01.02.03 · foundations / groups

Group action, orbit-stabiliser, class equation

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Anchor (Master): Burnside 1897 Theory of Groups of Finite Order; Sylow 1872; Dummit-Foote §4.1-4.3

Intuition [Beginner]

A group action is a way for the elements of a group to move objects around in a set. Think of the symmetries of a square: each symmetry rearranges the four corners. The group of symmetries acts on the set of corners by moving them to new positions.

An orbit is the set of places a single object can reach. Pick any corner of the square and apply all the symmetries: the rotations and reflections can send it to any of the other three corners. All four corners form one orbit because the symmetries connect every corner to every other corner.

A stabiliser is the set of symmetries that leave one specific object where it is. Pick the top-left corner. The only symmetries that keep it in place are the identity and the reflection across the diagonal through that corner: two elements. But the centre of the square is fixed by all eight symmetries, so its stabiliser is the whole group.

Why does this concept exist? Group actions connect abstract algebra to concrete objects you can see and count, turning questions about group structure into questions about rearrangements of a set.

Visual [Beginner]

The dihedral group $D_{4}$ acting on the four corners of a square. Eight symmetries — four rotations and four reflections — permute the corners. The arrows show how a $9 0^{\circ}$ rotation sends each corner to the next one clockwise.

Each corner belongs to the same orbit: any corner can be moved to any other by some symmetry in $D_{4}$ .

Worked example [Beginner]

Consider the dihedral group $D_{4}$ acting on the four corners of a square. The group has $8$ elements: the identity, rotations by $9 0^{\circ}$ , $18 0^{\circ}$ , and $27 0^{\circ}$ , and four reflections.

Step 1. Pick corner $1$ (top-left). Its orbit is the set of all positions reachable by symmetries: ${1, 2, 3, 4}$ . Every corner can be reached, so the orbit has size $4$ .

Step 2. The stabiliser of corner $1$ consists of the symmetries that fix corner $1$ . Only the identity and the reflection across the diagonal through corner $1$ keep it in place. So the stabiliser has size $2$ .

Step 3. The group has $8$ elements. The orbit size ( $4$ ) times the stabiliser size ( $2$ ) equals $8$ . This is the orbit-stabiliser relationship: the size of the group equals the size of the orbit times the size of the stabiliser.

What this tells us: counting where things go and what keeps them fixed gives a clean arithmetic relationship with the size of the whole symmetry group.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $G$ be a group and $X$ a set.

Definition (Group action). A left action of $G$ on $X$ is a map $G \times X \to X$ , written $(g, x) \mapsto g \cdot x$ , satisfying:

$e \cdot x = x$ for all $x \in X$ (identity fixes every point).
$(g h) \cdot x = g \cdot (h \cdot x)$ for all $g, h \in G$ and $x \in X$ (compatibility).

A set equipped with such an action is called a $G$ -set ^{[Dummit-Foote §4.1]}.

Definition (Orbit). For $x \in X$ , the orbit of $x$ is $$ \operatorname{Orb}(x) = {g \cdot x : g \in G}. $$

Definition (Stabiliser). For $x \in X$ , the stabiliser of $x$ is $$ \operatorname{Stab}(x) = {g \in G : g \cdot x = x}. $$

The orbits partition $X$ into disjoint subsets. The stabiliser $Stab (x)$ is a subgroup of $G$ for every $x \in X$ .

Definition (Fixed points). For $g \in G$ , the fixed-point set of $g$ is $X^{g} = {x \in X : g \cdot x = x}$ .

Definition (Centraliser and centre). When $G$ acts on itself by conjugation $g \cdot h = g h g^{- 1}$ , the stabiliser of $h$ is the centraliser $C_{G} (h) = {g \in G : g h = h g}$ . The centre of $G$ is $Z (G) = {g \in G : g x = xg for all x \in G}$ .

Counterexamples to common slips [Intermediate+]

A group action is not just any map. The constant map $(g, x) \mapsto x_{0}$ for a fixed $x_{0}$ fails the identity axiom unless $x_{0} = x$ for every $x$ . The map $(g, x) \mapsto x$ for all $g$ and $x$ satisfies both axioms but produces only singleton orbits.
The stabiliser depends on the point. Different points have different stabilisers. In $D_{4}$ acting on the corners, the stabiliser of a corner has $2$ elements, but the stabiliser of the centre (if included in the set) has $8$ elements.
Orbits need not all have the same size. The symmetric group $S_{4}$ acting on the set of ordered pairs ${1, 2, 3, 4}^{2}$ has orbits of sizes $12$ (off-diagonal pairs) and $4$ (diagonal pairs).

Key theorem with proof [Intermediate+]

Theorem (Orbit-Stabiliser). Let $G$ be a finite group acting on a set $X$ . For any $x \in X$ : $$ |G| = |\operatorname{Orb}(x)| \cdot |\operatorname{Stab}(x)|. $$

Proof. Define the map $φ : G \to Orb (x)$ by $φ (g) = g \cdot x$ . This map is surjective: every element of $Orb (x)$ has the form $g \cdot x$ for some $g$ .

Two elements $g, h \in G$ satisfy $φ (g) = φ (h)$ if and only if $g \cdot x = h \cdot x$ , which gives $h^{- 1} g \cdot x = x$ , meaning $h^{- 1} g \in Stab (x)$ . Equivalently, $g$ and $h$ lie in the same left coset of $Stab (x)$ in $G$ .

The fibres of $φ$ are precisely the left cosets of $Stab (x)$ . Each fibre has exactly $∣ Stab (x) ∣$ elements, and there are $∣ Orb (x) ∣$ distinct fibres (one for each element of the orbit). Therefore: $$ |G| = |\operatorname{Stab}(x)| \cdot |\operatorname{Orb}(x)|. $$ $□$

Bridge. The orbit-stabiliser theorem builds toward 01.02.04 (Sylow theorems), where group actions on sets of subsets produce the existence, conjugacy, and counting results for Sylow $p$ -subgroups. This result appears again in 01.02.02 (subgroups and cosets), since the coset partition underlying Lagrange's theorem is the same mechanism at work here: the fibres of $φ$ are cosets of the stabiliser. The foundational reason is that every group action produces a coset decomposition, and this is exactly the bridge between the algebraic structure of $G$ and the combinatorial structure of $X$ . The central insight is that the stabiliser measures how much symmetry fixes a point, and the orbit measures how far that point can travel.

Exercises [Intermediate+]

Exercise 7 (hard, symbolic).

Derive the class equation: for a finite group $G$ , $$ |G| = |Z(G)| + \sum_i [G : C_G(g_i)] $$ where the sum runs over representatives $g_{i}$ of the non-central conjugacy classes.

Hint

Apply the orbit-stabiliser theorem to the conjugation action. The orbits are conjugacy classes, and the stabilisers are centralisers. Elements in the centre have conjugacy class of size $1$ .

Answer

Let $G$ act on itself by conjugation. The orbits are the conjugacy classes. By Exercise 6, the stabiliser of $x$ is $C_{G} (x)$ .

By the orbit-stabiliser theorem, $∣ G ∣ = ∣ Orb (x) ∣ \cdot ∣ C_{G} (x) ∣$ , so $∣ Orb (x) ∣ = [G : C_{G} (x)]$ .

The conjugacy classes partition $G$ . The elements of $Z (G)$ are those with $∣ Orb (x) ∣ = 1$ (they commute with every element, so their conjugacy class is a singleton). Choosing one representative $g_{i}$ from each non-central conjugacy class: $$ |G| = \sum_{\text{classes}} |\text{class}| = |Z(G)| + \sum_i [G : C_G(g_i)]. $$

Exercise 8 (hard, symbolic).

Let $G$ act on a finite set $X$ . Prove Burnside's lemma: the number of distinct orbits is $$ \frac{1}{|G|}\sum_{g \in G} |X^g| $$ where $X^{g} = {x \in X : g \cdot x = x}$ is the fixed-point set of $g$ .

Hint

Count the pairs $(g, x)$ with $g \cdot x = x$ in two ways: fix $g$ first (sum $∣ X^{g} ∣$ ), then fix $x$ first (sum $∣ Stab (x) ∣$ and regroup by orbits using the orbit-stabiliser theorem).

Answer

Let $S = {(g, x) \in G \times X : g \cdot x = x}$ . Count $∣ S ∣$ in two ways.

Fixing $g$ : $∣ S ∣ = \sum_{g \in G} ∣ X^{g} ∣$ .

Fixing $x$ : $∣ S ∣ = \sum_{x \in X} ∣ Stab (x) ∣$ .

Group the second sum by orbits. Let $x_{1}, \dots, x_{k}$ be representatives of the $k$ distinct orbits. For any $x$ in the orbit of $x_{i}$ : $$ |\operatorname{Stab}(x)| = \frac{|G|}{|\operatorname{Orb}(x)|} = \frac{|G|}{|\operatorname{Orb}(x_i)|}. $$ Each orbit $Orb (x_{i})$ has $∣ Orb (x_{i}) ∣$ elements, each with stabiliser of size $∣ G ∣/∣ Orb (x_{i}) ∣$ . The total contribution of orbit $i$ is: $$ |\operatorname{Orb}(x_i)| \cdot \frac{|G|}{|\operatorname{Orb}(x_i)|} = |G|. $$ Therefore $\sum_{x \in X} ∣ Stab (x) ∣ = k \cdot ∣ G ∣$ . Equating the two counts: $$ \sum_{g \in G} |X^g| = k \cdot |G|, $$ and dividing by $∣ G ∣$ gives $k = \frac{1}{∣ G ∣} \sum_{g \in G} ∣ X^{g} ∣$ .

Advanced results [Master]

Theorem 1 (Burnside's Lemma). Let $G$ act on a finite set $X$ . The number of distinct orbits is $$ \frac{1}{|G|}\sum_{g \in G} |X^g| $$ where $X^{g} = {x \in X : g \cdot x = x}$ . Despite the name, this result is due to Cauchy (1845) and Frobenius (1887); Burnside included and popularised it in his 1897 textbook ^{[Burnside 1897]}.

Theorem 2 (Class Equation). For a finite group $G$ : $$ |G| = |Z(G)| + \sum_i [G : C_G(g_i)] $$ where $g_{i}$ are representatives of the non-central conjugacy classes. Each index $[G : C_{G} (g_{i})]$ exceeds $1$ and divides $∣ G ∣$ .

Theorem 3 (Cauchy via Group Actions). If $G$ is a finite group and a prime $p$ divides $∣ G ∣$ , then $G$ contains an element of order $p$ . The proof considers the set $X = {(g_{1}, \dots, g_{p}) \in G^{p} : g_{1} g_{2} \dots g_{p} = e}$ with the cyclic rotation action, and applies the orbit decomposition to show that some fixed-point orbit corresponds to an element of order $p$ ^{[Dummit-Foote §4.2]}.

Theorem 4 (Conjugacy Classes of $S_{n}$ ). The conjugacy classes of the symmetric group $S_{n}$ are indexed by partitions of $n$ . Two permutations are conjugate if and only if they have the same cycle type. The number of conjugacy classes equals the number of partitions of $n$ ^{[Dummit-Foote §4.3]}.

Theorem 5 (Fixed-Point-Free Automorphisms). If a finite group $G$ admits a fixed-point-free automorphism of prime order $p$ — meaning an automorphism $φ$ with $φ (g) = g$ only when $g = e$ — then $G$ is nilpotent. This result of Thompson (1959) relies on the structure of group actions and their fixed-point sets ^{[Rotman §3]}.

Theorem 6 (Centre of a $p$ -Group). If $G$ is a group of order $p^{n}$ for a prime $p$ , then $∣ Z (G) ∣ \geq p$ . The proof applies the class equation: each $[G : C_{G} (g_{i})]$ is a power of $p$ exceeding $1$ , so $p$ divides $\sum_{i} [G : C_{G} (g_{i})]$ . Since $p$ divides $∣ G ∣$ , it follows that $p$ divides $∣ Z (G) ∣$ , forcing $∣ Z (G) ∣ \geq p$ .

Synthesis. The orbit-stabiliser theorem is the foundational reason that counting arguments permeate finite group theory, and this is exactly the mechanism behind the class equation, Cauchy's theorem, and the Sylow theorems. The central insight is that conjugation makes the group act on itself, and putting these together with the orbit-stabiliser counting argument identifies the centre $Z (G)$ with the fixed points of this action. The bridge is between the combinatorial data of orbit sizes and the algebraic data of subgroup indices.

The pattern generalises from finite groups to compact Lie groups acting on manifolds, where orbits become homogeneous spaces and the class equation becomes the Weyl integration formula. Burnside's lemma appears again in 01.02.04 as the counting tool that underpins the Sylow congruence, and the pattern recurs in representation theory when counting irreducible characters via the class algebra constants.

Full proof set [Master]

Proposition 1 (Burnside's Lemma). Let the finite group $G$ act on the finite set $X$ . The number of orbits is $\frac{1}{∣ G ∣} \sum_{g \in G} ∣ X^{g} ∣$ .

Proof. Count the set $S = {(g, x) \in G \times X : g \cdot x = x}$ in two ways.

First, fixing $g \in G$ : the contribution is $∣ X^{g} ∣$ , the number of points fixed by $g$ . So $∣ S ∣ = \sum_{g \in G} ∣ X^{g} ∣$ .

Second, fixing $x \in X$ : the contribution is $∣ Stab (x) ∣$ . So $∣ S ∣ = \sum_{x \in X} ∣ Stab (x) ∣$ .

Regroup the second sum by orbits. Let $O_{1}, \dots, O_{k}$ be the distinct orbits with representatives $x_{1}, \dots, x_{k}$ . For any $x \in O_{i}$ , the orbit-stabiliser theorem gives $∣ Stab (x) ∣ = ∣ G ∣/∣ O_{i} ∣$ . Since $O_{i}$ contains $∣ O_{i} ∣$ elements, its total contribution to $\sum_{x} ∣ Stab (x) ∣$ is: $$ |\mathcal{O}i| \cdot \frac{|G|}{|\mathcal{O}i|} = |G|. $$ Summing over all orbits: $\sum{x \in X} |\operatorname{Stab}(x)| = k \cdot |G| $. E q u a t in g w i t h t h e f i r s t co u n t g i v es$ k = \frac{1}{|G|}\sum{g \in G} |X^g| $.$ \square$

Proposition 2 (Centre of a $p$ -Group). If $G$ is a group of order $p^{n}$ with $p$ prime, then $∣ Z (G) ∣ \geq p$ .

Proof. Apply the class equation: $$ p^n = |G| = |Z(G)| + \sum_i [G : C_G(g_i)] $$ where the sum runs over representatives of non-central conjugacy classes.

For each $g_{i} \in / Z (G)$ , the centraliser $C_{G} (g_{i})$ is a proper subgroup of $G$ , so $[G : C_{G} (g_{i})] > 1$ . Since $∣ G ∣ = p^{n}$ and $C_{G} (g_{i})$ is a subgroup, $[G : C_{G} (g_{i})] = p^{n} /∣ C_{G} (g_{i}) ∣$ is a power of $p$ exceeding $1$ , hence divisible by $p$ .

Therefore $p$ divides $\sum_{i} [G : C_{G} (g_{i})]$ , and since $p$ divides $∣ G ∣ = p^{n}$ , it follows that $p$ divides $∣ Z (G) ∣$ . Since $e \in Z (G)$ , we have $∣ Z (G) ∣ \geq 1$ , and divisibility by $p$ forces $∣ Z (G) ∣ \geq p$ . $□$

Proposition 3 (Cauchy's Theorem via Group Actions). If $G$ is a finite group and $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ .

Proof. Consider the set $X = {(g_{1}, \dots, g_{p}) \in G^{p} : g_{1} g_{2} \dots g_{p} = e}$ . For each choice of $g_{1}, \dots, g_{p - 1}$ , the element $g_{p} = (g_{1} \dots g_{p - 1})^{- 1}$ is uniquely determined, so $∣ X ∣ = ∣ G ∣^{p - 1}$ .

The cyclic group $⟨ σ ⟩ ≅ Z / p Z$ acts on $X$ by cyclic rotation: $σ \cdot (g_{1}, \dots, g_{p}) = (g_{2}, \dots, g_{p}, g_{1})$ . This is an action because rotating and then rotating again is the same as rotating twice, and the identity rotation fixes every tuple.

A tuple $(g_{1}, \dots, g_{p}) \in X$ is fixed by $σ$ if and only if $(g_{1}, \dots, g_{p}) = (g_{2}, \dots, g_{p}, g_{1})$ , meaning $g_{1} = g_{2} = \dots = g_{p}$ . Since $g_{1}^{p} = e$ , the fixed points correspond to elements of order dividing $p$ .

Let $k$ be the number of orbits. The tuple $(e, e, \dots, e)$ is a fixed point, so there is at least one orbit of size $1$ . By Burnside's lemma: $$ k = \frac{1}{p}\sum_{j=0}^{p-1} |X^{\sigma^j}|. $$ For $j = 0$ : $∣ X^{σ^{0}} ∣ = ∣ X ∣ = ∣ G ∣^{p - 1}$ . For $j \neq = 0$ : $∣ X^{σ^{j}} ∣$ counts tuples with $g_{1} = g_{2} = \dots = g_{p}$ and $g_{1}^{p} = e$ , so $∣ X^{σ^{j}} ∣$ is the number of elements $g \in G$ with $g^{p} = e$ .

Since $p$ divides $∣ G ∣$ , we have $p$ divides $∣ G ∣^{p - 1}$ , and since $k$ is an integer, $p$ must divide $∣ X^{σ^{0}} ∣ + \sum_{j = 1}^{p - 1} ∣ X^{σ^{j}} ∣$ . The number of elements with $g^{p} = e$ is at least $1$ (the identity), and $k \geq 1$ . For $p$ to divide $k \cdot p$ while $k$ is a positive integer, the fixed-point count under $σ^{j}$ ( $j \neq = 0$ ) must include at least one non-identity element. Any such element $g \neq = e$ with $g^{p} = e$ has order $p$ . $□$

Connections [Master]

Subgroup, coset, quotient group 01.02.02. The coset partition underlying Lagrange's theorem is the same structure that the orbit-stabiliser theorem exploits. The fibres of the map $φ : G \to Orb (x)$ are left cosets of $Stab (x)$ , and the bijection $G / Stab (x) \to Orb (x)$ is the concrete realisation of the orbit-stabiliser counting argument. The foundational reason that both Lagrange and orbit-stabiliser work is the same coset-decomposition mechanism.
Sylow theorems 01.02.04. The Sylow existence theorem is proved by letting $G$ act on the set of all subsets of $G$ of size $p^{k}$ , and the orbit-counting and stabiliser-size constraints force the existence of a subgroup of order $p^{k}$ . The class equation controls the number of Sylow $p$ -subgroups via the congruence $n_{p} \equiv 1 (mod p)$ . Burnside's lemma provides the orbit-counting machinery that makes the Sylow proofs work.
Group 01.02.01. The group axioms (identity and associativity) are precisely what make the action axioms well-behaved: the identity axiom for actions uses the group identity, and the compatibility axiom uses associativity. The action of a group on itself by left multiplication is the canonical free action, and the action by conjugation produces the class equation as its orbit decomposition.

Historical & philosophical context [Master]

Cauchy 1845 established the orbit-counting formula now called Burnside's lemma, in the context of permutation groups acting on finite sets ^{[Burnside 1897]}. Frobenius 1887 independently rediscovered the same counting result in Journal fur die reine und angewandte Mathematik 102. Burnside included the formula in the first edition of his Theory of Groups of Finite Order (1897, Cambridge University Press) and his textbook treatment made it widely known, which is the origin of the misattribution ^{[Burnside 1897]}.

Sylow 1872 proved the existence, conjugacy, and counting theorems for maximal $p$ -subgroups of finite groups in Mathematische Annalen 5, using the action of a group on the collection of its subsets ^{[Dummit-Foote §4.5]}. The orbit-stabiliser theorem in its modern abstract form developed with the axiomatisation of group theory in the early twentieth century, appearing in the treatments of Burnside (1911, second edition) and Speiser (1923, Die Theorie der Gruppen von endlicher Ordnung).

Bibliography [Master]

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}

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}

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}

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}

@article{Sylow1872,
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}

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}