Compute $\operatorname{Hom}_{\mathbb{Z}} ( \mathbb{Q}, \mathbb{Z})$

Solution 1:

Let $\varphi: \mathbb Q \rightarrow \mathbb Z$ suppose that $\varphi$ is non-trivial. Let $n$ be the smallest positive integer in $\mathrm{im}\; \varphi$. Pick $a/b \in \varphi^{-1}(n)$. Then

$$n=\varphi(a/b)=\varphi(a/2b+a/2b)=\varphi(a/2b)+\varphi(a/2b),$$

so $\varphi(a/2b)=n/2$, but this is impossible because $n$ was the smallest positive integer in the image. Thereby no non-trivial homomorphisms exist.

Solution 2:

$\,\Bbb Q\,$ is a divisible group , so is any of its homomorphic images. But $\,\Bbb Z\,$ is not a divisible group (why?) , so it must be that the image of any group homomorphism $\,\Bbb Q\to\Bbb Z\,$ is the trivial subgroup (which, BTW, is the only finite group which is divisible and the only subgroup of $\,\Bbb Z\,$ that is divisible), and thus $\,\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Z)=0\,$