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 $4\times 3$.

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The key word here is partition. See mathworld.wolfram.com/AbelianGroup.html – lhf Apr 27 '11 at 19:42
en.wikipedia.org/wiki/… – Qiaochu Yuan Apr 27 '11 at 19:57
@Qiaochu, unfortunately that Wikipedia page does not mention the number of isomorphism classes; it does not even mention partitions. Hence my link to MathWorld. I'd edit the Wikipedia page, but I couldn't find a nice place to add something on the number of isomorphism classes. (Ok, I didn't look too hard.) – lhf Apr 27 '11 at 20:11

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.

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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.

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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.

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