1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.