Are the following groups isomorphic? I have to find out if these groups are isomorphic:


*

*$( { \mathbb {Q} };+) $ and $(\mathbb{Z}\times\mathbb{Z};+)$ - (this exercise is the most important for me)

*$( { \mathbb {Q} };+)$ and $( { \mathbb {Q} \times \mathbb {Q} };+)$

*$(ℝ;+)$and $(\mathbb{R}\times \mathbb{R};+)$
I have tried to solve 1. and first of all I have tried to find some property which could show that these two groups can not be  isomorphic. But I have not find such property so I think that $( { \mathbb {Q} };+) $ and $(\mathbb{Z}\times\mathbb{Z};+)$ are isomorphic, but I can not to prove it, I can not find some isomorphism.
Thank you very much for any help.
 A: *

*Every homomorphism from $\mathbb Q$ to $C_2$ is trivial. Indeed, let $x \in \mathbb Q$. Write $x=2y$. If $y \mapsto g$, then $x \mapsto g^2 = e$. On the other hand, there is a nontrivial homomorphism from $\mathbb Z \times \mathbb Z$ to $C_2$ given by $(m,n) \mapsto g^m$.
A simpler reason is that every element in $\mathbb Q$ is a square, that is, $x \mapsto 2x$ is surjective. But not every element of $\mathbb Z \times \mathbb Z$  is a square, that is, of the form $(2m,2n)$.

*Consider $\mathbb Q$ and $\mathbb Q \times \mathbb Q$ as vector spaces over $\mathbb Q$. Every additive map $\mathbb Q \to \mathbb Q \times \mathbb Q$ is actually a linear transformation. But $\mathbb Q$ and $\mathbb Q \times \mathbb Q$ have different dimensions.

*The argument above does not apply to $\mathbb R$ and $\mathbb R \times \mathbb R$ as vector spaces over $\mathbb R$ because not every additive map is a linear transformation. Nevertheless, $\mathbb R$ and $\mathbb R \times \mathbb R$ have the same dimension as vector spaces over $\mathbb Q$ and so are isomorphic.
A: We know that $\mathbb{Q}$ is not a free-abelian group, see here:
How can we show that $\mathbb Q$ is not free?
On the other hand $\mathbb{Z}^2$ is free of rank $2$.
