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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'.

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    $\begingroup$ How about $(a,b) \mapsto (a-b,a)$? $\endgroup$ – Derek Holt May 12 '16 at 15:28
  • $\begingroup$ 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. $\endgroup$ – Helen Byrne May 12 '16 at 15:36
  • $\begingroup$ $\langle (3,3) \rangle$ denotes the set of all integer multiples of $(3,3)$, which is indeed $(3n,3n)$ for all $n \in {\mathbb Z}$. $\endgroup$ – Derek Holt May 12 '16 at 16:06
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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 $\mathbb{Z} \times \mathbb{Z}_3$

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