Let $G$ be an abelian group with generators $x, y, z$ and $t$ subject to the following relations:
$\begin{align*} 4x - 4y + 18z + 18t &= 0\\ 2x + 4z + 10t &= 0\\ x - 3y + 12z + 6t &= 0. \end{align*}$
My thought was to try to represent this as the matrix:
$$\begin{bmatrix}4 & -4 & 18 & 18\\ 2 & 0 & 4 & 10\\ 1 & -3 & 12 & 6 \end{bmatrix}$$
and calculate its Smith normal form. We know that $d_0 = 1$ and then $d_1$ will be the greatest common divisor of all the entries which is $1$ and then finally $d_2 = \operatorname{gcd}(8, -16, 108, -6, 12, -96) = 2.$
Then the Smith normal form of the matrix would be:
$$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 2 & 0\\ \end{bmatrix}$$
which would mean that $G \cong \mathbb{Z}/\langle 1 \rangle \oplus \mathbb{Z}_\langle 1 \rangle \oplus \mathbb{Z}/\langle 2 \rangle \cong \mathbb{Z}_2$. Is this correct? Any help is appreciated.