Generating function for groups of order n

Number Nine
Messages
813
Reaction score
25
I've done some searching and have thus far come up empty handed, so I'm hoping that someone here knows something that I don't.

I'm wondering if there has been any work on the enumeration of groups of order n (up to isomorphism); specifically, has anyone derived a generating function? Ideally someone would have one for all groups of order n, but I would imagine that there must at least be one for, say, finite abelian groups?
 
Physics news on Phys.org
Alright; I found a sort-of-answer to one half of my question that someone may find interesting, so I'll post it here.

Theorem: Let n be a positive integer with prime factorization \prod p_{k}^{e^{k}}, then the number of abelian groups of order n, up to isomorphism, is given by \prod \rho(e^{k}), where \rho(m) is the number of partitions of the integer m.

Useful note: The partition function \rho(n) is horrifically complicated, and is given to us courtesy of Ramanujan. It's easier to use the following generating function...
\sum_{n=0}^{\infty}\rho(n)q^{n} = \prod_{j=1}^{\infty}\frac{1}{1-q^{j}}\hspace{3 mm} where\hspace{2 mm} |q^{j}| \le 1
EDIT: Apparently the more general case (enumerating groups of order n) is an unsolved problem, which is driving me crazy enough that I've picked up a few books on finite group theory. The problem looks to be very closely tied with the distribution of prime numbers, so this might be difficult...
 
Last edited:
Cool fact. Thanks.
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...

Similar threads

Back
Top