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?

Solution 1:

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.

Solution 2:

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.