Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 1 down vote accepted

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 $P(a_1) \times P(a_2) \times \dots \times P(a_k)$.

share|improve this answer
1  
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
1  
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. –  Manasi Oct 27 '13 at 6:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.