Proving a quotient group $G/N$ is Isomorphic to $\mathbb{Z}$

Question

For $G=$$\mathbb{Z}$$\oplus$$\mathbb{Z}$ and $N=${$(x, y)$ | $3x=5y$}, prove that $G/N$$\cong$$\mathbb{Z}$.

I know that $G/N=${$aN$ | $a$$\in$$G$}, being $G\pmod N$.

As far as I can tell, this would be something along the lines of {$0(x, y), 1(x, y), 2(x, y)$...}

Thats as far as I got, its honestly the notation from $N$ that's confusing me the most, because to me, $(x, y)$ for $3x=5y$ would be {$(5, 3), (10, 6), (15, 9)$....}

any advice on how to proceed?

Answer 1

Note that $N$ can be written as $$N= \{ (x,y) : 3x-5y = 0 \}$$ hence it is the kernel of the group homomorphism $f: \Bbb{Z} \oplus \Bbb{Z} \to \Bbb{Z}$ $$f(x,y) = 3x-5y$$

Now, it is sufficient to show that $f$ is onto (or equivalently, that $1$ is in the range of $f$): this is clear since $3$ and $5$ are coprime integers.

Hence by the isomorphism theorem $$G/N = G/ \ker f \cong f(G) = \Bbb{Z}$$

Answer 2

This easily follows from the first isomorphism theorem. First, observes that $N=\{x(5,3): x\in \Bbb{Z}\}=(5,3)\Bbb{Z}$. Second, observe that $G/N\cong G/Z\}$ since the map $aN=a(5,3)\Bbb{Z}\mapsto a\Bbb{Z}$ is an isormophism. So the composition gives the desired result since $\phi:G\rightarrow Z$ given by $\phi(x,y)=3x-5y$ is a homomorphism with kernel $N$ .