Question

Prove that $\mathit{K}\mathbf{=}\left\{\overline{)\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)}n\in \mathbb{Z}\right\}$is a group under matrix multiplication.

Short answer

It is proved that$K=\{\overline{)\left(\begin{array}{cc}1-2\mathrm{n}& \mathrm{n}\\ -4\mathrm{n}& 1+2\mathrm{n}\end{array}\right)}\mathrm{n}\in \mathrm{\mathbb{Z}}\}$ 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 said to be isomorphic if there exists an isomorphism $\mathit{f}$from $\mathit{G}$onto $\mathit{H}$.

Prove that K={(1−2nn−4n1+2n)  n∈ℤ} is a group under matrix multiplication

Consider the given set,

$H=\left\{\overline{)\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)}n\in \mathbb{Z}\right\}$

Claim that set role="math" localid="1651568686284" $K$is a group under matrix multiplication

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

$\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right),\left(\begin{array}{cc}1-2m& m\\ -4m& 1+2m\end{array}\right)\in K$

Find the product of both the elements.

$$\begin{gathered} \left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right).\left(\begin{array}{cc}1-2m& m\\ -4m& 1+2m\end{array}\right)=\left(\begin{array}{cc}(1-2n)(1-2m)-4mn& m(1-2n)+n(1+2m)\\ -4n(1-2m)-4m(1+2n)& -4mn+(1+2n)(1+2m)\end{array}\right) \\ =\left(\begin{array}{cc}1-2m-2n+4mn-4mn& m-2mn+n+2mn\\ -4n+8mn-4m-8mn& -4mn+1+2m+2n+4mn\end{array}\right) \\ =\left(\begin{array}{cc}1-2(m+n)& (n+m)\\ -4(m+n)& 1+2(m+n)\end{array}\right) \end{gathered}$$

Therefore, $\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right),\left(\begin{array}{cc}1-2m& m\\ -4m& 1+2m\end{array}\right)\in K$

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

Prove that K=1−2nn−4n1+2n  n∈ℤ   is a group under matrix multiplication

Check the existence of identity element.

For $n=0$

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

And,

$$\begin{gathered} \left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right) \end{gathered}$$

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

Check the existence of the inverse element.

For $\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)\in K,n\in \mathbb{Z}$

The element $\left(\begin{array}{cc}1+2n& -n\\ 4n& 1-2n\end{array}\right)\in K,-n\in \mathbb{Z}$ is Such that,

$$\begin{gathered} \left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)\left(\begin{array}{cc}1+2n& -n\\ 4n& 1-2n\end{array}\right)=\left(\begin{array}{cc}(1-2n)(1+2n)+4n^2& -(1-2n)n+n(1-2n)\\ -4(1+2n)n+4n(1+2n)& 4n^2+(1-2n)(1+2n)\end{array}\right) \\ =\left(\begin{array}{cc}1-4n^2+4n^2& 0\\ 0& 4n^2+1-4n^2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

And,

$$\begin{gathered} \left(\begin{array}{cc}1+2n& -n\\ 4n& 1-2n\end{array}\right)\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)=\left(\begin{array}{cc}(1-2n)(1+2n)+4n^2& (1-2n)n-n(1-2n)\\ 4(1+2n)n-4n(1+2n)& 4n^2+(1-2n)(1+2n)\end{array}\right) \\ =\left(\begin{array}{cc}1-4n^2+4n^2& 0\\ 0& 4n^2+1-4n^2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Thus, the matrix $\left(\begin{array}{cc}1-2n& n\\ -4n& 1+2n\end{array}\right)$is multiplicative inverse of the element. $\left(\begin{array}{cc}1+2n& -n\\ 4n& 1-2n\end{array}\right)$.

Therefore, set $K$is a group under matrix multiplication.