Is there a onto group homomorphism from $\Bbb Q$ to $\Bbb Z$?

Question

I am learning group-homomorphisms. I have two questions:

1. Is there a onto group homomorphism from $\Bbb Z$ to $\Bbb Q$?
2. Is there a onto group homomorphism from $\Bbb Q$ to $\Bbb Z$?

I have the answer of the first one.

1. $\Bbb Z$ is cyclic and homomorphic image of a cyclic group is cyclic but $\Bbb Q$ is not.
2. I am stuck here. Please help me.

Answer 1

Your answer to the first question is correct.

For the second question, suppose that there were an onto homomorphism $f : \mathbb{Q} \to \mathbb{Z}$. Then there exists some $q \in \mathbb{Q}$ such that $f(q) = 1$. But then, $x = f(q/2)$ is an integer satisfying $$2x = x+x = f(q/2) + f(q/2) = f(q/2 + q/2) = f(q) = 1,$$ which is impossible. Therefore there is no such $f$.

Answer 2

The argument given for the first question is correct. Indeed, homomorphic image of a cyclic group is necessarily cyclic. $\mathbb Q$ is non-cyclic so there can't be an onto group homomorphism $\mathbb Z\to\mathbb Q$.

There's no onto group homomorphism $\mathbb Q\to\mathbb Z$ either, but you can prove something stronger i.e., the only group homomorphism $\mathbb Q\to\mathbb Z$ is the trivial one.

Let $f\colon\,\mathbb Q\to\mathbb Z$ be a group homomorphism. For each $n\in \mathbb Z^+$, we have $\begin{align}f(1)&=f\bigg(\underbrace{\frac{1}{n}+\ldots+\frac{1}{n}}_{\text{$n$ times}}\bigg)\\& = \underbrace{f\!\left(\frac{1}{n}\right)+\ldots+f\!\left(\frac{1}{n}\right)}_{\text{$n$ times}}\\ &= n\,f\!\left(\frac 1n\right)\end{align}\tag*{}$

Thus, for every $n\in \mathbb Z^+$, we have $n\mid f(1)$. It follows that $f(1)=0$.

Now, $0=f(1)=n\,f\!\left(\frac 1n\right)\implies f\!\left(\frac 1n\right)=0$ for each $n\in\mathbb Z^+$.

$\implies f\!\left(\frac mn\right) = m\times 0=0$ for all $m\in\mathbb Z$ and $n\in\mathbb Z^+$.

Thus, $f$ is the trivial map.