## 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:

- \(x=A^i, y=A^j\). This is easy.
- \(x = A^i, y = JA^j\). Then \(x^{-1}y = A^{-i}JA^j = JJA^{-i}JA^j = JA^i A^j = JA^{i+j} \in D_{2m}\).
- \(x=JA^i, y=A^j\). This is easy too.
- \(x=JA^i, y = JA^j\). Then \(x^{-1}y=JA^iJA^j = A^{-i}A^j=A^{j-i}\in D_{2m}\).

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