# Dihedral Group Computations

Let $n > 1$ be an integer and let $\theta = \dfrac{2\pi}{n}$. Let $P$ be the regular $n$-gon with vertices ($\cos i\theta$, $\sin i \theta$) for $i \in \mathbb Z_n$. The dihedral group $D_n$ is the symmetry group of $P$, which consists of rotations $R_i$ and reflections $F_i$ for $i \in \mathbb Z_n$.

$R_i$ is the counterclockwise rotation around the origin by angle $i$, and $F_i$ is the reflection across the line through the origin by angle $i\theta$ and $F_i$ is the reflection across ($\cos i\theta_2$, $\sin i\theta_2$).

How to give general formulas for $R_iR_j$ , $R_iF_j$ , $F_iR_j$ , and $F_iF_j$ . For example, $R_iR_j = R_{i+j}$ ,where the addition of indices is $\mod n$

And is $D_n$ a group? If not, what is missing?

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Have you considered trying to write this all down using matrices? Then you should hopefully see what is going on. – user1729 Feb 7 '12 at 10:55
(Meant to say that you can find the linear (matrix) representation of $D_n$ on the wikipedia page. However, you should try and understand how these matrices relate to what you are talking about here). – user1729 Feb 7 '12 at 11:52
What is $\theta_2$? Should that just be $\theta$? – Dylan Moreland Feb 7 '12 at 14:00

You seem to have two different definitions for $F_i$, but never mind.

Can you see that $R_iF_j$ is going to be a reflection, so it's going to be $F_k$ for some $k$? Then, by seeing where $(1,0)$ goes, can you figure out what $k$ is in terms of $i$ and $j$?

Can you do the same for $F_iR_j$?

Can you see that $F_iF_j$ is going to be a rotation $R_k$ for some $k$? and then work out $k$ in terms of $i$ and $j$?

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Not only is Dn a group, but it has only two 'generators' R and F so there is a simple formula for the possible products: The rules are $R^n$ = e (the identity transformation) and $F^2$ = e and the important rule relating R and F is FRF = $R^{-1}$.

Using this last relationship all the 2n elements of the group can be written as $R^iF^j$ where i = 0,1,..,n-1 and j = 0,1 , so the natural way to list them would be: $R^0$ (e), $R^1$, $R^2$, $R^3$,...,$R^{n-1}$, $R^0$F (=F) ,$R^1$F, $R^2$F,.. $R^{n-1}$F. Try it for a simple square and you will get 8 elements.

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