Is a given function a homomorphism?

I have a question that I have approached, but want to check if I'm on the right track.

Let G denote the group of symmetries of a circle. There are infinitely many reflections and rotations. There are no elements besides reflections and rotations. The identity element is the rotation by zero degrees. Let H denote the subgroup of H consisting of only the rotations.

Question: Define f: G--> {1,-1} by f(x) = 1 if x is a rotation and f(x) = -1 if x is a reflection. Is f a homomorphism?

What I did was check if f(xy)=f(x)f(y). So f((-1)*1)=f(-1)f(1)=-1, and f((1)*1)=f(1)f(1)=1, etc. It all checks out, so f is a homomorphism.

Did I do it right? Thanks!!

Re: Is a given function a homomorphism?

Let be a rotation and be a reflection. What are the relationships? Is it and ? Also, has elements that are neither reflections nor rotations, but products of both. Perhaps define so that for any word , if contains an even number of reflections and if contains an odd number of reflections. Now, you need some form of parity function to check that this is well defined. If there exists any two words, equal in G, with one word containing an even number of reflections and one word containing an odd number, then f is not a homomorphism (and is not even well-defined).

Re: Is a given function a homomorphism?

Hmm... sounds great. Can you help me construct the parity function? It is very challenging for me.

Re: Is a given function a homomorphism?

You can avoid checking parity if you can show that for any word with there exists a rotation and a reflection such that either or . This would essentially become your parity check. First, check that you can generate the entire group with a single reflection: and all rotations . If you can generate from just these elements, then the same way you show that any element of any other dihedral group consists only of elements of the form r^n or sr^n, now you would be trying to show G only contains elements of the form or .