Prove that $A/B$ is cyclic if and only if $\gcd(a,b,c,d) = 1$.

Question

Let $A$ be a free abelian group of rank 2. Let $\{e_1,e_2\}$ be a basis of A.
Let $\{a,b,c,d\}$ be integers, and $B = \langle \{ae_1+be_2,ce_1+de_2\} \rangle$ subgroup of A of rank two.
Prove that $A/B$ is cyclic if and only if $\gcd(a,b,c,d) = 1$.
I tried to decide what B looks like. I concluded that it looks like $a \mathbb{Z} \times (d-bc/a)\mathbb{Z}$ by Gaussian Elimination of the matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} Can I now finish with saying that $A/B$ is cylic if and only if $\gcd(a, d-bc/a) = 1$ and conclude the result? Can I even divide in $a$ at first?
Any help will be appreciated.

Answer 1

To elaborate on the comment of @TokenToucan. Since $A$ is free abelian of rank $2$ you know $A \cong \mathbb{Z}^2 = \langle e_1, e_2 \rangle$. Moreover $B$ is the sublattice generated by $(a, b)$ and $(c,d)$.

Notice that this sublattice is generated by the matrix \begin{equation*} M = \begin{pmatrix} a & c\\ b & d \end{pmatrix} \end{equation*}

Recall that we may put this matrix in Smith normal form by replacing $M$ by $SMT$ where $S$ and $T$ are invertible. So we may assume \begin{equation*} M = \begin{pmatrix} \alpha & 0\\ 0 & \beta \end{pmatrix} \end{equation*} I will leave it to you to check that $\alpha$ and $\beta$ are coprime if and only if $gcd(a, b, c, d) = 1$.