Register to reply

Prove reflections generate a the dihedral group Dn

by simmonj7
Tags: dihedral, generate, prove, reflections
Share this thread:
Nov3-11, 07:44 PM
P: 66
1. The problem statement, all variables and given/known data
Let l[itex]_{1}[/itex] and l[itex]_{2}[/itex] be the lines through the origin in [itex]\Re[/itex][itex]^{2}[/itex] that intersect in an angle [itex]\pi[/itex]/n and let r[itex]_{i}[/itex] be the reflection about l[itex]_{i}[/itex]. Prove the r[itex]_{1}[/itex] and r[itex]_{2}[/itex] generate a dihedral group D[itex]_{n}[/itex].

2. Relevant equations
D[itex]_{n}[/itex]: the dihedral group of order 2n generated by two elements: the rotation [itex]\rho[/itex][itex]_{\theta}[/itex], where [itex]\theta[/itex] = 2[itex]\pi[/itex]/n, and a reflection r' about a line l through the origin.
The product r[itex]_{1}[/itex]r[itex]_{2}[/itex] of these reflections preserves orientation and is a rotation about the origin. Its angle of rotation is [itex]\pm[/itex]2[itex]\theta[/itex].

3. The attempt at a solution
So I have spent a lot of time talking this problem out and I still don't feel like I completely grasp it. I had originally figured that I could just say that by the definition of the dihedral group (provided above) which was given in one of our theorems, we need to find a reflection and a rotation which fit the descriptions in that definition and then I would be set.

However, I have been told that doing so it not enough. It was suggested to me that I should create an n-gon centered at the origin with one vertex along the line l[itex]_{1}[/itex]. I was told that from here, I should be showing that the set generated by this rotation isn't too big (i.e. infinite) or too small and is therefore the whole of D[itex]_{n}[/itex]. Now here is where I am completely confused and am not sure if I am understanding anything right...I am being told that I needed to show that if I reflected my vertex lying along the line l[itex]_{1}[/itex] around l[itex]_{2}[/itex], I then get an angle of 2[itex]\pi[/itex]/n between my reflected point and the line l[itex]_{1}[/itex] (because the angle between l[itex]_{1}[/itex] and l[itex]_{2}[/itex] was [itex]\pi[/itex]/n so reflecting across the line l[itex]_{2}[/itex] would give me an angle twice as large i.e. 2[itex]\pi[/itex]/n). However, because the angular distance between two symmetries is 2[itex]\pi[/itex]/n, this means that the reflection I just got is another vertex of my n-gon. On the other hand, to show that the set generated is not too big (major confusion here again) I was told that because I have an original angle of [itex]\pi[/itex]/n, this means that the reflections generate a finite group because any angle which is a multiple of [itex]\pi[/itex]/n does so. This is what was suggested to me as the path I should take to prove this problem, however, I am not understanding how this is proving what I need to show.

If anyone as any insight as to why this works, please let me know. Thanks.
Phys.Org News Partner Science news on
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display

Register to reply

Related Discussions
Dihedral group - isomorphism Linear & Abstract Algebra 2
ABSTRACT- Dihedral Group Calculus & Beyond Homework 6
Dihedral and Symmetric Group Linear & Abstract Algebra 1
Dihedral group Calculus & Beyond Homework 1
Dihedral Group of Order 8 Linear & Abstract Algebra 4