# Smith normal form help

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?

• What makes you think the answer is $\mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}\oplus\mathbb{Z}$? That group requires at least four generators, and since $G$ is a quotient of a 3-generator group, it can be generated by 3 elements. It cannot equal the group you claim it equals. (Note also that $(4,-4,12) -2(2,0,10)=(0,-4,-8)$, so you can drop one of the generators of the normal subgroup you are moding out by). – Arturo Magidin Jun 9 '11 at 4:36
• I don't get it. We have a subgroup of $\mathbf{Z}^3$ generated by three elements. Shouldn't we be looking at the Smith normal form of a 3x3 matrix instead of 3x4? – Jyrki Lahtonen Jun 9 '11 at 5:12

You are correct that the decomposition of $G$ is $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}$. If the computer is telling you that the answer is $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z} \oplus \mathbb{Z}$, the most likely explanation is that the matrix for $G$ was somehow transposed at some point during your calculation. For example, if you originally entered the transpose of the original matrix for $G$, then you would get the transpose of the Smith normal form.