How to think about the presentation of $\Bbb Z \times\Bbb Z$

Question

I keep trying to understand this basic fact that $\mathbb{Z} \times \mathbb{Z}$ has a presentation $G = (x, y \vert xy = yx)$. I am stuck trying to understand the map between tuples $(x, y)$ in $\mathbb{Z} \times \mathbb{Z}$ and the scalars that result when I generate words from $G$. Clearly the tuples $(x, y)$ and $(y, x)$ are different in $\mathbb{Z} \times \mathbb{Z}$, so where does the commutative relation come from in the presentation?

I looked at this question but the explanation "since the exponents are just adding" really did not clarify it for me.

Answer 1

The components of this presentation notation are

$$\langle \underbrace{ {x, y}}_{\text{names for generators of the group}} \mid \underbrace{xy = yx}_\text{relations between the generators} \rangle$$

It has nothing to do with tuples.

Let $G$ be the group given by this presentation. To obtain an isomorphism $f : G \to \mathbb{Z} \times \mathbb{Z}$ you might:

• Choose two elements $u, v \in \mathbb{Z} \times \mathbb{Z}$ to be the image of $x$ and $y$ respectively; that is, to set $f(x) = u$ and $f(y) = v$.
• Verify that all of the relations hold after substituting $u$ for $x$ and $v$ for $y$; that is, check $f(xy) = f(yx)$, or equivalently, that $u + v = v + u$.
• Find a homomorphism $\mathbb{Z} \times \mathbb{Z} \to G$ that is inverse to $f$

The first bullet point is enough to define a homomorphism from the free group on two elements to $\mathbb{Z} \times \mathbb{Z}$. The second bullet point shows this homomorphism is well-defined on the group $G$. The third bullet point shows that it is an isomorphism.

Ultimately, it's not hard to guess at the choice of $u$ and $v$; the very first thing that springs to mind in response to the question "Give me a pair of elements of $\mathbb{Z} \times \mathbb{Z}$ that generate the group!" are probably going to work.