# Is a given function a homomorphism?

• Oct 7th 2013, 06:47 PM
abscissa
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!!
• Oct 7th 2013, 07:09 PM
SlipEternal
Re: Is a given function a homomorphism?
Let $r\in G$ be a rotation and $s \in G$ be a reflection. What are the relationships? Is it $rsrs = 1$ and $s^2 = 1$? Also, $G$ has elements that are neither reflections nor rotations, but products of both. Perhaps define $f$ so that for any word $w \in G$, $f(w) = 1$ if $w$ contains an even number of reflections and $f(w)=-1$ if $w$ 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).
• Oct 8th 2013, 03:49 PM
abscissa
Re: Is a given function a homomorphism?
Hmm... sounds great. Can you help me construct the parity function? It is very challenging for me.
• Oct 8th 2013, 04:12 PM
SlipEternal
Re: Is a given function a homomorphism?
You can avoid checking parity if you can show that for any word $g_1g_2\cdots g_n$ with $g_i \in G$ there exists a rotation $r$ and a reflection $s$ such that either $g_1g_2\cdots g_n = r$ or $g_1g_2\cdots g_n = sr$. This would essentially become your parity check. First, check that you can generate the entire group with a single reflection: $s: e^{i\theta} \mapsto e^{-i\theta}$ and all rotations $r_\delta: e^{i\theta} \mapsto e^{i(\theta + \delta)}$. If you can generate $G$ 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 $r_\delta$ or $sr_\delta$.