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

Question

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.

Answer 1

$(m,n)=l(a,b)$ 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?).

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