Generator of group $D_n$
Let be $D_n$ the group of symmetries of a regular n-sided polygon. Prove that $D_n$ is generated by a minimum rotation angle and a reflection (about a symmetry axis).
I really do not know how to start . Some help plis
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$\begingroup$Fix your rotation and reflection. Let us visualize the regular $n$-gon as a planer graph drawn on a plane with vertices labeled by numbers from $1, \ldots n$ increasing towards right until $n$. We will say a transformation in $D_n$ is "pure" is it preserves polarity that is the result of the transformation also has labels increasing towards right until $n$. And if it is not pure we will call it "impure".
Prove that a pure transformation is a rotation by an appropriate angle.
If a transformation is impure then apply your fixed reflection then prove that the resulting transformation is pure and hence is a rotation by a certain angle, then derive the result.
$\endgroup$ $\begingroup$The rotation group generated by the rotation through 2*pi/n will certainly preserve the polygon, hence is in the symmetric group. Now, we need to find what else apart from rotation is in Dn. Intuitively, they are reflections:
if n is odd (like triangle or pentagon):
n reflections are about the line that pass through a vertex and the midpoint of the opposite side. For each vertex, you get a reflection, hence n reflections.
if n is even(like square or hexagon)
in this case, n/2 reflections are through the midpoint of two opposite sides and remaining n/2 reflections are through the opposite vertices.
This is how you obtain your 2n elements. Now that you think about it, only generators you need is the rotation by 2*pi/n and one of the reflection.
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