This is the problem:

Let $G$ be the quotient of the free abelian group with $\mathbb{Z}$-basis $x_1, x_2, x_3$ by the subgroup $H = \langle x_1 + 3x_2, x_1 + 4x_2 + x_3, 2x_1 + 5x_2 + x_3\rangle$. Express $G/H$ as a direct sum of cyclic groups.

I would really appreciate an example using a different set of relations of what the procedure for doing this is.


As pointed out by i707107, the problem is essnetially equvalent to find the Smith normal form of the $3\times 3$ matrix which is defined by the relations. One can find the details of the algorithm to find the Smith normal form in Wikipedia as linked above. I just do the calculation and find the group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

  • $\begingroup$ When I did the steps for smith normal form I got the diagonal matrix with entries 1, 1, 2. If those are the elementary divisors, wouldn't that make it isomorphic to Z/2Z? Or did I misunderstand/do something wrong? $\endgroup$ – Xindaris Sep 23 '13 at 14:39
  • $\begingroup$ Dear Xindaris, I find that you are right after doing the calculation again. I am sorry to make the mistake. $\endgroup$ – Joy-Joy Sep 23 '13 at 15:53

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