Are Cyclic Groups with x^n = 1 the Only Finite Groups?

  • Thread starter Thread starter Identity
  • Start date Start date
  • Tags Tags
    Finite Groups
Identity
Messages
151
Reaction score
0
Is it true that cyclic groups with x^n = 1 the only finite groups (with order n)?

I've been experimenting with a few groups and I think this is true but I'm not sure.thanks
 
Physics news on Phys.org
Have you looked at the permutation groups Sn, dihedral groups Dn (symmetries of regular polygons), etc.?

In fact, almost all finite groups are non-abelian, so the cyclic groups are just very few in number.
 
The finite groups are completely categorized. See for example:

http://en.wikipedia.org/wiki/List_of_finite_simple_groups

Or, I strongly recommend Mark Ronan's book, "Symmetry and the Monster". It is a good read of this fascinating subject.
 
CompuChip said:
In fact, almost all finite groups are non-abelian, so the cyclic groups are just very few in number.
There are as many cyclic groups (\aleph_0) as finite groups. By, "almost all finite groups are non-abelian", do you mean \lim_{n\rightarrow \infty}a_n/g_n=0, where a_n is the number of abelian groups of order \leq n and g_n is the number of groups of order \leq n? (This doesn't seem likely.)
 
Given any n, there certainly exists a cylic group of order n.

It is also true that if p is a prime number then then the only group of order p is the cylic group.

But if n is NOT prime, then there exist other, non-cyclic, groups of order n.

The simplest example is for n= 4. The "Klein 4-group" is not cyclic.

The Klein 4-group has 4 members, e, a, b, c satifying
ee= e, ea= a, eb= b, ec= c (e is the identity)
ae= a, aa= e, ab= c, ac= b
be= b, ba= c, bb= e, bc= a
ce= c, ca= b, cb= a, cc= e

The fact that every element is its own inverse proves this group is not cyclic.

Any group of order 4 is isomorphic either to the cyclic group of order 4 or to the Klein four-group.
 
Martin Rattigan said:
...(This doesn't seem likely.)
Then again it doesn't seem unlikely. Has it been proved?

The non-abelian groups of orders 2n will probably arrange it by themselves (e.g. there are over ten times as many non-abelian groups of order 1024 as there are other groups up to and including 1024) so this may not be too difficult to prove.
 
Last edited:
##\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...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top