Question

Prove that $(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩\cong \mathbb{Z}$.Show that $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z},$ given by $f\left((a,b)\right)=a-b$, is a surjective homomorphism

Short answer

It is proved that,$(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩\cong \mathbb{Z}$.

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 $\mathit{G}\mathbf{/}\mathit{K}$with kernel K. Then, the quotient group is isomorphic to H.

Proving that  (ℤ×ℤ)/⟨(1,1)⟩≅ℤ

Let $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$where $f\left(\left(ab\right)\right)=a-b$.

For $(a,b)(c,d)\in \mathbb{Z}\times \mathbb{Z}$, find $f((a,b)-(c,d))$ as:

$\begin{array}{c}f((a,b)-(c,d))=f(a-c,b-d)\\ =(a-c)-(b-d)\\ =(a-b)-(c-d)\\ =f\left(ab\right)-f\left(cd\right)\end{array}$

Therefore, f is a homomorphism.

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

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

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

Therefore, f is surjective.

Thus, we can write ker f as:

$\begin{array}{c}\ker f=\{(a,b):f\left(ab\right)=0\}\\ =\{(a,b):a-b=0\}\end{array}$

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

If $a-b=0$ then $a=b$.

Therefore, let us assume$\ker f$= $⟨(a,a)⟩$and simplify as:

$\begin{array}{c}(a,a)=a(1,1)\\ =\underset{atimes}{(1,1)+(1,1)+\mathrm{...}+(1,1)}\end{array}$

This implies,$⟨(1,1)⟩\in \ker \text{\hspace{0.17em}}f$ .

Therefore, byFirst Isomorphism Theorem,$(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩\cong \mathbb{Z}$ .