Suppose that I have the free abelian group $\mathbb{Z}^3$, and I consider a subgroup $H$ which is generated by the elements $(2, 4, 6), (4, 5, 6), (0,4,8)$. I am trying to compute the rank of the quotient group $\mathbb{Z}^3/H$ and determine the invariant factors of $\mathbb{Z}^3/H$. I'm really not sure how to go about this problem, but here are my initial thoughts:
If I set $x = (2,4,6), y = (4,5,6)$ and $z = (0,4,8)$ then $H = \{ax + by + cz : a,b,c \in \mathbb{Z}\}$, in other words, I believe that $H$ is in some sense the $\mathbb{Z}$-span of $x,y$ and $z$. If I consider $x,y$ and $z$ to be column vectors then I would obtain a 3 by 3 matrix, whose row Echelon form would be (after augmenting with the zero column on the right)
$$ \begin{pmatrix} 2 & 4 & 0 & 0 \\ 4 & 5 & 4 & 0 \\ 6 & 6 & 8 & 0 \\ \end{pmatrix} \rightarrow_{REF} \begin{pmatrix} 3 & 0 & 8 & 0 \\ 0 & 3 & -4 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, $$
and the information I can pull from this is that $z$ is a free variable, $3y = 4z$, and $3x = -8z$ (I didn't go into fully reduced row Echelon form since the coefficients attached to $x,y$ and $z$ need to be integers). So, I think this would mean that the rank of this matrix would be $2$ (since there are two pivot columns), but I'm not sure how this would help me in solving for the rank of $\mathbb{Z}^3/H$ since there are infinite amount of elements in $\mathbb{Z}^3$.
As far as going about trying to find the invariant factors of the quotient group, I'm also unsure of where to start. Can someone assist me or verify that the first part of my efforts were correct? References to where I could look in Dummit and Foote would be most helpful.
Use Smith normal form. You can only add integer multiples of rows and columns to each other.
$$\begin{pmatrix}2\quad 4\quad 0\\4\quad 5\quad 4\\6\quad 6\quad 8\end {pmatrix}\to\begin {pmatrix}2\quad 0\quad 0\\4\quad-3\quad 4\\6\quad-6\quad 8\end {pmatrix}\to \begin {pmatrix}2\quad 0\quad 0\\0\quad-3\quad 4\\0\quad-6\quad 8\end {pmatrix}\to \begin {pmatrix}2\quad 0\quad 0\\0\quad 3\quad 1\\0\quad 0\quad 0\end {pmatrix}\to \begin {pmatrix}2\quad 0\quad 0\\0\quad 1\quad 0\\0\quad 0\quad 0\end{pmatrix}\to \begin {pmatrix}1\quad 0\quad 0\\0\quad 2\quad 0\\0\quad 0\quad 0\end {pmatrix}$$.
We get $\Bbb Z×\Bbb Z_2$, and the rank is $1$.