Let p be a prime number. How many Abelian groups (up to isomorphism) are there of order p^100.
I am stuck, Please Help
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?
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
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.
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. . I am scared!