I wish to classify $\mathbb{Z}_{12} \times \mathbb{Z}_3 \times \mathbb{Z}_6/\langle(8,2,4)\rangle$ according to the fundamental theorem of finitely generated abelian groups.

We have that it is of order $72$. Based on previous help received here, I have attempted to set it up as a matrix:

$$\begin{bmatrix}12 & 3 & 6\\8&2&4\end{bmatrix}$$

But I am unable to find any logical way to set it up in Smith Normal Form that yields a group of order $72$. My bag of tools (looking at collapsing elements etc.) are nothing but pitfalls, they all yield groups of a different order.

  • $\begingroup$ One way of seeing that the group has order 72 is noticing that your original group as order 216 and the subgroup generated by (8, 2, 4) has 3 elements. $\endgroup$ – Jack Yoon Apr 23 '14 at 10:49
  • $\begingroup$ I have already established that it has $72$ elements, that is not the problem at hand. The problem is the classification itself. $\endgroup$ – Andrew Thompson Apr 23 '14 at 10:53

The matrix you have obtained is actually incorrect.

$$\left(\begin{array}{cccc} 12 & 0 &0&8 \\ 0 & 3 &0&2 \\ 0 &0 &6 &4\\ \end{array}\right)$$

is in fact the matrix you want and if you do the Smith normal form of this you should get what you want.

  • 3
    $\begingroup$ It won't affect the answer, but it would agree better with the OP's notation if you used the transpose of that matrix. The point is, you are computing the structure of ${\mathbb Z}^3/\langle (12,0,0),(0,3,0),(0,0,6),(8,2,4) \rangle$. $\endgroup$ – Derek Holt Apr 23 '14 at 12:53
  • $\begingroup$ Oh, I see it so clearly now! Thanks a lot, Jack and Derek! $\endgroup$ – Andrew Thompson Apr 23 '14 at 15:16
  • $\begingroup$ Just to check that I got this correctly: we have an isomorphism to $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_6$ which is isomorphic again to $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_2$? $\endgroup$ – Andrew Thompson Apr 23 '14 at 16:11
  • $\begingroup$ From my memory I got $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$ which is isomorphic to what you got. If you put it into smith normal form that's the most natural answer you should have got rather than the answer you posted. $\endgroup$ – Jack Yoon Apr 23 '14 at 20:46

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