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How many mutually non-isomorphic Abelian groups whose order is $3^{2}*11^{4}*7$ are there?

Can anyone give a quick (obviously logical) solution?

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Do you know a structure theorem for finite abelian groups? I know of two - one looks at the components for each prime and then decomposes these (Sylow) subgroups into cyclic groups. Your question looks set up to use such a decomposition (the whole group is the direct product of the pieces). –  Mark Bennet Oct 22 '12 at 9:27
    
Just to go straight in the direction of the previous comment. Do you know the Theorem stated in this Wiki page? en.wikipedia.org/wiki/Finitely-generated_abelian_group –  Giovanni De Gaetano Oct 22 '12 at 10:10
    
I've noticed that you have 10 questions in two days. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. –  Martin Sleziak Oct 22 '12 at 13:42
    
@MartinSleziak I'm new here and thanks for your kindly remind. –  J.A.F Oct 24 '12 at 16:47

1 Answer 1

up vote 7 down vote accepted

The answer is $\,2\cdot 5=10\,$, but way more important and interesting is how to reach that answer, which is based in the Fundamental theorem for finitely generated abelian groups.

Definition: A partition of a number $\,n\in\Bbb N:=\{1,2,...\}\,$ is a finite sum of the form

$$a_1+a_2+...+a_r=n\,\,\,\,,\,\,a_i\in\Bbb N\;\;\;\forall\,r$$

We say two partitions of the same natural number are identical if they only difer in the order of the summands, and let us define $\,\cal P(n):=\,$number of different partitions of $\,n\,$

Theorem: If $\,n=p_1^{n_1}\cdot\ldots\cdot p_k^{n_k}\,$ is the prime decomposition of the natural number $\,n\,$ then there are $\,\prod_{m=1}^k\cal P(n_k)\,$ different abelian groups of order $\,n\,$ up to isomorphism

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