8
$\begingroup$

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?

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You missed the partition $\,5=5\,$ $\endgroup$ – DonAntonio Sep 24 '12 at 22:27
  • 1
    $\begingroup$ Yes I did. Thanks. $\endgroup$ – Dane Sep 24 '12 at 22:29
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.