The question is as followed: Consider $A,B \in M_{m\times n}(\Bbb Z)$. Show that there are $P \in GL_{m}(\Bbb Z)$ and $Q \in GL_{n}(\Bbb Z)$ such that $PAQ = B$ if and only if abelian groups $\Bbb Z^{m}/A\Bbb Z^{n}$ and $\Bbb Z^{m}/B\Bbb Z^{n}$ form an isomorphism.
I have shown the only if part by constructing a natural isomorphism, yet for the if part, I don't have much clue on it. It seems that considering the diagonal matrices corresponding to $A$ and $B$ may help to simplify things a bit, but I still have no clue on how to proceed.
(Minor question: does $\Bbb Z^{m}/A\Bbb Z^{n}\cong \Bbb Z^{m}/B\Bbb Z^{n}$ imply $A\Bbb Z^{n}\cong B\Bbb Z^{n}$?)
For the "if" part of the proof, you can proceed as follows. Suppose that $\Bbb Z^m/(A\Bbb Z^n) \cong \Bbb Z^m/(B \Bbb Z^n)$. Let $A = P_1DQ_1$ be a smith normal form decomposition, so that $D$ is rectangular and diagonal with diagonal entries equal to $a_1 \mid a_2 \mid \cdots \mid a_{k}$ (with $k = \min(m,n)$). It's easy to see that $\Bbb Z^m/(D \Bbb Z^n)$ is congruent to $\Bbb Z/(a_1 \Bbb Z) \oplus \cdots \oplus \Bbb Z^m/(a_k\Bbb Z)$. However, from the "only if" part of the statement, we know that $\Bbb Z^m/(A \Bbb Z^m) \cong \Bbb Z^m / (D\Bbb Z^m)$.
By the same line of reasoning conclude that $\Bbb Z^m/(B \Bbb Z^n)$ is congruent to $\Bbb Z/(b_1 \Bbb Z) \oplus \cdots \oplus \Bbb Z^m/(b_k\Bbb Z)$, where $b_1\mid \cdots \mid b_k$ are the invariant factors of $B$. Because $\Bbb Z^m/(A\Bbb Z^n) \cong \Bbb Z^m/(B \Bbb Z^n)$, we can conclude that these invariant factors are the same (i.e. $a_j = b_j$ for all $j$), from which it follows that $A$ and $B$ have a common Smith normal form, from which it follows that $PAQ = B$ for some $P \in \mathrm{GL}_m(\Bbb Z)$ and $Q \in \mathrm{GL}_n(\Bbb Z)$.