Color the four positions of a necklace black or white, identifying configurations that differ by rotation. There are sixteen labeled colorings, but dividing by four gives the wrong answer because some colorings have fewer than four distinct rotations. The all-black necklace has only one. Burnside’s lemma replaces uniform division by an average of fixed-point counts, automatically accounting for these unequal orbit sizes.
Predict
Predict what the quarter-turn fixes. All four positions lie in one cycle, so every position must have the same color: two choices. A half-turn has two position cycles and fixes four colorings. The identity fixes all sixteen. The two quarter-turns contribute the same count. Their average is (16+2+4+2)/4=6 necklaces. Reflections are not included in this equivalence relation; adding them changes the acting group.
Experiment
Build the fixed-point table explicitly.
Rotation
Position cycles
Fixed binary colorings
0 degrees
4
16
90 degrees
1
2
180 degrees
2
4
270 degrees
1
2
List the six orbits by number and arrangement of black positions. With two black positions, adjacent and opposite arrangements form different orbits even though they have the same color totals.
Understand
Let a finite group G act on a finite set X. Count the set of pairs (g, x) for which g fixes x. Summing over g gives Σ_g size (Fix (g)). Summing over x gives Σ_x size (Stab (x)). Within one orbit, orbit–stabilizer gives size (orbit)·size (stabilizer)=size (G), so each orbit contributes size (G) to this latter sum. Dividing by size (G) proves that the number of orbits is the average fixed-point count. This is a proof about a specified group action: one must check that the proposed symmetries preserve the object set and that composition behaves as claimed.
Apply
For colorings of positions, a permutation fixes a coloring exactly when each of its cycles is monochromatic. With k available colors and c cycles there are k^c fixed colorings. For rotation by r positions on an n-cycle, the number of cycles is gcd (n, r), including n cycles for the identity. Recording the lengths of these cycles in a cycle-index polynomial refines the count by color inventory; the substitution rule follows by choosing one color for each cycle and giving it the cycle’s full weight. This explains the mechanism behind Pólya-style enumeration without treating a symbolic substitution as magic.
Continue
Choose whether rotation alone or rotation plus reflection represents physical equivalence in an application. A printed bracelet that can be flipped differs from a one-sided cyclic arrangement whose orientation is meaningful. Constraints can also alter fixed-point counts: if exactly two positions must be black, replace k^c by a count of cycle unions of total size two. Burnside remains valid, but the easy unrestricted coloring formula no longer applies. Verify the identity contribution and one nontrivial symmetry independently before averaging.
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Count binary colorings of four cyclic positions up to rotation.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Count colorings of triangle vertices with three colors up to all six symmetries.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove Burnside’s lemma.
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Why does a permutation with c cycles fix k^c colorings?
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Why is total colorings divided by group size generally wrong?
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Derive the fixed-coloring count for rotation by r positions of an n-cycle.
Structure your argument and preserve revisions
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Reflect and continue
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