Presentation of the abelianization of $G$.

The abelianization of a group $G$ is an abelian group $A$ and a homomorphism $\varphi: G \to A$ such if $B$ is any abelian group, and $\phi: G \to B$ is any homomorphism, there is a unique homomorphism $\psi: A \to B$ (which might depend on $\phi$) such that $\psi\varphi = \phi$.

Now, I am reading some lecture notes, and the following is asserted.

If $G = \langle e_1, e_2, \ldots, e_n \mid w_1, w_2, \ldots, w_m\rangle$ is a finitely presented group, then$$A = \langle e_1, e_2, \ldots, e_n \mid w_1, \ldots, w_m, [e_1, e_2], \ldots, [e_i, e_j], \ldots, [e_{n - 1}, e_n]\rangle$$is a presentation of the abelianization of $G$, where the homomorphism $\varphi: G\to A$ sends the equivalence class of $w$ in $G$ to the equivalence class of $w$ in $A$ for each word $w \in G$.

To me, this is not a priori clear at all. Could anybody tell me why this is true?

• As you seem to be diving deeper and deeper in this, perhaps it'd be a good idea if you google and research a little about the commutator subgroup, or the derived subgroup of a group. – DonAntonio Oct 23 '16 at 23:04
• The idea is that if we have a homomorphism $G \to B$ to an abelian group $B$, then the kernel must contain all the different $[e_i, e_j]$. – Arthur Oct 23 '16 at 23:04
• Have you tried to prove that this $A$ is actually the abelianization of $G$? Meaning, did you take any abelian group $B$ and a group morphism $G \to B$, then tried to see why it should factor through $A$? In the course of the proof, it should become clear, a posteriori, why this is true. – Pece Oct 24 '16 at 7:49

Actually, in the category of Abelian groups, the same presentation $\langle e_1,\dots\,|\, w_1,\dots\rangle$ works for the abelianization of $G$.