I've been trying to figure out the correct way to do this and I'm not sure I've got it quite right.
$\mathbb{Z}^4/S$ where $S = \{(4, 12,20,8),(6,30,18,12),(4,16,16,8),(8,40,24,16) \}$
So, $M_0 = \begin{bmatrix} 6 & 30 & 18 & 12 \\ 4 & 12 & 20 & 8 \\ 4 & 16 & 16 & 8 \\ 8 & 40 & 24 & 16 \end{bmatrix}$, since $\gcd(4,6,4,8)=2$ then $$M_1 = \begin{bmatrix} 2 & 18 & -2 & 4 \\ 0 & -24 & 24 & 0 \\ 0 & -20 & 20 & 0 \\ 0 & -32 & 32 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & -24 & 24 & 0 \\ 0 & -20 & 20 & 0 \\ 0 & -32 & 32 & 0 \end{bmatrix}$$
Now $\gcd(-24,-20,-32)=4$, $M_2 = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & -4 & 4 & 0 \\ 0 & -20 & 20 & 0 \\ 0 & -32 & 32 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Then the rank of $\mathbb{Z}^4/S$ is 2 and $\mathbb{Z}^4 / S \simeq \mathbb{Z}^2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4 $.
I haven't been able to explicitly find the isomorphism between the two, which is one of the reasons I'm not sure if it's right. Is there a straightforward way of doing this?