Linear algebra about cyclic group and torsion factors Express the commutative group $Z^{3}/(f_{1}.f_{2},f_{3})$ as a direct sum of cyclic group where 
$f_{1}=(4,6,9), f_{2}=(2,4,12), f_{3}=(4,8,16)$
my answer is $Z[x]/(-11x^{2}+22x-128)$.
I wonder if my calculation is wrong, for this is not a direct sum of cyclic group, hope someone can give me a more clear answer.
How to find the number of non-isomorphic commutative group of a certain order, like $100$ and $p^{3}$ where $p$ is a prime. 
I know that I can use Kronecker Decomposition Theorem, finite abelian group theorem.
 A: The smith normal form for 
$$\begin{pmatrix} 4 & 6 &9\\ 2 & 4 & 12 \\ 4 & 8 & 16\end{pmatrix}$$ can be found by several steps. We first have by substracting two times middle row from top and bottom row:
$$\begin{pmatrix} 0 & -2 &-15\\ 2 & 4 & 12 \\ 0 & 0 & -8\end{pmatrix}$$ which is the same as 
$$\begin{pmatrix} 2 & 4 & 12 \\0 & -2 &-15\\ 0 & 0 & -8\end{pmatrix}$$
Then we substracting the first column from second and third column, and do the same for second column:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & -2 &-1\\ 0 & 0 & -8\end{pmatrix}$$
and then we have:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & 0 &-1\\ 0 & 16 & -8\end{pmatrix}$$
changing order we have:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & 16 &0\\ 0 & 0 & -1\end{pmatrix}$$
A: For the second question, find all the ways to write $100$ as a product of integers greater than $1$, each of which divides the next one: 
$100$; $2\times50$; $5\times20$; $10\times10$. 
To each, there corresponds a different commutative group of order $100$: 
$C_{100}$; $C_2\times C_{50}$; $C_5\times C_{20}$; $C_{10}\times C_{10}$, 
where I'm writing $C_n$ for the cyclic group of order $n$. 
Alternatively, factor $100$ into prime powers, $100=2^2\times5^2$. Then working on the prime powers individually, you get 
$C_4\times C_{25}$; $C_4\times C_5\times C_5$; $C_2\times C_2\times C_{25}$; $C_2\times C_2\times C_5\times C_5$. 
