Finding the correct pivot in Smith normal form I have been working through Smith normal form examples and I am wondering if I am finding the correct pivot in order to carry out the calculation.
Let $V \subset \mathbb{Z}$ be an Abelian group with relation matrix $A$.
$$
A = \begin{pmatrix}
2 & -6 & 0 \\
0 & 2 & -6 \\
-6 & 0 &  2 \end{pmatrix} 
$$
Question 1: For the case of entries in $\mathbb{Z}$, is the first step always to bring smallest integer to 1-1 position in the matrix? So we don't divide by 2 in this case right?
(I was trying to do something similar to the matrix A  in the Wikipedia article but I have no idea how to make the first pivot 1 and still get $SNF(xI-A) = \begin{pmatrix}1&0  \\0 &(x-1)^2 \end{pmatrix}$
Applying my logic this is what I get for the Smith normal form (SNF) of the original problem
$$\begin{pmatrix}
2 & -6 & 0 \\
0 & 2 & -6 \\
-6 & 0 &  2 \end{pmatrix}  \sim \begin{pmatrix}
2 & -6 & 0 \\
0 & 2 & -6 \\
0 & -18 &  2 \end{pmatrix} \sim \begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & -6 \\
0 & -18 &  2 \end{pmatrix} \sim \begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & -6 \\
0 & 0 &  52 \end{pmatrix}  \sim \begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 &  52 \end{pmatrix}$$
Question 2: Is $V = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z} / 52\mathbb{Z}$ based on the above Smith normal form?

To answer your Question 1:  We don't divide by $2$ or any other non-unit in $\mathbb{Z}$ because division by $2$ is not a $\mathbb{Z}$-map from $\mathbb{Z}$ to itself.  In fact, we are not permitted to multiply by a non-unit as well, because multiplication by $m$ is not an invertible $\mathbb{Z}$-map unless $m = \pm 1$.
Your SNF computation seems fine, and you get the correct invariant factors.
