Question

Is K isomorphic to$\mathrm{\mathbb{Z}}$ ?

Short answer

It proved that K isomorphic to$\mathrm{\mathbb{Z}}$ .

Step-by-step solution

 Step 1: Isomorphic Groups

Two groups $G$and $H$are said to be isomorphic if there exists an isomorphism $f$from $G$onto $H$

Check that K isomorphic to ℤ

Claim that set$K$is isomorphic to$\mathrm{\mathbb{Z}}$.

Define a function $f:\mathbb{Z}\to K$as,

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

Claim that the function$f:\mathbb{Z}\to K$is an isomorphism.

Check that the function $f:\mathbb{Z}\to K$is a homomorphism.

For $n,m\in \mathbb{Z}$

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

Therefore,$f(m+n)=f\left(n\right)f\left(m\right)$ for all$n,m\in \mathbb{Z}$ .

Hence, the function$f:\mathbb{Z}\to K$ is a homomorphism.

Check that K is isomorphic to ℤ

Check the function $f:\mathbb{Z}\to K$is an injective function.

For $n,m\in \mathbb{Z}$such that $f\left(n\right)=f\left(m\right)$,

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

Compare the entries of the two matrices.

$m=n$

Therefore, no two distinct elements of $\mathbb{Z}$are mapped to the same element of $K$.

Hence, the function $f:\mathbb{Z}\to H$is an injective function.

Check the function $f:\mathbb{Z}\to K$is a surjective function.

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

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

Thus, every element of $K$has a pre-image in $\mathrm{\mathbb{Z}}$.

Hence, the function $f:\mathbb{Z}\to K$is a surjective function.

Therefore, the function $f:\mathbb{Z}\to K$is a bijective homomorphism and hence, an isomorphism. This proves the claim.

Hence, it is proved that K isomorphic to$\mathrm{\mathbb{Z}}$ .