§ Clean way to write burnside lemma
Burnside lemma says that ∣Orb(G)∣≡1/∣G∣∑g∈Gfix(g). We prove this
as follows:
g∈G∑fix(g)=g∈G∑∣{x:g(x)=x}∣=∣{(g,x):g(x)=x}∣=x∈X∑∣{x:g(x)=x}∣=x∈X∑Stab(x)
- From orbit stabilizer, we know that ∣Orb(x)∣∣Stab(x)∣=∣G∣.
- Since ∣Orb(x) is the total cardinality of the orbit, each element in the orbit contributes 1/∣Orb(x)∣ towards cardinality of the full orbit.
- Thus, the sum over an orbit ∑x∈Orb(x)1/∣Orb(x)∣ will be 1.
- Suppose a group action has two orbits, O1 and O2. I can write the sum ∑x∈g1/∣Orb(x)∣ as: ∑x∈O11/∣O1∣+∑x∈O21/∣O2∣, which is equal to 2.
- I can equally write the sum as ∑o∈Orbits∑x∈o1/∣o∣. But this sum is equal to ∑o∈Orbits∑x∈o1/∣Orb(x)∣.
- This sum sums over the entire group, so it can be written as ∑x∈G1/∣Orb(x)∣.
- In general, the sum over the entire group ∑x∈g1/∣Orb(x)∣ will be the number of orbits, since the same argument holds for each orbit .
=x∈X∑Stab(x)=x∈X∑∣G∣/∣Orb(x)∣=∣G∣o∈orbits∑x∈o∑1/∣o∣=∣G∣num.orbits
So we have derived:
g∈G∑fix(g)=∣G∣num.orbits1/∣G∣(g∈G∑fix(g))=num.orbits
If we have a transformation that fixes many things, ie, fix(g) is large, then this g is not helping "fuse" orbits of x together, so the number of orbits will increase.