## 4.8 The dihedral groups

Given $$\theta \in \RR$$ we let $$A(\theta)$$ be the element of $$GL(2,\RR)$$ which represents a rotation about the origin anticlockwise through $$\theta$$ radians. So $A(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ Then $$A(\theta)^n = A(n\theta)$$, since a rotation by $$\theta$$ done n times is the same as a rotation of $$n\theta$$.

Fix a positive whole number n, and let $$A = A(2\pi /n)$$. Then A is an element of $$GL(2,\RR)$$ with order n. Let $J = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ so J represents a reflection in the x-axis of $$\RR^2$$.

Lemma 4.10 $$AJ = JA^{-1}$$
Proof. This is equivalent to $$AJA=J$$. Multiply the matrices and see what happens!
Corollary 4.5 For all m we have $$J A^mJ = A^{-m}$$.
Proof. Let I be the $$2\times 2$$ identity matrix. Multiplying the previous lemma on the left by J tells us $$JAJ=A^{-1}$$ (because $$J^2=I$$). Now $$JAJ = A^{-1}$$, so $$A^{-2} = JAJJAJ = JAIAJ=JA^2J$$, and so on.
Lemma 4.11 $$D_{2n} := \{ I, A, \ldots, A^{n-1}, J, JA, \ldots, JA^{n-1} \}$$ is a subgroup of $$GL(2,\RR)$$ with order $$2n$$.

Proof. Apply the subgroup test. To do this we’ll need to work case by case, and we need to know the inverse of an element that looks like $$JA^m$$. But $JA^m JA^m = A^{-m} A^m = I$ by the previous corollary, so $$JA^m$$ is its own inverse.

To use the subgroup test we have to pick $$x,y \in D_{2n}$$ and show $$x^{-1}y \in D_{2n}$$. There are four cases:

1. $$x=A^i, y=A^j$$. This is easy.
2. $$x = A^i, y = JA^j$$. Then $$x^{-1}y = A^{-i}JA^j = JJA^{-i}JA^j = JA^i A^j = JA^{i+j} \in D_{2m}$$.
3. $$x=JA^i, y=A^j$$. This is easy too.
4. $$x=JA^i, y = JA^j$$. Then $$x^{-1}y=JA^iJA^j = A^{-i}A^j=A^{j-i}\in D_{2m}$$.
Certainly $$|D_{2n}|$$ is at most $$2n$$ as its elements were $$A^i$$ and $$JA^j$$ for $$0 \leq i,j \leq n$$. To show that $$|D_{2n}|=2n$$, first note that all of the elements $$I, A, \ldots, A^{n-1}$$ are different since A has order n: these elements make up the cyclic subgroup $$\langle A \rangle$$ of $$D_{2n}$$. Now $$J \notin \langle A \rangle$$ since every element of that subgroup has determinant one, whereas J has determinant $$-1$$. So the coset $$J\langle A\rangle$$ is disjoint from $$\langle A \rangle$$. This means $$D_{2n}$$ has at least $$|\langle A \rangle| + |J \langle A \rangle| = 2n$$ elements, and we are done.

$$D_{2n}$$ is called the dihedral group of order $$2n$$. It consists of n rotations $$I, A, \ldots A^{n-1}$$ and n reflections $$J, JA, \ldots, JA^{n-1}$$. You can think of it as the symmetry group of the regular n-gon with vertices at $\begin{pmatrix} \cos(2\pi k/n) \\ \sin (2\pi k/n) \end{pmatrix} : k = 0,1,\ldots n-1.$