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Looked pretty simple at first sight but I am stuck with this problem now. I have to write $M=\mathbb{Z}^3/N$ as a direct sum of cyclic modules where $$N=\left\{(x,y,z)\in \mathbb{Z}^3\;\big|\; 2x+3y-5z=0\right\} \subset \mathbb{Z}^3.$$

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We're looking at $N$, the kernel of $\begin{pmatrix} 2 & 3 & -5 \end{pmatrix}$, by changing coordinates (as in the standard proof of Smith Normal Form), this matrix can be written as $\begin{pmatrix} 1 & 0 & 0 \end{pmatrix}$, so the kernel is a copy of $\mathbb{Z}^2$, and hence $\mathbb{Z}^3/N \simeq \mathbb{Z}$.

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Can you tell me the generating set for N? –  Waqas Ali Azhar Jan 21 '13 at 6:12

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