# Abelianization of Free Group is Free Abelian Group

## Theorem

Let $X$ be a set.

Let $\struct {F_X, \iota}$ be a free group on $X$.

Let $F_X^{\mathrm {ab} }$ be its abelianization.

Let $\pi : F_X \to F_X^{\mathrm {ab} }$ be the quotient group epimorphism.

Then $\struct {F_X^{\mathrm {ab} }, \pi \circ \iota}$ is a free abelian group on $X$.

## Proof

$\xymatrix { X \ar[d]_\iota \ar[rd]^{\forall f} & \\ F_X \ar[d]_\pi \ar[r]^{\exists ! g} & H \\ F_X^{\mathrm {ab} } \ar[ru]_{\exists ! h} }$

Lef $H$ be any abelian group and $f:X \to H$ be any mapping.

By Universal Property of Free Group on Set, there exists a unique group homomorphism $g:F_X \to H$ such that $g \circ \iota = f$.

By Universal Property of Abelianization of Group, there exists a unique group homomorphism $h:F_X^{\mathrm {ab} } \to H$ such that $h \circ \pi = g$.

Therefore, $h \circ (\pi \circ \iota) = f$.

Hence, $F_X^{\mathrm {ab} }$ satisfies the Universal Property of Free Abelian Group on Set.

Hence $F_X^{\mathrm {ab} }$ is a Free Abelian Group on $X$.

$\blacksquare$