# Dicyclic Group Dic3/Matrix Representation

## Matrix Representation of Dicyclic Group $\Dic 3$

Let $\omega$ denote the complex number $\map \exp {\dfrac {2 \pi i} 6}$, so that $\omega^6 = 1$.

Let $\mathbf X$ be the matrix defined as:

$\mathbf X = \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}$

Let $\mathbf Y$ be the matrix defined such that:

$\mathbf X \mathbf Y = \mathbf Y \mathbf X^{-1}$

and:

$\mathbf Y^2 = \mathbf X^3$

Then the set:

$G = \set {\mathbf X^i, \mathbf Y \mathbf X^j: 1 \le i, j \le 6}$

defines the dicyclic group $\Dic 3$.

## Proof

By calculation, we have:

 $\displaystyle \mathbf X^3$ $=$ $\displaystyle \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}^3$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^{-2} \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} \omega^3 & 0 \\ 0 & \omega^{-3} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} \map \exp {3 \times \dfrac {2 \pi i} 6} & 0 \\ 0 & \map \exp {-3 \times \dfrac {2 \pi i} 6} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} \map \exp {\pi i} & 0 \\ 0 & \map \exp {-\pi i} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

and it is seen that:

$\mathbf X^{-1} = \begin{bmatrix} \omega^{-1} & 0 \\ 0 & \omega \end{bmatrix}$

Now consider $\mathbf Y = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$

We have:

 $\displaystyle \mathbf X \mathbf Y$ $=$ $\displaystyle \begin{bmatrix} \omega^{-1} & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} 0 & i \omega^{-1} \\ i \omega & 0\end{bmatrix}$

and:

 $\displaystyle \mathbf Y \mathbf X^{-1}$ $=$ $\displaystyle \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} 0 & i \omega^{-1} \\ i \omega & 0\end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \mathbf X \mathbf Y$

Thus $\mathbf X$ and $\mathbf Y$ fulfil the criteria given.

$\blacksquare$