Q.Show that $\left\{\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}, \begin{pmatrix}\omega & 0\\0 & \omega^2\end{pmatrix}, \begin{pmatrix}\omega^2 & 0\\0 & \omega\end{pmatrix}, \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}, \begin{pmatrix}0 & \omega^2\\\omega & 0\end{pmatrix}, \begin{pmatrix}0 & \omega\\\omega^2 & 0\end{pmatrix}\right\}$, where $\omega^3 = 1$, $\omega \neq 1$ form a group with respect to matrix multiplication.

Verify the four group axioms — closure, associativity, identity, inverse — for the given set of six matrices under multiplication, using $\omega^3=1,\ \omega\neq 1$ (so $1+\omega+\omega^2=0$).

1. Label the elements. Let

   $I=\begin{pmatrix}1&0\\0&1\end{pmatrix},\ A=\begin{pmatrix}\omega&0\\0&\omega^2\end{pmatrix},\ B=\begin{pmatrix}\omega^2&0\\0&\omega\end{pmatrix},\ C=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ D=\begin{pmatrix}0&\omega^2\\\omega&0\end{pmatrix},\ E=\begin{pmatrix}0&\omega\\\omega^2&0\end{pmatrix},$and let $G=\{I,A,B,C,D,E\}$.

2. Closure — representative products (using $\omega^3=1$ throughout):

   A\cdot A = \begin{pmatrix}\omega^2&0\\0&\omega^4\end{pmatrix}=\begin{pmatrix}\omega^2&0\\0&\omega\end{pmatrix}=B, \qquad A\cdot B = \begin{pmatrix}\omega^3&0\\0&\omega^3\end{pmatrix}=I,$$$$C\cdot C = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I,\qquad A\cdot C = \begin{pmatrix}0&\omega\\\omega^2&0\end{pmatrix}=E, \qquad C\cdot A = \begin{pmatrix}0&\omega^2\\\omega&0\end{pmatrix}=D,$$$$D\cdot D = \begin{pmatrix}\omega^3&0\\0&\omega^3\end{pmatrix}=I,\qquad E\cdot E = I,\qquad D\cdot E = A,\qquad E\cdot D = B.Carrying this out for every pair (36 products) shows every product again lies in $G$ — closure holds. The complete multiplication table is:

(reading row $\times$ column, e.g. row $D$, column $A$ gives $D\cdot A = E$). Every entry lies in $G$.

3. Associativity. Matrix multiplication is associative in general (this follows from associativity of the underlying field/ring multiplication and addition), so $(XY)Z=X(YZ)$ for all $X,Y,Z\in G$ automatically. …