Abelian Group as the quotient of a free Abelian Group

Question

Is it true that every abelian group $G$ is the quotient of a free abelian group $F$?

I think so, since every abelian group $G$ is the quotient of a free group $H$ under some relations, but some of them are the commutators of the generators, thus if I quotient by those relations first, I get a free abelian group $F$, then I can take the quotient of the other relations getting the result as requested.

Thanks!

Answer 1

Yours is one possible way to show this.

One can also directly use the universal property of the free abelian group: Given an abelian group $G$ let $A(G)$ be the free abelian group on the underlying set of $G$. Then by the universal property of the free abelian group there exists a unique group homomorphism $\varphi \colon A(G) \to G$ with $\varphi(g) = g$ for all $g \in G$. Then $\varphi$ is surjective and therefore $G \cong A(G)/\ker \varphi$.