Theorem: Generator for Quaternion Group

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:
{{begin-eqn}}
{{eqn | l = \mathbf a^2
      | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
      | c = 
}}
{{eqn | r = \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}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \mathbf a^4
      | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
      | c = 
}}
{{eqn | r = \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}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
      | c = 
}}
{{eqn | r = \mathbf I
      | c = 
}}
{{end-eqn}}
Next we have:
{{begin-eqn}}
{{eqn | l = \mathbf b^2
      | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
      | c = 
}}
{{eqn | r = \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}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
      | c = 
}}
{{end-eqn}}
and so:
:$\mathbf b^2 = \mathbf a^2$
Then we have:
{{begin-eqn}}
{{eqn | l = \mathbf a \mathbf b \mathbf a
      | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
      | c = 
}}
{{eqn | r = \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}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
      | c = 
}}
{{eqn | r = \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}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
      | c = 
}}
{{eqn | r = \mathbf b
      | c = 
}}
{{end-eqn}}
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 $\Dic 2$:
{{:Quaternion Group/Group Presentation}}
Hence the result.
{{qed}}
Category:Quaternion Group
Category:Examples of Generators of Groups
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