The Isomorphism for Smith Normal Form

Question

Let $G$ be an abelian group that is generated by the elements $a$, $b$ and $c$ in it such that the relations $2a -4b =0$, $2b - 4c$ and $4a - 2c = 0$ generate all of the relations on $a$, $b$ and $c$. Then, a relation matrix $R$ for $G$ is $$ \begin{pmatrix} 2 & -4 & 0 \\ 0 & 2 & -4 \\ 4 & 0 & -2 \\ \end{pmatrix}. $$ Let the matrix $A$ and $B$ be $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & -4 & 1 \\ \end{pmatrix} \text{ and } \begin{pmatrix} 1 & 2 & 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{pmatrix}, $$ respectively. Then, the Smith normal form of $R$ is $A R B$, i.e., $$ \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 14 \\ \end{pmatrix}. $$ (Here, $A$ is the composition of row operations, and $B$ is the composition of column operations.) This shows that $G$ is isomophic to $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z \times \mathbb Z / 14 \mathbb Z$. How do I take an element in this latter group, such as $(1,0,0)$, and find what it is in terms of $a$, $b$ and $c$? If there were not any column operations, then I would try to map this element by $A^{-1}$. But, since there is also the matrix $B$, I am not sure what the isomorphism from $G$ to $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z \times \mathbb Z / 14 \mathbb Z$ is.

Answer 1

The rows of the matrix $B$ give the original generators $a,b,c$ in terms of the new generators - let's call them $x,y,z$. So $a=x+2y+4z$, $b=y+2z$, $c=z$.

So, by solving the equations or by computing $B^{-1}$, we get $x=a-2b$, $y=b-2c$, $z=c$.

Note that multiplying on the left by $A$ just replaces the original relations with equivalent relations - it does not change the geneting set of the abelian group.