Proving a group of rotaions is cyclic

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In summary, the conversation discusses how to prove that a finite group of rotations about the origin is cyclic. One approach is to show that the group must have a smallest nonzero rotation, which can form a cyclic subgroup. Another approach involves using the division algorithm to show that a rotation s is between nr and (n+1)r for some rotation r, and this leads to a contradiction if the group is assumed to be noncyclic. Ultimately, this contradiction demonstrates that the group must be cyclic.
  • #1
SNOOTCHIEBOOCHEE
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Homework Statement


Let G be a finite group of rotation of the plane about the origin. Prove that G is cyclic.


The Attempt at a Solution



What it means to be cyclic is that every element of the group can be written as a^n for some integer n.

I can see this is true if i take some examples. i.e. {0,pi/2, pi, 3pi/2} is cleraly cyclic. but i can't for the life of me figure out how to prove this.
 
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  • #2
There are lots of ways to prove this depending on what tools you happen to have around. The group must have a smallest nonzero rotation r. All multiples of r must also be in the group. That makes a nice cyclic subgroup, right? If it's noncyclic there must be an element s which is not a multiple of r. Now what?
 
  • #3
Dick said:
There are lots of ways to prove this depending on what tools you happen to have around. The group must have a smallest nonzero rotation r. All multiples of r must also be in the group. That makes a nice cyclic subgroup, right? If it's noncyclic there must be an element s which is not a multiple of r. Now what?

By division algorithm s= qr+ m . I am guessing we have to somehow show that m=0? Or show that if s is in a group then r is not.

dont know how to do that.
 
  • #4
s=qr+m. If r is in the group and s is in the group, s-qr=m is in the group too.The notation is a little bit non-rigorous here, but I think it gets the idea across
 
  • #5
SNOOTCHIEBOOCHEE said:
By division algorithm s= qr+ m . I am guessing we have to somehow show that m=0? Or show that if s is in a group then r is not.

dont know how to do that.

The idea is that the rotation s is between nr and (n+1)r for some n. Then, yes, s-nr must be in the group. Which is bigger s-nr or r? Then remember we picked r to be the smallest nonzero rotation.
 
  • #6
s-nr is smaller than r.

because we defined s to be between nr and nr+r. so s is at most (by most i mean slightly less than) nr+r. so s<nr+r ==> s-nr<r.

since we chose r to be the smallest non zero rotation we have a contradiction. But i have no clue what this contradiction tells us.
 
  • #7
It contradicts that the group is finite. An infinite group of group of rotations doesn't need to have a smallest member. And doesn't need to be cyclic.
 

1. What is a cyclic group of rotations?

A cyclic group of rotations is a mathematical concept in group theory that describes a group of rotations that can be generated by a single rotation. This means that all the other rotations in the group can be obtained by repeatedly applying the single rotation.

2. How do you prove that a group of rotations is cyclic?

To prove that a group of rotations is cyclic, you need to show that there exists a single rotation that can generate all the other rotations in the group. This can be done by showing that any rotation in the group can be expressed as a power of the generating rotation.

3. What is the significance of proving that a group of rotations is cyclic?

Proving that a group of rotations is cyclic is important in mathematics as it helps us understand the structure and properties of the group. It also allows us to simplify calculations and makes it easier to analyze the group.

4. Can a group of rotations be cyclic if it has more than one generating rotation?

No, a group of rotations can only be cyclic if it has a single generating rotation. If there are multiple generating rotations, then the group is not cyclic and is instead classified as a dihedral group.

5. How is the cyclic nature of a group of rotations related to the order of the group?

The cyclic nature of a group of rotations is directly related to the order of the group. If the group is cyclic, then the order of the group is equal to the order of the generating rotation. If the group is not cyclic, then the order of the group will be a multiple of the order of the generating rotation.

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