Dicyclic Group Dic3/Matrix Representation
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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:
\(\ds \mathbf X^3\) | \(=\) | \(\ds \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^{-2} \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \omega^3 & 0 \\ 0 & \omega^{-3} \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \map \exp {3 \times \dfrac {2 \pi i} 6} & 0 \\ 0 & \map \exp {-3 \times \dfrac {2 \pi i} 6} \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} \map \exp {\pi i} & 0 \\ 0 & \map \exp {-\pi i} \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\) |
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:
\(\ds \mathbf X \mathbf Y\) | \(=\) | \(\ds \begin{bmatrix} \omega^{-1} & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 & i \omega^{-1} \\ i \omega & 0\end{bmatrix}\) |
and:
\(\ds \mathbf Y \mathbf X^{-1}\) | \(=\) | \(\ds \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 0 & i \omega^{-1} \\ i \omega & 0\end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf X \mathbf Y\) |
Thus $\mathbf X$ and $\mathbf Y$ fulfil the criteria given.
$\blacksquare$
This needs considerable tedious hard slog to complete it. In particular: Yes there's a lot of work to be done yet on dicyclic groups. Not to be confused (like I did) with dihedral groups. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Exercise $8$