Prove reflections generate a the dihedral group Dn

[simmonj7]

Homework Statement

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$_{n}$.

Homework 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$.

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.

 

[mighty2000]

Thank you for bringing up this interesting problem. I am always intrigued by mathematical concepts and their applications. I will try my best to provide a clear explanation of the solution to this problem.

First, let's recall the definition of a dihedral group, which is the group of symmetries of a regular n-gon. This group is generated by two elements: a rotation, denoted by \rho_{\theta}, where \theta = 2\pi/n, and a reflection, denoted by r', about a line l through the origin. This means that any symmetry of the n-gon can be obtained by combining these two operations.

Now, let's consider the two given lines, l_{1} and l_{2}, which intersect at an angle \pi/n. We can create a regular n-gon centered at the origin with one of its vertices lying on l_{1}. This n-gon can be rotated by \rho_{\theta} to obtain all the other vertices. However, we also have the reflection r_{1} about l_{1}, which can be used to obtain the same set of vertices by reflecting them across l_{1}. Similarly, we have another reflection r_{2} about l_{2}.

Now, let's look at the product r_{1}r_{2}. This product can be thought of as first reflecting a point across l_{2} and then reflecting the result across l_{1}. This means that the final point will be rotated by 2\pi/n with respect to the original point (since the angle between l_{1} and l_{2} is \pi/n). This is exactly what a rotation by \rho_{\theta} does. Therefore, r_{1}r_{2} is a rotation by \pm2\theta.

Thus, we have shown that the two reflections r_{1} and r_{2} generate the dihedral group D_{n}. This is because any reflection or rotation of the n-gon can be obtained by combining these two operations. Moreover, since the angle between l_{1} and l_{2} is \pi/n, we can see that the set generated by these two reflections is not too big or too small, as it includes all the reflections and rotations by angles that are multiples of \pi/n.

I hope this explanation helps you understand the solution to this problem.