Free group on two generators and commutators. Why it's enough to add the relation ab=ba?

Question

I've looked through lots of question on this topics, but I cannot find what I want to prove:

I've seen in a lots of exercises sheets that the abelianization of a free group with two generators (let's call it $G$) is in fact the group given with the presentation $\langle a,b | ab=ba \rangle$. In fact I added the relationship $ab=ba$ forcing commutativity. Intuitively it seems clear to me.

So I tried proving it, let's call $N$ the smallest normal set containing $aba^{-1}b^{-1}$, $G/N$ is by definition $\langle a,b | ab=ba \rangle$

With the rule $ab=ba$ I can move all the powers of $a$ in a word to the left and all the power of $b$ to the right. So the quotient is abelian, and so we have $[G,G] \subseteq N$. For the other inclusion, we can argue in this way: $aba^{-1}b^{-1} \in [G,G]$ by definition, and so $N \subseteq [G,G]$.

Is it correct?

Then to prove the isomorphism between $G/N$ and $\mathbb{Z} \times \mathbb{Z}$ I would argue as this:

Consider $\phi \colon G \to \mathbb{Z} \times \mathbb{Z}$ defined in this way $\phi(a)=(1,0)$ and $\phi(b)=(0,1)$, and extend by linearity. Thanks to the fact that $\mathbb{Z} \times \mathbb{Z}$ is abelian, the map sum the coefficients of $a$ and the sum of the coefficient of $b$. Clearly it's surjective, and the kernel of such map is precisely the words that are killed if we pass to the quotient $G/N$ (because we would be allowed to move and simplify every power of $a$ and $b$. So the induced maps is injective and surjective and a morphism so we are done.

Is it correct again?

Sorry for posting elementary question, these are my first step with these arguments, so I want to be sure I'm doing it properly

Answer 1

First proof is correct and pretty "efficient". Second proof is correct in idea, but notions that were used not really correct.

1. "Consider $\phi:G→ℤ×ℤ$ defined in this way $\phi(a)=(1,0)$ and $\phi(b)=(0,1)$, and extend by linearity". Word linearity in abstract group is not correct, usually it is used in linear spaces. So in fact you can uniquely extend any map of generators by definition of the Free Group.

2. Secondly if you are considering some mapping $\phi:G→ℤ×ℤ$ and it is surjective then by theorem on homomorphisms $G/Ker(\phi) \simeq ℤ×ℤ$. So you have to find $Ker(\phi)$. After that make sure that $Ker(\phi) = [G, G]$ (in this problem) and the second item will be proved.