Reflections in Dihedral Group In Dihedral Groups, what is the meaning of reflection ?
A line needs to be specified for a reflection to take place, but, if you specify only one line how will $D_n$ give all the symmetries for a n-gon? as there might be more than one axis of symmetry in a n-gon?
 A: I can feel you... it's very frustrating that even this very simple but important question is somehow never explained explicitly everywhere!
The reflection could be defined completely and without any confusing, by the line of symmetry through vertex 1, and the center of the object.
It seems confusing as there are many lines of symmetry, but how could just one line of symmetry be enough to represent it all? You could see from the following picture, that no matter which line of symmetry we chose, the reflection would always turn the order of vertices from 1 to 4 clockwise to counter clockwise and vice versa.

Even though each line of symmetry above moves vertex 2 to a different position, it doesn't matter which line of symmetry we choose, we could always get the other configurations by simply rotating the n-polygon. As such, after a reflection there could be only two possibilities of the location of vertex 2, that is clockwise after vertex 1, or before vertex 1. Thus in all, for an n-polygon, there are $n$ configurations of rotations, and only $2$ configuration of reflections, make it all in total $2n$ configurations of symmetry.
It could also be understood that symmetries are rigid motions, so once the position of the ordered pair of vertices 1 and 2 are fixed, the action of the symmetry on all remaining vertices is completely determined. In the above picture, our convention chooses to always use line of symmetry through vertices 1 and 3.
A: There are many comments and answers and I will add a hint for a geometric approach. Draw a few regular polygons, for example with $n=3,4,5,6,7$ and $8$ vertices. 
The symmetry group consists of $n$ rotations and $n$ reflections: 


*

*If $n$ is even, a half of the symmetry axes pass through two opposite vertices, the other half through the midpoint of opposite sides.

*If $n$ is odd, all axes pass through a vertex and the midpoint of the opposite side


When you have one axis of symmetry, you get the other by rotating this axis with a certain angle. This can be done with $e^{i\phi}$ for certain angles $\phi$, when you see the problem in the complex plane. 
For example when $n=3$, you have to rotate with angle $120^{\circ}=\frac{360^{\circ}}{3}\equiv\frac{2\pi}{3}$. You get all symmetries by repeatedly rotation with this specified angle.
The representation of $D_{2n}$ is $\langle r,s|r^n=s^2=\mathrm{id},srs=r^{-1}\rangle$, the reflections are $s,sr,sr^2,\ldots,sr^{n-1}$. In the complex plane, $r$ can be viewed as $e^{2\pi i/n}$, thus you need $e^{2\pi i k/n}$ for $k=0,1,\ldots,n-1$, where you $k=0$ actually don't need.
