Chapter 08

Hard

Group Actions

00 · Symbol Glossary

$g\cdot x$G acts on x

Notation for the result of letting group element $g$ move a point $x$ in some set $X$ . Distinct from the group's own operation $gh$ — here $g$ and $x$ live in different sets.

$\mathrm{Orb}(x)$Orbit of x

$\mathrm{Orb}(x) = \{g\cdot x : g\in G\}$ — every point reachable from $x$ by acting with some group element. The set of everywhere $x$ can be "moved to."

$\mathrm{Stab}(x)$Stabilizer of x

$\mathrm{Stab}(x) = \{g\in G : g\cdot x = x\}$ — every group element that leaves $x$ fixed in place.

$C_G(x)$Centralizer of x (under conjugation)

$C_G(x) = \{g\in G : gxg^{-1}=x\} = \{g\in G : gx=xg\}$ — the stabilizer of $x$ specifically under the conjugation action. Everything commuting with $x$ .

$x^G$Conjugacy class of x

$x^G = \{gxg^{-1} : g\in G\}$ — the orbit of $x$ under conjugation. All elements "structurally equivalent" to $x$ by relabeling via some $g$ .

01 · What Is a Group Action?

Every example of a group so far has, at heart, been a group of motions: permutations moving points, matrices moving vectors, symmetries moving a polygon's vertices. A group action makes this idea of " $G$ moves the points of some set $X$ " completely precise — and, crucially, $X$ need not have any group structure of its own.

Definition — Group Action

A group $G$ acts on a set $X$ if there is a function $G\times X \to X$ , written $(g,x)\mapsto g\cdot x$ , satisfying:

A1. $e\cdot x = x$ for every $x\in X$ .

A2. $g\cdot(h\cdot x) = (gh)\cdot x$ for every $g,h\in G$ , $x\in X$ .

X is just a set — no group structure required

This is the key generalization beyond homomorphisms: $X$ can be a completely bare set — the vertices of a polygon, the elements of $G$ itself, the set of subgroups of $G$ — with no operation of its own. All the algebraic structure lives entirely in $G$ ; $X$ merely receives the motion.

02 · Actions Are Permutations, and Some Standard Examples

Theorem 8.1 — Each Group Element Acts as a Bijection

For fixed $g\in G$ , the map $\sigma_g: X\to X$ , $\sigma_g(x)=g\cdot x$ , is a bijection, with inverse $\sigma_{g^{-1}}$ .

Proof

Compute $\sigma_{g^{-1}}(\sigma_g(x)) = g^{-1}\cdot(g\cdot x) = (g^{-1}g)\cdot x = e\cdot x = x$ (using A2, then A1), and similarly $\sigma_g(\sigma_{g^{-1}}(x))=x$ . So $\sigma_{g^{-1}}$ is a two-sided inverse for $\sigma_g$ , meaning $\sigma_g$ is a bijection.

Example — G acting on itself by left multiplication

$X=G$ , action $g\cdot x := gx$ (the group's own operation). A1: $e\cdot x=ex=x$ . A2: $g\cdot(h\cdot x) = g(hx) = (gh)x = (gh)\cdot x$ , using associativity. Every group acts on itself this way — the very first example, and the source of Cayley's theorem (every group embeds into some $S_n$ , since left multiplication realizes $G$ as bijections of itself).

Example — $D_n$ acting on the vertices of a regular n-gon

$X=\{1,2,\ldots,n\}$ (vertex labels), and each symmetry $g\in D_n$ sends vertex $x$ to wherever that symmetry moves it. A1 (identity motion fixes every vertex) and A2 (composing two symmetries then applying to a vertex matches applying one after the other) both hold by the geometric meaning of composition. This is the action underlying every cycle-notation computation with $D_n$ from Chapter 02.

Example — G acting on itself by conjugation

$X=G$ , action $g\cdot x := gxg^{-1}$ . A1: $e\cdot x = exe^{-1}=x$ . A2: $g\cdot(h\cdot x) = g(hxh^{-1})g^{-1} = (gh)x(gh)^{-1} = (gh)\cdot x$ , using $(gh)^{-1}=h^{-1}g^{-1}$ (Theorem 2.3). This action will drive the rest of the chapter.

03 · Orbits and Stabilizers

Definition — Orbit and Stabilizer

For $x\in X$ : $\mathrm{Orb}(x) = \{g\cdot x : g\in G\} \subseteq X$ , and $\mathrm{Stab}(x) = \{g\in G : g\cdot x = x\} \subseteq G$ .

Theorem 8.2 — Orbits Partition X

The orbits of a group action partition $X$ : every element of $X$ lies in exactly one orbit.

Proof

Define $x\sim y \iff y=g\cdot x$ for some $g\in G$ . Reflexive: $x=e\cdot x$ (A1). Symmetric: if $y=g\cdot x$ , then $g^{-1}\cdot y = g^{-1}\cdot(g\cdot x) = (g^{-1}g)\cdot x = e\cdot x=x$ (A2, A1), so $x=g^{-1}\cdot y$ . Transitive: if $y=g\cdot x$ and $z=h\cdot y$ , then $z=h\cdot(g\cdot x) = (hg)\cdot x$ (A2), so $x\sim z$ . By Theorem 1.3, $\sim$ 's equivalence classes partition $X$ — and $[x]=\{y : y=g\cdot x\text{ some }g\}=\mathrm{Orb}(x)$ exactly.

Theorem 8.3 — Stabilizers Are Subgroups

$\mathrm{Stab}(x) \le G$ for every $x\in X$ .

Proof

Nonempty: $e\cdot x=x$ (A1), so $e\in\mathrm{Stab}(x)$ . Closure: if $g,h\in\mathrm{Stab}(x)$ , then $(gh)\cdot x = g\cdot(h\cdot x) = g\cdot x = x$ (A2, then $h$ fixes $x$ , then $g$ fixes $x$ ), so $gh\in\mathrm{Stab}(x)$ . Inverses: if $g\cdot x=x$ , apply $g^{-1}$ to both sides: $g^{-1}\cdot(g\cdot x) = g^{-1}\cdot x$ , and the left side simplifies to $x$ (A2, A1), so $x=g^{-1}\cdot x$ , meaning $g^{-1}\in\mathrm{Stab}(x)$ . By Theorem 3.1, $\mathrm{Stab}(x)\le G$ .

Example — Orbit and stabilizer under $D_4$ acting on a square's vertices

Fix vertex $1$ . $\mathrm{Orb}(1)$ : every symmetry sends vertex $1$ somewhere, and since the square is vertex-transitive (any vertex can be rotated/reflected to any other), $\mathrm{Orb}(1) = \{1,2,3,4\}$ , all four vertices. $\mathrm{Stab}(1)$ : the symmetries fixing vertex $1$ in place — the identity and the single reflection through vertex $1$ and the center. $|\mathrm{Stab}(1)|=2$ .

Notice $|\mathrm{Orb}(1)|\cdot|\mathrm{Stab}(1)| = 4\times2=8=|D_4|$ — not a coincidence, as the next theorem shows.

04 · The Orbit-Stabilizer Theorem

Theorem 8.4 — Orbit-Stabilizer Theorem

For any group action of a finite group $G$ on $X$ and any $x\in X$ :

|\mathrm{Orb}(x)| = [G : \mathrm{Stab}(x)] = \frac{|G|}{|\mathrm{Stab}(x)|}

Proof

Write $H=\mathrm{Stab}(x)$ . Define $\psi: G/H \to \mathrm{Orb}(x)$ by $\psi(gH) = g\cdot x$ .

Well-defined: if $gH=g'H$ , then by Theorem 4.1, $g^{-1}g'\in H$ , so $(g^{-1}g')\cdot x=x$ (definition of $H=\mathrm{Stab}(x)$ ). Applying $g$ to both sides: $g\cdot\big((g^{-1}g')\cdot x\big) = g\cdot x$ , and the left side is $(gg^{-1}g')\cdot x = g'\cdot x$ by A2. So $g'\cdot x = g\cdot x$ : the map doesn't depend on the representative.

Injective: if $\psi(gH)=\psi(g'H)$ , i.e. $g\cdot x = g'\cdot x$ , apply $g^{-1}$ : $x = g^{-1}\cdot(g'\cdot x) = (g^{-1}g')\cdot x$ , so $g^{-1}g'\in\mathrm{Stab}(x)=H$ , giving $gH=g'H$ by Theorem 4.1.

Surjective: every element of $\mathrm{Orb}(x)$ has the form $g\cdot x$ for some $g$ , which is $\psi(gH)$ .

$\psi$ is a bijection between $G/H$ (as a set of cosets — not requiring $H$ normal here, since we only need a bijection of sets, not a group isomorphism) and $\mathrm{Orb}(x)$ . So $|\mathrm{Orb}(x)| = |G/H| = [G:H]$ , and by Lagrange's Theorem, $[G:H]=|G|/|H|$ .

Common mistake — Stab(x) need not be normal

Unlike Chapter 05's quotient groups, $\psi$ here is only a bijection of sets, not a group isomorphism — $\mathrm{Orb}(x)$ generally has no group structure at all (it's just a subset of $X$ , and $X$ was never assumed to be a group). $G/\mathrm{Stab}(x)$ is used here purely as a counting device, valid whether or not $\mathrm{Stab}(x)$ happens to be normal. If $\mathrm{Stab}(x)$ is not normal, you cannot treat $G/\mathrm{Stab}(x)$ as a quotient group — only the orbit–stabilizer bijection for counting is valid.

05 · Conjugation and the Class Equation

Applying the general machinery of Sections 03–04 to the specific conjugation action from Section 02 gives one of the most powerful counting tools in finite group theory.

Definition — Conjugacy Class and Centralizer

Under the conjugation action, $\mathrm{Orb}(x) = x^G = \{gxg^{-1}:g\in G\}$ is the conjugacy class of $x$ , and $\mathrm{Stab}(x) = C_G(x) = \{g\in G : gxg^{-1}=x\}$ is the centralizer of $x$ .

Example — Central elements form their own conjugacy classes

If $x\in Z(G)$ , then $gxg^{-1}=x$ for every $g$ (that's exactly what commuting with everything means), so $x^G = \{x\}$ — a conjugacy class of size 1. Conversely, if $x^G=\{x\}$ , then $gxg^{-1}=x$ for every $g$ , meaning $x\in Z(G)$ . Conjugacy classes of size 1 are exactly the elements of $Z(G)$ .

Theorem 8.5 — The Class Equation

For a finite group $G$ with conjugacy classes $\{e\}=x_0^G,\ldots$ (writing out the size-1 classes from $Z(G)$ separately) and noncentral class representatives $x_1,\ldots,x_k$ :

|G| = |Z(G)| + \sum_{i=1}^{k} [G:C_G(x_i)]

Proof

By Theorem 8.2, conjugacy classes partition $G$ , so $|G|$ is the sum of all class sizes. By Theorem 8.4 (Orbit-Stabilizer applied to conjugation), each class $x^G$ has size $[G:C_G(x)]$ . Separating the size-1 classes (exactly the elements of $Z(G)$ , by the Example above) from the rest gives $|G| = \underbrace{|Z(G)|}_{\text{size-1 classes}} + \sum_{i=1}^k \underbrace{[G:C_G(x_i)]}_{\text{noncentral classes}}$ .

Step-by-step — The class equation for $S_3$

List conjugacy classes of $S_3$ :

\{e\}

(size 1), the three transpositions

\{(1\,2),(1\,3),(2\,3)\}

(one class, since conjugation permutes cycle shape, and all transpositions share the same shape), the two 3-cycles

\{(1\,2\,3),(1\,3\,2)\}

(one class).

Check sizes sum to $|S_3|=6$ :

1+3+2=6

. ✓

Identify $Z(S_3)$ : only the size-1 class contributes, so

Z(S_3)=\{e\}

, size 1 — consistent with

S_3

being nonabelian (Chapter 05's fact that

Z(G)=G

iff abelian).

Write the class equation:

6 = 1 + 3 + 2

, i.e.

|G|=|Z(G)| + [G:C_G((1\,2))] + [G:C_G((1\,2\,3))]

, giving

[G:C_G((1\,2))]=3

and

[G:C_G((1\,2\,3))]=2

Back out the centralizer sizes:

|C_G((1\,2))| = 6/3=2

and

|C_G((1\,2\,3))|=6/2=3

— matching direct computation (

(1\,2)

commutes only with itself and

e

;

(1\,2\,3)

commutes with all of

\{e,(1\,2\,3),(1\,3\,2)\}

06 · Application: Cauchy's Theorem

Lagrange's Theorem (Chapter 04) is one-directional: subgroup orders divide $|G|$ , but not every divisor is achieved (recall $A_4$ , order 12, with no subgroup of order 6). Remarkably, for divisors that are prime, the converse always holds — and the class equation is exactly the tool needed to prove it.

Theorem 8.6 — Cauchy's Theorem

If $G$ is a finite group and $p$ is a prime dividing $|G|$ , then $G$ has an element of order $p$ .

Proof

By strong induction on $|G|$ .

Case 1: $G$ is abelian. Pick any $a\neq e$ in $G$ , and let $N=\langle a\rangle$ . If $p$ divides $|a|=|N|$ , then $a^{|N|/p}$ has order exactly $p$ (a direct check: raising it to the $p$ -th power gives $a^{|N|}=e$ , and no smaller positive power works, by minimality of $|N|/p$ among divisors), and we're done.

Otherwise $p\nmid|N|$ . Since $G$ is abelian, $N\trianglelefteq G$ (Chapter 05), so $G/N$ is a group of order $|G|/|N| < |G|$ (strictly smaller, since $|N|>1$ ). Since $|G|=|N|\cdot|G/N|$ and $p$ divides $|G|$ but not $|N|$ , $p$ must divide $|G/N|$ (a prime dividing a product but not one factor divides the other). By the induction hypothesis, $G/N$ has an element $yN$ of order $p$ . Let $m=|y|$ in $G$ . Then $(yN)^m = y^mN = eN$ , so $p=|yN|$ divides $m$ (order of an element divides any exponent returning it to the identity — the same fact used in Theorem 7.2's proof). Then $y^{m/p}$ has order $p$ in $G$ .

Case 2: $G$ is nonabelian. If $p$ divides $|Z(G)|$ , then since $Z(G)$ is abelian and smaller than or equal to $G$ (with $Z(G)\neq G$ since $G$ is nonabelian, but we can still apply Case 1 directly to the abelian group $Z(G)$ regardless of its size), $Z(G)$ has an element of order $p$ by Case 1 — and that element also lives in $G$ .

Otherwise $p\nmid|Z(G)|$ . By the class equation (Theorem 8.5), $|G| = |Z(G)| + \sum_i [G:C_G(x_i)]$ . If $p$ divided every term $[G:C_G(x_i)]$ in the sum, it would divide the whole sum $\sum_i[G:C_G(x_i)] = |G|-|Z(G)|$ ; combined with $p\mid|G|$ , this would force $p\mid|Z(G)|$ too — contradicting our assumption. So some noncentral class representative $x_i$ has $p \nmid [G:C_G(x_i)]$ . Since $|G| = [G:C_G(x_i)]\cdot|C_G(x_i)|$ and $p$ divides $|G|$ but not the index factor, $p$ must divide $|C_G(x_i)|$ . Since $x_i$ is noncentral, $C_G(x_i)\neq G$ , so $|C_G(x_i)|<|G|$ — a strictly smaller group. By the induction hypothesis, $C_G(x_i)$ has an element of order $p$ , which is also an element of $G$ .

A genuine partial converse to Lagrange

Cauchy's Theorem says: for a prime divisor $p$ of $|G|$ , an element (equivalently, by Theorem 3.4, a cyclic subgroup) of order $p$ is guaranteed. Chapter 09's Sylow theorems extend this from single primes to full prime powers — guaranteeing subgroups of order $p^k$ whenever $p^k$ divides $|G|$ — using the exact same class-equation machinery built here, one level more refined.

07 · Exercises

EXERCISE 8.1

Check both action axioms directly: does the identity permutation fix everything, and does composing two permutations first vs. applying them in sequence give the same result?

$X=\{1,\ldots,n\}$ , action $\sigma\cdot k := \sigma(k)$ . A1: the identity permutation $e$ satisfies $e(k)=k$ for every $k$ , so $e\cdot k=k$ . A2: $\sigma\cdot(\tau\cdot k) = \sigma(\tau(k)) = (\sigma\circ\tau)(k) = (\sigma\tau)\cdot k$ , matching the composition convention from Chapter 02. Both axioms hold: $S_n$ acts on $\{1,\ldots,n\}$ by direct evaluation.

Verify that $S_n$ acts on $X=\{1,\ldots,n\}$ via $\sigma\cdot k := \sigma(k)$ , by checking both action axioms.

EXERCISE 8.2

Use the Orbit-Stabilizer Theorem: you're given $|G|=12$ and $|\mathrm{Orb}(x)|=4$ ; solve for $|\mathrm{Stab}(x)|$ .

By Theorem 8.4, $|\mathrm{Orb}(x)| = |G|/|\mathrm{Stab}(x)|$ , so $4 = 12/|\mathrm{Stab}(x)|$ , giving $|\mathrm{Stab}(x)| = 3$ .

A group of order 12 acts on a set, and some point $x$ has an orbit of size 4. What is $|\mathrm{Stab}(x)|$ ?

EXERCISE 8.3

Use the fact from Section 05: conjugacy classes of size 1 correspond exactly to central elements.

By the Example in Section 05, $x^G=\{x\}$ (a size-1 conjugacy class) if and only if $x\in Z(G)$ . If $G$ is abelian, $Z(G)=G$ (Chapter 05, Exercise 5.4), so every element's conjugacy class has size 1 — conjugation does nothing at all, consistent with $gxg^{-1}=x$ for all $g,x$ when $G$ is abelian.

If $G$ is abelian, what can you say about the size of every conjugacy class? Justify using the Section 05 fact about $Z(G)$ .

EXERCISE 8.4

$|A_4|=12=2^2\times3$ . Apply Cauchy's Theorem for each prime dividing 12.

$12 = 2^2\times3$ , so the prime divisors are $2$ and $3$ . By Theorem 8.6, $A_4$ must have an element of order $2$ and an element of order $3$ — but Cauchy's Theorem says nothing about elements of order $4$ (since $4$ is not itself prime, even though it divides $12$ ). This is consistent with $A_4$ 's well-known structure: it has elements of order $1,2,3$ but none of order $4$ or $6$ , showing Cauchy's Theorem is genuinely restricted to prime orders, not all divisors.

$|A_4|=12$ . Using Cauchy's Theorem, what element orders are guaranteed to exist in $A_4$ ? What does the theorem not guarantee?

08 · Chapter Summary

Concept	Statement
Group action	$G\times X\to X$ with $e\cdot x=x$ and $g\cdot(h\cdot x)=(gh)\cdot x$
Actions are bijections	Each $g$ acts as a bijection of $X$ (Thm 8.1)
Orbit	$\mathrm{Orb}(x)=\{g\cdot x:g\in G\}$ ; orbits partition $X$ (Thm 8.2)
Stabilizer	$\mathrm{Stab}(x)=\{g:g\cdot x=x\}$ ; always a subgroup (Thm 8.3)
Orbit-Stabilizer Theorem	$\lvert\mathrm{Orb}(x)\rvert = [G:\mathrm{Stab}(x)]$ (Thm 8.4)
Conjugation action	$g\cdot x = gxg^{-1}$ ; orbits = conjugacy classes; stabilizer = centralizer
Class equation	$\lvert G\rvert = \lvert Z(G)\rvert + \sum[G:C_G(x_i)]$ (Thm 8.5)
Cauchy's Theorem	$p$ prime, $p\mid\lvert G\rvert \implies G$ has an element of order $p$ (Thm 8.6)

Next: Chapter 09 — Sylow Theorems sharpens Cauchy's Theorem from single primes to full prime-power divisors, using the same orbit-counting and class-equation techniques developed here to describe the subgroup structure of any finite group almost completely.