# Number of abelian groups of order $p^n$

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 doing – yoyo Mar 17 '11 at 15:45
@yoyo: I should hope so, since we learned it the day these problems were assigned! – The Chaz 2.0 Mar 17 '11 at 15:58

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" ;) – The Chaz 2.0 Mar 17 '11 at 15:45
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! – The Chaz 2.0 Mar 17 '11 at 16:10
@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. :-) – Myself Mar 17 '11 at 16:40
@Myself: In that case, I should be leaving my exams blank! – The Chaz 2.0 Mar 17 '11 at 16:57
@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. – Arturo Magidin Mar 17 '11 at 17:25

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! Thanks – The Chaz 2.0 Mar 17 '11 at 15:46