Character group of torsion-free abelian group

Question

Let $G$ be a torsion-free abelian group. Consider the character group $\hat G :=Hom_\mathbb Z (G,\mathbb C^\times)$ which is the group of all group homomorphisms from $G$ to $\mathbb C^\times$. When can we say that $\hat G$ is also torsion-free ?

Answer 1

A nonzero torsion element of $\hat{G}$ is a nontrivial homomorphism $G\to\mathbb{Z}/(n)$ for some $n>0$. If such a nontrivial homomorphism exists, then a nontrivial homomorphism $G\to\mathbb{Z}/(p)$ exists for some prime $p$. Such a homomorphism would factor through $G/pG$. Since $G/pG$ is a vector space over $\mathbb{Z}/(p)$, it has a nontrivial homomorphism to $\mathbb{Z}/(p)$ iff it is nontrivial.

So we conclude that $\hat{G}$ is torsion-free iff $G/pG$ is trivial for all primes $p$. Equivalently, $G=pG$ for all $p$ meaning that $G$ is divisible.