# Character group of torsion-free abelian group

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 ?

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

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.