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?