Dihedral Group D4/Matrix Representation/Formulation 2/Examples of Generated Subgroups/D

Examples of Generated Subgroups of Dihedral Group $D_4$

Let the dihedral group $D_4$ be represented by the set of square matrices:

$D_4 = \set {\mathbf I, \mathbf A, \mathbf B, \mathbf C, \mathbf D, \mathbf E, \mathbf F, \mathbf G}$

under the operation of conventional matrix multiplication, where:

$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\qquad \mathbf A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \qquad \mathbf B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \qquad \mathbf C = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}$

$\mathbf D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

\qquad \mathbf E = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \qquad \mathbf F = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \qquad \mathbf G = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$

The generated subgroup $\gen {\mathbf D}$ is the (cyclic) group of order $2$:

$\set {\mathbf I, \mathbf D}$

Proof

The Cayley table of $D_4$ is as follows:

$\begin{array}{r|rrrrrrrr}

         & \mathbf I & \mathbf A & \mathbf B & \mathbf C & \mathbf D & \mathbf E & \mathbf F & \mathbf G \\

\hline \mathbf I & \mathbf I & \mathbf A & \mathbf B & \mathbf C & \mathbf D & \mathbf E & \mathbf F & \mathbf G \\ \mathbf A & \mathbf A & \mathbf B & \mathbf C & \mathbf I & \mathbf E & \mathbf F & \mathbf G & \mathbf D \\ \mathbf B & \mathbf B & \mathbf C & \mathbf I & \mathbf A & \mathbf F & \mathbf G & \mathbf D & \mathbf E \\ \mathbf C & \mathbf C & \mathbf I & \mathbf A & \mathbf B & \mathbf G & \mathbf D & \mathbf E & \mathbf F \\ \mathbf D & \mathbf D & \mathbf G & \mathbf F & \mathbf E & \mathbf I & \mathbf C & \mathbf B & \mathbf A \\ \mathbf E & \mathbf E & \mathbf D & \mathbf G & \mathbf F & \mathbf A & \mathbf I & \mathbf C & \mathbf B \\ \mathbf F & \mathbf F & \mathbf E & \mathbf D & \mathbf G & \mathbf B & \mathbf A & \mathbf I & \mathbf C \\ \mathbf G & \mathbf G & \mathbf F & \mathbf E & \mathbf D & \mathbf C & \mathbf B & \mathbf A & \mathbf I \end{array}$

We have:

\(\ds \mathbf D^2\)

\(=\)

\(\ds \mathbf I\)

Thus:

$\gen {\mathbf D} = \set {\mathbf I, \mathbf D}$

and its Cayley table is:

$\begin{array}{r|rr}

         & \mathbf I & \mathbf D \\

\hline \mathbf I & \mathbf I & \mathbf D \\ \mathbf D & \mathbf D & \mathbf I \\ \end{array}$

This is seen to be the (cyclic) group of order $2$.

$\blacksquare$