Put
$A=\begin{pmatrix} 1 & -5 & 4\\ 1 & -2 & 13\\ -2 & 13 & 7 \end{pmatrix}.$
The smith normal form of this matrix is \begin{pmatrix} 1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6 \end{pmatrix}
and now I want to find $a , b, c$ $\in$ $\Bbb Z_{\ge 0}$ such that $\Bbb Z^3/A\Bbb Z^3$ $\cong$ $\Bbb Z_a$$\oplus$$\Bbb Z_b$$\oplus$$\Bbb Z_c$ and $a$ | $b$ | $c$.
Please can anyone lend a hand here?
Recall that the Smith normal form defines matrices $P, Q$ respresenting automorphisms of $\Bbb Z^3$ (indeed they have determinant $1$) such that $QAP = D$ where the latter matrix is the diagonal one you mention. Now the set $A(\Bbb Z^3)$ represents a sublattice of $\Bbb Z^3$ (of volume $18 = \det(A)$). A first thing to take into account is that $A(\Bbb Z^3)$ and $QAQ^{-1}(\Bbb Z^3)$ represent the same lattice but in another basis. Moreover $QP(\Bbb Z^3) = \Bbb Z^3$ so that $QAQ^{-1}(QP(\Bbb Z^3)) = D(\Bbb Z^3)$ still represent the same lattice. Now this lattice corresponds to the subgroup consisting of the elements $(k, 3l, 6m)$ of $\Bbb Z^3$, so the quotient $\Bbb Z^3/D(\Bbb Z^3) \cong \Bbb Z_1 \oplus \Bbb Z_3 \oplus \Bbb Z_6$.
