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