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I've been trying to solve the following problem:

Suppose that $G$ is an abelian group, generated by $x_1, x_2, x_3, x_4$, and subject to the relations: $$4x_1 - 2x_2 - 2x_3 = 0; 8x_1 - 12x_3 + 20x_4 = 0; 6x_1 + 4x_2 - 16x_4 = 0.$$ Write $G$ as a direct product of cyclic groups.

$\textbf{My idea:}$ We can take the free $\mathbb{Z}$-module over $\{x_1, x_2, x_3, x_4\}$, and afterwards we can do quotient by the aforementioned relations. Then we can apply the Fundamental Theorem of Finitely Generated Modules over PIDs. The thing is that after doing quotient by the relations it is hard for me to "grab" the elements of this new abelian group. Any hint?

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    $\begingroup$ Are you familiar with Smith Normal Form? $\endgroup$ – Anurag A Aug 6 '19 at 20:33
  • $\begingroup$ Nope, but I can look it up. $\endgroup$ – user437748 Aug 6 '19 at 20:37
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To compute Smith Normal form we use the following:

  1. interchange two rows or two columns,
  2. multiply a row or column by $\pm 1$,
  3. add an integer multiple of row to another row (or an integer multiple of a column to another column). So, $$ \begin{bmatrix} 4&-2&-2&0\\ 8&0&-12&20\\ 6&4&0&16 \end{bmatrix} \rightarrow \begin{bmatrix} -2&4&-2&0\\ 0&8&-12&20\\ 4&6&0&16 \end{bmatrix}\rightarrow \begin{bmatrix} -2&0&0&0\\ 0&8&-12&20\\ 4&14&-4&16 \end{bmatrix}\rightarrow $$ $$\begin{bmatrix} -2&0&0&0\\ 0&8&-12&20\\ 0&14&-4&16 \end{bmatrix} \rightarrow \begin{bmatrix} -2&0&0&0\\ 0&8&-12&20\\ 0&6&8&-4 \end{bmatrix} \rightarrow \begin{bmatrix} -2&0&0&0\\ 0&2&-20&24\\ 0&6&8&-4 \end{bmatrix}\rightarrow $$ $$ \begin{bmatrix} 2&0&0&0\\ 0&2&-20&24\\ 0&0&68&-76 \end{bmatrix}\rightarrow \begin{bmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&68&-76 \end{bmatrix}\rightarrow \begin{bmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&-8&76 \end{bmatrix}\rightarrow $$ $$ \begin{bmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&8&-4 \end{bmatrix}\rightarrow \begin{bmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&4&0 \end{bmatrix} $$

So the given group is isomorphic to $\Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_4$.

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    $\begingroup$ I believe it should be $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}$ right? $\endgroup$ – user437748 Aug 7 '19 at 4:59
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    $\begingroup$ . . . n i c e ... $\endgroup$ – janmarqz Aug 11 '19 at 18:47

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