# Group generated by a nonzero translation

1. Apr 8, 2009

### zcdfhn

Let G1 be the group generated by a nonzero translation and G2 be the group generated by a glide reflection. Show that G1 and G2 are isomorphic.

Here is how I started:

G1 = <Tb> where b$$\in$$ C and Tb(z) = z+b

G2 = <ML $$\circ$$ Tc> where c and L are parallel to each other.

Let's define a function $$\Phi$$: G1 $$\rightarrow$$ G2

Then if $$\Phi$$ is a homomorphism and a bijection, it is an isomorphism.

But here lies my problem, I do not know what to make $$\Phi$$ equal to. Maybe this isn't the right way of approaching this problem.

2. Apr 8, 2009

### e(ho0n3

Re: Isomorphisms

How can you perform a translation using a glide reflection? How can you perform a glide reflection with a translation?

3. Apr 8, 2009

### zcdfhn

Re: Isomorphisms

A glide reflection composed with itself is a translation. And you get a glide reflection by composing a reflection with a translation.

4. Apr 8, 2009

### e(ho0n3

Re: Isomorphisms

So informally, translation = 2 * glide reflection and glide reflection = reflection + translation. Since you need a reflection to get a glide translation, I don't see any way of getting an isomorphism between the two groups.