Question

(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.

Answer 1

Solution: Given

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

a = - 66 -1), 1 0 0 -1 B= -1 0 0 1

Now,

(-1)(-1) = - 3 ) (32) 1) = (0 ) - 1

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

= -1 0 0 1) (0 9) = 1 0 0 1 1

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

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

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 .

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 .

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 .

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 .

Which is the required proof.

This complete the solution.