# How many Abelian groups are there

• Nov 16th 2008, 03:27 PM
mandy123
How many Abelian groups are there
Let p be a prime number. How many Abelian groups (up to isomorphism) are there of order p^100.

• Nov 16th 2008, 03:33 PM
ThePerfectHacker
Quote:

Originally Posted by mandy123
Let p be a prime number. How many Abelian groups (up to isomorphism) are there of order p^100.

If the problem was about $p^3$ then the Fundamental Theorem for Abelain groups would say it is one of the following:

$\mathbb{Z}_{p^3}$
$\mathbb{Z}_{p^2}\times \mathbb{Z}_p$
$\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p$

• Nov 16th 2008, 03:50 PM
mandy123
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, 04:08 PM
Jhevon
Quote:

Originally Posted by mandy123
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?

note that $100 = 5^2 \cdot 2^2$

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

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

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, 04:16 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
note that $100 = 5^2 \cdot 2^2$

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

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

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

I think mandy is talking about $p^{100}$.
• Nov 16th 2008, 04:16 PM
NonCommAlg
Quote:

Originally Posted by mandy123

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

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, 04:20 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
I think mandy is talking about $p^{100}$.

oh! oh yes, i see. i thought it was groups of order 100 as opposed to $p^{100}$
• Nov 16th 2008, 04:21 PM
Jhevon
Quote:

Originally Posted by NonCommAlg
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.

there should be some kind of combination formula for this right? similar to the one used in the multinomial theorem
• Nov 16th 2008, 04:25 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
there should be some kind of combination formula for this right? similar to the one used in the multinomial theorem

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!