Question

(a) Prove that $\mathbf{}\mathbf{}\mathbf{\{}\mathbf{}\overline{)\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)}\mathbf{n}\mathbf{\in }\mathbf{Z}\mathbf{\}}\mathbf{}$ is a group under matrix multiplication .

Short answer

It is proved that$\left\{\overline{)\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)}|n\in Z\right\}$ is a group under matrix multiplication.

Step-by-step solution

Step-by-Step Solution Step 1: Isomorphic Groups

Two groups $\mathit{G}$and$\mathit{H}$are isomorphic if there exists an isomorphism$\mathit{f}$from $\mathit{G}$onto$\mathit{H}$ .

Prove that {1−n−nn1+n n∈Z}  is a group under matrix multiplication.

Consider the given set.

$H=\left\{\overline{)\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)}n\in Z\right\}$

Claim that the set $H$is a group under matrix multiplication.

Set is closed under matrix multiplication. For any two elements,

$\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right),\left(\begin{array}{cc}1-m& -m\\ m& 1+m\end{array}\right)\in H$

Find the product of both the elements.

$$\begin{gathered} \left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right).\left(\begin{array}{cc}1-m& -m\\ m& 1+m\end{array}\right)=\left(\begin{array}{cc}(1-n)(1-m)-mn& -m(1-n)-n(1+m)\\ n(1-m)+m(1+n)& -mn+(1+n)(1+m)\end{array}\right) \\ =\left(\begin{array}{cc}1-m-n+mn-mn& -m+mn-n-mn\\ n+m& 1+(m+n)\end{array}\right) \\ =\left(\begin{array}{cc}1-(m+n)& -(m+n)\\ n+m& 1+(m+n)\end{array}\right) \end{gathered}$$

Therefore, $\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right),\left(\begin{array}{cc}1-m& -m\\ m& 1+m\end{array}\right)\in H$

Thus, set is closed under matrix, and associativity of the operation holds in as the matrix multiplication is associative.

Prove that {1−n−nn1+n  n∈Z}is a group under matrix multiplication

Check the existence of identity element.

For $n=0$,

$\left(\begin{array}{cc}1-0& -0\\ 0& 1+0\end{array}\right)\in H$

And,

$$\begin{gathered} \left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right) \end{gathered}$$

Therefore, the matrix $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$is the identity element of the set $H$.

Check the existence of the inverse element.

The element $\left(\begin{array}{cc}1+n& n\\ -n& 1-n\end{array}\right)\in H,-n\in \mathbb{Z}$is such that,

$$\begin{gathered} \left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)\left(\begin{array}{cc}1+n& n\\ -n& 1-n\end{array}\right)=\left(\begin{array}{cc}(1-n)(1+n)+n^2& (1-n)n-n(1-n)\\ (1+n)n-n(1+n)& n^2+(1-n)(1+n)\end{array}\right) \\ =\left(\begin{array}{cc}1-n^2+n^2& 0\\ 0& n^2+1-n^2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

And,

$$\begin{gathered} \left(\begin{array}{cc}1+n& n\\ -n& 1-n\end{array}\right)\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)=\left(\begin{array}{cc}(1-n)(1+n)+n^2& -n(1+n)+n(1+n)\\ -n(1-n)+(1-n)n& n^2+(1-n)(1+n)\end{array}\right) \\ =\left(\begin{array}{cc}1-n^2+n^2& 0\\ 0& n^2+1-n^2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Thus, the matrix $\left(\begin{array}{cc}1+n& n\\ -n& 1-n\end{array}\right)$is multiplicative inverse of the element$\left(\begin{array}{cc}1-n& -n\\ n& 1+n\end{array}\right)$.

Therefore, set $H$is a group under the matrix multiplication.