Isomorphism of finite abelian quotient groups

Question

Let $A,B,C,D$ be finitely generated free abelian groups (lattices), s.t. $B,C \subseteq D$ full rank sublattices and $A \subseteq B,C$ full rank sublattice. Moreover, let $f: D \to C$ be an isomorphism that restricts to an isomorphism $f|_B:B \to A$. In particular, this induces an isomorphism $D/B \cong C/A$. Do we also have $D/C \cong B/A$?

Answer 1

It turns out the answer is no. Here is a counter-example:

Take $D=\mathbb{Z}\times \mathbb{Z}$, $B=\mathbb{Z}\times 2\mathbb{Z}$, $C=4\mathbb{Z}\times 4\mathbb{Z}$ and $A=\langle (4,4), (0,8) \rangle$. Also take $f:D\to C:(a,b)\mapsto (4a, 4a+4b)$.

Then $f$ is an isomorphism, and $f(B)=A$. However, $D/C \cong \mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, whereas $B/A \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/8\mathbb{Z}$.

To compute this last isomorphism, note that if we fix the bases $\{(4,4), (0,8)\}$ of $A$ and $\{(1,0), (0,2)\}$ of $B$, then the matrix of the inclusion of $A$ into $B$ is $$ \begin{pmatrix} 4 & 0 \\ 2 & 4 \end{pmatrix}. $$ The Smith normal form of this matrix is $$ \begin{pmatrix} 2 & 0 \\ 0 & 8 \end{pmatrix}, $$ so the cokernel of the inclusion (in other words, $B/A$) is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/8\mathbb{Z}$.