# Homework Help: Proving a group of rotaions is cyclic

1. Nov 13, 2008

### SNOOTCHIEBOOCHEE

1. The problem statement, all variables and given/known data
Let G be a finite group of rotation of the plane about the origin. Prove that G is cyclic.

3. 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 cant for the life of me figure out how to prove this.

2. Nov 13, 2008

### Dick

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. Nov 13, 2008

### SNOOTCHIEBOOCHEE

By division algorithm s= qr+ m . Im 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. Nov 13, 2008

### Office_Shredder

Staff Emeritus
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. Nov 13, 2008

### Dick

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. Nov 13, 2008

### SNOOTCHIEBOOCHEE

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. Nov 13, 2008

### Dick

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.