I'm confused about exactly how a subgroup of $\Bbb Z \times \Bbb Z$ can be isomorphic to $\Bbb Z$

Question

I have read that there exists a subgroup $H$ of $\Bbb Z \times \Bbb Z$ such that $H \cong \Bbb Z$.

If $\Bbb Z \times \Bbb Z:=\{(a,b)|a,b \in \Bbb Z\}$

With the group operation $(a,b)+(c,d)=(a+c, b+d)$.

Then by defining a map

$\phi:\Bbb Z \times \Bbb Z \rightarrow \Bbb Z$

$\phi:(a,b) \rightarrow c $

Then I would assume that the only isomorphic subgroup would be $H:=\{(0,d)|d \in \Bbb Z \}$. Is that correct?

Answer 1

Think about the "lines" of $\mathbb{Z} \times \mathbb{Z}$, i.e

$$\{(an, bn) \mid n \in \mathbb{Z}\}$$

with $a, b \in \mathbb{N}_0$ fixed.

Answer 2

What about the diagonal subgroup?

$$H=\{(d,d):d \in \mathbb{Z}\}$$