Burnside's Lemma Derivation
📂Abstract AlgebraBurnside's Lemma Derivation
Overview
Burnside’s lemma is a representative application to the actions of a group and the isotropy subgroups that can be immediately used in combinatorics and other fields.
For a finite group G and a finite set X that is a G-set, let r be the number of orbits of X under G.
r∣G∣=g∈G∑∣Xg∣
Derivation
Let the cardinality of the set {(g,x)∈G×X∣gx=x} be N then
Xg={x∈X ∣ gx=x}
and, for Gx={g∈G ∣ gx=x}, the following holds.
N=g∈G∑∣Xg∣=x∈X∑∣Gx∣
Properties of the Isotropy Subgroup: If X is a G-set, then ∣Gx∣=(G:Gx). If G is a finite group, then ∣Gx∣ is a divisor of ∣G∣.
Since ∣G∣=∣Gx∣∣Gx∣, from N=x∈X∑∣Gx∣
N=x∈X∑∣Gx∣∣G∣=∣G∣x∈X∑∣Gx∣1
For any orbit O, ∣Gx∣ has the same value for x∈O,
x∈O∑∣Gx∣1=1
The number of orbits is r, so N=∣G∣r and summarizing, we obtain the following.
r∣G∣=g∈G∑∣Xg∣
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Examples
Circular Table Problem
Calculate the number of ways 7 people can be seated around a circular table.
The number of ways to arrange 7 people in a row is 7!.
The ‘action’ of rotating counter-clockwise is in total 7 ways. Therefore, the number of ways is
r=∣G∣1g∈G∑∣Xg∣=717!=720
Bead Problem
Calculate the number of ways to thread 7 distinct beads on a string.
The number of ways to arrange 7 beads in a row is 7!. The action of rotating counter-clockwise and flipping the string of beads is in total 7×2=14 ways. Therefore,
r=∣G∣1g∈G∑∣Xg∣=1417!=360