Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Burnside Lemma looks like it should have an intuitive explanation. Does anyone have one?

share|cite|improve this question
Probably there should be some probabilistic explanation? – Grigory M Jul 30 '10 at 9:31
up vote 5 down vote accepted

As an example, we consider the number of ways of colouring a cube with n colours with uniqueness up to rotation. We call each unique colouring where rotations are not allowed a static colouring and each unique one where they are allowed a dynamic colouring. We define the the set of orbits to be the (disjoint) static colourings that correspond to each dynamic colouring. We will use rotations to mean a rotation that makes the cube occupy the same space, and as being unique if it is a unique function from the cube to the cube. This includes the identity rotation. Intuitively, the lemma says:

Proposition 1. #Orbits * #Rotations = sum for each rotation r of #static colourings unchanged by this rotation

We will now consider each orbit O separately. Pick a static colouring c inside O. Suppose two (possibly equal) rotations p, q give the same static colouring, d, when applied on c. Then p^-1*q fixes d. Additionally, suppose r (possibly the identity) fixes d. p^-1*(p*r) will also fix d. So p*r will take c to d. Since p*r is different for each r, and p^-1*q is different for each q, the mapping functions are injective in both directions and there is a bijection between the q and r values.

So, for each O, the number of rotations is the sum over each static colorings x in O times the number of rotations producing x. This can be rewritten as the sum over each rotation r of the number of static colourings in O fixed by r (due to the bijection in the previous paragraph). We get proposition 1 by adding over all O.

The general proof is quite similar to this, except that it uses group theory.

share|cite|improve this answer
I created this solution while looking for a non-group theory answer to this question, but it didn't quite fit the question and I didn't want to just throw out the answer – Casebash Jul 29 '10 at 22:13

You can quickly reduce to the case of a transitive action, in which case we just want to explain why the total number of times that something gets fixed is exactly the size of the group. But in this case everything is symmetric at all points in the (unique) orbit. So to count all the times something gets fixed, we can just count how many times a particular x gets fixed, and multiply by the size of the orbit. Now we've reduced to the fact that the size of an orbit is the index of the stabilizer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.