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I was trying to solve the following problem:

Find number of abelian groups of order $10^5$, up to isomorphism ? Could someone point me in the right direction?

Thanks in advance for your time.

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    $\begingroup$ Every finite abelian group is a product of cyclic groups of prime power order. $\endgroup$ – Joe Johnson 126 Mar 10 '17 at 15:15
  • $\begingroup$ @JoeJohnson126 You mean "is isomorphic to" rather than just "is", right? $\endgroup$ – Bobson Dugnutt Mar 10 '17 at 15:18
  • $\begingroup$ The Structure Theorem for Finitely Generated Abelian Groups will tell you what these groups are. $\endgroup$ – Henning Makholm Mar 10 '17 at 15:18
  • $\begingroup$ @Lovsovs If you don't mean "is isomorphic to" then you have no hope of answering a question of the number of groups of a given size: there is a proper class of trivial groups. $\endgroup$ – Patrick Stevens Mar 10 '17 at 15:20
  • $\begingroup$ @Lovsovs Every finite abelian group can be expressed as an internal direct product of cyclic subgroups of prime power order, and is not merely isomorphic to such a group. $\endgroup$ – Bungo Mar 10 '17 at 15:35
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Hint : The number of abelian groups of order $$p_1^{a_1}\cdots p_n^{a_n}$$

upto isomorphism is $$p(a_1)\cdots p(a_n)$$ where $p(n)$ denotes the number of partitions of $n$

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If $n=\prod_{i=1}^rp_i^{k_i}$, then the number of distinct abelian groups of order $n$ is given by $$ \prod_{i=1}^rp(k_i), $$ where $p(k)$ denotes the number of partitions of $k$. Now $n=10^5=2^5\cdot 5^5$, and $p(5)^2=7^2=49$.

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