Dihedral groups explained by I N Herstein

Reference: Abstract Algebra, Third Edition, I N Herstein

First consider the following: Let S be the plane, that is, S= \{ (x,y): x, y \in R\} and consider f, g \in A(S) defined by f(x,y)=(-x,y) and g(x,y)=(-y,x); f is the reflection about the y-axis and g is the rotation through 90 degrees in a counterclockwise direction about the origin. We then define G = \{ f^{i}g^{j}: i=0,1; j=0,1,2,3\}, and let * in G be the product of elements in A(S). Clearly, f^{2}=g^{4}= identity mapping;

(f*g)(x,y)=(fg)(x,y)=f(g(x,y))=f(-y,x)=(y,x) and (g*f)(x,y)=g(f(x,y))=g(-x,y)=(-y,-x)

So g*f=f*g.

It is a good exercise to verify that g*f=f*g^{-1} and G is a non-abelian group of order 8. This group is called the dihedral group of order 8. [Try to find a formula for (f^{i}g^{j})*(f^{s}g^{t})=f^{a}g^{b} that expresses a, b in terms of i, j, s and t.

II) Let S be as in above example and f the mapping in above example. Let n>2 and let h be the rotation of the plane about the origin through an angle of 2\pi/n in the counterclockwise direction. We then define G=\{ f^{k}h^{j}: k=0,1 ; j=0,1,2, \ldots, n-1\} and define the product * in G via the usual product of mappings. One can verify that f^{2}=h^{n}= identity mapping and fh=h^{-1}f. These relations allow us to show that (with some effort) G is a non-abelian group of order 2n. G is called the dihedral group of order 2n.

More later,

Nalin Pithwa