Prove reflections generate a the dihedral group Dnby simmonj7 Tags: dihedral, generate, prove, reflections 

#1
Nov311, 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 ngon 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 ngon. 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. 


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