Order of Elements in Quaternion Group
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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:
and:
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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $14$