\(\mathbf{S}_3 = \mathbf{D}_3\)
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
S_{3} | e | C_{2} | C_{3} | notes |
# | 1 | 3 | 2 | |S_{3}| = 6 |
A_{0} | 1 | 1 | 1 | trivial rep |
A_{1} | 1 | -1 | 1 | alternating rep |
E | 2 | 0 | -1 |
- All the irreducible representations are absolutely irreducible and are defined over the integers Z.
Representation ring
⊗ | A_{0} | A_{1} | E |
A_{0} | A_{0} | A_{1} | E |
A_{1} | A_{1} | A_{0} | E |
E | E | E | A_{0}+A_{1}+E |
It can be seen from this that the representation ring for S_{3} 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 A_{1}, and Y is the 2-dimensional irreducible E.
Permutation representations
- The permutation representation on 3 points (vertices of an equilateral triangle) is A_{0} + E
- The "orientation permutation" representation on the set of 3 edges of the triangle is A_{1} + 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 | Z_{2} | Z_{3} | S_{3} | |
O_{4} = S_{3}/1 | 6 | - | - | - |
O_{3} = S_{3}/Z_{2} | 3 | 1 | - | - |
O_{2} = S_{3}/Z_{3} | 2 | 0 | 2 | - |
O_{1} = S_{3}/S_{3} | 1 | 1 | 1 | 1 |
Product structure in Burnside Ring Ω(G):
X | O_{1} | O_{2} | O_{3} | O_{4} |
O_{1} | O_{1} | O_{2} | O_{3} | O_{4} |
O_{2} | O_{2} | 2O_{2} | O_{4} | 2O_{4} |
O_{3} | O_{3} | O_{4} | O_{3}+O_{4} | 3O_{4} |
O_{4} | O_{4} | 2O_{4} | 3O_{4} | 6O_{4} |
Homomorphism β: Ω(G) → R(G)
- β(O_{1}) = A_{0}
- β(O_{2}) = A_{0}+A_{1}
- β(O_{3}) = A_{0}+E
- β(O_{4}) = A_{0}+A_{1}+2E
Note that this is not injective: indeed the kernel is generated by 2O_{1} - O_{2} - 2O_{3} + O_{4}. (This is a 'virtual set' of cardinality 0, as it must be.)