Determining the Smith Normal Form

Consider the integral matrix

$$R = \left(\begin{matrix} 2 & 4 & 6 & -8 \\ 1 & 3 & 2 & -1 \\ 1 & 1 & 4 & -1 \\ 1 & 1 & 2 & 5 \end{matrix}\right).$$

Determine the structure of the abelian group given by generators and relations.

$$A_r = \{a_1, a_2, a_3, a_4 | R \circ \vec{a} = 0\}$$

I know you have to row/column reduce the matrix however am unsure what to do next.

Solutions Collecting From Web of "Determining the Smith Normal Form"

We do want to use a kind of Gaussian elimination, but you have to be careful since you should not multiply a row by anything other than $1$ and $-1$, and you should not add non-integer multiples of one row to another row. So we can get started simply enough:
$$\begin{align*}
\left(\begin{array}{rrrr}
2 & 4 & 6 & -8 \\
1 & 3 & 2 & -1 \\
1 & 1 & 4 & -1 \\
1 & 1 & 2 & 5
\end{array}\right)
&\to
\left(\begin{array}{rrrr}
1 & 1& 2 & 5\\
1 & 3 & 2 & -1\\
1 & 1 & 4 & -1\\
2 & 4 & 6 & -8
\end{array}\right)
&&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 2 & 2 & -18
\end{array}\right)\\
&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 2 & -24
\end{array}\right)
&&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & -30
\end{array}\right)\\
&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & 30
\end{array}\right).
\end{align*}$$
This uses only elementary row operations.

From this we see that the relations on your group are equivalent to:
$$\begin{array}{rcccccccl}
r_1&+&r_2&+&2r_3&+&5r_4 &= & 0\\
& &2r_2 & & & – & 6r_4 &=& 0\\
& & & & 2r_3 & – &6r_4 & = & 0\\
& & & & & &30r_4 & = & 0
\end{array}$$
These elementary row operations replace the relations on our original set of generators with a new set of relations which are equivalent to the original, in the sense that if the generators satisfy these relations, then they satisfy the original relations and vice-versa.

We can now use elementary column operations, which also correspond to performing certain base changes. For example, subtracting five times the first column from the fourth column is equivalent to replacing $r_1$ with $r_1-5r_4$, which does not change the subgroup generated by $r_1,r_2,r_3,r_4$. Etc. Performing those elementary column operations, since the $(1,1)$ entry is the gcd of the entries on the first row (and similarly for the rest of the rows), we can eliminate all nondiagonal entries and end up with the diagonal matrix
$$\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 30
\end{array}\right),$$
from which you can read off the structure of the group in question.

The column operations corresponds to replacing our set of generators with a new set that generates the same group. For example, the operations we performed with the first column, subtracting the first column from the second, twice the first column from the third, and five times the first column from the fourth, correspond to replacing the original generator $r_1$ with the generator $r_1-r_2-r_3-5r_4$. This does not change the subgroup, because $$\langle r_1,r_2,r_3,r_4\rangle = \langle r_1-r_2-r_3-5r_4,r_2,r_3,r_4\rangle.$$
Similarly with the other operations. In the end we will have an abelian group generated by elements $a,b,c,d$, where, in terms of the original generators, we have
$$\begin{align*}
a & = r_1-r_2-r_3-5r_4\\
b &= r_2-3r_4\\
c &= r_3-3r_4\\
d &= r_4,
\end{align*}$$
which yields the relations
$$\begin{align*}
a&=0\\
2b&=0\\
2c&=0\\
30d&=0
\end{align*}$$
from which we can just read off the abelian group structure as well.