Show that there is no epimorphism from ($\mathbb{Q}$, +) to ($\mathbb{Z}$, +)

I can't figure out why such an epimorphism cannot exist.

I think the fact that a bijection exists between the $\mathbb{Q}$ and $\mathbb{Z}$ throws me off a little. What is it about having a homomorphic surjection that causes problems?

Solution 1:

Daniel's hint should suffice, but here's another approach:

1) Show $\;\Bbb Q\;$ is divisible, meaning: for any natural $\;n\;$ and any $\;g\in \Bbb Q\;$ there exists $\;x\in \Bbb Q\;$ s.t. $\;nx=g\;$ ;

2) If $\;G\;$ is divisible and $\;\phi:G\to H\;$ is a group homomorphism, then $\;\phi(G)\;$ is also divisible

3) $\;\Bbb Z\;$ is not divisible

Extra stuff not related to question: the only finite group that is divisible is the trivial one.

Solution 2:

Thanks to Daniel Fischer and Praphulla Koushik for providing the hints.

Suppose there exists an epimorphism $f:\mathbb{Q}\to\mathbb{Z}$. Thus, there's $x\in\mathbb{Q}$ such that $f(x)=1$.

$$1=f(x)=f\left(2\cdot\frac{x}{2}\right)=2f\left(\frac{x}{2}\right) \Rightarrow \frac{1}{2}=f\left(\frac{x}{2}\right)$$

Which is a contradiction, since $f\left(\frac{x}{2}\right)\in\mathbb{Z}$.