# Finite abelian groups

Find all finite abelian groups (up to isomorphism) of order 320.

So I found the prime factorization to be $2^6 \times 5$. I found the 11 groups to be

$\mathbb{Z}_{64} \times \mathbb{Z}_5$,

$\mathbb{Z}_{32} \times \mathbb{Z}_2 \times \mathbb{Z}_5$,

$\mathbb{Z}_{16} \times \mathbb{Z}_4 \times \mathbb{Z}_5$,

$\mathbb{Z}_{16} \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$,

$\mathbb{Z}_8 \times \mathbb{Z}_8 \times\mathbb{Z}_5$,

$\mathbb{Z}_8 \times \mathbb{Z}_4 \times\mathbb{Z}_2 \times\mathbb{Z}_5$,

$\mathbb{Z}_8 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$,

$\mathbb{Z}_4 \times \mathbb{Z}_4 \times\mathbb{Z}_4 \times\mathbb{Z}_5$,

$\mathbb{Z}_4 \times \mathbb{Z}_4 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$,

$\mathbb{Z}_4 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times \mathbb{Z}_5$,

$\mathbb{Z}_2 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2\times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$.

This is all I have to show right? I listed all the groups.

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Whether this is all you have to show would depend on whom you have to show it to, and on what she expects to see. She might expect to see a proof that these all work, and/or a proof that they are pairwise nonisomorphic, and/or a proof that there aren't any other ones --- but only she knows what she wants to see, so I'd advise you to ask her. – Gerry Myerson Oct 27 '13 at 6:04
Well I have notes that listed just the groups and the number of groups we had. I also see a mention of invariant factors but nothing more in depth than that. Well actually my prof is a male lol. – Ruth Gutierrez Oct 27 '13 at 6:10

You have found the prime factorization to be $2^6 \times 5^1$, so the number of groups would be $P(6) \times P(1)$ where $P(i)$ is the partition function of i i.e the number of ways of expressing natural number $i$ in a distinct manner.

6=6

6=5+1

6=4+2

6=4+1+1

6=3+3

6=3+2+1

6=3+1+1+1

6=2+2+2

6=2+2+1+1

6=2+1+1+1+1

6=1+1+1+1+1+1

Infact if the prime factorization of your order $n$ is $p_1^{a_1} \times p_2^{a_2} \times \dots p_k^{a_k}$ then the total number of abelian groups possible upto isomorphism of order $n$ is $P(a_1) \times P(a_2) \times \dots \times P(a_k)$.

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Yes, I was able to get that. I did that so I could use it to find my groups. I moreless just need to know if I finished answering the question. – Ruth Gutierrez Oct 27 '13 at 6:47
Yes,you have! Great job! Actually, there is a proof for my last statement. You can find it in I.N Herstein or any standard Algebra textbook for undergraduates. – wannadeleteacct Oct 27 '13 at 6:52