Finding an explicit isomorphism from $\mathbb{Z}^{4}/H$ to $\mathbb{Z} \oplus \mathbb{Z}/18\mathbb{Z}$ There was a past qualifying exam problem, I was having trouble with, it is stated below as follows:

In the group $G= \mathbb{Z} \times \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}=\mathbb{Z}^{4}$, let $H$ be the subgroup generated by $[0,0,3,1], [0,6,0,0], [0,1,0,1]$. Find an explicit isomorphism between $G/H$ and a product of cyclic groups.

I truly do not fully understand how to define such a map on $G/H$. I have given some thought to this.
We have that $H$ consists of integer combinations of the generators above, that is, if $\xi \in H$, then $\xi=a[0,0,3,1]+b[0,6,0,0]+c[0,1,0,1]$. In particular, we have that $\xi$ is in the image of a module homomorphism $\mathbb{Z}^{3} \rightarrow \mathbb{Z}^{4}$ whose matrix with respect to a standard basis $(e_{1},e_{2}, e_{3})$ and $(f_{1}, f_{2}, f_{3}, f_{4})$ for $\mathbb{Z}^{3}$ and $\mathbb{Z}^{4}$ respectively, will be
$$A=\begin{bmatrix}
0&0&0\\
0&6&1\\
3&0&0\\
1&0&1\\
\end{bmatrix}.$$
We can perform row operations by multiplying on the left by the matrix $P$ and column operations by multiplying  on the right by the matrix $Q$
$$P=
\begin{bmatrix}
0&0&0&1\\
0&1&0&0\\
0&3&1&-3\\
1&0&0&0\\
\end{bmatrix},$$
$$Q=\begin{bmatrix}
1&6&-1\\
0&1&0\\
0&-6&1\\
\end{bmatrix}$$
to obtain the matrix
$$PAQ=\begin{bmatrix}
1&0&0\\
0&0&1\\
0&18&0\\
0&0&0\\
\end{bmatrix}.$$
I believe this tells us if $\xi \in H$, then $\xi=[a,6b,18c,0]$ for $[a,b,c] \in \mathbb{Z}^{3}$. From here, I think we can conclude that
$$\mathbb{Z}^{4}/H \cong \mathbb{Z} \times  \mathbb{Z}/18\mathbb{Z}.$$
I do not see how to explicitly write a function between these two objects.
Do I think of $\mathbb{Z}^{4}/H$ as cosets?
The matrices $P$ and $Q$ above, tell us exactly the change of basis in the domain and codomain, I was wondering if I could use that.
Thanks
 A: This problem hinges on interpreting the Smith normal form of a matrix.
As you have said, we need to study the homomorphism $\varphi: \mathbb{Z}^3 \to \mathbb{Z}^4$ mapping the standard basis vectors to generators of the subgroup $H$.  With respect to the standard bases $\mathcal{E}$ and $\mathcal{F}$ for $\mathbb{Z}^3$ and $\mathbb{Z}^4$, $\varphi$ has matrix $A$.  We will find new bases $\mathcal{B}$ and $\mathcal{C}$ with respect to which $\varphi$ is represented by a (much simpler) diagonal matrix.
Computing the Smith normal form by row and column operations, I find that
$$
PAQ =
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 18\\
0 & 0 & 0
\end{pmatrix}
$$
where
$$
P =
\begin{pmatrix}
0 & 0 & 0 & 1\\
0 & 1 & 0  & 0\\
0 & -15 & 1 & -3\\
1 & 0 & 0 & 0
\end{pmatrix}
\qquad
\text{and}
\qquad
Q =
\begin{pmatrix}
1 & 5 & 6\\
0 & 1 & 1\\
0 & -5 & -6
\end{pmatrix} \, .
$$
Note that $P$ and $Q$ both have determinant $-1$, hence are invertible over $\mathbb{Z}$.  Interpreting $P$ and $Q$ as change of basis matrices, then
$$
PAQ = {_\mathcal{C} [\text{id}]_\mathcal{F}} \, {_\mathcal{F}[\varphi]_\mathcal{E}} \, {_\mathcal{E}[\text{id}]_\mathcal{B}}
$$
for some bases $\mathcal{B}$ and $\mathcal{C}$ as above.  Then $\mathcal{B} = \left\{b_1 = e_1, b_2 = 5e_1 + e_2 - 5e_3, b_3 = 6e_1 + e_2 - 6e_3 \right\}$, and since
$$
{_\mathcal{F} [\text{id}]_\mathcal{C}} = P^{-1}= 
\begin{pmatrix}
0 & 0 & 0 & 1\\
0 & 1 & 0 & 0\\
3 & 15 & 1 & 0\\
1 & 0 & 0 & 0
\end{pmatrix}
$$
we find $\mathcal{C} = \{c_1 = 3f_3 + f_4, c_2 = f_2 + 15f_3, c_3 = f_3, c_4 = f_1\}$.  By the form of $PAQ$, then $\varphi(b_1) = c_1$, $\varphi(b_2) = c_2$, $\varphi(b_3) = 18c_3$ and $\varphi(b_4) = 0$, so $H = \text{img}(\varphi) = \mathbb{Z}c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z}18 c_3$.  Thus
\begin{align*}
\frac{\mathbb{Z}^4}{H} = \frac{\mathbb{Z} c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z} c_3 \oplus \mathbb{Z} c_4}{\mathbb{Z}c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z}18 c_3} \cong \frac{\mathbb{Z}}{18\mathbb{Z}} \oplus \mathbb{Z} \, .
\end{align*}
More explicitly, the isomorphism is induced by the map
$$
\alpha_1 c_1 + \alpha_2 c_2 + \alpha_3 c_3 + \alpha_4 c_4 \mapsto (\overline{\alpha_3},\alpha_4)
$$
where the bar indicates the residue mod $18$.
