subgroup of rational numbers under addition is isomophic to integers Let $G =\mathbb Q$ be the rational numbers, with addition as the group operation. Let $x$, $y$ be non-zero elements of $G$, and let $N < G$ be the subgroup they generate. Show that $N$ is isomorphic to $\mathbb Z$. 
Find non-zero elements $x,y \in \mathbb R$ which generate a subgroup that is not isomorphic to $\mathbb Z$.
Thoughts about: to prove isomorphism we need to show homomorphism and bijection. So if we define a map $f: N \to \mathbb Z$. We need to find a number such that $f(a) =1$ and make sure $f(0)=0$ since $f(e)=e$. Since $1$ or $-1$ are generators of $\mathbb Z$, so it is natural to map $x$, $y$ to $1$ or $-1$. But $x+y$ is not necessarily equal to $0$, however by group homomorphism $f(x+y)=f(x)+f(y)=0$. So it is where I get stuck. Greatly appreciated if anyone could help!

This question is worded in quite a misleading way, as I don't think there is an "obvious" isomorphism $N\to\mathbb Z$. Here's a way to rephrase it:

Let $N$ be the subgroup of $(\mathbb Q, +)$ generated by the non-zero elements $x,y\in \mathbb Q$. Then there exists some $z\in\mathbb Q$ such that $N$ is generated by $z$ (i.e. $N$ is cyclic).

Since the additive group $\mathbb Z$ is characterised as the free cyclic group on $1$ generator, there is now an obvious isomorphism $N\to\mathbb Z$ which sends $z\mapsto 1$.
To prove the restated version of the problem, let $x=\frac ab$, $y=\frac cd$ be written in lowest terms. Can you find some $r$ such that $\langle x,y\rangle \subset \langle r\rangle$? This need not be an equality: since a subgroup of a cyclic group is cyclic we can use this to find a generator.
More generally, any finitely generated subgroup of $(\mathbb Q,+)$ is cyclic - the above is the inductive step in a proof.
