Edit: $a,b,c,d$ are fixed in advance. Otherwise it is clearly $\mathbb{Z}^2$. Apologies.
Given $a,b,c,d \in \mathbb{Z}$ such that $ad-bc\neq 0$. Set $\Gamma:=\langle (a,b),(c,d)\rangle$, what is its index in $\mathbb{Z}^2$?
I know this group has finite index in $\mathbb{Z}^2$, as you can quickly see that $(ad-bc,0)$ and $(0,ad-bc)$ are elements of $\Gamma$, and the index of $\langle (ad-bc,0),(0,ad-bc)\rangle$ in $\mathbb{Z}^2$ is $(ad-bc)^2$.
Best Answer
The index of this subgroup is $n=|ad-bc|$. The quickest argument I can think of right now is that $n$ is the total area of the torus $T=\mathbf R^2/\Gamma$ (the parallelogram spanned by $(a,b)$ and $(c,d)$ is a fundamental domain for $\Gamma$), and that the projection $T\to\mathbf R^2/\mathbf Z^2$ is a locally area-preserving covering of a torus of area $1$, so that it must be an $n$-fold covering; the size of the preimage of $(0,0)$ is the index of $\Gamma$ in $\mathbf Z^2$.
As suggested in the comment by Steve D, there is also a way to see this that avoids using real numbers and area considerations. One can start with a matrix $$ \begin{pmatrix}a&c\\b&d\end{pmatrix} $$ whose columns express a pair of generators of $\Gamma$ in the standard basis of $\mathbf Z^2$, and repeatedly modify it by either right-multiplying by a matrix of $\mathbf{GL}(2,\mathbf Z)$ (integer entries, determinant $\pm1$) which corresponds to choosing a different pair of generators of $\Gamma$, or left-multiplying by a matrix of $\mathbf{GL}(2,\mathbf Z)$, which corresponds to changing the basis of $\mathbf Z^2$ that the generators are expressed in. These operations do not change the absolute value of the determinant, and one can arrange for the resulting matrix to become diagonal (the Smith normal form is an even more special form which one does not need here, but the theory of Smith normal forms at least shows this is possible). Now each generator is just a multiple of a basis vector of $\mathbf Z^2$, and it is now clear that the index of $\Gamma$ is the absolute value of the product of those multiples (diagonal entries), which is the absolte value of the determinant, which hasn't changed. so it is $|ad-bc|$.