I want to find the structure of the abelian group: $G=\frac{\mathbb{Z}^{3}}{\langle (2,0,10),(0,4,8),(4,-4,12) \rangle}$ The Smith normal form of the matrix associated to $G$ is: $P= \left( \begin{array}{ccc} 2 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0\end{array} \right),$ which is correct (verified it with software). Thus the decomposition of $G$ as a direct sum of cyclic groups is $\mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}$ yes? why the answer is $\mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z} \oplus \mathbb{Z}$, i.e why the extra summand $\mathbb{Z}$?. I thought that we only look at the elements of the diagonal of $P$ which in this case there are only three: $2,4,0$.
What am I doing incorrect? Can you please explain?