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I'm trying to resolve the next one:

Describe, as a direct sum of cyclic groups, the cokernel of the map $\phi: \mathbb{Z}^{3} \longrightarrow \mathbb{Z}^{3}$ given by left multiplication by the matrix

$$ \left(\begin{matrix} 15 & 6 & 9 \\ 6 & 6 & 6 \\ -3 & -12 & -12 \end{matrix}\right) $$

So, the cokernel is $\mathbb{Z}^{3}/\phi(\mathbb{Z}^{3})$, I know that I can only get at most two groups of order 3, but I'm not able to describe it. Any hint?

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    $\begingroup$ You have to put the matrix into Smith Normal Form. $\endgroup$ – Derek Holt May 23 '16 at 19:25
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    $\begingroup$ Relevant links: 1, 2 $\endgroup$ – André 3000 May 24 '16 at 0:35
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Thanks to Derek Holt and SpamIAM for the recomendations and useful links, after a while to read and understand Modules over a PID, I finally got an answer.

Let $\phi$ be a $\mathbb{Z}$-linear map such that can be determined by $\phi(e_{1}) = f_{1}, \dots, \phi(e_{n}) = f_{n}$, where $e_{1}, \dots, e_{n}$ be the basis of $\mathbb{Z}^{n}$. Then $\phi(e_{j}) = \sum_{i=1}^{n} = c_{ij}e_{i}$ for $j = 1, \dots, n$, so $(c_{ij})$ is the matrix representation of $\phi$ with respect to the basis. Then

$$ \phi(\mathbb{Z}^{n}) = \mathbb{Z}\phi(e_{1}) \oplus \dots \oplus \mathbb{Z}\phi(e_{n}) = \mathbb{Z}f_{1} \oplus \dots \oplus \mathbb{Z}f_{n}, $$

By aligned bases for $\mathbb{Z}^{n}$ and its modulo $\phi(\mathbb{Z}^{n})$, we can say that

$$ \mathbb{Z}^{n} = \mathbb{Z}v_{1} \oplus \dots \oplus \mathbb{Z}v_{n}, \hspace{1 em} \phi(\mathbb{Z}^{n}) = \mathbb{Z}a_{1}v_{1} \oplus \dots \oplus \mathbb{Z}a_{n}v_{n} $$

where $a_{i}$'s are nonzero integers. Then

$$ \mathbb{Z}^{n}/\phi(\mathbb{Z}^{n}) \cong \bigoplus_{i=1}^{b} \mathbb{Z}/a_{i}\mathbb{Z} $$

Now, for our solution we need to get the Smith Normal Form, since each $a_{i}$ is the $M_{i,i}$ element of the matrix, the Smith Normal Form of the cokernel is:

$$ \left(\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 18 \end{matrix}\right) $$

So, we can describe the cokernel as the sum of cyclic groups:

$$ \mathbb{Z}^{3}/\phi(\mathbb{Z}^{3}) \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/18\mathbb{Z} $$

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