Dihedral Groups
The dihedral group D_{n} or Dih(2n) is of order 2n. It is the symmetry group of the regular ngon.
(Some denote this group D_{2n} because its order is 2n, but I prefer D_{n}  after all, one doesn't denote the symmetric group S_{n} by S_{n!}  nor J_{1} by J_{175560}. On the other hand, Dih(2n) is fine as there's no conflict of notation.)
On this page:
D_{2}
D_{3}
D_{4}
D_{5}
D_{6}
 The Schoenflies notation for D_{n} depends on which 3dimensional representation is used: C_{nv} refers to the the 3dimensional representation acting trivially in the 'vertical', while D_{n} denotes the representation in 3 dimensions where the reflections act by (1) in the 'vertical'.
 A fixed generator of order n (rotation by \(2\pi/n\) is denoted ρ).
 For n even, there are two conjugacy classes of reflection, denoted κ and κ', with κ acting as reflections in a line joining two vertices, and κ' reflection in a line joining the midpoints of two opposite edges. There is an (outer) automorphism of D_{n} exchanging the two reflections.
 For n odd, all reflections are conjugate;
 Rotations through θ and θ are conjugate for all n.
 This is close to the theory of Fourier series, and symmetric circulant matrices. See Notes for details.
D_{2} = Dih(4)
\(D_2 \simeq \mathbb{Z}_2\times\mathbb{Z}_2\) with generators κ and κ'. It is the symmetry group of the rectangle.
D_{2}
 e
 κ
 κ'
 ρ=κκ'

A_{0}
 1
 1
 1
 1

A_{1}
 1
 1
 1
 1

A_{2}
 1
 1
 1
 1

A_{3}
 1
 1
 1
 1

D_{3} = Dih(6)
D_{3}
 e
 ρ
 κ

#
 1
 2
 3

A_{0}
 1
 1
 1

A_{1}
 1
 1
 1

E
 2
 1
 0

D_{4} = Dih(8)
D_{4}
 e
 ρ
 ρ^{2}
 κ
 κ'

#
 1
 2
 1
 2
 2

A_{0}
 1
 1
 1
 1
 1

A_{1}
 1
 1
 1
 1
 1

B_{1}
 1
 1
 1
 1
 1

B_{2}
 1
 1
 1
 1
 1

E
 2
 0
 2
 0
 0

 The permutation representation on the 4 vertices of the square is A_{0} + B_{1} + E
 The permutation representation on the 4 edges of the square is A_{0} + B_{2} + E
 The orientation permutation representation on the 4 edges of the square is A_{1} + B_{1} + E
D_{5} = Dih(10)
D_{5}
 e
 \(\rho\)
 \(\rho^2\)
 \(\kappa\)
 notes

#
 1
 2
 2
 5
 D_{5}=10

A_{0}
 1
 1
 1
 1
 trivial rep

A_{1}
 1
 1
 1
 1
 "orientation" rep

E_{1}
 2
 \(\gamma\)
 \(\gamma\)_{2}
 0
 symmetry of pentagon

E_{2}
 2
 \(\gamma\)_{2}
 \(\gamma\)
 0


 \(\gamma = 2\cos(2\pi/5) = \frac12(\sqrt51)\) (= golden ratio)
 \(\gamma_2 = 2\cos(4\pi/5) = \frac12(\sqrt5+1)\)
 There is an outer automorphism of D_{5}, taking \(\rho\) to \(\rho^2\). This interchanges E_{1} and E_{2}.
 The permutation representation on the 5 vertices of the pentagon is A_{0}+ E_{1}+ E_{2}
D_{6} = Dih(12)
D_{6}
 e
 ρ
 ρ^{2}
 ρ^{3}
 κ
 κ'
 notes

#
 1
 2
 2
 1
 3
 3
 D_{6}=12

A_{0}
 1
 1
 1
 1
 1
 1
 trivial rep

A_{1}
 1
 1
 1
 1
 1
 1
 "orientation rep"

B_{1}
 1
 1
 1
 1
 1
 1
 alternating rep

B_{2}
 1
 1
 1
 1
 1
 1


E_{1}
 2
 1
 1
 2
 0
 0
 symmetry of hexagon

E_{2}
 2
 1
 1
 2
 0
 0


 The permutation representation on the 6 vertices of the hexagon is A_{0}+ B_{1}+ E_{1}+ E_{2}
 B_{1} is the alternating rep because it is the rep obtained if the vertices of the hexagon are weighted successively with +1,1,+1,1,+1,1 (alternating signs)
 The permutation representation on the 3 diagonals joining opposite vertices of the hexagon is A_{0} + E_{2}
 The 'oriented permutation' representation on the 3 oriented diagonals is B_{1}+ E_{1}
And so the pattern goes on ...
 n even: D_{n} has four 1dimensional representations and ½(n2) 2dimensional representations.
 n odd: D_{n} has two 1dimensional representation and ½(n1) 2dimensional representations.
 All the irreducible reps are absolutely irreducible.