TITLE: Homomorphisms from abelian groups to $S^1$ separate points
QUESTION [5 upvotes]: It is stated in the book Natural Dualities for the Working Algebraist by Clark and Davey (and also in the context of Pontryagin Duality) that for every abelian group $A$ and elements $a \neq b$ of $A$, there exists a homomorphism $f: A \rightarrow S^1$ such that $f(a) \neq f(b)$. Here, $S^1 = \{z \in \mathbb C \mid |z| = 1\}$ is the (multiplicative) circle group. Is there an easy argument we're missing?
In the case of $A$ being finitely generated it is easy to see using the classification of finitely generated abelian groups. This shows that it is enough to show that homomorphisms into finitely generated abelian groups separate points.

REPLY [8 votes]: Note that $S^1$ is a divisible group. This is equivalent with the fact that for every abelian group $A$ and every subgroup $B\subseteq A$ every homomorphism $f:B\rightarrow S^1$ can be extended $f$ to $\tilde{f}:A\rightarrow S^1$.
Now this property implies that it suffices to show that $S^1$ separates elements of finitely generated groups. Indeed, pick any abelian group $A$ and two of its elements $a_1$, $a_2$. Clearly the subgroup $B$ generated by $a_1$ and $a_2$ is finitely generated. Take the separating homomorphism $f:B\rightarrow S^1$ and extend it to $\tilde{f}:A\rightarrow S^1$.
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