Presentation of the additive group of the rational numbers We know that $\mathbb{Q}\cong\mathbb{Z}\times\mathbb{Z}/\sim$, where the isomorphism is a ring isomorphism and the equivalence relation is defined as
$$(a,b)\sim(c,d)\Longleftrightarrow ad=bc$$
Then the relation is stated in terms of the integers' multiplication. So, if I want to give a presentation for the additive group of $\mathbb{Q}$, how should I modify the presentation
$$\mathbb{Z}\times\mathbb{Z}\cong\, \langle a,b|aba^{1}b^{-1} \rangle$$
as I can't use the multiplication? The equivalence relation that defines $\mathbb{Q}$ can be stated in additive terms as
$$(a,b)\sim(c,d)\Longleftrightarrow \begin{cases}a=nc\\b=nd\end{cases}$$
where of course $nx$ stands for $\sum_{i=1}^{n}x$, but I have no idea as how to insert it in the presentation. I'm in doubt that the additive group of $\mathbb{Q}$ is different from that of $\mathbb{Z}\times\mathbb{Z}$, but it seems to me that fractions should be identified on the (additive) group structure independently of the subsequent definition of the multiplication.
 A: By the way, a presentation of $(\mathbb{Q},+)$ is
$$\langle x_1,x_2,\ldots \mid x_n^n=x_{n-1}, \ n \geq 2 \rangle.$$
To understand why, it is possible to consider the morphism induced by $x_n \mapsto \frac{1}{n!}$.
For more information, see Johnson's book, Presentations of groups, Chapter 5.7.
A: $\mathbb{Q}$ is not a quotient of $\mathbb{Z}\times \mathbb{Z}$, or indeed any finite product of copies of $\mathbb{Z}$. By the classification of finitely generated abelian groups, a quotient of $\mathbb{Z}^N$ has a free component (which is a product of finitely many $\mathbb{Z}$'s) and a torsion component. 
If $\mathbb{Q}$ were isomorphic to such a quotient, it couldn't have a torsion component because $\mathbb{Q}$ doesn't have any torsion (there is no integer $n$ and nonzero element $x$ such that $nx=0$). However, $\mathbb Q$ is not free because for any element $x$ and any integer $n$ there is an element $y$ such that $ny=x$; this is not true for free abelian groups. We can see a simple example where this fails in a free abelian group in the element $x=(1,0,0,\ldots,0)$; there is no element $y$ such that $ny=x$ for any $n>1$. 
Any abelian group is a quotient of a free abelian group, but for $\mathbb{Q}$ we would need a free abelian group with infinitely many generators. In particular, $\mathbb{Z}\times\mathbb{Z}$ won't work.
