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Let p and q be distinct primes. How many mutually non-isomorphic Abelian groups are there of order $p^2q^4$. I think there are 6 of them:

$p^2q^4$ $q, qp, q^2p$ $q^2, q^2p^2$ $p, pq^3$ $pq, pq^3$ $q, q^3p^2$

in order that the former divides latter ones. The solution says 10. Any ideas?

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    $\begingroup$ Your list is very hard to read. Are you missing $q,q,qp,qp$ and $q,q,q,qp^2$, and $q,q,q^2p^2$? $\endgroup$ – Derek Holt Oct 22 '14 at 6:37
  • $\begingroup$ If $p(n)$ denotes the partition function and $n=p_1^{a_1}...p_n^{a_n}$, then the number of abelian groups of order $n$ is $p(a_1)...p(a_n)$. $\endgroup$ – Peter Dec 22 '15 at 17:11
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Hint: How many of order $p^2$? How many of order $q^4$? Multiply.

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  • $\begingroup$ No idea what you are talking about. I use my own way, the former divides the latter ones, and got 6. Anything wrong? $\endgroup$ – gter Oct 22 '14 at 4:22
  • $\begingroup$ Every nsuch group is isomorphic to a direct sum of an Abelian group of order $p^2$ and an Abelian group of order $q^4$, and non-isomorphic such groups give us non-isomorphic results, hence the hint. $\endgroup$ – André Nicolas Oct 22 '14 at 4:27
  • $\begingroup$ For $q^4$ they are (i) direct sum of $4$ copies of $\mathbb{Z}_q$; (ii) $2$ of $\mathbb{Z}_q$ and $1$ of $\mathbb{Z}_{q^2}$; (iii) $1$ of $\mathbb{Z}_q$ and $1$ of $\mathbb{Z}_{q^3}$; (iv) $2$ copies of $\mathbb{Z}_{q^2}$; (v) $\mathbb{Z}_{q^4}$. $\endgroup$ – André Nicolas Oct 22 '14 at 4:35

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