# 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:

$\begin{array}{r|rrrrrrrr} & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ \hline e & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ a & a & a^2 & a^3 & e & a b & a^2 b & a^3 b & b \\ a^2 & a^2 & a^3 & e & a & a^2 b & a^3 b & b & a b \\ a^3 & a^3 & e & a & a^2 & a^3 b & b & a b & a^2 b \\ b & b & a^3 b & a^2 b & a b & a^2 & a & e & a^3 \\ a b & a b & b & a^3 b & a^2 b & a^3 & a^2 & a & e \\ a^2 b & a^2 b & a b & b & a^3 b & e & a^3 & a^2 & a \\ a^3 b & a^3 b & a^2 b & a b & b & a & e & a^3 & a^2 \end{array}$

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.

$\blacksquare$