Necklaces There are 8 possible
bit strings of length 3, but tying together the string ends gives only four distinct 2-colored
necklaces of length 3, given by the canonical forms 000, 001, 011, 111: the other strings 100 and 010 are equivalent to 001 by rotation, while 110 and 101 are equivalent to 011. That is, rotation equivalence splits the set X of strings into four orbits: X = \{{000}\}\cup \{{001}, {010}, {100}\}\cup \{{011}, {101}, {110}\}\cup \{{111}\}.The Burnside formula uses the number of rotations, which is 3 including the null rotation, and the number of bit strings left unchanged by each rotation. All eight vectors are unchanged by the null rotation, and two (000 and 111) are unchanged by the other two rotations. Thus the number of orbits is: 4 = \frac{1}{3}(8+2+2). For length 4, there are 16 possible bit strings; 4 rotations; the null rotation leaves all 16 strings unchanged; the 1-rotation and 3-rotation each leave two strings unchanged (0000 and 1111); the 2-rotation leaves 4 bit strings unchanged (0000, 0101, 1010, 1111). The number of distinct necklaces is thus: 6 = \tfrac{1}{4}(16+2+4+2), represented by the canonical forms 0000, 0001, 0011, 0101, 0111, 1111. The general case of
n bits and
k colors is given by a
necklace polynomial.
Colorings of a cube Burnside's lemma can compute the number of rotationally distinct colourings of the faces of a
cube using three colours. Let X be the set of 36 possible face color combinations that can be applied to a fixed cube, and let the rotation group
G of the cube act on X by moving the colored faces: two colorings in X belong to the same orbit precisely when one is a rotation of the other. Rotationally distinct colorings correspond to group orbits, and can be found by counting the sizes of the
fixed sets for the 24 elements of
G, the colorings left unchanged by each rotation: • the
identity element fixes all 36 colorings • six 90-degree face rotations each fix 33 colorings • three 180-degree face rotations each fix 34 colorings • eight 120-degree vertex rotations each fix 32 colorings • six 180-degree edge rotations each fix 33 colorings. A detailed examination may be found
here. The average fixed-set size is thus: : |X/G| = \frac{1}{24}\left(3^6+6\cdot 3^3 + 3 \cdot 3^4 + 8 \cdot 3^2 + 6 \cdot 3^3 \right) = 57. There are 57 rotationally distinct colourings of the faces of a cube in three colours. In general, the number of rotationally distinct colorings of the faces of a cube in
n colors is: : \frac{1}{24}\left(n^6+3n^4 + 12n^3 + 8n^2\right).
Conjugacy classes Let G be a finite group. Consider the group action of G on itself given by the
conjugation map g \mapsto \phi_g where \phi_g(h) = g h g^{-1} . The orbits are the
conjugacy classes of G and the set of fixed points of an element g is the
centralizer C_G(g) . Thus, by Burnside's lemma, the number of conjugacy classes of G is equal to \frac{1} \sum_g |C_G(g)|, that is, the average size of the centralizer. == Proof ==