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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!

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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 $G \cong A(G)/\ker \varphi$.

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  • $\begingroup$ I feel like using the universal property here kind of obscures things. Just take $A(G)$ and mod out by the relations $a+b-c$ for every $a, b, c\in G$ such that $a+b=c$. $\endgroup$
    – Noah Schweber
    Commented Jan 20, 2016 at 1:00
  • $\begingroup$ I do prefer the construction using the universal property of the free abelian group, because I fell that it is more rigorous and is hiding less formally needed things. When moding out the corresponding relations we (strictly speaking) still have to check that we resulting group is not too small, i.e. that we did not accidently mod out anything we needed. And I don’t see how one would prove this without using the universal property. To formally prove that your quotient is isomorphic to $G$ we also need the universal property. $\endgroup$
    – Jendrik Stelzner
    Commented Jan 20, 2016 at 1:24

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