Question

Show that $\left\{\begin{array}{c}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right],\left[\begin{array}{cc}{\omega }^2& 0\\ 0& \omega \end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& {\omega }^2\\ \omega & 0\end{array}\right],\left[\begin{array}{cc}0& \omega \\ {\omega }^2& 0\end{array}\right]\end{array}\right\}$ where

${\omega }^2=1,\omega \ne 1$ form a group with respect to matrix multiplication.

Answer 1

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}w& 0\\ 0& w^2\end{array}\right],\left[\begin{array}{cc}{w}^2& 0\\ 0& w\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& w^2\\ w& 0\end{array}\right],\left[\begin{array}{cc}0& w\\ w^2& 0\end{array}\right]$

$\left(1\right)$ $\left[\begin{array}{cc}w& 0\\ 0& w^2\end{array}\right]\left[\begin{array}{cc}0& w\\ w^2& 0\end{array}\right]=\left[\begin{array}{cc}0& w^2\\ w^4& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

$\left[\begin{array}{cc}w& 0\\ 0& w^2\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& w\\ w^2& 0\end{array}\right]$

Similarly, Multiplication of any two matrix within the set also lies in the set

$\therefore$ Matrix Multiplication satisfies closure.

$\left(2\right)$ Identity

$\because \left[\begin{array}{cc}{w}^2& 0\\ 0& w\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{w}^2& 0\\ 0& w\end{array}\right]$

i.e. $AI=A$

$\therefore$ Identity property is also satisfied.

$\left(3\right)$ Inverse

$\left[\begin{array}{cc}0& w\\ w^2& 0\end{array}\right]\left[\begin{array}{cc}0& w^2\\ w& 0\end{array}\right]=\left[\begin{array}{cc}{w}^2& 0\\ 0& w^4\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore$ Inverse also exists.

Since all three properties are satisfied

$\therefore$ They form a group.