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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?

azimut
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Ronald
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    Yes one may calculate the Smith normal form. – Myself Jul 08 '13 at 17:09
  • @Myself: As I have seen, this method works when $G=\langle X| R\rangle$, such that $|X|-|R|\le 0$. – Mikasa Jul 08 '13 at 17:34
  • @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 $[x, y]$. – user1729 Jul 08 '13 at 17:48
  • i think i found similar example which used Smith normal form to find out the group. http://books.google.se/books?id=BecLeCWOjI4C&pg=PA73&lpg=PP1&dq=Symmetries++By+D.+L.+Johnson page 73 – Ronald Jul 08 '13 at 17:51
  • (@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 $G\cong\langle X; R\rangle$.) – user1729 Jul 08 '13 at 18:01
  • @user1729: No, you noted right. In fact, your remark reminded me the points which I saw in Todd-Coxeter Algorithm. There; sometimes, I added some redundant relations. I got my bad mistake. Thanks. – Mikasa Jul 08 '13 at 18:05

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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 $$\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z\times\mathbb Z\times\mathbb Z.$$

azimut
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