The dual for Abelian group on Q/Z

Question

Let $A$ be any Abelian group. Take $A^\star:=\operatorname{Hom}(A,\Bbb{Q}/\Bbb{Z})$ be the dual of $A$. Then is $A=0$ equivalent to $A^\star=0$?

Answer 1

Yes. Note that $\mathbb Q / \mathbb Z$ is a divisible group, thus an injective object in the category of abelian groups, thus any homomorphism into it can be lifted from subgroups.

Take any $x\in A$. If it has finite order $n$, define $f(x)={1\over n}$, otherwise define $f(x)$ to be any nonzero element of $\mathbb Q / \mathbb Z$. This defines a nonzero homomorphism $f: \langle x \rangle \rightarrow \mathbb Q / \mathbb Z$. By injectivity of $\mathbb Q / \mathbb Z$ this can be extended to a nonzero homomorphism $A \rightarrow \mathbb Q / \mathbb Z$. Thus, if $A$ is nonzero, $A^*$ is nonzero as well.

Answer 2

Yes, every non-zero Abelian group has a homomorphism onto either a non-zero cyclic group or $\Bbb Q$ or a Prufer group. All these groups have homomorphisms onto non-zero subgroups of $\Bbb{Q}/\Bbb Z$. So the dual of every bon-zero Abelian group is not zero.