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The problem is to find a $\mathbb{Z}$-submodule $H$ of $\mathbb{Z}^2$ such that $\mathbb{Z}^2/H\cong\mathbb{Z}_{15}\oplus \mathbb{Z}_{12}\oplus \mathbb{Z}_{10}$.

To solve this problem, I'm trying to apply technique described in Chapter 14.5 of Artin's book (2nd edition).

In general, given any ring $R$, a presentation matrix for an $R$-module $V$ is a matrix $A$ such that $R^m/AR^n\rightarrow V$ is an isomorphism; here $AR^n$ stands for the image of the map $R^n\rightarrow R^m$ given by left multiplication by $A$. If $R=\mathbb{Z}$ and we know such a matrix $A$, then we can determine the structure of the abelian group $V$ by diagonalizing $A$ via row and column operations.

Now the situation in question is converse. As far as I understand, I am given a presentation matrix for the $\mathbb{Z}$-module $\mathbb{Z}^2/H$, which is, in the above terminology, $$A=\begin{bmatrix}15&0 & 0\\0&12 & 0 \\ 0&0&10\end{bmatrix}.$$ Also, in the above terminology, $R=\mathbb{Z}$, $V=\mathbb{Z}^2/H$, and $m=2$. I need to find $n$ and $A\mathbb{Z}^n$ (the latter will be the answer). But the matrix $A$ is $3\times 3$ and it does not contain $e_j$ as one of its colums or the zero column, so its size cannot be changed (I mean if it contained the zero column, we could have deleted this column). And such a matrix only determines a homomorphism $\mathbb{Z}^3\rightarrow \mathbb{Z}^3$. But we need a homomorphism $\mathbb{Z}^n\rightarrow \mathbb{Z}^2$ for some $n$.

Do I miss something?

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  • $\begingroup$ I would look for a set of two generators of $G := \mathbb{Z}_{15} \oplus \mathbb{Z}_{12} \oplus \mathbb{Z}_{10}$ then find the kernel of the corresponding map $\mathbb{Z}^2 \to G$. I think $x=(1,1,0)$ and $y=(0,1,1)$ would work: $15x = (0,3,0)$ and $10y = (0,10,0)$ so $(0,1,0) = 15x + 10y$. $\endgroup$ – Daniel Schepler May 31 '17 at 18:07
  • $\begingroup$ My calculation would seem to show the kernel of the map is generated by $(60, 0)$ and $(30, 30)$. $\endgroup$ – Daniel Schepler May 31 '17 at 18:20
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The only finite groups that are quotients of $\Bbb Z^2$ are those of the form $\Bbb Z_m\oplus\Bbb Z_n$. So you have to write your group in that form. Now $$\Bbb Z_{15}\oplus\Bbb Z_{12}\oplus\Bbb Z_{10}\cong\Bbb Z_3\oplus\Bbb Z_5\oplus\Bbb Z_4\oplus\Bbb Z_3\oplus\Bbb Z_2\oplus\Bbb Z_5.$$ Now you have to reassemble this in the form $\Bbb Z_m\oplus\Bbb Z_n$.....

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  • $\begingroup$ So you mean I can replace $A$ with $$A'=\begin{bmatrix}60&0 & 0\\0&30 & 0 \end{bmatrix}$$ which defines a map $\mathbb{Z}^3\rightarrow \mathbb{Z}^2$ with image generated by $(60, 0)^t$ and $(0, 30)^t$? And so the answer is the subgroup generated by $(60, 0)^t$ and $(0, 30)^t$, right? $\endgroup$ – Cary May 31 '17 at 18:47
  • $\begingroup$ @Cary Exactly!. $\endgroup$ – Lord Shark the Unknown May 31 '17 at 18:47

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