# How to find isomorphic group of the quotient $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle$?

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....

• 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! Mar 18 at 4:04
• Here in your case,,you have shown $\mathbb{Z}\times \mathbb{Z}$ is isomorphic to itself,,,but ask for $\mathbb{Z}\times \mathbb{Z} / \langle(2,2)\rangle$ Mar 18 at 4:14
• 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. Mar 18 at 4:26
• @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 $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$.... Mar 18 at 4:47

$$(m,n)=l(a,b)$$ 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?).
$$\Bbb{Z^2}/\Bbb{Z}(m,n)\cong \Bbb{Z^2}U/\Bbb{Z}(m,n)U=\Bbb{Z}^2/\Bbb{Z}(l,0)$$
• Which part? ${}{}$ Mar 18 at 4:44
• 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 $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$. Mar 18 at 4:48