Group homomorphism between $\Bbb{Q}$ and $\Bbb{Q}\times\Bbb{Q}$ I am trying to find out the number of homomorphisms from $(\mathbb{Q},+)$ to $(\mathbb{Q}\times \mathbb{Q},+)$. All I understood is that if $f$ is a homomorphism, then it must satisfy
$f(0)=f(0,0)$. Moreover,
How do I make the conclusion that there exists no isomorphisms between them?
 A: There's a lot more we can tell about a homomorphism $f:\mathbb{Q}\to G$ for any abelian group $(G,+)$. If $n\in\mathbb{N}$ then
$$f(n)=f(1+\cdots+1)=f(1)+\cdots+f(1)=nf(1)$$
Now if $q\in\mathbb{N}$ is nonzero, then
$$f(1)=f\bigg(\frac{q}{q}\bigg)=f\bigg(\frac{1}{q}+\cdots+\frac{1}{q}\bigg)=f\bigg(\frac{1}{q}\bigg)+\cdots+f\bigg(\frac{1}{q}\bigg)=qf\bigg(\frac{1}{q}\bigg)$$
and thus $f\big(\frac{1}{q}\big)=\frac{1}{q}f(1)$ if $G$ is a divisible group (like $\mathbb{Q}\times\mathbb{Q}$). Combining both together, and extending to negatives, we obtain that if $G$ is a divisible group, then
$$f\bigg(\frac{p}{q}\bigg)=\frac{p}{q}f(1)$$
Note the subtlety: on the right side we don't have a multiplication, but rather a solution to $qf(p/q)=pf(1)$.
In the case when $G=\mathbb{Q}\times\mathbb{Q}$ we obtain that
$$f\bigg(\frac{p}{q}\bigg)=\bigg(\frac{p}{q}v_1, \frac{p}{q}v_2\bigg)$$
for some fixed $v_1,v_2\in\mathbb{Q}$. Here on the right side we have the standard rationals multiplication (which corresponds to the solution to the equation mentioned earlier).
So can $f$ be an isomorphism? If $v_1=v_2=0$ then it is the zero morphism. On the other hand if $v_1\neq 0$, then $(0,1)$ does not belong to the image of $f$. If $v_2\neq 0$ then $(1,0)$ does not belong to the image of $f$. Hence $f$ is never surjective.
As an alternative we can utilize linear algebra. The formula $f(k)=kf(1)$ also means that every group homomorphism $f:\mathbb{Q}\to\mathbb{Q}^n$ is also a linear map over $\mathbb{Q}$. And thus we can simply use the dimension argument to conclude that it cannot be an isomorphism when $n>1$.
