# The Isomorphism for Smith Normal Form

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.

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.