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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.

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Alright, so it follows that $x = -14y -16z$. I understand that this means $G$ is generated by $y$ and $z$ since $x$ is a linear comb. of $y$ and $z$. What more do I need to know to determine the abstract structure of $G$? –  Jack Rousseau Oct 14 '12 at 1:55
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You can describe the abstract structure of the group directly from the SNF, without doing any more calculations. I would advise you to learn how to do that in general rather than in a specific example. –  Derek Holt Oct 14 '12 at 9:49
    
In each case I obtain $19y + 23z = 0$ In my lecture notes, there is a section that states for the SNF diag(1,2,0) we should have $G \simeq \mathbb{Z}/1\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$ but I find the derivation hard to follow. –  Jack Rousseau Oct 14 '12 at 10:00
    
Each step in the calculation of the SNF corresponds to a change of generators in your presentation. So you have actually shown that your group is generated by elements $x',y',z'$ with the relations $x'=2y'=0$. You can use one of the transforming matrices to get $x',y',z'$ in terms of $x,y,z$, but you have not been asked to do that. –  Derek Holt Oct 14 '12 at 16:39

1 Answer 1

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 $\mathbb{Z}_2\oplus \mathbb{Z}$.

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