So im asked to calculate $Hom_{Grp}(\Bbb{Q},\Bbb{Z})$, where $Hom_{Grp}$ stands for group homomorphisms. However, im not being able to translate the fact that a group homomorphism should comply with $f(xy)=f(x)f(y)$ into calculating $Hom_{Grp}(\Bbb{Q},\Bbb{Z})$. That is, lets take $f \in Hom_{Grp}(\Bbb{Q},\Bbb{Z})$. Then $f(0)=0$ and given $\frac{p}{q}, \frac{p'}{q'} \in \Bbb{Q}$, $$f(\frac{p}{q} + \frac{p'}{q'})=f(\frac{p}{q})+f(\frac{p'}{q'})$$
And now? how can I extract from there how my $f$ should more or less look like?
We claim that this group is the trivial group. (!)
To see this, let $n$ be the smallest positive integer in the image of some homomorphism $\phi$ with the desired properties. We can choose any pre-image of $n$, let's call it $a/b$. Then we have $n = \phi(a/b) = \phi(a/2b + a/2b) = \phi(a/2b) + \phi (a/2b) = 2\phi(a/2b).$ Thus $n/2$ is in the image, contradicting $n$ being the smallest positive integer.