

Proof
From Identity is Only Group Element of Order 1, the identity element $e$, and only $e$, is of order $1$.

From here, we inspect the Cayley table:

It is immediately seen that:
 * $\paren {a^2}^2 = e$

and so by definition $a^2$ is of order $2$.

As can be seen from inspection of the main diagonal, for all other $x \in Q$, we have:
 * $x^2 = a^2 \ne e$

and so:
 * $x^4 = \paren {a^2}^2 = e$

There are $6$ such elements.

Hence the result by definition of order of group element.