Burnside's Lemma

$|X/G|\; =\; \backslash frac\{1\}$>

/dd> Thus the number of orbits (a natural number or infinity) is equal to the average number of points fixed by an element of G (which consequently is also a natural number or infinity). Example application The number of rotationally distinct colourings of the faces of a cube using three colours can be determined from this formula as follows. Let X be the set of 36 fixed coloured cubes, and let the rotation group G of the cube act on X in the natural manner. Then two elements of X belong to the same orbit precisely when one is simply a rotation of the other. The number of rotationally distinct colourings is thus the same as the number of orbits and can be found by counting the sizes of the fixed sets for the 24 elements of G. Cube one identity element fixing all 36 elements of X six 90-degree face rotations fixing 33 elements of X three 180-degree face rotations fixing 34 elements of X eight 120-degree vertex rotations fixing 32 elements of X six 180-degree edge rotations fixing 33 elements of X The average fix size is thus $\backslash frac\{1\}\{24\}\backslash left(3^6+6\backslash times\; 3^3\; +\; 3\; \backslash times\; 3^4\; +\; 8\; \backslash times\; 3^2\; +\; 6\; \backslash times\; 3^3\; \backslash right)\; =\; 57.$ Hence there are 57 rotationally distinct colourings of the faces of a cube in three colours. Proof The proof uses the orbit-stabilizer theorem and the fact that X is the disjoint union of the orbits: $\backslash sum\_\{g\; \backslash in\; G\}|X^g|\; =\; |\backslash \{(g,x)\backslash in\; G\backslash times\; X\; \backslash mid\; g.x\; =\; x\backslash \}|\; =\; \backslash sum\_\{x\; \backslash in\; X\}\; |G\_x|$ $=\; \backslash sum\_\{x\; \backslash in\; X\}\; \backslash frac\{|G|\}\{|Gx|\}\backslash \; =\backslash \; \backslash sum\_\{A\backslash in\; X/G\}\; |G|\backslash sum\_\{x\backslash in\; A\}\; \backslash frac\{1\}\{|A$

\ = \ |G| \sum_{A\in X/G} 1

$=\; |G|\; \backslash cdot\; |X/G|.\backslash ,$

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