Generator for Quaternion Group
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Theorem
The Quaternion Group can be generated by the matrices:
- $\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
\qquad \mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
where $i$ is the imaginary unit:
- $i^2 = -1$
Proof
Note that:
- $\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
is the identity for (conventional) matrix multiplication of order $2$.
We have:
\(\ds \mathbf a^2\) | \(=\) | \(\ds \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 \times 0 + 1 \times \paren {-1} & 0 \times 1 + 1 \times 0 \\ \paren {-1} \times 0 + 0 \times \paren {-1} & \paren {-1} \times 1 + 0 \times 0 \end{bmatrix}\) | Definition of Matrix Product (Conventional) | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) |
and so:
\(\ds \mathbf a^4\) | \(=\) | \(\ds \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \paren {-1} \times \paren {-1} + 0 \times 0 & \paren {-1} \times 0 + 0 \times \paren {-1} \\ 0 \times \paren {-1} + \paren {-1} \times 0 & 0 \times 0 + \paren {-1} \times \paren {-1} \end{bmatrix}\) | Definition of Matrix Product (Conventional) | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf I\) |
Next we have:
\(\ds \mathbf b^2\) | \(=\) | \(\ds \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 \times 0 + i \times i & 0 \times i + i \times 0 \\ \paren i \times 0 + 0 \times i & i \times i + 0 \times 0 \end{bmatrix}\) | Definition of Matrix Product (Conventional) | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) |
and so:
- $\mathbf b^2 = \mathbf a^2$
Then we have:
\(\ds \mathbf a \mathbf b \mathbf a\) | \(=\) | \(\ds \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \paren {-1} \times 0 + 0 \times i & \paren {-1} \times i + 0 \times 0 \\ \paren 0 \times 0 + \paren {-1} \times i & 0 \times i + \paren {-1} \times 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) | Definition of Matrix Product (Conventional) | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 \times \paren {-1} + \paren {-i} \times 0 & 0 \times 0 + \paren {-i} \times \paren {-1} \\ \paren {-i} \times \paren {-1} + 0 \times 0 & \paren {-i} \times 0 + 0 \times \paren {-1} \end{bmatrix}\) | Definition of Matrix Product (Conventional) | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf b\) |
Thus $\gen {\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} }$ fulfils the conditions for the group presentation of $\Dic 2$:
- $\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
Hence the result.
$\blacksquare$