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It's easy when $\langle(m,n)\rangle=\langle(1,2)\rangle$

Just define the homomorphism $f\colon \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} $ by $f(a,b)=2a-b$

Then $f$ onto

And, $kerf=\langle (1,2) \rangle $

So, $\mathbb{Z}\times \mathbb{Z} / \langle (1,2) \rangle \cong \mathbb{Z}$

if $ \langle (m,n) \rangle $( where $m,n$ prime to each other)

Then , $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle \cong \mathbb{Z}$

What if $ \langle (m,n) \rangle = \langle (2,2) \rangle $

What if $ \langle (m,n) \rangle = \langle (2,4) \rangle $

And others arbitrarily....

Please help.

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    $\begingroup$ In general this problem can be approached with Smith normal forms. A manual way to find eg $\Bbb Z \times \Bbb Z / \langle (2, 2) \rangle$ is to define $\phi: \Bbb Z \times \Bbb Z \to \Bbb Z \times \Bbb Z$ by $\phi(a, b) = (a - b, a)$. Show that $\phi$ is an isomorphism and calculate the image of this subgroup under $\phi$. You should find that this makes the quotient group easier to identify! $\endgroup$ Mar 18 at 4:04
  • $\begingroup$ Here in your case,,you have shown $\mathbb{Z}\times \mathbb{Z}$ is isomorphic to itself,,,but ask for $\mathbb{Z}\times \mathbb{Z} / \langle(2,2)\rangle $ $\endgroup$
    – A learner
    Mar 18 at 4:14
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    $\begingroup$ The idea is that if $\phi$ is an isomorphism, then $G/N \equiv \phi(G) / \phi(N)$, and in this case $\phi(N)$ is much easier to work with. This is exactly the same as what reuns' lovely answer below is doing, except they are representing the isomorphism by a matrix. $\endgroup$ Mar 18 at 4:26
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    $\begingroup$ @Izaak van Dongen Ohh, thanks. Just assure me here ,, $\mathbb{Z}\times \mathbb{Z} / \langle (2,2) \rangle $ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{2}$.......and hence $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle $ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$.... $\endgroup$
    – A learner
    Mar 18 at 4:47
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$(m,n)=l(a,b)$ with $\gcd(a,b)=1$. Then $ad-bc=1$ and $\pmatrix{a&b\\c&d}$ has an integer matrix inverse $U$ (can you find it?).

$$\Bbb{Z^2}/\Bbb{Z}(m,n)\cong \Bbb{Z^2}U/\Bbb{Z}(m,n)U=\Bbb{Z}^2/\Bbb{Z}(l,0)$$

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  • $\begingroup$ I didn't understand your last line,,,how do you get into that? Can you please explain a little bit more? Thank you... $\endgroup$
    – A learner
    Mar 18 at 4:19
  • $\begingroup$ Which part? ${}{}$ $\endgroup$
    – reuns
    Mar 18 at 4:44
  • $\begingroup$ Just assure me here ,, $\mathbb{Z}\times \mathbb{Z} / \langle (2,2) \rangle $ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{2}$.......and hence $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle $ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$. $\endgroup$
    – A learner
    Mar 18 at 4:48
  • $\begingroup$ Am I wrong? Please respond. $\endgroup$
    – A learner
    Mar 18 at 4:55
  • $\begingroup$ Which part is unclear to you? $\endgroup$
    – reuns
    Mar 18 at 4:59

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