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✓ 28 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 28 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Group Actions, Orbits, Stabilisers and Cayley's Theorem

1 · Prerequisites

2 · Summary

A group action turns the elements of a group into symmetries of a set. Groups, homomorphisms, kernels, quotient groups and isomorphisms supply the declared algebraic prerequisites, while symmetric groups turn actions into permutation representations. Subgroups and cosets provide stabilisers and orbit models; normal subgroups and the isomorphism theorems control action kernels. Finite cardinality, index, Lagrange's theorem and elementary counting support the finite results.

Equivariant maps, free actions and fixed-point sets lead to orbit–stabiliser, coset actions, cores and the classification of transitive actions. Conjugation then yields centralisers, normalisers, conjugacy-class counts, the non-covering theorem and the class equation. For finite $p$ -groups, orbit decomposition gives the fixed-point congruence, index- $p$ normality, nontrivial centres, the order- $p^{2}$ theorem, central intersection for normal subgroups and Cauchy's theorem. Double-counting fixed pairs closes the page with Cauchy–Frobenius orbit counting and Jordan's derangement theorem, while Cayley's theorem remains the regular-action case.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Left group actions, transitive actions, and faithful actions

Definition

Let $G$ be a group with identity $e$ and let $X$ be a set. A left action of $G$ on $X$ is a function $G \times X \to X$ , written $(g, x) \mapsto g \cdot x$ , such that

$e \cdot x = x, (g h) \cdot x = g \cdot (h \cdot x)$

for all $g, h \in G$ and $x \in X$ . Then $X$ is a $G$ -set. The action is transitive when every $x, y \in X$ satisfy $g \cdot x = y$ for some $g \in G$ . It is faithful when $g \cdot x = x$ for every $x \in X$ implies $g = e$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$

Statement

For groups $G$ and a set $X$ , left actions of $G$ on $X$ are in bijection with group homomorphisms $ρ : G \to Sym (X)$ . The action attached to $ρ$ is $g \cdot x : = ρ (g) (x)$ ; the homomorphism attached to an action sends $g$ to the permutation $x \mapsto g \cdot x$ .

Facts & Assumptions

Given: A group $G$ with identity $e$ and a set $X$ .

[L1]

A left action satisfies $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L2]

$Sym (X)$ is the group of bijections $X \to X$ under composition (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L3]

A group homomorphism preserves multiplication and sends the identity to the identity (Monoid homomorphism and group homomorphism, A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

Proof

technique · direct

1.1

Given an action, define $ρ (g) (x) = g \cdot x$ . The maps $ρ (g)$ are bijective: $ρ (g^{- 1})$ is a two-sided inverse because the action laws give $g^{- 1} \cdot (g \cdot x) = x = g \cdot (g^{- 1} \cdot x)$ .

L1givenalgebra

1.2

Conversely, let $ρ : G \to Sym (X)$ be a homomorphism and set $g \cdot x = ρ (g) (x)$ . Then $e \cdot x = ρ (e) (x) = x$ , and $(g h) \cdot x = ρ (g) (ρ (h) (x)) = g \cdot (h \cdot x)$ .

L2L3given

2.1

The action law gives $ρ (g h) (x) = g \cdot (h \cdot x) = (ρ (g) \circ ρ (h)) (x)$ for every $x$ ; thus $ρ (g h) = ρ (g) \circ ρ (h)$ , so $ρ$ is a homomorphism into $Sym (X)$ .

step 1.1L1L2L3given

3.1

The two constructions recover their input pointwise, so they are mutually inverse correspondences.

step 2.1step 1.2∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action

Definition

For a left action of $G$ on $X$ and $x \in X$ , the orbit of $x$ is

$G \cdot x : = {g \cdot x : g \in G},$

and the stabilizer of $x$ is

$G_{x} : = {g \in G : g \cdot x = x} .$

The subgroup claim implicit in the word “stabilizer” is proved in The stabilizer $G_{x}$ is a subgroup of $G$ ↗.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The stabilizer $G_{x}$ is a subgroup of $G$

Statement

For every left action of a group $G$ on a set $X$ and every $x \in X$ , the stabilizer $G_{x}$ is a subgroup of $G$ .

Facts & Assumptions

Given: A left action of $G$ on $X$ and $x \in X$ .

[L1]

$G_{x} = {g \in G : g \cdot x = x}$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L2]

The action satisfies $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L3]

A nonempty subset $H$ of a group is a subgroup exactly when $a b^{- 1} \in H$ for all $a, b \in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

Proof

technique · direct

1.1

The identity lies in $G_{x}$ , since $e \cdot x = x$ . Thus $G_{x}$ is nonempty.

L1L2given

1.2

If $a, b \in G_{x}$ , then $(a b^{- 1}) \cdot x = a \cdot (b^{- 1} \cdot x) = a \cdot x = x$ : indeed $b \cdot x = x$ implies $b^{- 1} \cdot x = b^{- 1} \cdot (b \cdot x) = x$ . Therefore $a b^{- 1} \in G_{x}$ .

L1L2givenalgebra

2.1

The subgroup criterion now gives $G_{x} \leq G$ .

step 1.1step 1.2L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set

Statement

For a left action of $G$ on $X$ , define $x \sim y$ when $y = g \cdot x$ for some $g \in G$ . This is an equivalence relation, its equivalence class at $x$ is $G \cdot x$ , and the distinct orbits partition $X$ .

Facts & Assumptions

Given: A left action of a group $G$ on a set $X$ .

[L1]

The action laws are $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L2]

The orbit at $x$ is $G \cdot x = {g \cdot x : g \in G}$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L3]

Equivalence classes of an equivalence relation partition the underlying set (Equivalence relation, equivalence class, and the quotient set $A / \sim$ , The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

Proof

technique · direct

1.1

The relation is reflexive: $x = e \cdot x$ , so $x \sim x$ .

L1given

1.2

If $y = g \cdot x$ , then $x = g^{- 1} \cdot y$ , so $x \sim y$ implies $y \sim x$ .

L1givenalgebra

1.3

If $y = g \cdot x$ and $z = h \cdot y$ , then $z = (h g) \cdot x$ , so $x \sim y$ and $y \sim z$ imply $x \sim z$ .

L1givenalgebra

2.1

Steps 1.1–1.3 show that $\sim$ is an equivalence relation. Its class at $x$ is precisely the set of $y = g \cdot x$ , namely $G \cdot x$ .

step 1.1step 1.2step 1.3L2L3

3.1

Therefore the distinct orbits partition $X$ .

step 2.1L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Cayley's theorem: every group $G$ is isomorphic to a subgroup of $Sym (G)$

Statement

Every group $G$ is isomorphic to the subgroup of $Sym (G)$ formed by its left translations $λ_{g} : x \mapsto g x$ .

Facts & Assumptions

Given: A group $G$ with identity $e$ .

[L1]

A left action of $G$ on a set gives a homomorphism into its symmetric group (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ , Left group actions, transitive actions, and faithful actions).

[L2]

The image of a group homomorphism is a subgroup, and a homomorphism is injective exactly when its kernel is trivial (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, A group homomorphism is injective if and only if its kernel is trivial).

[L3]

A bijective group homomorphism is an isomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1

Define $g \cdot x = g x$ on the set underlying $G$ . Then $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ , so this is a left action.

L1givenalgebra

2.1

By [L1], the action yields a homomorphism $λ : G \to Sym (G)$ with $λ (g) (x) = g x$ .

step 1.1L1

3.1

If $λ (g)$ is the identity permutation, then evaluating it at $e$ gives $g = λ (g) (e) = e$ . Hence $\ker λ = {e}$ and $λ$ is injective.

step 2.1L2given

4.1

The image $λ [G]$ is a subgroup of $Sym (G)$ , and the injective homomorphism $λ : G \to λ [G]$ is bijective.

step 2.1step 3.1L2

5.1

Thus $λ$ is an isomorphism from $G$ to a subgroup of $Sym (G)$ .

step 4.1L3∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Equivariant maps and isomorphisms of group actions

Definition

Let $X$ and $Y$ be $G$ -sets (Left group actions, transitive actions, and faithful actions). A function $f : X \to Y$ (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain) is $G$ -equivariant when

$f (g \cdot x) = g \cdot f (x)$

for every $g \in G$ and $x \in X$ . An isomorphism of $G$ -sets is an equivariant bijection (Injection, surjection, bijection). Two actions are equivariantly isomorphic when such a bijection exists between their $G$ -sets.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

A free group action has no nonidentity element fixing a point

Definition

A left action of a group $G$ on a set $X$ (Left group actions, transitive actions, and faithful actions) is free when

$g \cdot x = x ⟹ g = e$

for every $g \in G$ and $x \in X$ . Equivalently, no nonidentity element of $G$ fixes any point of $X$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

The fixed-point sets $X^{g}$ and $X^{G}$ of a group action

Definition

Let a group $G$ act on a set $X$ (Left group actions, transitive actions, and faithful actions). For $g \in G$ , the fixed-point set of $g$ is

$X^{g} : = {x \in X : g \cdot x = x} .$

The global fixed-point set is

$X^{G} : = {x \in X : g \cdot x = x for every g \in G} = ⋂_{g \in G} X^{g} .$

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection

Statement

Let $G$ act on $X$ and let $x \in X$ . The rule

$Φ : G / G_{x} ⟶ G \cdot x, Φ (g G_{x}) = g \cdot x,$

is well-defined and bijective. Thus every orbit is naturally in bijection with the left cosets of its stabilizer.

Facts & Assumptions

Given: A left action of a group $G$ on a set $X$ and a point $x \in X$ .

[L1]

The orbit and stabilizer are $G \cdot x = {g \cdot x : g \in G}$ and $G_{x} = {g \in G : g \cdot x = x}$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L2]

The stabilizer $G_{x}$ is a subgroup of $G$ (The stabilizer $G_{x}$ is a subgroup of $G$ ).

[L3]

For a subgroup $H \leq G$ , the left coset represented by $g$ is $g H = {g h : h \in H}$ (Left and right cosets $g H$ and $H g$ of a subgroup).

[L4]

For $H \leq G$ , one has $g H = h H$ exactly when $g^{- 1} h \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L5]

A function is bijective exactly when it is injective and surjective (Injection, surjection, bijection).

Proof

technique · constructive

1.1

Define $Φ (g G_{x}) = g \cdot x$ . If $g G_{x} = h G_{x}$ , then $g^{- 1} h \in G_{x}$ by [L4], so $(g^{- 1} h) \cdot x = x$ by [L1], and the action law gives $h \cdot x = g \cdot ((g^{- 1} h) \cdot x) = g \cdot x$ ; hence $Φ$ is well-defined.

L1L2L3L4L5construct

2.1

Every $y \in G \cdot x$ has the form $y = g \cdot x = Φ (g G_{x})$ by [L1], so $Φ$ is surjective.

step 1.1L1

3.1

If $Φ (g G_{x}) = Φ (h G_{x})$ , then $g \cdot x = h \cdot x$ , so $(g^{- 1} h) \cdot x = x$ and $g^{- 1} h \in G_{x}$ ; [L4] gives $g G_{x} = h G_{x}$ , so $Φ$ is injective and therefore bijective.

step 1.1L1L4L5discharge-construct∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$

Statement

For an action of $G$ on $X$ and $x \in X$ ,

$∣ G \cdot x ∣ = [G : G_{x}]$

whenever either side is finite. In particular, if $G$ is finite, then

$∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣ .$

Facts & Assumptions

Given: A left action of $G$ on $X$ and a point $x \in X$ .

[L1]

The map $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a bijection (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L2]

The index is the finite cardinality $[G : H] = ∣ G / H ∣$ when the coset set is finite (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L3]

Finite cardinality is preserved by a bijection (The cardinality $∣ A ∣$ of a finite set).

[L4]

If $G$ is finite and $H \leq G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

By [L1], the sets $G / G_{x}$ and $G \cdot x$ are bijective; [L2] and [L3] therefore give $∣ G \cdot x ∣ = ∣ G / G_{x} ∣ = [G : G_{x}]$ whenever they are finite.

L1L2L3

2.1

If $G$ is finite, [L4] applied to $G_{x} \leq G$ gives $∣ G ∣ = [G : G_{x}] ∣ G_{x} ∣ = ∣ G \cdot x ∣ ∣ G_{x} ∣$ .

step 1.1L4algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

If $y = g \cdot x$ , then $G_{y} = g G_{x} g^{- 1}$

Statement

Let $G$ act on $X$ . If $y = g \cdot x$ , then

$G_{y} = g G_{x} g^{- 1} .$

In particular, stabilizers of points in the same orbit are conjugate and hence isomorphic.

Facts & Assumptions

Given: A left action of $G$ on $X$ , points $x, y \in X$ , and $g \in G$ with $y = g \cdot x$ .

[L1]

The stabilizer is $G_{x} = {h \in G : h \cdot x = x}$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L2]

A left action satisfies $(a b) \cdot z = a \cdot (b \cdot z)$ and $e \cdot z = z$ (Left group actions, transitive actions, and faithful actions).

[L3]

Conjugation $h \mapsto g h g^{- 1}$ is an automorphism of $G$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

Proof

technique · direct

1.1

For $h \in G$ , one has $h \in G_{y}$ exactly when $h \cdot (g \cdot x) = g \cdot x$ , which by [L2] is equivalent, after applying $g^{- 1}$ , to $(g^{- 1} h g) \cdot x = x$ , that is, to $g^{- 1} h g \in G_{x}$ .

L1L2L3

2.1

The last condition is equivalent to $h \in g G_{x} g^{- 1}$ , so $G_{y} = g G_{x} g^{- 1}$ ; [L3] also makes conjugation an isomorphism from $G_{x}$ onto $G_{y}$ .

step 1.1L3algebra∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup

Definition

Let $H \leq G$ be a subgroup (Subgroup). Its core in $G$ is

${Core}_{G} (H) : = ⋂_{g \in G} g H g^{- 1} .$

Each $g H g^{- 1}$ is a subgroup because conjugation is an automorphism (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism). The facts that the displayed intersection is normal in $G$ and is the largest normal subgroup of $G$ contained in $H$ are proved in ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ↗.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$

Statement

For $H \leq G$ , the core $K = {Core}_{G} (H)$ is a normal subgroup of $G$ , satisfies $K \leq H$ , and contains every normal subgroup of $G$ that is contained in $H$ . Thus it is the largest normal subgroup of $G$ contained in $H$ .

Facts & Assumptions

Given: A group $G$ , a subgroup $H \leq G$ , and $K : = {Core}_{G} (H)$ .

[L1]

The core is $K = ⋂_{g \in G} g H g^{- 1}$ (The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup).

[L2]

A subgroup $N$ is normal when $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

[L3]

Normality is equivalent to $g N g^{- 1} \subseteq N$ for every $g \in G$ (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

[L4]

A nonempty subset $S$ of a group is a subgroup if $a b^{- 1} \in S$ for every $a, b \in S$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

Proof

technique · direct

1.1

The identity belongs to every $g H g^{- 1}$ , and if $a, b$ belong to every such conjugate then $a b^{- 1}$ does too; [L4] makes $K$ a subgroup. The factor for $g = e$ is $H$ , so $K \leq H$ .

L1L4

1.2

For $t \in G$ , conjugation sends the family ${g H g^{- 1} : g \in G}$ to ${(t g) H (t g)^{- 1} : g \in G}$ , the same family because $g \mapsto t g$ is a bijection; hence $t K t^{- 1} = K$ , and [L2] gives $K ⊴ G$ .

L1L2L3

2.1

If $N ⊴ G$ and $N \leq H$ , then $N = g N g^{- 1} \leq g H g^{- 1}$ for every $g \in G$ , so $N \leq ⋂_{g} g H g^{- 1} = K$ .

L1L2L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$

Statement

Let $H \leq G$ . Left multiplication defines a transitive action of $G$ on the coset set $G / H$ by

$g \cdot (a H) : = (g a) H .$

The stabilizer of the point $H$ is $H$ . The corresponding homomorphism $ρ : G \to Sym (G / H)$ has

$\ker ρ = {Core}_{G} (H) .$

Facts & Assumptions

Given: A group $G$ and a subgroup $H \leq G$ .

[L1]

A left action satisfies the identity and multiplication laws and is transitive when some group element carries any chosen point to any other (Left group actions, transitive actions, and faithful actions).

[L2]

Every action yields a homomorphism into the symmetric group of the acted-on set (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ).

[L3]

The elements of $G / H$ are the left cosets $a H$ (Left and right cosets $g H$ and $H g$ of a subgroup, The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L4]

One has $a H = b H$ exactly when $a^{- 1} b \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L5]

The kernel of a homomorphism consists of the elements mapped to the identity (The kernel and image of a group homomorphism).

[L6]

The core is ${Core}_{G} (H) = ⋂_{a \in G} a H a^{- 1}$ (The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup).

[L7]

The core is a normal subgroup contained in $H$ ( ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ).

Proof

technique · direct

1.1

If $a H = b H$ , then $a^{- 1} b \in H$ by [L4], and $(g a)^{- 1} (g b) = a^{- 1} b \in H$ , so $(g a) H = (g b) H$ and the rule is well-defined. It satisfies $e \cdot (a H) = a H$ and $g \cdot (k \cdot (a H)) = (g k) a H = (g k) \cdot (a H)$ ; moreover $a \cdot H = a H$ , so the action is transitive, and $g \cdot H = H$ exactly when $g \in H$ .

L1L3L4

2.1

By [L2], the action defines $ρ : G \to Sym (G / H)$ . By [L5], an element $k$ lies in $\ker ρ$ exactly when $k \cdot (a H) = a H$ for every $a \in G$ , that is, when $(k a) H = a H$ for every $a$ .

step 1.1L2L5

3.1

By [L4], $(k a) H = a H$ is equivalent to $a^{- 1} k a \in H$ , or to $k \in a H a^{- 1}$ . Requiring this for every $a$ gives $k \in ⋂_{a} a H a^{- 1} = {Core}_{G} (H)$ , so $\ker ρ = {Core}_{G} (H)$ , which is normal by [L7].

step 2.1L4L6L7∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

If $[G : H] = n < \infty$ , then ${Core}_{G} (H) ⊴ G$ , $[G : {Core}_{G} (H)] ∣ n!$ , and only finitely many subgroups contain $H$

Statement

Let $H \leq G$ have finite index $[G : H] = n$ , and put $N = {Core}_{G} (H)$ . Then $N ⊴ G$ , the index $[G : N]$ divides $n!$ , and there are only finitely many subgroups $K$ with $H \leq K \leq G$ .

Facts & Assumptions

Given: A group $G$ , a subgroup $H \leq G$ of finite index $n$ , and $N : = {Core}_{G} (H)$ .

[L1]

The action on $G / H$ gives a homomorphism $ρ : G \to Sym (G / H)$ whose kernel is $N$ (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

[L2]

The core $N$ is normal in $G$ , lies in $H$ , and contains every normal subgroup of $G$ lying in $H$ ( ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ).

[L3]

First isomorphism gives $G / \ker ρ ≅ im ρ$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L4]

The symmetric group of a set is its group of bijections (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L5]

A set with $n$ elements has exactly $n!$ bijections to itself (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality, The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L6]

The order of a subgroup of a finite group divides the order of the group (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L7]

The canonical projection $π : G \to G / N$ is a surjective homomorphism (The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L8]

The power set of a finite set is finite ( $∣ P (A) ∣ = 2^{∣ A ∣}$ for finite $A$ ).

[L9]

Every subset of a finite set is finite (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L10]

For a finite-index subgroup, the index is the cardinality of its coset set (The coset set $G / H$ and the index $[G : H]$ of a subgroup, The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

By [L1] and [L2], the coset action has kernel $N ⊴ G$ with $N \leq H$ , and its image is a subgroup of $Sym (G / H)$ .

L1L2L4

2.1

By [L3], $G / N ≅ im ρ$ . The set $G / H$ has $n$ elements by [L10], so [L5] gives $∣ Sym (G / H) ∣ = n!$ ; [L6] therefore gives $∣ G / N ∣ = ∣ im ρ ∣ ∣ n!$ , that is, $[G : N] ∣ n!$ by [L10].

step 1.1L3L4L5L6L10

3.1

Every subgroup $K$ containing $H$ also contains $N$ by [L2]. If $π (x) \in π [K]$ , then $π (x) = π (k)$ for some $k \in K$ , so $k^{- 1} x \in N \leq K$ and hence $x \in K$ ; thus $K = π^{- 1} [π [K]]$ . Therefore $K \mapsto π [K]$ injects the set of such overgroups into the power set of the now known finite set $G / N$ , which is finite by [L8] and [L9].

step 2.1L2L7L8L9

4.1

Thus the core is normal and finite-index, its index divides $n!$ , and the collection of subgroups containing $H$ is finite.

step 1.1step 2.1step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every transitive $G$ -set is equivariantly isomorphic to $G / G_{x}$ for any chosen point $x$

Statement

Let $X$ be a transitive $G$ -set and choose $x \in X$ . Then the orbit map

$G / G_{x} ⟶ X, g G_{x} ⟼ g \cdot x,$

is an equivariant isomorphism from the left-coset action on $G / G_{x}$ to the given action on $X$ .

Facts & Assumptions

Given: A transitive action of a group $G$ on a nonempty set $X$ and a point $x \in X$ .

[L1]

An isomorphism of $G$ -sets is an equivariant bijection (Equivariant maps and isomorphisms of group actions).

[L2]

Transitivity means that for every $y \in X$ there is $g \in G$ with $g \cdot x = y$ (Left group actions, transitive actions, and faithful actions).

[L3]

The map $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a bijection (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L4]

The left-coset action is $a \cdot (g G_{x}) = (a g) G_{x}$ (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

Proof

technique · constructive

1.1

Define $Φ (g G_{x}) = g \cdot x$ . By [L3], this is a bijection from $G / G_{x}$ onto $G \cdot x$ .

L1L3construct

2.1

For $a, g \in G$ , one has $Φ (a \cdot (g G_{x})) = Φ ((a g) G_{x}) = (a g) \cdot x = a \cdot (g \cdot x) = a \cdot Φ (g G_{x})$ , so $Φ$ is equivariant.

step 1.1L1L4

3.1

By transitivity [L2], $G \cdot x = X$ . Thus $Φ : G / G_{x} \to X$ is an equivariant bijection and hence an isomorphism of $G$ -sets.

step 1.1step 2.1L1L2discharge-construct∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

A transitive action is faithful exactly when a point stabiliser is core-free

Statement

Let $G$ act transitively on a nonempty set $X$ , and let $x \in X$ . The action is faithful if and only if

${Core}_{G} (G_{x}) = {e} .$

Equivalently, faithful transitive $G$ -sets are precisely the coset actions $G / H$ for core-free subgroups $H$ .

Facts & Assumptions

Given: A transitive action of $G$ on a nonempty set $X$ and a point $x \in X$ .

[L1]

An action is faithful when the only group element fixing every point is the identity (Left group actions, transitive actions, and faithful actions).

[L2]

The action on $X$ is equivariantly isomorphic to the left-coset action on $G / G_{x}$ (Every transitive $G$ -set is equivariantly isomorphic to $G / G_{x}$ for any chosen point $x$ ).

[L3]

The kernel of the action on $G / H$ is ${Core}_{G} (H)$ (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

[L4]

The core is the intersection of all conjugates of the subgroup (The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup).

Proof

technique · direct

1.1

By [L2], the equivariant isomorphism identifies the given action with the action on $G / G_{x}$ .

2.1

A group element fixes every point of $X$ exactly when it fixes every point of the equivariantly isomorphic coset set; by [L3] and [L4], the set of such elements is ${Core}_{G} (G_{x})$ .

step 1.1L3L4

3.1

By [L1], the action is faithful exactly when this kernel is ${e}$ , proving both directions.

step 2.1L1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element

Definition

Let $G$ be a group and $x \in G$ (Group and abelian group). The conjugacy class of $x$ is

${Cl}_{G} (x) : = {g x g^{- 1} : g \in G} .$

The centralizer of $x$ is

$C_{G} (x) : = {g \in G : g x = x g} = {g \in G : g x g^{- 1} = x} .$

The subgroup property implicit in the notation is proved in $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ↗.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup

Definition

Let $H \leq G$ be a subgroup (Subgroup). The normalizer of $H$ in $G$ is

$N_{G} (H) : = {g \in G : g H g^{- 1} = H} .$

Thus $g \in N_{G} (H)$ exactly when the conjugation automorphism $c_{g}$ preserves $H$ setwise (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism). The subgroup property is proved in $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ↗.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

$C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$

Statement

For every group $G$ , element $x \in G$ , and subgroup $H \leq G$ , both the centralizer $C_{G} (x)$ and the normalizer $N_{G} (H)$ are subgroups of $G$ .

Facts & Assumptions

Given: A group $G$ , an element $x \in G$ , and a subgroup $H \leq G$ .

[L1]

The centralizer is $C_{G} (x) = {g \in G : g x = x g}$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[L2]

The normalizer is $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L3]

A nonempty subset $S$ is a subgroup if $a b^{- 1} \in S$ whenever $a, b \in S$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

Proof

technique · direct

1.1

The identity commutes with $x$ . If $a, b \in C_{G} (x)$ , then $b^{- 1}$ commutes with $x$ , and hence $(a b^{- 1}) x = a (b^{- 1} x) = a (x b^{- 1}) = x (a b^{- 1})$ ; [L3] gives $C_{G} (x) \leq G$ .

L1L3

1.2

The identity normalizes $H$ . If $a, b \in N_{G} (H)$ , then $b^{- 1} H b = H$ , and therefore $(a b^{- 1}) H (a b^{- 1})^{- 1} = a (b^{- 1} H b) a^{- 1} = a H a^{- 1} = H$ .

2.1

Applying [L3] to step 1.2 gives $N_{G} (H) \leq G$ , so both asserted sets are subgroups.

step 1.1step 1.2L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

$G / C_{G} (x) \to {Cl}_{G} (x)$ is a bijection, so $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever these cardinalities are finite

Statement

For a group $G$ and $x \in G$ , the map

$G / C_{G} (x) ⟶ {Cl}_{G} (x), g C_{G} (x) ⟼ g x g^{- 1},$

is a well-defined bijection. Consequently

$∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$

whenever these cardinalities are finite, in particular when $G$ is finite.

Facts & Assumptions

Given: A group $G$ and an element $x \in G$ .

[L1]

Orbit-stabiliser gives a bijection from the cosets of a point stabilizer to its orbit (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L2]

The finite cardinality of an orbit is the index of its stabilizer (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L3]

The conjugacy class is ${Cl}_{G} (x) = {g x g^{- 1} : g \in G}$ and the centralizer is $C_{G} (x) = {g : g x g^{- 1} = x}$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[L4]

The centralizer $C_{G} (x)$ is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L5]

The maps $x \mapsto g x g^{- 1}$ form the conjugation homomorphism (The map $g \mapsto (x \mapsto g x g^{- 1})$ is a homomorphism $G \to Aut (G)$ with kernel $Z (G)$ and image $Inn (G)$ ).

[L6]

A homomorphism into a symmetric group defines a group action (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ).

Proof

technique · direct

1.1

By [L5] and [L6], $G$ acts on itself by conjugation. By [L3], the orbit of $x$ is ${Cl}_{G} (x)$ and its stabilizer is $C_{G} (x)$ , which is a subgroup by [L4].

L3L4L5L6

2.1

Applying [L1] to this action gives the displayed well-defined bijection $g C_{G} (x) \mapsto g x g^{- 1}$ .

step 1.1L1

3.1

Applying [L2] to the same orbit gives $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever finite.

step 1.1step 2.1L2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$

Statement

Let $H \leq G$ . The rule

$G / N_{G} (H) ⟶ {g H g^{- 1} : g \in G}, g N_{G} (H) ⟼ g H g^{- 1},$

is a well-defined bijection. If $G$ is finite, the number of distinct conjugates of $H$ is $[G : N_{G} (H)]$ .

Facts & Assumptions

Given: A group $G$ and a subgroup $H \leq G$ .

[L1]

Orbit-stabiliser identifies an orbit with the cosets of its stabilizer (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L2]

The cardinality of a finite orbit is the index of its stabilizer (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L3]

The normalizer is $N_{G} (H) = {g : g H g^{- 1} = H}$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L4]

The normalizer is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L5]

Conjugation by each $g \in G$ is an automorphism of $G$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

Proof

technique · direct

1.1

Let $G$ act on the set of subgroups of $G$ by $g \cdot K = g K g^{- 1}$ . By [L5], conjugation sends subgroups to subgroups, and the conjugation identities give the action laws.

2.1

The orbit of $H$ is its set of conjugates, while [L3] says that its stabilizer is $N_{G} (H)$ , a subgroup by [L4].

step 1.1L3L4

3.1

Applying [L1] gives the displayed bijection, and [L2] gives the finite count $[G : N_{G} (H)]$ .

step 2.1L1L2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

The conjugates of a proper subgroup do not cover a finite group

Statement

If $H$ is a proper subgroup of a finite group $G$ , then

$⋃_{g \in G} g H g^{- 1} \neq G .$

Thus some element of $G$ lies in no conjugate of $H$ .

Facts & Assumptions

Given: A finite group $G$ and a proper subgroup $H < G$ .

[L1]

There are $m = [G : N_{G} (H)]$ distinct conjugates of $H$ (The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$ ).

[L2]

The normalizer is $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L3]

The normalizer is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L4]

Conjugation is an automorphism, so every conjugate of $H$ has cardinality $∣ H ∣$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[L5]

For a finite group and subgroup, $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The coset set $G / H$ and the index $[G : H]$ of a subgroup, The cardinality $∣ A ∣$ of a finite set).

[L6]

A subset of a finite set is finite, has no larger cardinality, and has equal cardinality only when it is the whole set (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L7]

The cardinality of a finite disjoint union is the sum of the cardinalities of its parts (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L8]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Proof

technique · direct

1.1

Let $K_{1}, \dots, K_{m}$ be the distinct conjugates of $H$ , where $m = [G : N_{G} (H)]$ by [L1]. Each $K_{i}$ has $∣ H ∣$ elements by [L4] and contains $e$ . For $h \in H$ , subgroup closure gives $h H h^{- 1} = H$ , so [L2] gives $H \leq N_{G} (H)$ .

L1L2L3L4

2.1

Add the sets successively after removing elements already counted. The common identity contributes once and each $K_{i} ∖ {e}$ contributes at most $∣ H ∣ - 1$ , so [L6], [L7], and [L8] give $∣ ⋃_{i} K_{i} ∣ \leq 1 + m (∣ H ∣ - 1)$ .

step 1.1L6L7L8

2.2

Put $n = [G : H]$ . Properness gives $n \geq 2$ . Since $H \leq N_{G} (H)$ , [L5] gives $m \leq n$ , and [L5] also gives $∣ G ∣ = n ∣ H ∣$ .

step 1.1L2L3L5L6

3.1

Therefore $∣ ⋃_{i} K_{i} ∣ \leq 1 + n (∣ H ∣ - 1) = ∣ G ∣ - n + 1 < ∣ G ∣$ , so the union is a proper subset of $G$ .

step 2.1step 2.2algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

The class equation $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i} [G : C_{G} (x_{i})]$ for a finite group

Statement

Let $G$ be finite, and let $x_{1}, \dots, x_{r}$ contain one representative from each conjugacy class having more than one element. Then

$∣ G ∣ = ∣ Z (G) ∣ + \sum_{i = 1}^{r} [G : C_{G} (x_{i})] .$

Facts & Assumptions

Given: A finite group $G$ and representatives $x_{1}, \dots, x_{r}$ of its non-singleton conjugacy classes.

[L1]

The orbits of an action partition the acted-on set (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set).

[L2]

Under conjugation, $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ ( $G / C_{G} (x) \to {Cl}_{G} (x)$ is a bijection, so $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever these cardinalities are finite).

[L3]

Conjugacy classes and centralizers are as in The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element.

[L4]

The center is $Z (G) = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L5]

A finite partition has cardinality equal to the sum of its block cardinalities (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L6]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Proof

technique · direct

1.1

Let $G$ act on itself by conjugation. By [L1] and [L3], its orbits are the conjugacy classes and they partition $G$ .

L1L3

2.1

The class of $x$ is a singleton exactly when $g x g^{- 1} = x$ for every $g$ , equivalently when $x \in Z (G)$ by [L4]. Thus the singleton classes contribute $∣ Z (G) ∣$ .

step 1.1L3L4

3.1

Applying the finite partition sum rule to the singleton classes and to the classes represented by $x_{1}, \dots, x_{r}$ gives $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i = 1}^{r} ∣ {Cl}_{G} (x_{i}) ∣$ .

step 1.1step 2.1L5L6

4.1

Replacing each remaining class size by [L2] yields $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i = 1}^{r} [G : C_{G} (x_{i})]$ .

step 3.1L2∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$

Definition

Let $p$ be a prime natural number (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ). A finite $p$ -group is a finite group $P$ (Group and abelian group, The cardinality $∣ A ∣$ of a finite set) whose order has the form

$∣ P ∣ = p^{n}$

for some $n \in N$ , with natural exponentiation as in Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ . The case $n = 0$ permits the trivial group. A finite $p$ -group is nontrivial exactly when $n \geq 1$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every subgroup of a finite $p$ -group has order a power of $p$

Statement

If $P$ is a finite $p$ -group and $H \leq P$ , then $H$ is finite and

$∣ H ∣ = p^{k}$

for some $k \in N$ . If $∣ P ∣ = p^{n}$ , then $k \leq n$ .

Facts & Assumptions

Given: A finite $p$ -group $P$ with $∣ P ∣ = p^{n}$ and a subgroup $H \leq P$ .

[L1]

A finite $p$ -group has order $p^{n}$ for a prime $p$ and a natural $n$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

Lagrange gives $∣ P ∣ = [P : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L3]

Positive integers have unique prime factorisations (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1

By [L2], the finite set $H$ has positive order dividing $p^{n}$ .

L1L2

2.1

By uniqueness in [L3], no prime other than $p$ can divide $∣ H ∣$ , so $∣ H ∣ = p^{k}$ for some natural $k \leq n$ .

step 1.1L3

3.1

This includes the trivial subgroup, whose order is $1 = p^{0}$ , and proves that every subgroup of $P$ is a finite $p$ -group.

step 2.1algebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every subgroup of index $p$ in a finite $p$ -group is normal

Statement

Let $P$ be a finite $p$ -group and let $H \leq P$ . If $[P : H] = p$ , then $H ⊴ P$ .

Facts & Assumptions

Given: A finite $p$ -group $P$ and a subgroup $H \leq P$ with $[P : H] = p$ .

[L1]

For $K = {Core}_{P} (H)$ , one has $K ⊴ P$ , $K \leq H$ , and $[P : K] ∣ p!$ (If $[G : H] = n < \infty$ , then ${Core}_{G} (H) ⊴ G$ , $[G : {Core}_{G} (H)] ∣ n!$ , and only finitely many subgroups contain $H$ , ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ).

[L2]

Every subgroup of $P$ has order a power of $p$ (Every subgroup of a finite $p$ -group has order a power of $p$ ).

[L3]

If $K \leq H \leq P$ , then $[P : K] = [P : H] [H : K]$ (For $K \leq H \leq G$ with $G$ finite, $[G : K] = [G : H] [H : K]$ ).

[L4]

The factorial is the product of the positive natural numbers at most $p$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L5]

If a prime divides a finite product, it divides one of the factors (If a prime $p$ divides a finite product $\prod_{i < n} a_{i}$ of integers then $p ∣ a_{i}$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold).

[L6]

Prime factorisation is unique (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

[L7]

A prime $p$ is greater than $1$ and has no positive divisors other than $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L8]

For a subgroup of a finite group, the subgroup order divides the group order and the quotient is the index (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Put $K = {Core}_{P} (H)$ . By [L1] and [L3], $K ⊴ P$ , $K \leq H$ , $[P : K] ∣ p!$ , and $[P : K] = p [H : K]$ .

L1L3

2.1

By [L2] and [L8], the orders of $P$ and $K$ are powers of $p$ and their quotient is $[P : K]$ ; [L6] therefore makes $[P : K]$ a positive power of $p$ , and step 1.1 makes it divisible by $p$ .

step 1.1L2L6L8

3.1

Among the factors $1, \dots, p$ in $p!$ , only $p$ is divisible by $p$ by [L7]. If $p^{2}$ divided $p!$ , cancellation of the factor $p$ and [L5] would make $p$ divide one of $1, \dots, p - 1$ , impossible. Thus the positive power of $p$ in step 2.1 that divides $p!$ is exactly $p$ .

step 1.1step 2.1L4L5L7

4.1

Step 1.1 now gives $p = p [H : K]$ , so $[H : K] = 1$ and $H = K$ . Since $K$ is normal in $P$ , so is $H$ .

step 1.1step 3.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$

Statement

If a finite $p$ -group $P$ acts on a finite set $X$ , then

$∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p) .$

Facts & Assumptions

Given: A finite $p$ -group $P$ acting on a finite set $X$ .

[L1]

A finite $p$ -group has prime-power order (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

The global fixed-point set is $X^{P} = {x : g \cdot x = x for every g \in P}$ (The fixed-point sets $X^{g}$ and $X^{G}$ of a group action).

[L3]

The $P$ -orbits partition $X$ (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set).

[L4]

An orbit has size $∣ P \cdot x ∣ = [P : P_{x}]$ (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L5]

Every subgroup of $P$ has prime-power order (Every subgroup of a finite $p$ -group has order a power of $p$ ).

[L6]

The congruence $a \equiv b (m o d p)$ means that $p$ divides $a - b$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L7]

A finite partition has total cardinality equal to the sum of its block cardinalities (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L8]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Proof

technique · direct

1.1

By [L3], $X$ is the disjoint union of its $P$ -orbits. An orbit is a singleton exactly when its point is fixed by every element of $P$ , so the singleton orbits are indexed by $X^{P}$ .

L2L3

1.2

For a non-singleton orbit $P \cdot x$ , the stabilizer $P_{x}$ is proper. By [L1], [L4], and [L5], its index is a positive power of $p$ , so $p$ divides $∣ P \cdot x ∣$ .

L1L4L5

2.1

Applying [L7] and [L8] to the orbit partition, every non-singleton orbit contributes a multiple of $p$ and the singleton orbits contribute $∣ X^{P} ∣$ . Thus $p$ divides $∣ X ∣ - ∣ X^{P} ∣$ , which is the asserted congruence by [L6].

step 1.1step 1.2L6L7L8∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

A finite $p$ -group action on $X$ has a global fixed point whenever $p ∤ ∣ X ∣$

Statement

Let a finite $p$ -group $P$ act on a finite set $X$ . If $p ∤ ∣ X ∣$ , then $X^{P} \neq \emptyset$ ; equivalently, the action has a point fixed by every element of $P$ .

Facts & Assumptions

Given: A finite $p$ -group $P$ acting on a finite set $X$ , with $p ∤ ∣ X ∣$ .

[L1]

The fixed-point congruence gives $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L2]

The set $X^{P}$ consists of the points fixed by every element of $P$ (The fixed-point sets $X^{g}$ and $X^{G}$ of a group action).

[L3]

Congruence modulo $p$ means divisibility of the difference by $p$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L4]

A prime is positive and greater than $1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $X^{P} = \emptyset$ . Then $∣ X^{P} ∣ = 0$ .

assume-contraL2

2.1

By [L1] and [L3], $p$ divides $∣ X ∣ - ∣ X^{P} ∣ = ∣ X ∣$ , contradicting the hypothesis.

step 1.1L1L3

3.1

Therefore $X^{P}$ is nonempty, and any of its elements is a global fixed point by [L2].

step 2.1L2L4discharge-contradiction∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$

Statement

If $P$ is a nontrivial finite $p$ -group, then

$p ∣ ∣ Z (P) ∣ .$

In particular, the center $Z (P)$ contains a nonidentity element.

Facts & Assumptions

Given: A nontrivial finite $p$ -group $P$ .

[L1]

Nontriviality means $∣ P ∣ = p^{n}$ with $n \geq 1$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

For an action of $P$ on a finite set $X$ , $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L3]

Conjugation gives a homomorphism $P \to Aut (P)$ (The map $g \mapsto (x \mapsto g x g^{- 1})$ is a homomorphism $G \to Aut (G)$ with kernel $Z (G)$ and image $Inn (G)$ ).

[L4]

A homomorphism into a symmetric group defines an action (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ).

[L5]

The center is the set of elements commuting with every element of $P$ (The center $Z (G)$ of a group).

[L6]

Congruence modulo $p$ means divisibility of the difference by $p$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

Proof

technique · direct

1.1

By [L3] and [L4], $P$ acts on itself by conjugation. An element is fixed by all conjugations exactly when it lies in $Z (P)$ by [L5].

L3L4L5

2.1

Applying [L2] to this action gives $∣ P ∣ \equiv ∣ Z (P) ∣ (m o d p)$ .

step 1.1L2

3.1

By [L1], $p$ divides $∣ P ∣$ . Hence [L6] and step 2.1 show that $p$ divides $∣ Z (P) ∣$ . Since $Z (P)$ contains the identity and its cardinality is a positive multiple of $p > 1$ , it also contains a nonidentity element.

step 2.1L1L5L6algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

If $G / Z (G)$ is cyclic, then $G$ is abelian

Statement

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian.

Facts & Assumptions

Given: A group $G$ such that $G / Z (G)$ is cyclic.

[L1]

The center $Z (G)$ consists of the elements commuting with every element of $G$ (The center $Z (G)$ of a group).

[L2]

The center is a normal subgroup of $G$ (The center of a group is a normal subgroup).

[L3]

Multiplication in $G / Z (G)$ is $(g Z (G)) (h Z (G)) = g h Z (G)$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L4]

A cyclic group is generated by one element (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L5]

The subgroup generated by an element is the set of its integer powers ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L6]

Powers of one group element commute with one another (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L7]

Equality of left cosets $a H = b H$ is equivalent to $a^{- 1} b \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

Proof

technique · direct

1.1

Choose $a Z (G)$ generating $G / Z (G)$ . By [L3]--[L6], arbitrary $x, y \in G$ satisfy $x Z (G) = a^{m} Z (G)$ and $y Z (G) = a^{n} Z (G)$ for some integers $m, n$ . By [L7], $a^{- m} x$ and $a^{- n} y$ lie in $Z (G)$ ; setting $z = a^{- m} x$ and $w = a^{- n} y$ gives $x = a^{m} z$ and $y = a^{n} w$ .

L1L2L3L4L5L6L7choose

1.2

The elements $z, w$ commute with every element by [L1], and $a^{m}$ commutes with $a^{n}$ by [L6].

L1L6

2.1

Therefore $x y = a^{m} z a^{n} w = a^{m + n} z w = a^{n + m} w z = a^{n} w a^{m} z = y x$ . Since $x, y$ were arbitrary, $G$ is abelian.

step 1.1step 1.2L6algebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every group of order $p^{2}$ , for prime $p$ , is abelian

Statement

If $p$ is prime and $G$ is a group of order $p^{2}$ , then $G$ is abelian.

Facts & Assumptions

Given: A prime $p$ and a finite group $G$ with $∣ G ∣ = p^{2}$ .

[L1]

A nontrivial finite $p$ -group has nontrivial center (Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$ ).

[L2]

If $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L3]

For finite $G$ , $∣ G / Z (G) ∣ = ∣ G ∣ / ∣ Z (G) ∣$ (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[L4]

A group of prime order is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it).

[L5]

A group of order $p^{2}$ is a finite $p$ -group (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L6]

Every subgroup of a finite $p$ -group has prime-power order (Every subgroup of a finite $p$ -group has order a power of $p$ ).

[L7]

The center $Z (G)$ is a normal subgroup, hence in particular a subgroup, of $G$ (The center of a group is a normal subgroup).

[L8]

A finite subset with the same cardinality as its ambient finite set is the whole set (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

Proof

technique · direct

1.1

By [L1] and [L7], $Z (G)$ is a nontrivial subgroup of $G$ ; [L5] and [L6] therefore show that it has order $p$ or $p^{2}$ .

L1L5L6L7

2.1

If $∣ Z (G) ∣ = p^{2} = ∣ G ∣$ , then [L8] gives $Z (G) = G$ , so $G$ is abelian. If $∣ Z (G) ∣ = p$ , then [L3] gives $∣ G / Z (G) ∣ = p$ , so [L4] makes $G / Z (G)$ cyclic.

step 1.1L3L4L8

3.1

In the second case [L2] makes $G$ abelian, and the first case already did so. Hence every group of order $p^{2}$ is abelian.

step 2.1L2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Every nontrivial normal subgroup of a finite $p$ -group meets the center nontrivially

Statement

Let $P$ be a finite $p$ -group and let $N ⊴ P$ be nontrivial. Then

$N \cap Z (P) \neq {e} .$

Facts & Assumptions

Given: A finite $p$ -group $P$ and a nontrivial normal subgroup $N ⊴ P$ .

[L1]

A finite $p$ -group has prime-power order (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

A finite $p$ -group action satisfies the fixed-point congruence (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L3]

Normality means $g N g^{- 1} = N$ for every $g \in P$ (Normal subgroup: invariance under conjugation, Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

[L4]

Conjugation by $g$ is an automorphism (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[L5]

The center consists of the elements fixed by every conjugation (The center $Z (G)$ of a group).

[L6]

A nontrivial subgroup of a finite $p$ -group has order $p^{k}$ for some $k \geq 1$ (Every subgroup of a finite $p$ -group has order a power of $p$ ).

Proof

technique · direct

1.1

By [L3] and [L4], conjugation restricts to an action of $P$ on the finite set $N$ .

L3L4

2.1

A point of $N$ is fixed by every element of $P$ exactly when it lies in $N \cap Z (P)$ by [L5].

step 1.1L5

3.1

By [L6], $p$ divides $∣ N ∣$ . Applying [L2] to the action in step 1.1 therefore shows that $p$ divides $∣ N \cap Z (P) ∣$ .

step 1.1step 2.1L1L2L6

4.1

The intersection contains $e$ , and its cardinality is a positive multiple of the prime $p > 1$ ; hence it contains a nonidentity element.

step 3.1L5algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$

Statement

Let $G$ be a finite group and let $p$ be prime. If $p ∣ ∣ G ∣$ , then $G$ contains an element of order $p$ .

Facts & Assumptions

Given: A finite group $G$ and a prime $p$ dividing $∣ G ∣$ .

[L1]

A finite $p$ -group acting on a finite set satisfies the fixed-point congruence (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L2]

The additive group $Z / p$ is a group with $p$ elements and hence is a finite $p$ -group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L3]

If $A$ and $B$ are finite, the set of functions $B \to A$ has cardinality $∣ A ∣^{∣ B ∣}$ (The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $∣ A^{B} ∣ = ∣ A ∣^{∣ B ∣}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ ).

[L4]

A prime is greater than $1$ and has only $1$ and itself as positive divisors (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L5]

A congruence $a \equiv b (m o d p)$ means that $p$ divides $a - b$ , and divisibility means existence of an integer factor (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L6]

The order of an element is the least positive exponent giving the identity (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · constructive

1.1

Let $Ω$ be the set of $p$ -tuples $(g_{0}, \dots, g_{p - 1}) \in G^{p}$ whose ordered product is $e$ . The first $p - 1$ coordinates determine the last uniquely as $(g_{0} \dots g_{p - 2})^{- 1}$ , so [L3] gives $∣ Ω ∣ = ∣ G ∣^{p - 1}$ ; since $p ∣ ∣ G ∣$ and $p - 1 \geq 1$ , one has $p ∣ ∣ Ω ∣$ .

L3L4L5construct

2.1

Let $1 \in Z / p$ act on $Ω$ by cyclic rotation. If $g_{0} \dots g_{p - 1} = e$ , then $g_{1} \dots g_{p - 1} g_{0} = g_{0}^{- 1} (g_{0} \dots g_{p - 1}) g_{0} = e$ , so rotation preserves $Ω$ ; $p$ rotations are the identity, and [L2] therefore gives an action of the finite $p$ -group $Z / p$ .

step 1.1L2

3.1

A tuple is fixed by every rotation exactly when it is constant, say $(g, \dots, g)$ , and it lies in $Ω$ exactly when $g^{p} = e$ .

step 2.1L6

3.2

By [L1], $∣ Ω ∣ \equiv ∣ Ω^{Z / p} ∣ (m o d p)$ . Step 1.1 makes the left side divisible by $p$ , so the number of fixed tuples is divisible by $p$ . The constant identity tuple is fixed, and a positive multiple of $p > 1$ cannot equal $1$ , so there is another fixed tuple.

step 1.1step 2.1L1L4L5

4.1

By step 3.1, this second tuple is $(g, \dots, g)$ for some $g \neq e$ with $g^{p} = e$ . By [L6], the positive order of $g$ divides the prime $p$ and is not $1$ , so it is $p$ .

step 3.1step 3.2L4L6discharge-construct∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Cauchy-Frobenius orbit counting: $∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣$ for a finite group action

Statement

Let a finite group $G$ act on a finite set $X$ , and let $X / G$ denote the set of orbits. Then

$∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣ .$

Equivalently, the number of orbits is the average number of fixed points of an element of $G$ .

Facts & Assumptions

Given: A finite group $G$ acting on a finite set $X$ .

[L1]

The fixed-point set of $g$ is $X^{g} = {x \in X : g \cdot x = x}$ (The fixed-point sets $X^{g}$ and $X^{G}$ of a group action).

[L2]

The orbits partition $X$ (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set).

[L3]

For every $x \in X$ , $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L4]

A finite incidence relation can be counted by either family of fibres (Double counting: $\sum_{x \in X} ∣ R_{x} ∣ = ∣ R ∣ = \sum_{y \in Y} ∣ R^{y} ∣$ for a relation between finite sets).

[L5]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

[L6]

A finite sum splits along a finite partition of its index set (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

Proof

technique · direct

1.1

Let $R = {(g, x) \in G \times X : g \cdot x = x}$ . Counting its fibres over $g$ and using [L1], [L4], and [L5] gives $∣ R ∣ = \sum_{g \in G} ∣ X^{g} ∣$ .

L1L4L5

1.2

Counting the same relation over $x$ gives $∣ R ∣ = \sum_{x \in X} ∣ G_{x} ∣$ .

L4L5

2.1

Split the second sum along the orbit partition using [L2] and [L6]. On an orbit $O = G \cdot x$ , [L3] gives $∣ G_{y} ∣ = ∣ G ∣ / ∣ O ∣$ for every $y \in O$ , so that orbit contributes $∣ O ∣ (∣ G ∣ / ∣ O ∣) = ∣ G ∣$ .

step 1.2L2L3L5L6

3.1

There is one contribution for each orbit in $X / G$ , hence $∣ R ∣ = ∣ G ∣ ∣ X / G ∣$ . Combining this with step 1.1 gives the stated identity.

step 1.1step 2.1L2L5∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Jordan's derangement theorem: every transitive action of a finite group on a finite set with more than one element has a nonidentity element with no fixed points

Statement

Let a finite group $G$ act transitively on a finite set $X$ with $∣ X ∣ > 1$ . Then some nonidentity $g \in G$ is a derangement:

$X^{g} = \emptyset .$

Facts & Assumptions

Given: A transitive action of a finite group $G$ on a finite set $X$ with $∣ X ∣ > 1$ .

[L1]

A transitive action has exactly one orbit (Left group actions, transitive actions, and faithful actions).

[L2]

The fixed-point set is $X^{g} = {x : g \cdot x = x}$ (The fixed-point sets $X^{g}$ and $X^{G}$ of a group action).

[L3]

Cauchy-Frobenius gives $∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣$ (Cauchy-Frobenius orbit counting: $∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣$ for a finite group action).

[L4]

Finite cardinalities add over disjoint unions (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition, The cardinality $∣ A ∣$ of a finite set).

[L5]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Proof

technique · contradiction

1.1

By transitivity [L1], $∣ X / G ∣ = 1$ , so [L3] gives $\sum_{g \in G} ∣ X^{g} ∣ = ∣ G ∣$ .

L1L3

1.2

The identity fixes every point, so $∣ X^{e} ∣ = ∣ X ∣$ ; splitting its term from the finite sum gives $\sum_{g \in G} ∣ X^{g} ∣ = ∣ X ∣ + \sum_{g \neq e} ∣ X^{g} ∣$ .

L2L4L5

2.1

Suppose, for contradiction, that every nonidentity $g$ fixes a point. Then every term in the remaining sum is at least $1$ , so step 1.2 gives $\sum_{g} ∣ X^{g} ∣ \geq ∣ X ∣ + (∣ G ∣ - 1) > ∣ G ∣$ , since $∣ X ∣ > 1$ .

assume-contrastep 1.2L2L4L5

3.1

This contradicts step 1.1. Therefore some $g$ has $X^{g} = \emptyset$ ; the identity fixes all of $X$ , so this $g$ is nonidentity.

step 1.1step 2.1L2discharge-contradiction∎

5 · Examples, counterexamples and false statements

None yet.

Group Actions, Orbits, Stabilisers and Cayley's Theorem

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Left group actions, transitive actions, and faithful actions

Definition

Actions of GG on XX correspond exactly to homomorphisms G→Sym⁡(X)G→Sym(X)

Statement

Facts & Assumptions

Proof

The orbit G⋅xG⋅x and stabilizer GxGx​ of a point in a group action

Definition

The stabilizer GxGx​ is a subgroup of GG

Statement

Facts & Assumptions

Proof

The orbits of a group action are the equivalence classes of x∼yx∼y iff y=g⋅xy=g⋅x for some gg, and hence partition the acted-on set

Statement

Facts & Assumptions

Proof

Cayley's theorem: every group GG is isomorphic to a subgroup of Sym⁡(G)Sym(G)

Statement

Facts & Assumptions

Proof

Equivariant maps and isomorphisms of group actions

Definition

A free group action has no nonidentity element fixing a point

Definition

The fixed-point sets XgXg and XGXG of a group action

Definition

Orbit-stabiliser: G/Gx→G⋅xG/Gx​→G⋅x, gGx↦g⋅xgGx​↦g⋅x, is a well-defined bijection

Statement

Facts & Assumptions

Proof

Orbit-stabiliser cardinality: ∣G⋅x∣=[G:Gx]∣G⋅x∣=[G:Gx​] whenever either side is finite, and ∣G∣=∣Gx∣ ∣G⋅x∣∣G∣=∣Gx​∣∣G⋅x∣ for finite GG

Statement

Facts & Assumptions

Proof

If y=g⋅xy=g⋅x, then Gy=gGxg−1Gy​=gGx​g−1

Statement

Facts & Assumptions

Proof

The core Core⁡G(H)=⋂g∈GgHg−1CoreG​(H)=⋂g∈G​gHg−1 of a subgroup

Definition

Core⁡G(H)CoreG​(H) is the largest normal subgroup of GG contained in HH

Statement

Facts & Assumptions

Proof

Left multiplication on G/HG/H is transitive, has stabiliser HH at HH, and has kernel Core⁡G(H)CoreG​(H)

Statement

Facts & Assumptions

Proof

If [G:H]=n<∞[G:H]=n<∞, then Core⁡G(H)⊴GCoreG​(H)⊴G, [G:Core⁡G(H)]∣n![G:CoreG​(H)]∣n!, and only finitely many subgroups contain HH

Statement

Facts & Assumptions

Proof

Every transitive GG-set is equivariantly isomorphic to G/GxG/Gx​ for any chosen point xx

Statement

Facts & Assumptions

Proof

A transitive action is faithful exactly when a point stabiliser is core-free

Statement

Facts & Assumptions

Proof

The conjugacy class Cl⁡G(x)ClG​(x) and centralizer CG(x)CG​(x) of an element

Definition

The normalizer NG(H)={g∈G:gHg−1=H}NG​(H)={g∈G:gHg−1=H} of a subgroup

Definition

CG(x)CG​(x) and NG(H)NG​(H) are subgroups of GG

Statement

Facts & Assumptions

Proof

G/CG(x)→Cl⁡G(x)G/CG​(x)→ClG​(x) is a bijection, so ∣Cl⁡G(x)∣=[G:CG(x)]∣ClG​(x)∣=[G:CG​(x)] whenever these cardinalities are finite

Statement

Facts & Assumptions

Proof

The conjugates of HH are in bijection with G/NG(H)G/NG​(H) and, for finite GG, number [G:NG(H)][G:NG​(H)]

Statement

Facts & Assumptions

Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$

The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action

The stabilizer $G_{x}$ is a subgroup of $G$

The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set

Cayley's theorem: every group $G$ is isomorphic to a subgroup of $Sym (G)$

The fixed-point sets $X^{g}$ and $X^{G}$ of a group action

Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection

Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$

If $y = g \cdot x$ , then $G_{y} = g G_{x} g^{- 1}$

The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup

${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$

Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$

If $[G : H] = n < \infty$ , then ${Core}_{G} (H) ⊴ G$ , $[G : {Core}_{G} (H)] ∣ n!$ , and only finitely many subgroups contain $H$

Every transitive $G$ -set is equivariantly isomorphic to $G / G_{x}$ for any chosen point $x$

The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element

The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup

$C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$

$G / C_{G} (x) \to {Cl}_{G} (x)$ is a bijection, so $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever these cardinalities are finite

The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$

The class equation $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i} [G : C_{G} (x_{i})]$ for a finite group

A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$

Every subgroup of a finite $p$ -group has order a power of $p$

Every subgroup of index $p$ in a finite $p$ -group is normal

If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$

A finite $p$ -group action on $X$ has a global fixed point whenever $p ∤ ∣ X ∣$

Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$

If $G / Z (G)$ is cyclic, then $G$ is abelian

Every group of order $p^{2}$ , for prime $p$ , is abelian

Every nontrivial normal subgroup of a finite $p$ -group meets the center nontrivially

Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$

Cauchy-Frobenius orbit counting: $∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣$ for a finite group action