Proof of the universal property of the abelianisation

Proof

We show that $\varphi$ kills the commutator subgroup, and then apply the universal property of the quotient group.

1. Let $G$ be a group, let $A$ be an abelian group with identity $e_A$, and let $\varphi \colon G \to A$ be a group homomorphism.
2. By the definition of the abelianisation, $G^{\mathrm{ab}} = G/[G, G]$, and $\pi(g) = g[G, G]$.
3. Let $g, h \in G$. Since homomorphisms preserve inverses, $\varphi([g, h]) = \varphi(g)^{-1} \varphi(h)^{-1} \varphi(g) \varphi(h)$, where $[g, h] = g^{-1}h^{-1}gh$ is the commutator.
4. Since $A$ is abelian, we may reorder the factors in step 3. This gives $\varphi([g, h]) = \varphi(g)^{-1} \varphi(g) \varphi(h)^{-1} \varphi(h) = e_A$.
5. So the set $C$ of all commutators lies in the kernel $\ker \varphi$.
6. By the kernel is a normal subgroup, $\ker \varphi$ is a subgroup of $G$ that contains $C$.
7. The commutator subgroup $[G, G]$ is the subgroup generated by $C$, the intersection of all subgroups that contain $C$. By step 6, $[G, G] \subseteq \ker \varphi$.
8. By step 7 and the universal property of the quotient group with $N = [G, G]$, there is exactly one homomorphism $\bar{\varphi} \colon G^{\mathrm{ab}} \to A$ with $\bar{\varphi} \circ \pi = \varphi$.