Group Actions. The Fundamental Counting Principal.

Definitions. Let X be a set and G be a group. A left group action of G on X is a map G x X → X given by (g, x) → g·x such that the following axioms hold:

a) Identity: ∀x ∈ X, e·x = x

b) Compatibility: ∀x ∈ X, ∀g, h ∈ G, g·(h·x) = (gh)·x

The group G is said to act on the set X. A set X equipped with an action by G is called a G-set.

Let X be a set and G be a group. A right group action of G on X is a map X x G→ X given by (x, g) → x·g such that the following axioms hold:

a) Identity: ∀x ∈ X, x·e = x

b) Compatibility: ∀x ∈ X, ∀g, h ∈ G, (x·g)·h = x·(gh)

Examples.

• Let G = GL_n(ℝ), and X = ℝⁿ. Then G acts on X by left multiplication, that is, A·x = Ax for A ∈ G and x ∈ X.

a) ∀x ∈ ℝⁿ, I_nx = x

b) ∀x ∈ ℝⁿ, ∀A, B ∈ G = GL_n(ℝ), A(Bx) = (AB)x.

This is called the matrix action.

• Let G = S_n and X = {1, 2, ···, n}. Then G acts on X by function evaluation, that is, σ·x = σ(x), for σ ∈ S_n and x ∈ X.

a) ∀x ∈ X, 1·x = Id(x) = x b) ∀x ∈ X, ∀σ, τ ∈ S_n: σ·(τ·x) = σ·(τ(x)) = σ(τ(x)) = (σ∘τ)(x) = (στ)·x

This is called the permutation action.

• Let G act on X and H ≤ G. Then the action G x X → X, naturally restricts to an action H x X → X, which makes X also into an H-set, e.g., D_n acts on the regular n-gon geometrically by rotations and reflects. Identifying the regular n-gon with X = {1, 2, ···, n} and noting that D_n ≤ S_n, then the action by S_n restricts to an action by D_n.

Cayley’s Theorem states that every group is isomorphic to a permutation group, i.e., a subgroup of a symmetric group, therefore up to isomorphism every group action is a permutation action.

• A group G acts on itself by left multiplication, that is, g·x = gx. This is called the left regular action. Viewing the regular action as a permutation action, we have g·x = λ_g(x), where λ_g: G → G is the left multiplication map defined as x → gx

• A group G acts on itself by conjugation, that is, g, x ∈ G, and define g·x = gxg^-1.

a) e·x = exe^-1 = x.

b) ∀ g, h ∈ G. g·(h·x)= g·(hxh^-1) = g(hxh^-1)g^-1= [shoe-sock theorem] (gh)x(gh)^-1 = (gh)·x

This is called the conjugation action.

• Let G be a group, let H ≤ G, and let X = G/H the set of left cosets of H. Then G acts on G/H by the rule g·xH = (gx)H

  Notice that X does not need to be a subgroup, so H does not need to be a normal subgroup.

1. is it well defined? Suppose xH = yH ⇒ x^-1y ∈ H ⇒ x^-1g^-1gy ∈ H ⇒ (x^-1g^-1)(gy) ∈ H ⇒ [Shoe-Sock Theorem] (gx)^-1(gy) ∈ H ⇒ gxH = gyH
2. e·xH = (ex)H = xH.
3. g·(h·xH) = g·(hxH) = (gh)xH = (gh)·xH

Theorem. Let X be a G-set. Then define a relation ~ on X, by the rule x ~ y if ∃g ∈ G: g·x = y. Then ~ is an equivalence relation on X, called G-equivalence. The equivalence classes of this relation are called orbits, where O_x is the orbit containing x ∈ X.

Proof.

Recall, a set X equipped with an action by G is called a G-set. Claim: ~ is an equivalence relation on X

• Reflexive. Let x ∈ X, e·x = x where e is the group identity ⇒ ∀x ∈ X, x ~ x
• Symmetric. Let x, y ∈ X such that x ~ y ⇒ ∃g ∈ G: g·x = y.

[G is a group ⇒ ∃g^-1∈G: g^-1g = gg^-1 = e] g^-1·y = g^-1·(g·x) = (g^-1g)·x = e·x = x ⇒ y ~ x.

• Transitive. Let x, y, z ∈ X such that x ~ y, y ~ z ⇒ ∃g, h ∈ G: h·x = y and g·y = z ⇒ (gh)·x = g·(h·x) = g·y = z ⇒ x ~ z.

Example. Let G the permutation group G = ⟨(123), (45)⟩ = ⟨(123)⟩⟨(45)⟩ = {1, (123), (132), (45), (123)(45), (132)(45)} ≤ S₅, and let X = {1, 2, 3, 4, 5}. Then, G acts a permutation on X (σ·x = σ(x) for x ∈S_n)

1. O₁ = O₂ = O₃ = {1, 2, 3} e.g., let σ = (123), x= 1, 1 ~ 2 because σ∈G: σ·1=σ(1)= 2; 2 ~ 3 because σ∈G: σ·2=σ(2)=3. τ = (132), τ·1 = τ(1) = 3, so 1 ~ 3. Therefore, [1]_~ = O₁ = O₂ = O₃ = {1, 2, 3}

2. O₄ = O₅ = {4, 5}.

The Fundamental Counting Principle

The Fundamental Counting Principle. Let G be a group acting on X and x an element of X. Then, the cardinality of an orbit is equal to the index of its stabilizer, that is, |O_x| = [G:G_x]

Recall: The stabilizer of x is the set G_x = { g ∈ G ∣ g ⋅ x = x} , the set of elements of G which leave x unchanged under the action.

Proof.

Reclaim: O_x = {y ∈ X | y = g·x}, G_x = {g ∈ X | g·x = x}

It suffices to show a bijection between O_x and G/G_x. Notice that if y ∈ O_x ⇒ ∃g ∈ X | y = g·x ⇒ g·x denotes an arbitrary element of O_x

Let’s define the map f: O_x → G/G_x by the rule, f(g·x) = gG_x

• f is well-defined. ∀y, y’∈ O_x, y = y’ ⇒ ∃g, g’ such that g·x = h·x ⇒ h^-1·(g·x) = h^-1·(h·x) ⇒ (h^-1g)·x = (h^-1h)·x = e·x = x ⇒ (h^-1g)·x = x ⇒ h^-1g ∈ G_x ⇒ gG_x = hG_x
• f is one-to-one. Suppose f(g·x) = f(h·x) ⇒ gG_x = hG_x ⇒ h^-1g ∈ G_x ⇒ (h^-1g)·x = x ⇒ h·(h^-1g)·x = h·x ⇒ (hh^-1)·(g·x) = h·x ⇒ g·x = h·x
• f is onto. Let gG_x ∈ G/G_x, then f(g·x) = gG_x. Therefore, f: O_x → G/G_x is a bijection and |O_x| = |G/G_x| = [G:G_x]

Remark. If G is a finite group, then |O_x| = [G:G_x] = |G|/|G_x| and therefore, |G| = |O_x||G_x|, and the cardinality of the orbit divides the cardinality of the group.

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