Classifying $\mathbb{Z}_{12} \times \mathbb{Z}_3 \times \mathbb{Z}_6/\langle(8,2,4)\rangle$

Question

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.

Answer 1

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.