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(4) Consider the 2 x 2 matrices 1-69). -:- (-1). «-6 -1) :-(1) (a) Prove that...

(4) Consider the 2 x 2 matrices 1-69). -:- (-1). «-6 -1) :-(1) (a) Prove that {1,-1}, {1,a}, and {1,3} are finite abelian gro

(4) Consider the 2 x 2 matrices 1-69). -:- (-1). «-6 -1) :-(1) (a) Prove that {1,-1}, {1,a}, and {1,3} are finite abelian groups of order 2. (b) Prove that {1,-1,a,ß} is a finite abelian group of order 4, and compute the multiplication table for this group.
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Solution: Given

I=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix},\ \ \ -I=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}

\alpha =\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\ \ \ \beta=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}

Now,

(-I)(-I)=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=I

\alpha^2=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=I

\beta^2=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=I

\alpha \beta=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}=-I

\beta\alpha=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}=-I

a).

Now , the composition tables for \left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} are respectively below

\begin{matrix} * & {\color{Red} I} & {\color{Red} -I}\\ {\color{Red} I} & I & -I\\ {\color{Red} -I} & -I &I \end{matrix}\ \ \ \ \ \ \begin{matrix} * & {\color{Red} I} & {\color{Red} \alpha}\\ {\color{Red} I} & I & \alpha\\ {\color{Red} \alpha } & \alpha &I \end{matrix}\ \ \ \ \ \ \ \ \ \ \begin{matrix} * & {\color{Red} I} & {\color{Red} \beta}\\ {\color{Red} I} & I & \beta\\ {\color{Red} \beta} & \beta &I \end{matrix}

From above composition tables it is clear that

\left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} are closed.

\left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} are associative.

\left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} all have identity element I .

and inverse of each element is itself.

also all elements in composition tables commute to each other.

Thus, \left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} are abelian groups with 2 elements.

Hence, \left \{ I,-I \right \},\left \{ I,\alpha \right \},\left \{ I,\beta \right \} are finite abelian groups of order 2 .

b).

Now, the composition table for \left \{ I,-I,\alpha,\beta \right \} is below

\begin{matrix} * & {\color{Red} I} &{\color{Red} -I} & {\color{Red}\alpha } & {\color{Red} \beta} \\ {\color{Red}I } &I & -I & \alpha &\beta \\ {\color{Red} -I} & -I & I & -\alpha & -\beta\\ {\color{Red} \alpha} & \alpha & -\alpha & I & -I\\ {\color{Red}\beta } & \beta & -\beta & -I &I \end{matrix}

From above composition tables it is clear that

\left \{ I,-I,\alpha,\beta \right \} is closed.

\left \{ I,-I,\alpha,\beta \right \} is associative.

\left \{ I,-I,\alpha,\beta \right \}  has identity element I .

and inverse of each element is itself.

also all elements in composition tables commute to each other.

Thus, \left \{ I,-I,\alpha,\beta \right \} are abelian groups with 4 elements.

Hence, \left \{ I,-I,\alpha,\beta \right \} are finite abelian groups of order 4 .

Which is the required proof.

This complete the solution.

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(4) Consider the 2 x 2 matrices 1-69). -:- (-1). «-6 -1) :-(1) (a) Prove that...
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