Universal Property of Abelianization of Group

Theorem
Let $G$ be a group.

Let $G^{\operatorname{ab}}$ be its abelianization and $\pi : G \to G^{\operatorname{ab}}$ the quotient group epimorphism.

Let $H$ be an abelian group.

Let $f : G \to H$ be a group homomorphism.

Then there exists a unique group homomorphism $g : G^{\operatorname{ab}} \to H$ such that $g \circ \pi = f$:
 * $\xymatrix{

G \ar[d]_\pi \ar[r]^{\forall f} & H\\ G^{\operatorname{ab}} \ar[ru]_{\exists ! g} }$

Proof
By Universal Property of Quotient Group, it suffices to show that the kernel $\ker f$ contains the commutator subgroup $[G, G]$.

By definition of generated subgroup, this is equivalent to showing that $\ker f$ contains all commutators of $G$.

Let $g, h \in G$.

Then by Group Homomorphism Preserves Inverses:
 * $f(g^{-1}h^{-1}gh) = f(g)^{-1}f(h)^{-1}f(g)f(h) = 1 \in H$.

Thus the commutator $[g, h] = g^{-1}h^{-1}gh$ is in $\ker f$.