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Show that { $\bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} \omega & 0 \\ 0 & \omega^2\end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} \omega^2 & 0 \\ 0 & \omega\end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} 0 & 1 \\ 1 & 0\end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} 0 & \omega^2\\ \omega & 0\end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} 0 & \omega\\ \omega^2 & 0\end{smallmatrix} \bigr) $} where $\omega^3 = 1, \omega \neq 1$ form a group with respect to matrix multiplication.

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Toolbox:
  • A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
  • (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
  • (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
  • (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a ^{-1} G.$
  • (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
  • e is called the identity element of $G$ and $a^{-1}$ is called the inverse of a in $G.$
Step 1:
Let $G=\{I,A,B,C,D,E\}$
Where $I=\bigl(\begin{smallmatrix}1 &0\\0 &1\end{smallmatrix} \bigr)$
$\qquad A=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)$
$\qquad B=\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega^2\end{smallmatrix} \bigr)$
$\qquad C=\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)$
$\qquad D=\bigl(\begin{smallmatrix}0 &\omega^2\\\omega &0\end{smallmatrix} \bigr)$
$\qquad E=\bigl(\begin{smallmatrix}0&\omega \\\omega^2&0\end{smallmatrix} \bigr)$
Where $\omega^3=1$
Step 2:
$IA=A,IB=B,IC=C,ID=D,IE=E$
$AI=A,AA=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega^4\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega\end{smallmatrix} \bigr)$=B
$AB=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega\end{smallmatrix} \bigr)=I$
$AC=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega\\\omega^2 &0\end{smallmatrix} \bigr)=E$
$AD=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &\omega^2\\\omega &0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^3\\\omega^3 &0\end{smallmatrix} \bigr)=C$
$AE=\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &\omega\\\omega^2 &0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^2\\\omega^4 &0\end{smallmatrix} \bigr)=D$
BI=B,BA=I,BB=A,$BC=\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^2\\\omega &0\end{smallmatrix} \bigr)=D$
$BD=\bigl(\begin{smallmatrix}\omega^2 &0\\0 &\omega\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &\omega^2\\\omega &0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^4\\\omega^2 &0\end{smallmatrix} \bigr)=E$
$BE=C$
$CI=C,CA=\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega &0\\0 &\omega^2\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^2\\\omega &0\end{smallmatrix} \bigr)=D$
$CB=\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega^2&0\\0&\omega\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0&\omega\\\omega^2&0\end{smallmatrix} \bigr)=E$
$CE=B$
$CC=I$
$DI=D,DA=\bigl(\begin{smallmatrix}0&\omega^2\\\omega&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega&0\\0&\omega^2\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^4\\\omega^2 &0\end{smallmatrix} \bigr)=E$
$DB=\bigl(\begin{smallmatrix}0&\omega^2\\\omega&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}\omega^2&0\\0&\omega\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}0 &\omega^3\\\omega^3&0\end{smallmatrix} \bigr)=C$
$DC=\bigl(\begin{smallmatrix}0&\omega^2\\\omega&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\omega^2&0\\0&\omega\end{smallmatrix} \bigr)=B$
$DD=\bigl(\begin{smallmatrix}0&\omega^2\\\omega&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0&\omega^2\\\omega^2&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\omega^3&0\\0&\omega^3\end{smallmatrix} \bigr)=I$
$DE=A$
$EI=E,EA=C,EB=D,EC=A,ED=B,EE=I$
Step 3:
Drawing up cayley's Table,we have
Step 4:
Closure :All the products belong to $G$.Closure is satisfied.
Associativity: Matrix multiplication is associative.
Existence of identity : $I$ is the identity.
Existence of inverse : From the table it is evident that using element has an inverse.
The four group axioms being satisfied,$G$ is a group under matrix multiplication.
answered Sep 14, 2013 by sreemathi.v
edited Sep 14, 2013 by sreemathi.v
 

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