Let p be a prime number. How many Abelian groups (up to isomorphism) are there of order p^100.

I am stuck, Please Help(Crying)

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- Nov 16th 2008, 02:27 PMmandy123How many Abelian groups are there
Let p be a prime number. How many Abelian groups (up to isomorphism) are there of order p^100.

I am stuck, Please Help(Crying) - Nov 16th 2008, 02:33 PMThePerfectHacker
If the problem was about $\displaystyle p^3$ then the Fundamental Theorem for Abelain groups would say it is one of the following:

$\displaystyle \mathbb{Z}_{p^3}$

$\displaystyle \mathbb{Z}_{p^2}\times \mathbb{Z}_p$

$\displaystyle \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p$

What about your more general problem? - Nov 16th 2008, 02:50 PMmandy123
I don't have a more general problem, this is all I am given. Is that bad?

So if I follow what you did, would there be 100 Abelian groups? - Nov 16th 2008, 03:08 PMJhevon
note that $\displaystyle 100 = 5^2 \cdot 2^2$

so that if we let $\displaystyle p = 5$ and $\displaystyle q = 2$, then in the most basic cases you have

$\displaystyle \mathbb{Z}_{p^2} \times \mathbb{Z}_{q^2}$

now you want all possible combinations of those two in direct products. the number of combinations you come up with is the number of Abelian groups - Nov 16th 2008, 03:16 PMThePerfectHacker
- Nov 16th 2008, 03:16 PMNonCommAlg
by the fundamental theorem of finite abelian groups, the answer is the number of partitions of 100, which according to this website is equal to 190569292.

- Nov 16th 2008, 03:20 PMJhevon
- Nov 16th 2008, 03:21 PMJhevon
- Nov 16th 2008, 03:25 PMThePerfectHacker
I do not think there is a partitions formula.

The # of partitions is a very complicated combinatorics problem.

There are ways to get them using recurrence relations and all that stuff. But as far as a formula that give you an answer it is does not exist. At least I never seen one. (Worried). I am scared!