Group Actions, Orbits, Stabilisers and Cayley's Theorem

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

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

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 $G\to\operatorname{Sym}(X)$

Statement

For groups $G$ and a set $X$ , left actions of $G$ on $X$ are in bijection with group homomorphisms $\rho:G\to\operatorname{Sym}(X)$ . The action attached to $\rho$ is $g\cdot x:=\rho(g)(x)$ ; 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 $(gh)\cdot x=g\cdot(h\cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L2]

$\operatorname{Sym}(X)$ is the group of bijections $X\to X$ under composition (The symmetric group $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ 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 $n \in \mathbb{Z}$ ; for monoid homomorphisms preservation of the identity must be assumed).

Proof

technique · direct

1.1

Given an action, define $\rho(g)(x)=g\cdot x$ . The maps $\rho(g)$ are bijective: $\rho(g^{-1})$ 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 $\rho:G\to\operatorname{Sym}(X)$ be a homomorphism and set $g\cdot x=\rho(g)(x)$ . Then $e\cdot x=\rho(e)(x)=x$ , and $(gh)\cdot x=\rho(g)(\rho(h)(x))=g\cdot(h\cdot x)$ .

L2L3given

2.1

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

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 $(gh)\cdot x=g\cdot(h\cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L3]

A nonempty subset $H$ of a group is a subgroup exactly when $ab^{-1}\in H$ for all $a,b\in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ 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 $(ab^{-1})\cdot x=a\cdot(b^{-1}\cdot x)=a\cdot x=x$ : indeed $b\cdot x=x$ implies $b^{-1}\cdot x=b^{-1}\cdot(b\cdot x)=x$ . Therefore $ab^{-1}\in G_x$ .

L1L2givenalgebra

2.1

The subgroup criterion now gives $G_x\le G$ .

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 $(gh)\cdot x=g\cdot(h\cdot x)$ (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 $z=(hg)\cdot x$ , 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 $\operatorname{Sym}(G)$

Statement

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

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 $G\to\operatorname{Sym}(X)$ , 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 $\operatorname{Aut}(G)$ ).

Proof

technique · direct

1.1

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

L1givenalgebra

2.1

By [L1], the action yields a homomorphism $\lambda:G\to\operatorname{Sym}(G)$ with $\lambda(g)(x)=gx$ .

step 1.1L1

3.1

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

step 2.1L2given

4.1

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

step 2.1step 3.1L2

5.1

Thus $\lambda$ is an isomorphism from $G$ to a subgroup of $\operatorname{Sym}(G)$ .

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

$g\cdot x=x\quad\Longrightarrow\quad g=e$

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

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

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

Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a well-defined bijection

Statement

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

$\Phi:G/G_x\longrightarrow G\cdot x,\qquad \Phi(gG_x)=g\cdot x,$

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

[L4]

For $H\le G$ , one has $gH=hH$ exactly when $g^{-1}h\in H$ ( $x\in aH$ iff $a^{-1}x\in H$ , and $aH=bH$ 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 $\Phi(gG_x)=g\cdot x$ . If $gG_x=hG_x$ , 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 $\Phi$ is well-defined.

L1L2L3L4L5construct

2.1

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

step 1.1L1

3.1

If $\Phi(gG_x)=\Phi(hG_x)$ , then $g\cdot x=h\cdot x$ , so $(g^{-1}h)\cdot x=x$ and $g^{-1}h\in G_x$ ; [L4] gives $gG_x=hG_x$ , so $\Phi$ 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 $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a bijection (Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , 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 $\lvert A\rvert$ of a finite set).

[L4]

If $G$ is finite and $H\le G$ , 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 $G_x\le G$ 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 $G_y=gG_xg^{-1}$

Statement

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

$G_y=gG_xg^{-1}.$

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 $(ab)\cdot z=a\cdot(b\cdot z)$ and $e\cdot z=z$ (Left group actions, transitive actions, and faithful actions).

[L3]

Conjugation $h\mapsto ghg^{-1}$ is an automorphism of $G$ (Conjugation $x\mapsto gxg^{-1}$ 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 $(g^{-1}hg)\cdot x=x$ , that is, to $g^{-1}hg\in G_x$ .

L1L2L3

2.1

The last condition is equivalent to $h\in gG_xg^{-1}$ , so $G_y=gG_xg^{-1}$ ; [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 $\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}$ of a subgroup

Definition

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

$\operatorname{Core}_G(H):=\bigcap_{g\in G}gHg^{-1}.$

Each $gHg^{-1}$ is a subgroup because conjugation is an automorphism (Conjugation $x\mapsto gxg^{-1}$ 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 $\operatorname{Core}_G(H)$ is the largest normal subgroup of $G$ contained in $H$ ↗.

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

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

Statement

For $H\le G$ , the core $K=\operatorname{Core}_G(H)$ is a normal subgroup of $G$ , satisfies $K\le H$ , 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 $H\le G$ , and $K:=\operatorname{Core}_G(H)$ .

[L1]

The core is $K=\bigcap_{g\in G}gHg^{-1}$ (The core $\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}$ of a subgroup).

[L2]

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

[L3]

Normality is equivalent to $gNg^{-1}\subseteq N$ 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 $ab^{-1}\in S$ for every $a,b\in S$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ 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 $gHg^{-1}$ , and if $a,b$ belong to every such conjugate then $ab^{-1}$ does too; [L4] makes $K$ a subgroup. The factor for $g=e$ is $H$ , so $K\le H$ .

L1L4

1.2

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

L1L2L3

2.1

If $N\mathrel{\trianglelefteq}G$ and $N\le H$ , then $N=gNg^{-1}\le gHg^{-1}$ for every $g\in G$ , so $N\le\bigcap_g gHg^{-1}=K$ .

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 $\operatorname{Core}_G(H)$

Statement

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

$g\cdot(aH):=(ga)H.$

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

$\ker\rho=\operatorname{Core}_G(H).$

Facts & Assumptions

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

[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 $G\to\operatorname{Sym}(X)$ ).

[L3]

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

[L4]

One has $aH=bH$ exactly when $a^{-1}b\in H$ ( $x\in aH$ iff $a^{-1}x\in H$ , and $aH=bH$ 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 $\operatorname{Core}_G(H)=\bigcap_{a\in G}aHa^{-1}$ (The core $\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}$ of a subgroup).

[L7]

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

Proof

technique · direct

1.1

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

L1L3L4

2.1

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

step 1.1L2L5

3.1

By [L4], $(ka)H=aH$ is equivalent to $a^{-1}ka\in H$ , or to $k\in aHa^{-1}$ . Requiring this for every $a$ gives $k\in\bigcap_a aHa^{-1}=\operatorname{Core}_G(H)$ , so $\ker\rho=\operatorname{Core}_G(H)$ , 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 $\operatorname{Core}_G(H)\trianglelefteq G$ , $[G:\operatorname{Core}_G(H)]\mid n!$ , and only finitely many subgroups contain $H$

Statement

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

Facts & Assumptions

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

[L1]

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

[L2]

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

[L3]

First isomorphism gives $G/\ker\rho\cong\operatorname{im}\rho$ (First isomorphism theorem for groups: $G/\ker f\cong\operatorname{im}f$ ).

[L4]

The symmetric group of a set is its group of bijections (The symmetric group $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ 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 $\lvert A\rvert = n$ 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 $\mathbb{N}$ ).

[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 $\pi:G\to G/N$ is a surjective homomorphism (The canonical projection $\pi:G\to G/N$ , $\pi(g)=gN$ , is a surjective group homomorphism, The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L8]

The power set of a finite set is finite ( $\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert}$ for finite $A$ ).

[L9]

Every subset of a finite set is finite (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$ , 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 $\lvert A\rvert$ of a finite set).

Proof

technique · direct

1.1

By [L1] and [L2], the coset action has kernel $N\mathrel{\trianglelefteq}G$ with $N\le H$ , and its image is a subgroup of $\operatorname{Sym}(G/H)$ .

L1L2L4

2.1

By [L3], $G/N\cong\operatorname{im}\rho$ . The set $G/H$ has $n$ elements by [L10], so [L5] gives $|\operatorname{Sym}(G/H)|=n!$ ; [L6] therefore gives $|G/N|=|\operatorname{im}\rho|\mid n!$ , that is, $[G:N]\mid n!$ by [L10].

step 1.1L3L4L5L6L10

3.1

Every subgroup $K$ containing $H$ also contains $N$ by [L2]. If $\pi(x)\in\pi[K]$ , then $\pi(x)=\pi(k)$ for some $k\in K$ , so $k^{-1}x\in N\le K$ and hence $x\in K$ ; thus $K=\pi^{-1}[\pi[K]]$ . Therefore $K\mapsto\pi[K]$ 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

$G/G_x\longrightarrow X,\qquad gG_x\longmapsto g\cdot x,$

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 $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a bijection (Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a well-defined bijection).

[L4]

The left-coset action is $a\cdot(gG_x)=(ag)G_x$ (Left multiplication on $G/H$ is transitive, has stabiliser $H$ at $H$ , and has kernel $\operatorname{Core}_G(H)$ ).

Proof

technique · constructive

1.1

Define $\Phi(gG_x)=g\cdot x$ . By [L3], this is a bijection from $G/G_x$ onto $G\cdot x$ .

L1L3construct

2.1

For $a,g\in G$ , one has $\Phi(a\cdot(gG_x))=\Phi((ag)G_x)=(ag)\cdot x=a\cdot(g\cdot x)=a\cdot\Phi(gG_x)$ , so $\Phi$ is equivariant.

step 1.1L1L4

3.1

By transitivity [L2], $G\cdot x=X$ . Thus $\Phi:G/G_x\to X$ 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

$\operatorname{Core}_G(G_x)=\{e\}.$

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

[L4]

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

Proof

technique · direct

1.1

By [L2], the equivariant isomorphism identifies the given action with the action on $G/G_x$ .

L2

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 $\operatorname{Core}_G(G_x)$ .

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 $\operatorname{Cl}_G(x)$ 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

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

The centralizer of $x$ is

$C_G(x):=\{g\in G:gx=xg\}=\{g\in G:gxg^{-1}=x\}.$

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 $N_G(H)=\{g\in G:gHg^{-1}=H\}$ of a subgroup

Definition

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

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

Thus $g\in N_G(H)$ exactly when the conjugation automorphism $c_g$ preserves $H$ setwise (Conjugation $x\mapsto gxg^{-1}$ 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 $H\le G$ , 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 $H\le G$ .

[L1]

The centralizer is $C_G(x)=\{g\in G:gx=xg\}$ (The conjugacy class $\operatorname{Cl}_G(x)$ and centralizer $C_G(x)$ of an element).

[L2]

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

[L3]

A nonempty subset $S$ is a subgroup if $ab^{-1}\in S$ whenever $a,b\in S$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ 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 $(ab^{-1})x=a(b^{-1}x)=a(xb^{-1})=x(ab^{-1})$ ; [L3] gives $C_G(x)\le G$ .

L1L3

1.2

The identity normalizes $H$ . If $a,b\in N_G(H)$ , then $b^{-1}Hb=H$ , and therefore $(ab^{-1})H(ab^{-1})^{-1}=a(b^{-1}Hb)a^{-1}=aHa^{-1}=H$ .

L2

2.1

Applying [L3] to step 1.2 gives $N_G(H)\le G$ , so both asserted sets are subgroups.

step 1.1step 1.2L3∎

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

$G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever these cardinalities are finite

Statement

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

$G/C_G(x)\longrightarrow\operatorname{Cl}_G(x),\qquad gC_G(x)\longmapsto gxg^{-1},$

is a well-defined bijection. Consequently

$|\operatorname{Cl}_G(x)|=[G:C_G(x)]$

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: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , 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 $\operatorname{Cl}_G(x)=\{gxg^{-1}:g\in G\}$ and the centralizer is $C_G(x)=\{g:gxg^{-1}=x\}$ (The conjugacy class $\operatorname{Cl}_G(x)$ 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 $x\mapsto gxg^{-1}$ form the conjugation homomorphism (The map $g\mapsto(x\mapsto gxg^{-1})$ is a homomorphism $G\to\operatorname{Aut}(G)$ with kernel $Z(G)$ and image $\operatorname{Inn}(G)$ ).

[L6]

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

Proof

technique · direct

1.1

By [L5] and [L6], $G$ acts on itself by conjugation. By [L3], the orbit of $x$ is $\operatorname{Cl}_G(x)$ 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 $gC_G(x)\mapsto gxg^{-1}$ .

step 1.1L1

3.1

Applying [L2] to the same orbit gives $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ 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 $H\le G$ . The rule

$G/N_G(H)\longrightarrow\{gHg^{-1}:g\in G\},\qquad gN_G(H)\longmapsto gHg^{-1},$

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 $H\le G$ .

[L1]

Orbit-stabiliser identifies an orbit with the cosets of its stabilizer (Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , 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 $N_G(H)=\{g:gHg^{-1}=H\}$ (The normalizer $N_G(H)=\{g\in G:gHg^{-1}=H\}$ 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 $x\mapsto gxg^{-1}$ is an automorphism).

Proof

technique · direct

1.1

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

L5

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

$\bigcup_{g\in G}gHg^{-1}\ne G.$

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 $N_G(H)=\{g\in G:gHg^{-1}=H\}$ (The normalizer $N_G(H)=\{g\in G:gHg^{-1}=H\}$ 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 $x\mapsto gxg^{-1}$ 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 $\lvert A\rvert$ 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 $\lvert B\rvert \le \lvert A\rvert$ , 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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , 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 $K_1,\ldots,K_m$ 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 $hHh^{-1}=H$ , so [L2] gives $H\le N_G(H)$ .

L1L2L3L4

2.1

Add the sets successively after removing elements already counted. The common identity contributes once and each $K_i\setminus\{e\}$ contributes at most $|H|-1$ , so [L6], [L7], and [L8] give $|\bigcup_iK_i|\le 1+m(|H|-1)$ .

step 1.1L6L7L8

2.2

Put $n=[G:H]$ . Properness gives $n\ge2$ . Since $H\le N_G(H)$ , [L5] gives $m\le n$ , and [L5] also gives $|G|=n|H|$ .

step 1.1L2L3L5L6

3.1

Therefore $|\bigcup_iK_i|\le1+n(|H|-1)=|G|-n+1<|G|$ , 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 $x_1,\ldots,x_r$ 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 $x_1,\ldots,x_r$ 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, $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ ( $G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever these cardinalities are finite).

[L3]

Conjugacy classes and centralizers are as in The conjugacy class $\operatorname{Cl}_G(x)$ and centralizer $C_G(x)$ of an element.

[L4]

The center is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$ (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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , 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 $gxg^{-1}=x$ 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 $x_1,\ldots,x_r$ gives $|G|=|Z(G)|+\sum_{i=1}^{r}|\operatorname{Cl}_G(x_i)|$ .

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 $n\in\mathbb N$

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 $\lvert A\rvert$ of a finite set) whose order has the form

$|P|=p^n$

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

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 $H\le P$ , then $H$ is finite and

$|H|=p^k$

for some $k\in\mathbb N$ . If $|P|=p^n$ , then $k\le n$ .

Facts & Assumptions

Given: A finite $p$ -group $P$ with $|P|=p^n$ and a subgroup $H\le P$ .

[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 $n\in\mathbb N$ ).

[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 $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ 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 $k\le n$ .

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 $H\le P$ . If $[P:H]=p$ , then $H\mathrel{\trianglelefteq}P$ .

Facts & Assumptions

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

[L1]

For $K=\operatorname{Core}_P(H)$ , one has $K\mathrel{\trianglelefteq}P$ , $K\le H$ , and $[P:K]\mid p!$ (If $[G:H]=n<\infty$ , then $\operatorname{Core}_G(H)\trianglelefteq G$ , $[G:\operatorname{Core}_G(H)]\mid n!$ , and only finitely many subgroups contain $H$ , $\operatorname{Core}_G(H)$ 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 $K\le H\le P$ , then $[P:K]=[P:H][H:K]$ (For $K\le H\le G$ 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 $\mathbb{N}$ ).

[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 $p \mid a_i$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold).

[L6]

Prime factorisation is unique (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ 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 $K=\operatorname{Core}_P(H)$ . By [L1] and [L3], $K\mathrel{\trianglelefteq}P$ , $K\le H$ , $[P:K]\mid p!$ , 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 $1,\ldots,p$ 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 $1,\ldots,p-1$ , 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 $|X|\equiv|X^P|\pmod p$

Statement

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

$|X|\equiv|X^P|\pmod p.$

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 $n\in\mathbb N$ ).

[L2]

The global fixed-point set is $X^P=\{x:g\cdot x=x\text{ for every }g\in P\}$ (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]

[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 $a\equiv b\pmod p$ means that $p$ divides $a-b$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , 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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , 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 $p\nmid|X|$

Statement

Let a finite $p$ -group $P$ act on a finite set $X$ . If $p\nmid|X|$ , then $X^P\ne\varnothing$ ; 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 $p\nmid|X|$ .

[L1]

The fixed-point congruence gives $|X|\equiv|X^P|\pmod p$ (If a finite $p$ -group $P$ acts on a finite set $X$ , then $|X|\equiv|X^P|\pmod p$ ).

[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: $a\equiv b\pmod n$ when $n\mid(a-b)$ , 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 $X^P=\varnothing$ . 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

$p\mid |Z(P)|.$

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 $n\ge1$ (A finite $p$ -group has order $p^n$ for a prime $p$ and some $n\in\mathbb N$ ).

[L2]

For an action of $P$ on a finite set $X$ , $|X|\equiv|X^P|\pmod p$ (If a finite $p$ -group $P$ acts on a finite set $X$ , then $|X|\equiv|X^P|\pmod p$ ).

[L3]

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

[L4]

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

[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: $a\equiv b\pmod n$ when $n\mid(a-b)$ , 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 $|P|\equiv|Z(P)|\pmod p$ .

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 $(gZ(G))(hZ(G))=ghZ(G)$ (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L4]

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

[L5]

The subgroup generated by an element is the set of its integer powers ( $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , 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 $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L7]

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

Proof

technique · direct

1.1

Choose $aZ(G)$ generating $G/Z(G)$ . By [L3]--[L6], arbitrary $x,y\in G$ satisfy $xZ(G)=a^mZ(G)$ and $yZ(G)=a^nZ(G)$ 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 $x=a^mz$ and $y=a^nw$ .

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 $xy=a^mza^nw=a^{m+n}zw=a^{n+m}wz=a^nwa^mz=yx$ . 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 $n\in\mathbb N$ ).

[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 $\lvert B\rvert \le \lvert A\rvert$ , 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 $N\mathrel{\trianglelefteq}P$ be nontrivial. Then

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

Facts & Assumptions

Given: A finite $p$ -group $P$ and a nontrivial normal subgroup $N\mathrel{\trianglelefteq}P$ .

[L1]

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

[L2]

A finite $p$ -group action satisfies the fixed-point congruence (If a finite $p$ -group $P$ acts on a finite set $X$ , then $|X|\equiv|X^P|\pmod p$ ).

[L3]

Normality means $gNg^{-1}=N$ 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 $x\mapsto gxg^{-1}$ 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 $k\ge1$ (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 $p\mid|G|$ , 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 $|X|\equiv|X^P|\pmod p$ ).

[L2]

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

[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 $\lvert A^{B}\rvert = \lvert A\rvert^{\lvert B\rvert}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $\mathbb{R}$ ).

[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 $a\equiv b\pmod p$ means that $p$ divides $a-b$ , and divisibility means existence of an integer factor (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ , Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ 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 $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · constructive

1.1

Let $\Omega$ be the set of $p$ -tuples $(g_0,\ldots,g_{p-1})\in G^p$ whose ordered product is $e$ . The first $p-1$ coordinates determine the last uniquely as $(g_0\cdots g_{p-2})^{-1}$ , so [L3] gives $|\Omega|=|G|^{p-1}$ ; since $p\mid|G|$ and $p-1\ge1$ , one has $p\mid|\Omega|$ .

L3L4L5construct

2.1

Let $1\in\mathbb Z/p$ act on $\Omega$ by cyclic rotation. If $g_0\cdots g_{p-1}=e$ , then $g_1\cdots g_{p-1}g_0=g_0^{-1}(g_0\cdots g_{p-1})g_0=e$ , so rotation preserves $\Omega$ ; $p$ rotations are the identity, and [L2] therefore gives an action of the finite $p$ -group $\mathbb Z/p$ .

step 1.1L2

3.1

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

step 2.1L6

3.2

By [L1], $|\Omega|\equiv|\Omega^{\mathbb Z/p}|\pmod p$ . 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 $(g,\ldots,g)$ for some $g\ne e$ 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: $\sum_{x \in X}\lvert R_x\rvert = \lvert R\rvert = \sum_{y \in Y}\lvert R^y\rvert$ 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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , 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:

$X^g=\varnothing.$

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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , and a sum over a finite index set splits along a partition, The cardinality $\lvert A\rvert$ 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 $\sum_{g\in G}|X^g|=|X|+\sum_{g\ne e}|X^g|$ .

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 $\sum_g|X^g|\ge |X|+(|G|-1)>|G|$ , since $|X|>1$ .

assume-contrastep 1.2L2L4L5

3.1

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

step 1.1step 2.1L2discharge-contradiction∎

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 GGG on XXX correspond exactly to homomorphisms G→Sym⁡(X)G\to\operatorname{Sym}(X)G→Sym(X)

Statement

Facts & Assumptions

Proof

The orbit G⋅xG\cdot xG⋅x and stabilizer GxG_xGx​ of a point in a group action

Definition

The stabilizer GxG_xGx​ is a subgroup of GGG

Statement

Facts & Assumptions

Proof

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

Statement

Facts & Assumptions

Proof

Cayley's theorem: every group GGG is isomorphic to a subgroup of Sym⁡(G)\operatorname{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 XgX^gXg and XGX^GXG of a group action

Definition

Orbit-stabiliser: G/Gx→G⋅xG/G_x\to G\cdot xG/Gx​→G⋅x, gGx↦g⋅xgG_x\mapsto g\cdot xgGx​↦g⋅x, is a well-defined bijection

Statement

Facts & Assumptions

Proof

Orbit-stabiliser cardinality: ∣G⋅x∣=[G:Gx]|G\cdot x|=[G:G_x]∣G⋅x∣=[G:Gx​] whenever either side is finite, and ∣G∣=∣Gx∣ ∣G⋅x∣|G|=|G_x|\,|G\cdot x|∣G∣=∣Gx​∣∣G⋅x∣ for finite GGG

Statement

Facts & Assumptions

Proof

If y=g⋅xy=g\cdot xy=g⋅x, then Gy=gGxg−1G_y=gG_xg^{-1}Gy​=gGx​g−1

Statement

Facts & Assumptions

Proof

The core Core⁡G(H)=⋂g∈GgHg−1\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}CoreG​(H)=⋂g∈G​gHg−1 of a subgroup

Definition

Core⁡G(H)\operatorname{Core}_G(H)CoreG​(H) is the largest normal subgroup of GGG contained in HHH

Statement

Facts & Assumptions

Proof

Left multiplication on G/HG/HG/H is transitive, has stabiliser HHH at HHH, and has kernel Core⁡G(H)\operatorname{Core}_G(H)CoreG​(H)

Statement

Facts & Assumptions

Proof

If [G:H]=n<∞[G:H]=n<\infty[G:H]=n<∞, then Core⁡G(H)⊴G\operatorname{Core}_G(H)\trianglelefteq GCoreG​(H)⊴G, [G:Core⁡G(H)]∣n![G:\operatorname{Core}_G(H)]\mid n![G:CoreG​(H)]∣n!, and only finitely many subgroups contain HHH

Statement

Facts & Assumptions

Proof

Every transitive GGG-set is equivariantly isomorphic to G/GxG/G_xG/Gx​ for any chosen point xxx

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)\operatorname{Cl}_G(x)ClG​(x) and centralizer CG(x)C_G(x)CG​(x) of an element

Definition

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

Definition

CG(x)C_G(x)CG​(x) and NG(H)N_G(H)NG​(H) are subgroups of GGG

Statement

Facts & Assumptions

Proof

G/CG(x)→Cl⁡G(x)G/C_G(x)\to\operatorname{Cl}_G(x)G/CG​(x)→ClG​(x) is a bijection, so ∣Cl⁡G(x)∣=[G:CG(x)]|\operatorname{Cl}_G(x)|=[G:C_G(x)]∣ClG​(x)∣=[G:CG​(x)] whenever these cardinalities are finite

Statement

Facts & Assumptions

Proof

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

Statement

Facts & Assumptions

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

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 $\operatorname{Sym}(G)$

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

Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , 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 $G_y=gG_xg^{-1}$

The core $\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}$ of a subgroup

$\operatorname{Core}_G(H)$ 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 $\operatorname{Core}_G(H)$

If $[G:H]=n<\infty$ , then $\operatorname{Core}_G(H)\trianglelefteq G$ , $[G:\operatorname{Core}_G(H)]\mid n!$ , 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 $\operatorname{Cl}_G(x)$ and centralizer $C_G(x)$ of an element

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

$C_G(x)$ and $N_G(H)$ are subgroups of $G$

$G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ 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 $n\in\mathbb N$

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 $|X|\equiv|X^P|\pmod p$

A finite $p$ -group action on $X$ has a global fixed point whenever $p\nmid|X|$

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