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If $p$ is prime, determine the number of abelian groups of order $p^n$ for each $1\leq n\leq8$

(I assume that "up to isomorphism" should be included somewhere in the question for the sake of precision...)

Could someone please review/confirm my work?
n = 1: $\mathbb{Z}_p $
n = 2: $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$
n = 3: $\mathbb{Z}_{p^3}$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p$
n = 4: $\mathbb{Z}_{p^4}$, $\mathbb{Z}_{p^3} \times \mathbb{Z}_p$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^2}$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_p \times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p \times \mathbb{Z}_p$

et cetera

I am simply considering all the options for when the largest exponent of $p$ is $n$, then $n-1$, and so on. How does this look? Thanks!
(Apparently I don't know how to "end a quote"...)

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    @yoyo: I should hope so, since we learned it the day these problems were assigned!2011-03-17

2 Answers 2

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Your work is correct, except that you aren't answering the question asked (they asked you for the number of (nonisomorphic) groups, not for a list of the groups). So for $n=1$, the answer should be "1"; for $n=2$ the answer should be "2"; for $n=3$ the answer should be "3"; for $n=4$ the answer should be "5", etc.

The magic words you are looking for are "partitions of $n$." You should verify that there is a bijection between the isomorphism types of abelian groups of order $p^n$ and the partitions of $n$.

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    @Arturo: I see. Standby!2011-03-17
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It's equal to the number of partitions of $n$. See http://mathworld.wolfram.com/AbelianGroup.html

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    Wow. They actually list the case n = 8 ... no way I'm LaTexing **that**! Thanks2011-03-17