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I am given a group $G=\text{Span}(w,x,y,z)$ with relations defined by $$\begin{bmatrix}0&0&1&3\\-2&1&1&3\\-2&4&1&3\\0&-3&1&5\end{bmatrix}\begin{bmatrix}w\\x\\y\\z\end{bmatrix}=0.$$ I was wondering if anyone could give me any hints, or explain a simply way to do this. I have figured out how to solve it, but the method seems slow. Maybe my method can be improved, or maybe there is another method entirely. Any help would be so welcomed.

I know that we can perform Gaussian Elimination* (we can only multiply rows by integer units) to the matrix to get an equivalent set of relations. So the above system after a hairy computation boils down to

\begin{equation}\begin{bmatrix}-2&1&0&0\\0&3&0&0\\0&0&1&3\\0&0&0&2\end{bmatrix}\begin{bmatrix}w\\x\\y\\z\end{bmatrix}=0.\end{equation}

As we cannot deduce $z=0$ from our relations, and $2z=0$ is a relation it follows that $z$ has order $2.$ Then since $$y+3z=y+z=0$$ it follows that $y=-z=z.$

As we cannot deduce $x=0$ from our relations, and $3x=0$ is a relation it follows that $x$ has order $3.$ Well then the relation $-2w+x=0$ tells us that$$-6w+3x=-6w=0.$$ Then $6w=0$. As $x\in\text{Span}(w)$ we see that $\text{Ord}(w)=3,6$. If $\text{Ord}(w)=3,$ then $$-2w+x=w+x=0,$$ so we would have the relation $w+x=0.$ As $$\begin{bmatrix}1&1&0&0\end{bmatrix}$$ is not in the span of the rows of
$$\begin{bmatrix}-2&1&0&0\\0&3&0&0\\0&0&1&3\\0&0&0&2\end{bmatrix}$$ we do not have the relation $w+x=0,$ hence $\text{Ord}(w)=6.$

Note that we've reduced the generating set to $w,z$, hence $G=\left<w\right>+\left<z\right>.$ If the sum were not direct, then since $\text{Ord}(z)=2$ we would have $z\in\left<w\right>.$ Since $\text{Ord}(w)=6$ this would imply $3w=z,$ or $3w+z=0.$ As $$\begin{bmatrix}3&0&0&1\end{bmatrix}$$ is not in the span of the rows of $$\begin{bmatrix}-2&1&0&0\\0&3&0&0\\0&0&1&3\\0&0&0&2\end{bmatrix}$$ this cannot happen, hence the sum is direct, that is $$G=\left<w\right>\oplus\left<z\right>\cong\mathbb{Z}_6\oplus\mathbb{Z}_2\cong\mathbb{Z}_3\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2.$$

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    $\begingroup$ Check out the Smith Normal Form. I got the same answer as you with much less computation. mathworld.wolfram.com/SmithNormalForm.html $\endgroup$ – SKYejin May 6 at 1:28
  • $\begingroup$ @nilpo10, When we get the matrix in upper triangular form, this immediately gives us that the characteristic poly is $(x-1)(x-2)(x+2)(x-3).$ By properties of invariant factors this would give us the only invariant factor of $a(x):=(x-1)(x-2)(x+2)(x-3).$ This leads to a Smith Normal Form of $$\begin{bmatrix}1\\&1\\&&1\\&&&a(x)\end{bmatrix}.$$ From here can we immediately conclude that the cyclic decomposition is $$\mathbb{Z}_1\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{-2}\oplus\mathbb{Z}_3?$$ This is the end result I got (upto isomorphism). Is that all the work we need to do? $\endgroup$ – Melody May 6 at 2:23

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