Is a Discrete Group of Rotations Cyclic?

  • Thread starter Thread starter SNOOTCHIEBOOCHEE
  • Start date Start date
  • Tags Tags
    Discrete Groups
SNOOTCHIEBOOCHEE
Messages
141
Reaction score
0

Homework Statement



Prove that a discrete group G cosisting of rotations about the origin is cyclic and is generated by \rho_{\theta} where \theta is the smallest angle of rotation in G

The Attempt at a Solution



since G is by definition a discrete group we know that if \rho is a rotation in G about some point through a non zero angle \theta the the angle \theta is at least \epsilon:|\theta|\geq\epsilon

But i don't know how to apply this definition to show that G is cyclic. Is this definition even useful?
 
Physics news on Phys.org
Ok, if you pick theta to be the smallest angle (which you can do since the rotations are discrete), then all of the rotations n*theta for n an integer are in the group. If that's the whole group, then you are done since it's cyclic. If not there a rotation phi in the group that isn't equal to n*theta for any n. Can you take the next step?
 
with that i can show that there is a non zero positive rotation less than theta, a contradiction. Is that it?
 
SNOOTCHIEBOOCHEE said:
with that i can show that there is a non zero positive rotation less than theta, a contradiction. Is that it?

It sure is.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top