Does $B/A \cong D/C$ imply $C/A \cong D/B$ for fin. gen. free $\mathbb{Z}$-modules?

Question

Let $A,B,C,D$ be finitely generated free $\mathbb{Z}$-modules, s.t. $A \subseteq B,C$ and $B,C \subseteq D$. Moreover, all quotients should be finite abelian groups and $B/A \cong D/C$. Does this imply $C/A \cong D/B$?

Answer 1

This is true, and is a consequence of the Nine lemma.

Consider the following commutative diagram:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & 0 && 0 && 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ 0 & \ra{} & A & \ra{} & B & \ra{} & B/A & \ra{} & 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ 0 & \ra{} & C & \ra{} & D & \ra{} & D/C & \ra{} & 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ 0 & \ra{} & C/A & \ra{} & D/B & \ra{} & 0 & \ra{} & 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ & & 0 && 0 && 0 \\ \end{array} $$ The two left columns and two upper rows are exact by construction, and the right-most column is exact by your hypothesis. By the nine lemma, the bottom row has to be exact. Thus $C/A$ and $D/B$ are isomorphic.