Question

Prove that$\mathbb{Z}\times \mathbb{Z}/⟨(2,2)⟩\cong \mathbb{Z}\times {\mathbb{Z}}_2$ .$\mathbf{[}\mathbf{H}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{:}$ Show that ,$h:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}\times {\mathbb{Z}}_2$

given by $h\left((a,b)\right)=(a-b,{\left[b\right]}_2)$ is a surjective homomorphism$\mathbf{.}\mathbf{]}$

Short answer

It is proved that,$\mathbb{Z}\times \mathbb{Z}/⟨(2,2)⟩\cong \mathbb{Z}\times {\mathbb{Z}}_2$ .

Step-by-step solution

First Isomorphism Theorem

Theorem 8.20

Let $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$ be a surjective homomorphism of a group with kernel K. Then, the quotient group $\mathit{G}\mathbf{/}\mathit{K}$ is isomorphic to H.

Proving that  (ℤ×ℤ)/⟨(2,2)⟩≅ℤ×ℤ2

Let $h:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}\times {\mathbb{Z}}_2$ where $h\left(\left(ab\right)\right)=(a-b,{\left[b\right]}_2)$.

For$(a,b)(c,d)\in \mathbb{Z}\times \mathbb{Z}$ , prove h is a homomorphism as follows:

$\begin{array}{c}h((a,b)-(c,d))=h(a-c,b-d)\\ =((a-c)-(b-d),[b_2-d_2])\\ =((a-b)-(c-d),\left[b_2\right]-\left[d_2\right])\\ =h\left(ab\right)-h\left(cd\right)\end{array}$

Therefore, h is a homomorphism.

Now for$a\in \mathbb{Z}$ , find $h(a,0)$ as:

$\begin{array}{c}h(a,0)=a-0\\ =n\end{array}$

It shows that one-to-one mapping is possible.

Therefore, h is surjective.

Thus, we can write ker h as:

$\begin{array}{c}\ker h=\{(a,b):h\left(ab\right)=0\}\\ =\{(a,b):a-b=0,b\equiv 0\left(\mathrm{mod}2\right)\}\end{array}$

Which implies$(a,b)\in \ker h$ .

If $a-b=0$ then $a=b$ and b= 2x, $x\in \mathbb{Z}$.

Therefore, let us assume $\ker h$= $⟨(2x,2x)⟩$and simplify as:

$\begin{array}{c}(2x,2x)=x(2,2)\\ =\underset{xtimes}{(2,2)+(2,2)+\mathrm{...}+(2,2)}\end{array}$

This implies, $⟨(2,2)⟩\in \ker \text{\hspace{0.17em}}h$.

Therefore, by First Isomorphism Theorem,$\mathbb{Z}\times \mathbb{Z}/⟨(2,2)⟩\cong \mathbb{Z}\times {\mathbb{Z}}_2$ .