Counting the number of isomorphism types of abelian groups of order 2000 Number of isomorphism abelian groups of order 2000?
$2000 = 2^4 5^3$, and the non-isomorphic groups is just $4\times 3$.
 A: The isomorphism classes of abelian groups of order $p^m$ ($p$ a prime, $m\gt 0$) correspond to the number of ways in which we can partition $m$, since such a group must be of the form
$$C_{p^{a_1}}\oplus\cdots\oplus C_{p^{a_k}}$$
with $0\lt a_1\leq a_2\leq\cdots\leq a_k$ and $a_1+\cdots+a_k = m$. So you want to count the number of ways to partition $4$ (to get all the isomorphism types of abelian groups of order $2^4$) and all the ways to partition $3$ (to get all the isomorphism types of abelian groups of order $5^3$. Then multiply them together, since the group is isomorphic to the product of its $p$-parts, and they can be chosen independently.
A: GAP tells me there are 15 of them, so there is something wrong with your method.
Interestingly, according to GAP there are $963$ groups of alder $2000$ in all.
A: I'll be a little more explicit
You need to count all the ways you could group the four 2's in the direct sum decomposition (2*2*2*2), (2*2*2)*2, (2*2)*2*2, (2*2)(2*2), 2*2*2*2 
Note: The ordering of the groups (partition cells) doesn't matter, only their sizes eg $Z/2Z \times Z/4Z \times Z/2Z$ is isomorphic $Z/2Z \times  Z/2Z \times Z/4Z$ but $Z/16Z$ isn't isomorphic to $Z/4Z \times Z/4Z$.
And you also need to count all the ways you could group the three 5's
(5*5*5),(5*5)*5, 5*5*5
Then as mentioned above you need to pick one grouping (partition) for the 2's and another for the 5's
All told there are $5 \times 3 = 15$ possibilities.
