How can $\mathbb{Z} \times \mathbb{Z}$ be generated by $(1,1)$ and $(0, 1)$?

Question

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 $(0, 1)$?

Answer 1

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 $$(m,0)=m(1,1)+(-m)(0,1).$$

Answer 2

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 $(1,1)$.

Answer 3

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.)

Answer 4

\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.