# 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.

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This does not make any sense. How could you take quotient of $\mathbb{Z}$ with a polynomial? – Bombyx mori Dec 24 '12 at 15:57
I am sorry that should be $Z[x]$ – user53800 Dec 24 '12 at 16:14
But what are the $f_{i}$s? This still does not make sense to me. – Bombyx mori Dec 24 '12 at 16:16
yes, you are right, I see this question on a note, I think the teacher assuming that they can form a matrix, actually I just wonder if there is such a expression saying that $z^{3}/(f_{1},f_{2},f_{3})$ – user53800 Dec 24 '12 at 16:20
But $(f_{1},f_{2},f_{3})$ does not give you $(-11x^{2}+22x-128)$. This does not make any sense. – Bombyx mori Dec 24 '12 at 16:26

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

$C_4\times C_{25}$; $C_4\times C_5\times C_5$; $C_2\times C_2\times C_{25}$; $C_2\times C_2\times C_5\times C_5$.

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can you tell me the relation between the groups in two different ways? – user53800 Dec 24 '12 at 17:40
I'm sure you can work that out --- all you need to know is that if $m$ and $n$ have no prime factors in common then $C_m\times C_n=C_{mn}$. So, for example, $C_{50}=C_2\times C_{25}$. – Gerry Myerson Dec 25 '12 at 6:38

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:

$$\begin{pmatrix} 2 & 0 & 0 \\0 & 16 &0\\ 0 & 0 & -1\end{pmatrix}$$

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thanks I think that there is a typo in the above matrix, that should be -8 in the second matrix. – user53800 Dec 24 '12 at 17:37
:( embarassing... – Bombyx mori Dec 24 '12 at 17:44
haha:) , just for kidding, well I know that the statement in the problem is not clear now , and thanks very much for your help. – user53800 Dec 24 '12 at 17:46
:D hopefully fixed now. Thanks. – Bombyx mori Dec 24 '12 at 17:50
In the 4th matrix, you have $-8$ where you want zero, right? – Gerry Myerson Dec 25 '12 at 6:39