I'm new on stackexange, so I hope the way I'm asking this question, by putting it on the title, is ok. I didn't see how to do it differently.
This is just a detail in a proof and I'm sure the answer isn't that difficult (or I didn't get it right), but I can't find the isomorphism. I'm sorry, I'm not asking for the answer but maybe a tip to help me get to the idea.
Thanks a lot
I'm pretty sure, this is what you want: Given two fixed $x,y\in \mathbb{Z}$ such that $x,y\neq 0$, consider the set $$ A_{x,y}=\{(a,b)\in \mathbb{Z}^2: ax+by=0\} $$ Then if $(a,b)$ and $(a',b')$ are in $A_{x,y}$ then so is $(a+a', b+b')$, meaning $A_{x,y}$ has a natural addition structure (also $(0,0)$ is the additive identity and $(-a,-b)$ is the inverse of $(a,b)$), meaning $A_{x,y}$ is an additive group.
Now let $k$ be the greatest common divisor of $x,y$ and $x=kx'$ and $y=ky'$. Then if $(a,b)\in A_{x,y}$, then $(a,b)\in A_{x',y'}$ too since $$ 0=ax+by=k(ax'+by')\Longrightarrow ax'+by'=0 $$ Conversely if $(a,b)\in A_{x',y'}$ then $(a,b)\in A_{x,y}$ too obviously. So $A_{x,y}= A_{x',y'}$. So without loss of generality, let us assume $x,y$ are relatively prime from now on.
Note that if $x,y$ are relatively prime. Suppose $(a,b)\in A_{x,y}$, i.e. $ax=-by=t$. Note that $t$ is a common multiple of $x,y$ and the least common multiple of $x,y$ is $xy$ (since $x,y$ are coprime). Therefore $a=ny$ and $b=-nx$ for some $n\in \mathbb{Z}$. Now define $\phi:\mathbb{Z}\to A_{x,y}$ via $n\mapsto (ny,-nx)$. The discussion above shows that this is an isomorphism.