Showing that the matrix of the generator has nonzero determinant means the group is finite 
Let $G=\mathbb{Z}\times\mathbb{Z}$ and $H$ a subgroup generated by $(a,b)$, and $(c,d)$. Then I'm trying to show if:
The determinant of matrix
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
is non zero, then the group $G/H$ is finite.

I have a hint to assume that the determinant is $n$ and then to show $H$ contains both $(n,0),(0,n)$. But I don't see how this is useful.
It makes sense that then $H\ge\langle (n,0),(0,n)\rangle $ but I don't see how this is useful.
 A: If we want to determine what elements are in $H$, we need to solve
$$
(x,y)=p(a,b)+q(c,d)
$$
that is,
$$
\begin{cases} x=pa+qc \\[1ex] y=pb+qd \end{cases}
$$
and Cramer's rule yields
$$
p=\frac{xd-yc}{n}\qquad q=\frac{ya-xb}{n}
$$
where I used $n$ with the meaning you assigned.
In particular the system has integer solutions for $x=n,y=0$ or $x=0,y=n$.
As a consequence, every element of the form $(kn,0)$ or $(0,kn)$ belongs to $H$, so $H\supseteq n\mathbb{Z}\times n\mathbb{Z}$ and therefore there exists a surjective homomorphism
$$
(\mathbb{Z}\times\mathbb{Z})/(n\mathbb{Z}\times n\mathbb{Z})\to G/H
$$
Can you finish?
A: Let $n = ad - bc$ be the determinant of the matrix $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$ and suppose that $n \neq 0$. First we consider the case when $n > 0$.
Note that $d(a,b) - b(c,d) = (n,0)$ and $-c(a,b) + a(c,d) = (0,n)$. Therefore, $(n,0), (0,n) \in H$, and consequently the group $n\mathbb{Z} \times n\mathbb{Z}$ generated by $(n,0)$ and $(0,n)$ is a subgroup of $H$.
Now, consider the projection $$\varphi : \mathbb{Z} \times \mathbb{Z} \longrightarrow \mathbb{Z}_{n} \times \mathbb{Z}_{n}$$ defined by $\varphi (x,y) = (\overline{x},\overline{y})$. It is easy to verify that $\varphi$ is surjective and its kernel is the subgroup $n\mathbb{Z} \times n\mathbb{Z}$, so it follows from the homomorphism theorem that $(\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z}) \simeq \mathbb{Z}_{n} \times \mathbb{Z}_{n}$, and in particular $(\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z})$ is a finite group. (Here we used the fact that $n \neq 0$.)
Finally, since $n\mathbb{Z} \times n\mathbb{Z} \leqslant H$, we may define the map $$\psi : (\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z}) \longrightarrow (\mathbb{Z} \times \mathbb{Z})/H$$ by $\psi ((x,y) + n\mathbb{Z} \times n\mathbb{Z}) = (x,y) + H$. It is easy to see that $\psi$ is well defined and that it is surjective, which implies that $(\mathbb{Z} \times \mathbb{Z})/H$ is finite, since $(\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z})$ is finite.
For the case $n < 0$ we may consider $m = -n > 0$ and proceed as above, using $\mathbb{Z}_{m}$ instead of $\mathbb{Z}_{n}$. (Observe that $n\mathbb{Z} \times n\mathbb{Z}$ = $m\mathbb{Z} \times m\mathbb{Z}$.)
