## Characterization of free abelian groups of finite rank

Let $S$ be a set. The group presented by $(S, R)$ where $R = \{ [s,t] \ |\ s,t \in S \}$ 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 $\varphi S \rightarrow G$ is a set function, then there is a unique group homomorphism $\Phi : A(S) \rightarrow G$ such that $\Phi|_S = \varphi$. Deduce that if $A$ is a free abelian group on a set of (finite) cardinality $n$, then $A \cong \mathbb{Z}^n$.

Recall from the universal property of free groups that there is a unique group homomorphism $\Psi : F(S) \rightarrow G$ such that $\Psi|_S = \varphi$. Moreover, clearly $R \leq \mathsf{ker}\ \Psi$. Then there is a unique group homomorphism $\Psi : A(S) \rightarrow G$ such that $\Psi = \Phi \circ \pi$. Identifying $s$ in $S$ with $sR$ in $A(S)$, we have $\Phi|_S = \varphi$. This gives existence.

To see uniqueness, note that any homomorphism $\Phi : A(S) \rightarrow G$ extending $\varphi$ is uniquely determined by $\Psi$, which is uniquely determined by $\varphi$.

Now let $n \in \mathbb{N}$ and let $S = \{1, \ldots, n \}$. If $n = 0$, then $A(S)$ is a quotient of $F(S) = 1$, so that in fact $A(S) = 1$. Suppose $n > 0$, and for each $1 \leq k \leq n$, define $\varphi(k)$ to be the $n$-tuple with 1 in the $k$th coordinate and 0 in all other coordinates. By the universal property, there exists a unique group homomorphism $\Phi : A(S) \rightarrow \mathbb{Z}^n$ extending $\varphi$; $\Phi$ is surjective since $\Phi[S]$ generates $\mathbb{Z}^n$. Now if $w = s_1^{e_1} \cdots s_n^{e_n} \in \mathsf{ker}\ \Phi$, note that since $A(S)$ is abelian, we can collect terms so that $s_i = i$. Now $\Phi(w) = (e_1, \ldots, e_n) = (0, \ldots, 0)$, and in fact $w = 1$. Thus $\Phi$ is injective, and we have $A(S) \cong \mathbb{Z}^n$.