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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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    the fundamental theorem of finitely generated abelian groups justifies what youre doing2011-03-17
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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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    Right. I intend to count them when I'm done! Thanks again. I will look into this *so-called* "partition" ;)2011-03-17
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    When you say "...should verify", is that for the sake of the question asked, or for my overall education, or ... ? I don't want to leave any stone uncovered on this assignment!2011-03-17
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    @Arturo Magidin: In fact the question asks to determine the number only, not to return it to the person asking the question. Therefore counting them and not answering anything should be sufficient. :-)2011-03-17
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    @Myself: In that case, I should be leaving my exams blank!2011-03-17
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    @The Chaz: My point was that if you want to count the number of non-isomorphic groups by *instead* looking at the partitions, then in order to justify that you need to show there is a bijection. If someone asks you to count the number of cows on the field, and you instead count the number of hoofs and divide by four, you need to show that what you counted actually gives the answer you are looking for.2011-03-17
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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