How to find $k$ such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$ In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that 
$$
(\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2
$$ 
One does this with the first isomorphism theorem. With the same idea I proved for example that 
$$
(\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2
$$
So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by
$$
\phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right)
$$
is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.

What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?

Thanks in advance for any help/hint/comment!
 A: Another approach is to solve $mx-ny=k$ and then show that $r(m/k,n/k)+s(x,y)$ is all of $\mathbb Z\times \mathbb Z$, and each element of $\mathbb Z\times\mathbb Z$ is expressible in this way only one way.
Then it is clear that $r(m/k,n/k)+s(x,y)\sim r'(m/k,n/k)+s'(x,y)$ if and only if $k\mid r-r'$ and $s=s'$.
So, how do we show the above? If $mx-ny=k$ then $$\begin{pmatrix}m/k&n/k\\y&x\end{pmatrix}$$ has determinant $1$, and thus has an inverse with integer coefficients.
A: As SpamIAm mentioned, Smith Normal Form is a useful tool here. With $(m,n)=(4,6)$ we get $(4,6)\rightarrow (4,2)\rightarrow (2,2)\rightarrow (2,0)$ which means we get $\mathbb{Z} \times \mathbb{Z}_2$. The steps are to turn your $m$ and $n$ into a matrix and you can perform operations such as subtracting an integer multiple of a column from another and re-ordering the columns. Try this for the cases you used above to get a sense of the pattern. Then see if you can get a general homomorphism from these examples. Best of luck!
