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Let $L \subset Z$ be the subgroup of $\mathbb{Z}^3$ generated by the elements $(−1, −1, 4)$, $(2, 4, 0)$ and $(3, 3, 8)$.

Write $\mathbb{Z}^3/L$ as a direct sum of cyclic groups.

Since L is spanned by $(-1,-1,4),(2,4,0)$, and $(3, 3, 8)$, and they are linear independent, I think L = $\mathbb{Z}^3$, but I am not sure about it. So I don't know how to solve this problem, please help me to figure it out! Thank you!

Quinn Greicius
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Nhay
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  • Here are some relevant links: one, two and three. – Viktor Vaughn Mar 28 '16 at 01:59
  • The short answer is to compute the Smith normal form of the matrix whose columns are these vectors. – Viktor Vaughn Mar 28 '16 at 02:01
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    Just because they are linearly independent does not mean that you can associate $L$ to $\Bbb Z^3$. Could you somehow get $(1,0,0)$ from integer linear combinations? Obviously this is the case with rational linear combinations, but need not be the case for integer linear combinations. – Cameron Williams Mar 28 '16 at 02:20
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    So I got the Smith normal form $\begin{pmatrix} 1 & 0&0\0&2&0\0&0&20\end{pmatrix}$. Is it correct? Then the answer should be $\frac{\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}}{\mathbb{Z}\oplus 2\mathbb{Z} \oplus 20 \mathbb{Z}} = \mathbb{Z}2 \oplus \mathbb{Z}{20}$ – Nhay Mar 29 '16 at 23:44
  • @NoboruHayashi You should post this as an answer. – Viktor Vaughn Mar 30 '16 at 01:48
  • Related: https://math.stackexchange.com/questions/1910883/ – Watson Oct 21 '16 at 16:43

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