\(\mathbf{S}_3 = \mathbf{D}_3\)

This is the symmetric group on three things. It can be realized as the group of symmetries of an equilateral triangle (in its incarnation as \(\mathbf{D}_3\)).

Character table

S3 e C2 C3 notes
# 1 3 2 |S3| = 6
A0 1 1 1 trivial rep
A1 1 -1 1 alternating rep
E 2 0 -1
  • All the irreducible representations are absolutely irreducible and are defined over the integers Z.
  • The permutation representation on 3 points (vertices of an equilateral triangle) is A0 + E
  • The "orientation permutation" representation on the set of 3 oriented edges of the triangle is A1 + E

Representation ring

⊗ A0 A1 E
A0 A0 A1 E
A1 A1 A0 E
E E E A0+A1+E

It can be seen from this that the representation ring for S3 satisfies $$R(S_3) \simeq \mathbb{Z}[X, Y] / \left<XY-Y,\,X^2-1,\,Y^2-X-Y-1\right>,$$ where X is the alternating representation A1, and Y is the 2-dimensional irreducible E.

Burnside ring

Table of marks:

The rows are the orbit types, the columns are the subgroups. The entries represent #Fix(H, G/K) - the number of elements in the orbit G/K fixed by the subgroup H.

1 Z2 Z3 S3
O4 = S3/1 6 - - -
O3 = S3/Z2 3 1 - -
O2 = S3/Z3 2 0 2 -
O1 = S3/S3 1 1 1 1

Product structure in Burnside Ring Ω(G):

X O1 O2 O3 O4
O1 O1 O2 O3 O4
O2 O2 2O2 O4 2O4
O3 O3 O4 O3+O4 3O4
O4 O4 2O4 3O4 6O4

Homomorphism β: Ω(G) → R(G)

  • β(O1) = A0
  • β(O2) = A0+A1
  • β(O3) = A0+E
  • β(O4) = A0+A1+2E

Note that this is not injective: indeed the kernel is generated by 2O1 - O2 - 2O3 + O4. (This is a 'virtual set' of cardinality 0, as it must be.)