So far a group has been a self-contained gadget: a set of elements with a way to combine them. But groups earn their keep when they do something — when they shuffle, rotate, reflect or permute the members of some other set. A rotation group spins the corners of a square; a symmetry group rearranges the atoms of a molecule; the group of card shuffles rearranges a deck. This is the idea of a group action: a group G reaching out and moving the points of a set X around.

Once a group acts on a set, two beautiful structures fall out for free. Each point traces out an orbit — everywhere it can be sent — and each point pins down a stabiliser — the elements that leave it exactly where it is. These two are locked together by one of the most useful counting facts in all of algebra, the orbit–stabiliser theorem, and its companion Burnside's lemma lets us count genuinely-different configurations — distinct necklaces, distinct painted dice — by the surprisingly gentle method of averaging.

What is a group action?

A (left) action of a group G on a set X is a rule that takes a group element g and a point x \in X and returns another point g \cdot x \in X, subject to just two rules:

The first says the identity does nothing. The second says that acting by h and then by g is the same as acting once by the product gh — the group's own multiplication and the shuffling of X agree. That's the whole definition.

There is a slicker way to say the same thing. Each fixed g gives a map x \mapsto g \cdot x from X to itself, and the two axioms force this map to be a bijection — a permutation of X. So an action is exactly a homomorphism

sending each group element to the permutation it performs. "Group action" and "homomorphism into a symmetric group" are two names for one thing — the second axiom is precisely the statement that \rho(gh) = \rho(g)\rho(h). This is why permutation groups sit at the heart of the subject: every group action is secretly a way of realising G as permutations.

Orbits and stabilisers

Fix a point x \in X. Two questions immediately suggest themselves: where can x be sent, and what leaves it alone?

The stabiliser really is a subgroup: the identity fixes x, so e \in \operatorname{stab}(x); if g and h both fix x then so does gh; and if g \cdot x = x then applying g^{-1} to both sides gives g^{-1} \cdot x = x. Closure, identity, inverses — it passes the subgroup test.

A crucial structural fact: the orbits partition X. Every point lies in its own orbit (take g = e), and if two orbits share a single point they are identical — so X breaks cleanly into disjoint orbits, with no overlaps and nothing left out. "Being in the same orbit" is an equivalence relation, and orbits are its equivalence classes.

A picture: a group rotating a necklace

Here is the friendliest example. Take n beads arranged in a ring and let the cyclic group C_n act by rotating the whole ring. The set X is the beads (or, later, the colourings of the beads); a group element is "rotate by k steps." Watch a single colouring get carried around its orbit — the rotations that bring it back to something you've already seen are exactly its stabiliser. A fresh random necklace appears each time you press Refresh.

Notice the two extremes. A necklace with all beads the same colour is fixed by every rotation — its stabiliser is the whole group and its orbit has size 1. A necklace with no rotational symmetry at all is only fixed by doing nothing — stabiliser \{e\} — and its orbit contains all n distinct rotations. The orbit and the stabiliser trade off against each other, which is exactly what the next theorem makes precise.

The orbit–stabiliser theorem

The bigger the stabiliser (the more symmetry pinning x down), the smaller the orbit — and their product is always |G|. The proof is a clean bijection: two group elements g and h send x to the same place exactly when h^{-1}g fixes it, i.e. when g and h lie in the same coset of \operatorname{stab}(x). So points of the orbit correspond one-to-one with cosets of the stabiliser, and there are |G| / |\operatorname{stab}(x)| of those by Lagrange's theorem.

Worked example — a square's corners. Let G = C_4, the four rotations of a square, act on its four corners X = \{1,2,3,4\}. Any corner can be rotated to any other, so the orbit of a corner is all four corners: |\operatorname{orb}(x)| = 4. Which rotations fix a chosen corner? Only the identity — a genuine turn always moves every corner. So |\operatorname{stab}(x)| = 1, and indeed 4 = |G| / 1 = 4/1. The theorem checks out. (Include the four reflections as well, giving the dihedral group D_4 of order 8, and each corner is now fixed by the identity plus one reflection through it — stabiliser of size 2, orbit 8/2 = 4. Still all four corners.)

Counting by averaging: Burnside's lemma

Now the payoff. Suppose you want to count configurations up to symmetry: how many truly different bracelets can you make from beads of a few colours, if two bracelets that differ only by a rotation count as the same? "The same up to symmetry" means "in the same orbit," so what you're really asking is: how many orbits are there? Counting orbits directly is fiddly. Burnside's lemma replaces it with an average that is often easy to compute.

Read that carefully: you go through the group element by element, count how many configurations each one leaves unchanged, add those counts up, and divide by the size of the group. The word "average" is exact — the number of orbits is literally the mean of the fixed-point counts. It comes from double-counting the pairs (g, x) with g \cdot x = x: summing over g gives \sum_g |\operatorname{Fix}(g)|, while summing over x gives \sum_x |\operatorname{stab}(x)|, and orbit–stabiliser turns the latter into |G| times the number of orbits.

Worked example: counting necklaces

How many distinct necklaces have n = 4 beads, each coloured one of 2 colours, where rotations count as the same? Without symmetry there are 2^4 = 16 colourings. The group is C_4 = \{e, r, r^2, r^3\}. Count the colourings fixed by each rotation:

• e (rotate by 0) fixes all 2^4 = 16 colourings.
• r (rotate by 1) cycles all four beads together, forcing them equal: 2^1 = 2 fixed.
• r^2 (rotate by 2) swaps beads in two opposite pairs: 2^2 = 4 fixed.
• r^3 (rotate by 3) again cycles all four: 2^1 = 2 fixed.

Six genuinely different two-colour necklaces of length four — and you can list them by hand to check: the general rule is that a rotation by k fixes c^{\gcd(n,k)} colourings when there are c colours, because it splits the beads into \gcd(n,k) cycles that must each be a single colour. Averaging those counts over the group is Burnside in action.