LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1L1L2given

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

1.2L1L2givenalgebra

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 $a b^{- 1} \in G_{x}$ .

2.1step 1.1step 1.2L3∎

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

Depends on

Dependency tree · 2 levels

Sources

Brosnan, Orbits and stabilizers