# Prove reflections generate a the dihedral group Dn

by simmonj7
Tags: dihedral, generate, prove, reflections
 P: 66 1. The problem statement, all variables and given/known data Let l$_{1}$ and l$_{2}$ be the lines through the origin in $\Re$$^{2}$ that intersect in an angle $\pi$/n and let r$_{i}$ be the reflection about l$_{i}$. Prove the r$_{1}$ and r$_{2}$ generate a dihedral group D$_{n}$. 2. Relevant equations D$_{n}$: the dihedral group of order 2n generated by two elements: the rotation $\rho$$_{\theta}$, where $\theta$ = 2$\pi$/n, and a reflection r' about a line l through the origin. The product r$_{1}$r$_{2}$ of these reflections preserves orientation and is a rotation about the origin. Its angle of rotation is $\pm$2$\theta$. 3. The attempt at a solution So I have spent a lot of time talking this problem out and I still don't feel like I completely grasp it. I had originally figured that I could just say that by the definition of the dihedral group (provided above) which was given in one of our theorems, we need to find a reflection and a rotation which fit the descriptions in that definition and then I would be set. However, I have been told that doing so it not enough. It was suggested to me that I should create an n-gon centered at the origin with one vertex along the line l$_{1}$. I was told that from here, I should be showing that the set generated by this rotation isn't too big (i.e. infinite) or too small and is therefore the whole of D$_{n}$. Now here is where I am completely confused and am not sure if I am understanding anything right...I am being told that I needed to show that if I reflected my vertex lying along the line l$_{1}$ around l$_{2}$, I then get an angle of 2$\pi$/n between my reflected point and the line l$_{1}$ (because the angle between l$_{1}$ and l$_{2}$ was $\pi$/n so reflecting across the line l$_{2}$ would give me an angle twice as large i.e. 2$\pi$/n). However, because the angular distance between two symmetries is 2$\pi$/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 $\pi$/n, this means that the reflections generate a finite group because any angle which is a multiple of $\pi$/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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