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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
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    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
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    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 07:49

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

But stepping back to general groups, we have to encode the Abelianness, i.e. that each pair of elements commute. It is enough to pose that each pair of generating elements commute.
And that's exactly the extra data in in the formula of the presentation of abelianization among groups.

Berci
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