How one can find the abelian group which has a presentation $$\langle x,y,z,w\mid6x+8y+10z+14w, 4x+4y+4z+4w\rangle$$ Is there any way indicates the steps to find such a group? Or just by guesswork and experience?

Edit: Can one proves directly whether it is the group $\mathbb Z_2 \times \mathbb Z \times \mathbb Z \times \mathbb Z$ or not?

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    $\begingroup$ Yes one may calculate the Smith normal form. $\endgroup$ – Myself Jul 8 '13 at 17:09
  • $\begingroup$ @Myself: As I have seen, this method works when $G=\langle X| R\rangle$, such that $|X|-|R|\le 0$. $\endgroup$ – mrs Jul 8 '13 at 17:34
  • $\begingroup$ @BabakS. What you say does not make sense. One can always reduce $|X|-|R|$ by simply introducing redundant relators. For example, $[x, y], [x, y^{-1}], [x^{-1}, y], [x^{-1}, y^{-1}], [y, x], [y, x^{-1}], \ldots$ all follow from $[x, y]$. $\endgroup$ – user1729 Jul 8 '13 at 17:48
  • $\begingroup$ i think i found similar example which used Smith normal form to find out the group. books.google.se/… page 73 $\endgroup$ – Ronald Jul 8 '13 at 17:51
  • $\begingroup$ (@BabakS. Perhaps you mean the deficiency of a group $G$? This is defined to be the maximum which $|X|-|R|$ can be for any presentation $G\cong\langle X; R\rangle$.) $\endgroup$ – user1729 Jul 8 '13 at 18:01

The smith normal form of the matrix $$\begin{pmatrix}2 & 4 & 6 & 8 \\ 4 & 4 & 4 & 4\end{pmatrix}$$ is $$\begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0\end{pmatrix},$$ so your group is isomorphic to $$\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z\times\mathbb Z\times\mathbb Z.$$


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