Prove that $\frac{\Bbb{Z} \times \Bbb{Z}}{\langle(3,3)\rangle}$ is isomorphic to $\Bbb{Z} \times \Bbb{Z_3}$

Question

I must prove that $\frac{\Bbb{Z} \times \Bbb{Z}}{\langle(3,3)\rangle}$ is isomorphic to $\Bbb{Z} \times \Bbb{Z_3}$.

I am trying to do this using the first isomorphism theorem ie. for $\phi: G \to H$ a homomorphism, we have that:

$$\frac{G}{\operatorname{Ker}(\phi)} \cong \operatorname{Im}(\phi) \subset H$$

However, I just can't seem to be able to find the right homomorphism that will give me the kernel of $\phi$ that I need. If someone can point out what the homomorphism should be, it would help greatly.

Thanks!

Just in case these differ in any way, here are some of the definitions we have:

$\langle(3,3)\rangle$ is the generating set of $(3,3)$

$\Bbb{Z}_3$ is the integers modulo $3$

Product of groups, kernel and image defined as 'normal'.

Answer 1

The first thing to do to make this isomorphism obvious is to find a more convenient basis for $\mathbb{Z} \times \mathbb{Z}$; how about using $\{(0,1),(1,1)\}$ as opposed to the 'natural' choice $\{(1,0),(0,1)\}$.

Now, express your quotient in terms of generators and relations:

$\frac{\mathbb{Z} \times \mathbb{Z}}{\langle (3,3)\rangle} \cong \{\langle (0,1),(1,1) \rangle | (3,3)=(0,0)\}$

This presentation is clearly $\mathbb{Z} \times \mathbb{Z}_3$