Question

Show that
$\{\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right).\left(\begin{array}{cc}{\omega }^2& 0\\ 0& {\omega }^2\end{array}\right).\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right).\left(\begin{array}{cc}0& {\omega }^2\\ \omega & 0\end{array}\right).\left(\begin{array}{cc}0& \omega \\ {\omega }^2& 0\end{array}\right).\}$, where ${\omega }^2=1,\omega =1$ form a group with respect to matrix multiplication

Answer 1

Let $G=\{\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right),\left(\begin{array}{cc}{\omega }^2& 0\\ 0& \omega \end{array}\right),\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),\left(\begin{array}{cc}0& {\omega }^2\\ \omega & 0\end{array}\right),\left(\begin{array}{cc}0& \omega \\ {\omega }^2& 0\end{array}\right)\}$

and $\ast =$matrix multiplication.

Let $e=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),a=\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right),b=\left(\begin{array}{cc}{\omega }^2& 0\\ 0& \omega \end{array}\right),c=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),d=\left(\begin{array}{cc}0& {\omega }^2\\ \omega & 0\end{array}\right),f=\left(\begin{array}{cc}0& \omega \\ {\omega }^2& 0\end{array}\right)$

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

$e\ast a=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right)=a$

Similarly $e\ast b=b,e\ast c=c,e\ast d=d,e\ast f=f$

$a\ast a=\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right)\left(\begin{array}{cc}\omega & 0\\ 0& {\omega }^2\end{array}\right)=\left(\begin{array}{cc}{\omega }^2& 0\\ 0& \omega \end{array}\right)=b$

Similarly finding all possibilities we generate the following table

$\ast$	$e$	$a$	$b$	$c$	$d$	$f$
$e$	$e$	$a$	$b$	$c$	$d$	$f$
$a$	$a$	$b$	$e$	$f$	$c$	$d$
$b$	$b$	$e$	$a$	$d$	$f$	$c$
$c$	$c$	$d$	$f$	$e$	$a$	$b$
$d$	$d$	$f$	$c$	$b$	$e$	$a$
$f$	$f$	$c$	$d$	$a$	$b$	$e$

From the table, it is seen that $G$ is closed under $\ast$, matrix multiplication is associative and $e$ is the identity.

Also satisfies inverse property.

• Inverse of $e$ is $e$
• Inverse of $a$ is $b$
• Inverse of $b$ is $a$
• Inverse of $c$ is $c$
• Inverse of $d$ is $d$
• Inverse of $f$ is $f$

Hence $(G,\ast )$ is a group.