Computing the number of nonisomorphic finite abelian groups of order $n$ Let's say I'm given a number $n = p^{3}q^{4}r^{2}s$ and I want to find the number of (non-isomorphic) abelian groups of order $n$. How I'm computing the number is basically partitioning, that is, I'm just looking at it and saying that I have $\mathbb{Z}/p^{3}q^{4}r^{2}s\mathbb{Z}$, $\mathbb{Z}/p^{2}q^{4}r^{2}s\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, and so on. Then I count the number of groups I've written down. However, I've found that I'm quite prone to error when I do this. Is there a faster way of doing this? If not, is there a systematic trick/way I could use to make this easier?
 A: You only need to be able to compute the partitions of a single number, and then the rest follows from the Fundamental Counting Principle.
We know there are three partitions of 3, five partitions of 4, two partitions of 2 and one partition of 1. Then there are 3*5*2*1 distinct mixtures of partitions. So there are 30 total outcomes.
A: Here's a systematic way of doing it.
If $n = \prod p_i^{a_i}$ is a prime factorization, then for each $p_i$, find all partitions of the number $a_i$ into positive integers.  For example, if $a_i$ is 5, then these will be 
\begin{align*}
&5\\
&4+1, \\
&3+2, \\
&3+1+1, \\
&2+2+1, \\
&2+1+1+1, \\
&1+1+1+1+1.
\end{align*}
You see how these have been written down in a systematic pattern.  Then the Sylow $p_i$-subgroup of your group will be isomorphic to a product of cyclic groups, where you can choose any one of these partitions as the exponents of $p_i$ in the order of each cyclic factor.  For example, in the situation above, the Sylow $p_i$-subgroup would be one of
\begin{align*}
&Z_{p_i^5} \\
&Z_{p_i^4}+Z_{p_i}, \\
&Z_{p_i^3}+Z_{p_i^2}, \\
&Z_{p_i^3}+Z_{p_i}+Z_{p_i}, \\
&Z_{p_i^2}+Z_{p_i^2}+Z_{p_i}, \\
&Z_{p_i^2}+Z_{p_i}+Z_{p_i}+Z_{p_i}, \\
&Z_{p_i}+Z_{p_i}+Z_{p_i}+Z_{p_i}+Z_{p_i}. 
\end{align*}
Then you can take a product of any combination of these Sylow $p_i$-subgroups to form your group.
