Let $G$ be the Abelian group generated by $x,y,z$ with relations: \begin{equation*} -5x+6y+12z=3x+4y+2z=11x+2y-8z=0 \end{equation*} Describe the abstract structure of $G$.
I have never encountered a problem of this type before. I know that the above is equivalent to the matrix equation: \begin{equation*} \left[ \begin{matrix} -5 & 6 & 12 \\ 3 & 4 & 2 \\ 11 & 2 & -8 \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \end{equation*}
I have previously shown that: \begin{equation*} \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 5 & 0 \\ -1 & 2 & -1 \end{matrix} \right]\left[ \begin{matrix} -5 & 6 & 12 \\ 3 & 4 & 2 \\ 11 & 2 & -8 \end{matrix} \right]\left[ \begin{matrix} 1 & 4 & -18 \\ 0 & -6 & 23 \\ 0 & 5 & -19 \end{matrix} \right] = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \end{equation*}
Where the matrix on the RHS is the Smith Normal form. Write this as $UAV=D$, then we need to solve $A{\bf x} = {\bf 0}$ or equivalently $U^{-1}DV^{-1}{\bf x} = {\bf 0}$ so $DV^{-1}{\bf x} = 0$.
Setting: \begin{equation*} {\bf y} = V^{-1}{\bf x} = \left[\begin{matrix} u \\ v \\ w \end{matrix}\right] \end{equation*} We have that: \begin{equation*} D{\bf y} = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{matrix} \right]\left[\begin{matrix} u \\ v \\ w \end{matrix}\right] = \left[\begin{matrix} u \\ 2v \\ 0 \end{matrix}\right] =\left[\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right] \end{equation*}
So $u=v=0$ and $w$ is undetermined. Thus, \begin{equation*} {\bf x} = V{\bf y} = \left[ \begin{matrix} 1 & 4 & -18 \\ 0 & -6 & 23 \\ 0 & 5 & -19 \end{matrix} \right] \left[\begin{matrix} 0 \\ 0 \\ w \end{matrix}\right] = w\left[\begin{matrix} -18 \\ 23 \\ -19 \end{matrix}\right] \end{equation*}
How do I relate all of this to an abelian group $G$!? I really have no idea what to do next. Any help would be greatly appreciated.