TITLE: Number of abelian groups of order 108
QUESTION [7 upvotes]: What is the number  of  abelian groups of  order 108 upto isomorphism ?
To answer  this I wrote explicitly the possible abelian groups of order 108 as follows : 
$$\Bbb Z_{108}$$
$$\Bbb Z_{4}\times\Bbb Z_{3}\times\Bbb Z_{9}$$
$$\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{27}$$
$$\Bbb Z_{4}\times\Bbb Z_{3}\times\Bbb Z_{3}\times\Bbb Z_{3}$$
  $$\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{3}\times\Bbb Z_{9}$$
$$\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{3}\times\Bbb Z_{3}\times\Bbb Z_{3}$$
And I  found  the  answer  to be 6. But  my problem  is  that  what  if  I  was  given  a much bigger  number? Is this  the  only  way  to  find  abelian  groups  of  a  certain  order? If there  are  better  ways  to find  the  exact  answer  to  such  question please  let  me  know.

REPLY [9 votes]: Let $n=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}$ be a prime factorization of $n\in\mathbb{N}$.  For $m\in\mathbb{N}$, $C_m$ denotes the cyclic group of order $m$.  An abelian group of order $n$ will be a direct sum $\bigoplus_{i=1}^k\bigoplus_{j=1}^{s_i}\,C_{p_i^{t_{i,j}}}$, where $t_{i,j}$, for $j=1,2,\ldots,s_i$ and $i=1,2,\ldots,k$, is a positive integer such that $\sum_{j=1}^{s_i}\,t_{i,j}=r_i$.  We may assume that $t_{i,1}\leq t_{i,2}\leq \ldots \leq t_{i,s_i}$.  Hence, the number of such abelian groups (up to isomorphism) is the product $p\left({r_1}\right)\,p\left({r_2}\right)\,\cdots \,p\left({r_k}\right)$, where $p$ is the partition function.
In your example, $108=2^2\cdot 3^3$.  Now, $p(2)=2$ because $2=2$ and $2=1+1$, whereas $p(3)=3$ because $3=3$, $3=1+2$, and $3=1+1+1$.  Hence, the number of abelian groups of order $108$ up to isomorphism is $p(2)\,p(3)=2\cdot 3=6$.