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If $G$ is a nonzero abelian group show that

$$\operatorname{Hom}_{\Bbb Z}(G,\frac{\Bbb Q}{\Bbb Z}) \neq \{0\}.$$

user26857
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erfan
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1 Answers1

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Let $f: n \mathbb{Z} \rightarrow \mathbb{Q}/ \mathbb{Z}$ be any homomorphism of groups. Then we may extend this homomorphism to $\tilde{f}: \mathbb{Z} \rightarrow \mathbb{Q}/ \mathbb{Z}$ by putting $\tilde{f}(1)=f(n)/n$. Thus by Baers Criterion $\mathbb{Q}/ \mathbb{Z}$ is an injective group.

Now pick any nonzero element in $g\in G$ and define a nontrivial homomorphism $f: <g> \rightarrow \mathbb{Q}/ \mathbb{Z}$( we can set $f(g)$ to be any nonzero element of $\mathbb{Q}/ \mathbb{Z}$ if the order of $g$ is infinite. If the order of $g$ is $n$, then we set $f(g)=1/n$). Since $\mathbb{Q}/ \mathbb{Z}$ is injective we can extend $f$ to a nonzero morphism $\tilde{f}: G \rightarrow \mathbb{Q}/ \mathbb{Z}$.

math635
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