abelian groups of order $10^5$

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?

• Every finite abelian group is a product of cyclic groups of prime power order. – Joe Johnson 126 Mar 10 '17 at 15:15
• @JoeJohnson126 You mean "is isomorphic to" rather than just "is", right? – Bobson Dugnutt Mar 10 '17 at 15:18
• The Structure Theorem for Finitely Generated Abelian Groups will tell you what these groups are. – Henning Makholm Mar 10 '17 at 15:18
• @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. – Patrick Stevens Mar 10 '17 at 15:20
• @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. – Bungo Mar 10 '17 at 15:35

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 $n$
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 $p(5)^2=7^2=49$.