first, we use Cartesian coordinate to define the matrix representation of the Mirror, 180 degree Rotation and Parity.

$M_x= \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$

$R_x= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

$P= \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

i guess the reader can think the other matrix.

the group table is:

$\left( \begin{array}{cccccccc} i & M_x & M_y & M_z & R_x & R_x & R_z & P \\ M_x & i & R_z & R_x & P & M_z & M_y & R_x \\ M_y & R_z & i & R_x & M_z & P & M_x & R_x \\ M_z & R_x & R_x & i & M_y & M_x & P & R_z \\ R_x & P & M_z & M_y & i & R_z & R_x & M_x \\ R_x & M_z & P & M_x & R_z & i & R_x & M_y \\ R_z & M_y & M_x & P & R_x & R_x & i & M_z \\ P & R_x & R_x & R_x & M_x & M_y & M_z & i \end{array} \right)$

where i is identity matrix. we can see it is a group of order 8. there is a sub-group of $(i, R_x, R_y, R_z)$. since the group table are diagonal symmetry, each element from its own class.

since a rotational group can be represented by 2X2 matrix instead of 3X3 by introducing complex number. those matrix is Pauli’s spinor.

$\sigma_x= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

the transform from 3X3 matrix to 2X2 matrix is that…… then we can map the mirror matrix and parity into 2X2 matrix.