Presentations: fewer relations than generators implies $G$ infinite. Take $r=0$. Let $G=\langle a_1\rangle \times ...\times \langle a_n\rangle, \langle a_i\rangle$ infinite cyclic. $G$ is generated by $n$ elements and all relations in $A$ are relations in $G$. Therefor by von  Dyck's theorem there is an epimorphism from $A$ to $G$. But $G$ is infinite and so, $A$ is infinite too. 

For the case $r&gt;0$ I have been unable to find a proof. I don't want a solution. I only ask for a hint which allows me to begin working in the problem.
Hint: $A$ is an abelian group, so think of it as a $\Bbb Z$-module.  $A$ can be presented as the quotient of $\Bbb Z^n$ by the (free) submodule generated by the $r$ relations.
I get it. I have $G=\langle m_1,...,m_n\mid s_i(m)\rangle$ where the $s_i$ are the same relations as in $A$. For instance, if $2x_1+x_2=0$ is a relation in $A$ then $2m_1+m_2=0$ is a relation in $G$. So $G$ is a subgroup of $Z$ and, as such, it is infinite. Now von Dyck gives an epimorphism $\beta:A\to G$ and $A$ is infinite. 

However, how do I define the homomorphism $\alpha:A\to \langle t\rangle$?