# Intersection of kernels of elements of $\operatorname{hom}_\mathbb{Z}(G,\mathbb{Q}/\mathbb{Z})$ is $0$?

Suppose $G$ is an abelian group, and $G^\ast=\operatorname{hom}_\mathbb{Z}(G,\mathbb{Q}/\mathbb{Z})$ it's Pontrjagin dual/character module. If $a\in G$ is in the kernel of every element of $G^\ast$, why is $a=0$ necessarily?

I ask because the existence of a homomorphism not annihilating a nonzero element of $G$ is used in providing that a sequence of modules is exact iff the sequence of character modules is exact.

• Are you familiar with the result that $\Bbb{Q}/\Bbb{Z}$ is an injective object in the category of abelian groups? Meaning that a homomorphism $f:H\to\Bbb{Q}/\Bbb{Z}$ from any subgroup $H\le G$ can be extended to a homomorphism $F:G\to\Bbb{Q}/\Bbb{Z}$ (that is, $F(h)=f(h)$ for any $h\in H$). Apply that to the case where $H$ is the subgroup generated by $a$, and $f$ is a homomorphism that sends $a$ to a suitable element of $\Bbb{Q}/\Bbb{Z}$. – Jyrki Lahtonen Mar 2 '17 at 19:21
• @JyrkiLahtonen Oh right, thanks! Because it's divisible over a PID. If $a$ has finite order $n$, we could map $a$ to the class of $1/n$ and extend homomorphically. What if $a$ has infinite order? Wouldn't it have to be mapped to $0$ since $\mathbb{Q}/\mathbb{Z}$ is torsion? – Ankita Desari Mar 2 '17 at 19:29
• Oops, I think it's fine to just map $a$ anywhere since $\langle a\rangle$ is free on one generator? – Ankita Desari Mar 2 '17 at 19:32
• Correct. Well done. If you feel like fleshing that out to an answer, please do so. That way you may get more feedback. – Jyrki Lahtonen Mar 2 '17 at 20:11