How to compute $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$? I am trying to compute the factor group  $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$. Here is what i did:
First i expand  $\langle(2,1,2)\rangle$, it is equal to {(2,1,2),(0,0,4),(2,1,6),(0,0,0)}. Then, since order of $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}$ is 64, we have 64/4=16. So the order of factor group is 16 and so it must be isomorphic to one of the following: $\mathbb{Z}_{16}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{8}$,  $\mathbb{Z}_{4} \times\mathbb{Z}_{4}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{2} \times\mathbb{Z}_{2}$. But from here on i do not know how to continue. Can anyone help? How can i eliminate  $\mathbb{Z}_{16}$ etc.?
Edit: Actually, i have a solution, saying that consider 4(a,b,c)=(4a,4b,4c)=(0,0,4c). 4c is either 0 or 4, in both cases both (0,0,0) and (0,0,4) are in $\langle(2,1,2)\rangle$ so any coset in the factor group has order at most 4. I do not understand why this is the case. Can anyone explain it?
Thanks
 A: An alternate approach is to convert the problem to an integer problem: $\mathbb{Z}/4 \times \mathbb{Z}/2 \times \mathbb{Z}/8$ is isomorphic to the quotient of $\mathbb{Z}^3$ by the (integer) row space of the matrix
$$ \left( \begin{matrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{matrix} \right) $$
The quotient group you are interested in, then, is obtained as a quotient of $\mathbb{Z}^3$ by the rowspace of
$$ \left( \begin{matrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \\ 2 & 1 & 2 \end{matrix} \right) $$
We can use elementary integer row operations to perform row reduction (in the spirit of the Euclidean algorithm!) to simplify this matrix without changing its rowspace. However, we can do more: elementary integer column operations have the effect of a change of basis on $\mathbb{Z}^3$, so we can do those too in our quest to find the quotient group.
Doing both column and row operations to simplify the matrix will result in a diagonal matrix, which makes it easy to see the structure of the quotient group.
A: The original group was generated by $\langle1,0,0\rangle, 
\langle0,1,0\rangle, $ and $\langle0,0,1\rangle$, with orders 4, 2, and 8, respectively.  A good place to begin would be to investigate the orders of these elements of the new group, and to see if they generate the entire new group.
