# Dihedral Groups

The dihedral group Dn or Dih(2n) is of order 2n. It is the symmetry group of the regular n-gon.
(Some denote this group D2n because its order is 2n, but I prefer Dn - after all, one doesn't denote the symmetric group Sn by Sn! - nor J1 by J175560. On the other hand, Dih(2n) is fine as there's no conflict of notation.)

• The Schoenflies notation for Dn depends on which 3-dimensional representation is used: Cnv refers to the the 3-dimensional representation acting trivially in the 'vertical', while Dn 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 mid-points of two opposite edges. There is an (outer) automorphism of Dn 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.

### D2 = 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.

 D2 e κ κ' ρ=κκ' A0 1 1 1 1 A1 1 1 -1 -1 A2 1 -1 1 -1 A3 1 -1 -1 1

### D3 = Dih(6)

 D3 e ρ κ # 1 2 3 A0 1 1 1 A1 1 1 -1 E 2 -1 0

### D4 = Dih(8)

 D4 e ρ ρ2 κ κ' # 1 2 1 2 2 A0 1 1 1 1 1 A1 1 1 1 -1 -1 B1 1 -1 1 1 -1 B2 1 -1 1 -1 1 E 2 0 -2 0 0
• The permutation representation on the 4 vertices of the square is A0 + B1 + E
• The permutation representation on the 4 edges of the square is A0 + B2 + E
• The permutation representation on the 2 diagonals of the square is A0 + A1
• The orientation permutation representation on the 4 oriented edges of the square is A1 + B1 + E
• The orientation permutation representation on the 2 oriented diagonals of the square is E

### D5 = Dih(10)

 D5 e $$\rho$$ $$\rho^2$$ $$\kappa$$ notes # 1 2 2 5 |D5|=10 A0 1 1 1 1 trivial rep A1 1 1 1 -1 "orientation" rep E1 2 $$\gamma$$ $$\gamma$$2 0 symmetry of pentagon E2 2 $$\gamma$$2 $$\gamma$$ 0
• $$\gamma = 2\cos(2\pi/5) = \frac12(\sqrt5-1)$$ (= golden ratio)
• $$\gamma_2 = 2\cos(4\pi/5) = -\frac12(\sqrt5+1)$$
• There is an outer automorphism of D5, taking $$\rho$$ to $$\rho^2$$. This interchanges E1 and E2.
• The permutation representation on the 5 vertices of the pentagon is A0+ E1+ E2

### D6 = Dih(12)

 D6 e ρ ρ2 ρ3 κ κ' notes # 1 2 2 1 3 3 |D6|=12 A0 1 1 1 1 1 1 trivial rep A1 1 1 1 1 -1 -1 "orientation rep" B1 1 -1 1 -1 1 -1 alternating rep B2 1 -1 1 -1 -1 1 E1 2 1 -1 -2 0 0 symmetry of hexagon E2 2 -1 -1 2 0 0
• The permutation representation on the 6 vertices of the hexagon is A0+ B1+ E1+ E2
• B1 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 A0 + E2
• The 'oriented permutation' representation on the 3 oriented diagonals is B1+ E1

And so the pattern goes on ...

• n even: Dn has four 1-dimensional representations and ½(n-2) 2-dimensional representations.
• n odd: Dn has two 1-dimensional representation and ½(n-1) 2-dimensional representations.
• All the irreducible reps are absolutely irreducible.