For $x,y \in \mathbb{Q}\setminus\{ 0\}$, prove that $\mathbb{Q}/\langle x \rangle \simeq \mathbb{Q}/ \langle y \rangle$. For $x$, $y$ nonzero rationals, I need to prove that $\mathbb{Q}/\langle x \rangle \simeq \mathbb{Q}/ \langle y \rangle$.
To do this, I was thinking about using the Third Isomorphism Theorem: 

Let $M, N \trianglelefteq G$ and $M \leq N$. Then, $(G/M)/(N/M)\simeq G/N$.  

However, this would only make sense if $\langle y \rangle$ was a normal subgroup of $\mathbb{Q}$. I know it is cyclic, and hence abelian, but is it also normal? 
Then, I could let $M = e_{G}$ (the identity element in $G$), but then $\langle x \rangle = N/M = N/e = N = \langle y \rangle$, which doesn't make any sense.
So, suffice it to say, I am at a loss as to how to prove this result! :(
Could somebody please help me figure out how to prove this? Thank you.

I would suggest you look no further than the first isomorphism theorem. 
Take some nonzero $a \in \mathbb{Q}$ and convince yourself of the fact that if we show that $\mathbb{Q}/\langle a \rangle \cong \mathbb{Q}/\langle 1 \rangle$ we are done. Define the map $$f: \mathbb{Q} \to \mathbb{Q}/\langle 1 \rangle,\quad x \mapsto \frac{1}{a} \cdot x$$ and do the following:


*

*Show that $f$ is an homomorphism.

*Show that $f$ is surjective.

*Show that $\ker{(f)} = \langle a \rangle$.

*Apply the first isomorphism theorem to find $\mathbb{Q}/\ker{(f)} \cong \text{im}{(f)}$.

