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