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.)

On this page:

D2 D3 D4 D5 D6

  • 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 orientation permutation representation on the 4 edges of the square is A1 + B1 + 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.