Universal Property of Abelianization of Group

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Theorem

Let $G$ be a group.

Let $G^{\operatorname {ab} }$ be its abelianization.

Let $\pi : G \to G^{\operatorname {ab} }$ be 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 $\sqbrk {G, G}$.

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:

$\map f {g^{-1} h^{-1} g h} = \map f g^{-1} \, \map f h^{-1} \, \map f g \, \map f g = 1 \in H$

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

$\blacksquare$