Abelian group generated by $x,y,z$ with relations $-5x+6y+12z = 3x+4y+2z=11x+2y-8z=0$ 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.
 A: So what you're actually doing when you compute the Smith normal form is finding a new set of generators which still generates the same group.  (More abstractly, you've shown with your $U$ and $V$ that the free group module $F_{x,y,z}$ defined by the relations in your problem is isomorphic to the free group module $F_{u,v,w}$ under the relations you then obtain.)  When you subtract column $2$ from column $1$ for example you're replacing $x$ with $x-y$.
So, you ended up with $$\begin{array}{ccc}u&=&0\\ 2v&=&0\end{array}$$ and no relation on $w$.  You've concluded that this means $x=y=0$ but what it means is that $x$ has order $1$, $y$ has order $2$, and $w$ has infinite order (since there didn't end up being any relations on it).  Thus the group is isomorphic to $$\frac{\mathbb{Z}}{\mathbb{Z}}\oplus \frac{\mathbb{Z}}{2\mathbb{Z}} \oplus \mathbb{Z}.$$  You can see each of the group summands there comes from the relation you found on the generators... $\mathbb{Z}/\mathbb{Z}$ is the trivial group, generated by your guy $u$ of order $1$, $\mathbb{Z}/2\mathbb{Z}$ comes from $y$ who has order $2$, and $\mathbb{Z}$ represents $w$.  Anyway, this is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}$.
