The dihedral group is a particular instance of one of the two distinct abstract groups of group order 6. Unlike the cyclic group (which is Abelian), is nonAbelian. In fact, is the nonAbelian group having smallest group order.
Examples of include the point groups known as , , , , the symmetry group of the equilateral triangle (Arfken 1985, p. 246), and the permutation group of three objects (Arfken 1985, p. 249).
The cycle graph of is shown above. has cycle index given by
(1)

Its multiplication table is illustrated above and enumerated below, where 1 denotes the identity element. Equivalent but slightly different forms are given by (Arfken 1985, p. 247) and Cotton (1990, p. 12), the latter of which denotes the abstract group of by .
1  
1  1  
1  
1  
1  
1  
1 
Like all dihedral groups, a reducible twodimensional representation using real matrices has generators given by and , where is a rotation by radians about an axis passing through the center of a regular gon and one if its vertices and is a rotation by about the center of the gon. The multiplication table above corresponds to the following matrices:
(2)
 
(3)
 
(4)
 
(5)
 
(6)
 
(7)

The elements , , , and of satisfy , the elements , , and satisfy , the elements , , , and satisfy , and all elements satisfy .
The conjugacy classes are , , and . There are 6 subgroups of : , , , , , and . Of these, the subgroups , , and are normal
To find the irreducible representation, note that there are three conjugacy classes. The fifth rule of irreducible representations requires that there be three irreducible representations, and the second rule requires that
(8)

so it must be true that
(9)

By rule 6, we can let the first representation have all 1s.
1  
1  1  1  1  1  1 
To find a representation orthogonal to the totally symmetric representation, we must have three and three group characters. We can also add the constraint that the components of the identity element 1 be positive. The three conjugacy classes have 1, 2, and 3 elements. Since we need a total of three s and we have required that a occur for the conjugacy class of order 1, the remaining +1s must be used for the elements of the conjugacy class of order 2, i.e., and .
1  
1  1  1  1  1  1  
1  1  1 
Using group rule 1, we see that
(10)

so the final representation for 1 has group character 2. Orthogonality with the first two representations (group rule 3) then yields the following constraints:
(11)
 
(12)

Solving these simultaneous equations by adding and subtracting (12) from (11), we obtain , . The full character table is then
1  
1  1  1  1  1  1  
1  1  1  
2  0  0  0 
Since there are only three conjugacy classes, this table is conventionally written simply as
1  
1  1  1  
1  1  
2  0 
Writing the irreducible representations in matrix form then yields
(13)
 
(14)
 
(15)
 
(16)
 
(17)
 
(18)
