Free group is equal to $\mathbf{Z}^2$

Question

I would like to show that the free abelian group $\langle a, b : aba^{-1}b^{-1} = 1 \rangle$ is equal to $\mathbf{Z}^2$. A friend of mine suggested to use the Van-Kampen-theorem. However, I was hoping for a more elementary way to show this (e.g. give the explicit isomorphism).

Thanks!

Answer 1

The isomorphism should be mostly obvious: To specify a homomorphism $f\colon \langle\, a,b:aba^{-1}b^{-1}=1\,\rangle\to\Bbb Z^2$ it is necessary and sufficient to specify elements $f(a),f(b)$ with $f(a)+f(b)-f(a)-f(b)=0$. Since $\Bbb Z^2$ is abelian, the latter is not a restriction. This allows us to let $f(a)=(1,0)$ and $f(b)=(0,1)$ to obtain an epimorphism. Now verify that $(n,m)\mapsto a^nb^m$ is a homomorphism, and is inverse to $f$, so that we have an isomorphism.

Answer 2

Denote by $F = \langle a,b : ab=ba \rangle$.

Then it is easily checked (but it is boring) that the following $f: \Bbb{Z}^2 \to F$ is a group isomorphism $$(n,m) \mapsto a^nb^m$$

Answer 3

Given that $\langle a\rangle \simeq \mathbb{Z}$ by definition, it follows that $\langle a,b\mid ab=ba\rangle \simeq \mathbb{Z}\times \mathbb{Z}$. So the group is not free, but free-abelian of rank $2$.