Is there a surjective homomorphism from $(\Bbb{Q},+)$ to $(\Bbb{Z},+)$?

Question

Consider the groups $G = (\Bbb{Q},+)$ and $H = (\Bbb{Z},+)$. Is there a surjective homomorphism from $G$ to $H$? If not, how can I prove there isn't?

I considered a homomorphism that rounds up or down but I saw these operations are not "friendly" with the addition.

Answer 1

Let $\varphi: G\to H$ be a homomorphism.

If $\varphi(g)=1$, then

$$\begin{align} 1&=\varphi\left(\frac{g}{2}+\frac{g}{2}\right)\\ &=\varphi\left(\frac{g}{2}\right)+\varphi\left(\frac{g}{2}\right)\\ &=2\varphi\left(\frac{g}{2}\right), \end{align}$$

but then $\varphi\left(\frac{g}{2}\right)\notin\Bbb Z$.

Answer 2

Not only is there no surjection, there is no nontrivial additive map $\mathbb{Q}\to\mathbb{Z}$.

Note that for every $q\in\mathbb{Q}$ and every positive integer $n$, there exists an element $r\in\mathbb{Q}$ such that $nr=q$. This fact respected under a homomorphism, since $nf(r)=f(rn)=f(q)$. (This means that the additive group of rationals is divisible; homomorphic images of divisible groups are divisible).

So the image of any $q\in\mathbb{Q}$ must be an integer $a$ with the property that for every positive integer $n$, there is an integer $b$ such that $nb=a$. The only such integer is $0$ (in which case $b=0$ as well), so an additive map $f\colon\mathbb{Q}\to\mathbb{Z}$ must send everything to $0$.