How can $\mathbb{Z} \times \mathbb{Z}$ be generated by $(1,1)$ and $(0, 1)$? I read somewhere that $\mathbb{Z} \times \mathbb{Z}$ be generated by $(1,1)$ and $(0, 1)$ (taken together, I presume). I am trying to understand how this is true. I found this answer, based on which I developed the following argument:
The subgroup generated by $(1,1)$ is of the form $\{(m, n)\}$ where $m= n$. The subgroup generated by $(0, 1)$ is of the form $\{(0, n)\}$. But $\mathbb{Z} \times \mathbb{Z}$ also has elements of the form $\{(m, 0)\}$. How are they being generated by $(1,1)$ and $(0, 1)$?
 A: Consider the following: $(m,n)=m(1,1)+(n-m)(0,1)$. This shows that $\Bbb{Z} \times \Bbb{Z} \subseteq \langle (1,1), (0,1)\rangle$. The other inclusion is obvious.
To answer your second question:  If you want $(m,0)$, then we can have $$(m,0)=m(1,1)+(-m)(0,1).$$
A: One way to view $\Bbb Z\times\Bbb Z$ is by way of the presentation
$$\langle a, b\mid ab=ba\rangle.\tag{1}$$
Think of $a=(1,0)$ and $b=(0,1)$.
Introduce a new generator and relation (by way of Tietze transformations) in $(1)$ as follows:
$$\langle a, b, c\mid ab=ba, c=ab\rangle,\tag{2i}$$
which is equivalent to
$$\langle a, b, c\mid c=ba, cb^{-1}=a\rangle,\tag{2ii}$$
i.e.,
$$\langle b, c\mid c=bcb^{-1}\rangle.\tag{2iii}$$
This gives $\Bbb Z\times \Bbb Z$ as the group given by 
$$\langle b, c\mid cb=bc\rangle.\tag{3}$$
Now think of $c=ab$ as $(1,1)$.
A: If you think geometrically you may see why with integer combinations of the vectors $(1,1)$ and $(1,0)$ you can reach any vector with integer coordinates. In particular,
$$
(0,1) =  (1,1) - (1,0).
$$
(This is essentially @AnuragA 's correct answer, in another light.)
A: \begin{alignat}{1}
\mathbb{Z}\times\mathbb{Z} &= \{(m,n)\mid m,n\in\mathbb{Z}\} \\
&= \{m(1,1)+(n-m)(0,1)\mid m,n\in\mathbb{Z}\} \\
&= \{m(1,1)+l(0,1)\mid m,l\in\mathbb{Z}\} \\
&= \langle (1,1),(0,1)\rangle
\end{alignat}
Note that the last but one equality holds because, for every $n\in \mathbb{Z}$, the map $f_n\colon \mathbb{Z} \to \mathbb{Z}$, $m\mapsto n-m$ is surjective.
