Let $S$ be a set. The group with presentation $(S, R)$ , where $R = \{\ [s , t] \mid s,t\in S\ \}$ is called the free abelian group on $S$ -- denote it by $A (S)$ . Prove that $A (S)$ has the following universal property: if $G$ is any abelian group and $\varphi:S\to G$ is any set map, then there is a unique group homomorphism $\phi : A (S) \to G$ such that $\phi\mid_S=\varphi$ . Deduce that if $A$ is a free abelian group on a set of cardinality $n$ then $$ > A\cong\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}\ \ (n\text{ factors})\ . > $$
For all $N\unlhd G$ containing $R$ , we have $[s,t]\in R\subseteq N$ , then $[s,t]N=N$ , and so $$ \begin{aligned} \,[s,t]\langle R\rangle&=[s,t]\bigcap_{R\subseteq N\unlhd F(S)}N \\&=\bigcap_{R\subseteq N\unlhd F(S)}N \\&=\langle R\rangle\ , \end{aligned} $$ hence $$ \begin{aligned} st\langle R\rangle&=ts\langle R\rangle \\s\langle R\rangle\,t\langle R\rangle&=t\langle R\rangle\,s\langle R\rangle \end{aligned} $$ implies $F(S)/\langle R\rangle$ is abelian.
Denote $F(S)'$ the commutator subgroup of $F(S)$ . Let $\pi':F(S)\to F(S)/F(S)'$ and $\pi_R:F(S)\to F(S)/\langle R\rangle$ be the natural projection with $\ker\pi_R=\langle R\rangle$ . Then by the universal property of commutator subgroups, we have the following commutative diagram:
Then by the universal property of free groups, the diagram below also commutes:
Combining them I get
But I do not know how to continue this work to obtain a group homomorphism as desired between the two abelian groups $F(S)/\langle R\rangle$ and $G$ .
Lemma. The normal subgroup of $F(S)$ generated by $R$, $\langle R\rangle^{F(S)}$, is precisely $[F(S),F(S)]$.
Proof. Since $R$ is generated by elements of $[F(S),F(S)]$, then we have $\langle R\rangle^{F(S)}\leq [F(S),F(S)]$.
Conversely, using the commutator identities (I use the convention $[a,b]=a^{-1}b^{-1}ab$; similar identities hold for the other convention), we have: $$\begin{align*} [x,zy] &= [x,y][x,z]^y\\ [xz,y] &= [x,y]^z[z,y]\\ [x,y^{-1}] &= [y,x]^{y^{-1}}\\ [x^{-1},y] &= [y,x]^{x^{-1}} \end{align*}$$ it follows that any element of $[F(S),F(S)]$ can be written as a product of conjugates of elements of the form $[s,t]$ with $s,t\in S$, so $[F(S),F(S)]\leq \langle R\rangle^{F(S)}$, giving equality.
(Alternatively, you already proved that $[F(S),F(S)]\leq \langle R\rangle^{F(S)}$ using the universal property of the abelianization, but I figured one universal property at a time is enough). $\Box$
Proof of the Universal Property.
Existence. Let $G$ be an abelian group, and let $\varphi\colon S\to G$ be a set map. Then there exists a (unique) morphism $\Phi\colon F(S)\to G$ such that $\Phi\circ i = \varphi$, where $i\colon S\to F(S)$ is the inclusion of $S$ to the free group on $S$.
Now we have a morphism $\Phi\colon F(S)\to G$, where $G$ is abelian. Therefore, there exists $\Psi\colon F(S)/[F(S),F(S)]\to G$ such that $\Phi=\Psi\circ\pi$, where $\pi\colon F(S) \to F(S)/[F(S),F(S)]$ is the canonical projection. But $[F(S),F(S)]=\langle R\rangle^{F(S)}$, so the domain of $\Psi$ is actually $A(S)$. Note that the inclusion of $S$ into $A(S)$ is $\pi\circ i$. Thus, we have a morphism $\Psi\colon A(S)\to G$ such that $$\Psi|_S = \Psi\circ (\pi\circ i) = (\Psi\circ \pi)\circ i = \Phi\circ i = \varphi,$$ as desired.
Uniqueness. If $F\colon A(S)\to G$ is such that $F\circ (\pi\circ i) = \varphi$, then $F$ and $\Psi$ agree on $S$; since $A(S)$ is generated by $S$, it follows that $F=\Psi$, as required. $\Box$