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I am trying to classify abelian groups of order $19^5$ up to isomorphism. Can anyone provide any approaches or hints?

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    Do you know the structure theorem for finite abelian groups? If you don't, you can consider the endomorphism $\phi=-\times 19$ that sends an element $g$ to $\phi(g)=g+\cdots+g$ where $g$ is being summed $19$ times. What can you say about this endomorphism? Careful study of this will tell you the structure of your group.2012-07-21
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    Hint:Every finite Abelian group is a direct product of cyclic groups2012-07-21
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    Can you think of *any* abelian groups of order $19^5$?2012-07-21
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    I must ask you @Patience, how long have you worked on this question? And more generally, what is your backgroud? You are putting out an awful lot of questions on a broad range of topics; unless you've spent some time on them you won't profit from the solution. You'll know the basic framework of how to solve them, but if my experience has taught me anything, you'll quickly forget, and in the long run you might not retain very much of it at all. For this question in particular: do you know the structure theorem for finite(ly generated) abelian groups?2012-07-21
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    I forgot but I just saw this:every finite abelian group is isomorphic to the direct product of cyclic group of orime power and ofcourse the product of the primes is equal to the order of the abelian group. so there will be exactly one abelian group of order $19^5$2012-07-21
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    @Patience: No. Because two such products are isomorphic if and only if they are identical (up to order of the factors), and there are many ways of writing $19^5$ as a product of prime powers. In fact, there are *seven* nonisomorphic abelian groups of order $19^5$.2012-07-21
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    Ah!!!!! got it :)2012-07-21
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    So $(\Bbb Z_{19})^5$ and $\Bbb Z_{19^5}$ are isomorphic?2012-07-21
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    No, they are non isomorphic2012-07-21
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    @Sade: If you'd like, I can give you the codes in which you are able to classify these groups by using GAP. :)2013-11-18

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From the Fundamental Theorem for Fin. Gen. Abelian groups, it follows that we must take the partitions of 5 (all this can be googled easily): $$\begin{align*}5=&5\\5=&4+1\\5=&3+2\\5=&3+1+1\\5=&2+2+1\\5=&2+1+1+1\\5=&1+1+1+1+1\end{align*}$$ Since there are 7 such partitions, there are 7 non-isomorphic groups of order $\,19^5\,$, which are (notation: $\,C_k=\,$ the cyclic group of order $\,k\,$): $$C_{19^5}\,\,,\,\,C_{19^4}\times C_{19}\,\,,\,\,C_{19^3}\times C_{19^2}\,\,,\,\,C_{19^3}\times C_{19}\times C_{19}\,\,,...\text{you've got the idea}$$

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Hint. $19$ is prime; consider the primary decomposition of such an abelian group.