4.7 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 4.4 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 \[ \begin{pmatrix} \cos(2\pi k/n) \\ \sin (2\pi k/n) \end{pmatrix} : k = 0,1,\ldots n-1. \]