Unique morphism from the additive group $\mathbb Q$ to $\mathbb Z$ I am trying to prove that the only group homormorphism from $\mathbb Q$ to $\mathbb Z$ is the trivial one but I couldn't
Suppose there is $x \in \mathbb Q$ : $f(x)=z \neq 0$. We can write $x=\dfrac{m}{n}$ and since $nz=f(\dfrac{m}{n}n)=mf(1)$, we deduce $f(1) \neq 0$. We also have $mf(\dfrac{1}{n})=f(\dfrac{m}{n})=z$, so $f(\dfrac{1}{n})=\dfrac{z}{m} \in \mathbb Z$. It follows $m=-1,1,z$.
I don't know how to continue from here, any hint would be appreciated.
 A: Here is a simpler approach. Suppose $f(x) = n \ne 0$. Then
$$2n f\left(\frac{x}{2n}\right) = \underbrace{f\left(\frac{x}{2n}\right) + \ldots + f\left(\frac{x}{2n}\right)}_{2n \text{ times}} = f\left(\underbrace{\frac{x}{2n} + \ldots + \frac{x}{2n}}_{2n \text{ times}}\right) = f(x) = n,$$
which is impossible because no integer multiplied by $2n$ is $n$.
A: More generally, the group $\mathbb{Q}$ has the property that, for all $x\in\mathbb{Q}$ and all integers $n>0$, there exists $y\in\mathbb{Q}$ such that $ny=x$.
Let's stick to abelian groups, written additively.
An abelian group $G$ with this property is called divisible; more formally,

the abelian group $G$ is divisible if, for all $x\in G$ and all integers $n>0$, there exists $y\in G$ such that $ny=x$.

Now you can prove that, if $f\colon G\to G'$ is a group homomorphism and $G$ is divisible, then $\operatorname{im}f$ (the image of $f$) is a divisible subgroup of $G'$. What are the divisible subgroups of $\mathbb{Z}$? Can you conclude now?
There is another related concept. Consider an abelian group $G$; then $nG=\{nx:x\in G\}$ is a subgroup of $G$, for all integers $n>0$.

An abelian group $G$ is reduced if $\bigcap_{n>0}nG=\{0\}$.

For example $\mathbb{Z}$ is reduced.
Theorem. If $G$ is divisible and $G'$ is reduced, then the only homomorphism $f\colon G\to G'$ is the trivial homomorphism: $f(x)=0$ for all $x\in G$.
