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

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'.

• How about $(a,b) \mapsto (a-b,a)$? May 12, 2016 at 15:28
• Hang on, is the generating set of $(3,3)$ just $(3n,3n)$ for all $n \in \Bbb{Z}$? I thought that originally and then I confused myself and thought it was $(3n,3m)$. It is just $(3n,3n)$, right? I was trying to get a map where the kernel was the set of $(3n, 3m)$ but I think this is impossible while retaining the image that I want. Thanks for your help. May 12, 2016 at 15:36
• $\langle (3,3) \rangle$ denotes the set of all integer multiples of $(3,3)$, which is indeed $(3n,3n)$ for all $n \in {\mathbb Z}$. May 12, 2016 at 16:06

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 $$\{(1,0),(0,1)\}$$.
$$\frac{\mathbb{Z} \times \mathbb{Z}}{\langle (3,3)\rangle} \cong \{\langle (0,1),(1,1) \rangle | (3,3)=(0,0)\}$$
This presentation is clearly $$\mathbb{Z} \times \mathbb{Z}_3$$