TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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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∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Brosnan, Cayley's theorem (standard reference, not scraped)

Cayley's theorem: every group GGG is isomorphic to a subgroup of Sym⁡(G)\operatorname{Sym}(G)Sym(G)