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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 ?

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  • $\begingroup$ Are characters in $\hat G$ required to be continuous? In the f.g. case, the answer is never: $\hat{\mathbb{Z}} = \mathbb{R}/\mathbb{Z}$. $\endgroup$ – anomaly Feb 1 at 4:07
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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.

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