Pascal's Triangle and Its Relatives

Exercises

Symmetries of regular polygons

A regular polygon with n sides has two types of symmetries:

rotations about the center reflections across an axis

These symmetries form the dihedral group Dn of order 2n.

The order is 2n because there are n rotations, by (m/n)360 deg for 0 <= m <= n-1, and n reflections.

The number of rotations that preserve the figure is evident. A bit more thought is needed to study the reflections.

1. Using a pentagon and a hexagon template, find all the reflections that are symmetries of these polygons. Describe the reflections on terms of sides or angles of the polygons. You may need different descriptions for polygons with an odd or an even number of sides. Answer

2. Certainly, all the rotations can be generated by compositions of rotation by (1/n)360 deg. Perhaps less obviously, all reflections are compositions of any given reflection and a rotation. Illustrate this by showing reflection across a diagonal of a square is the same as a rotation followed by reflection across the perpendicular bisector of two opposite sides. Can you prove this in general? (Hint: any orientation-preserving symmetry of a regular n-gon is one of the rotations. Suppose c and d are two reflection symmetries of an n-gon. Then cd is orientation-preserving, hence is one of the rotations.) Answer

Consequently, the group D_n is generated by rotation a through (1/n)360 deg, and a reflection b. Denote by 1 the identity symmetry, and observe that aⁿ = 1 and b² = 1. Are there other relations?

3. By studying the pentagon and hexagon templates, show bab = a^-1. Show this relation is equivalent to (ab)² = 1. Answer

4. Show aⁿ = 1, b² = 1, and (ab)² = 1 are the only relations. That is, if some other combination w of a and b satisfies w = 1, then the three relations can be used to show w is equivalent to one of the relations. Answer

In abstract terms, the dihedral group of order 2n is written

D_n = < a,b: aⁿ = b² = (ab)² = 1>

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