TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

Blocks containing a point correspond to intermediate subgroups

Statement

Let $G$ act transitively on a set $Ω$ , fix $α \in Ω$ , and write $G_{α} : = {g \in G : g \cdot α = α} .$

Then the assignments $H ⟼ H \cdot α : = {h \cdot α : h \in H}, B ⟼ G_{B} : = {g \in G : g \cdot B = B}$ give mutually inverse bijections between

subgroups $H$ with $G_{α} \leq H \leq G$ , and
blocks $B \subseteq Ω$ with $α \in B$ .

Facts & Assumptions

Given: A transitive left action of $G$ on $Ω$ and a point $α \in Ω$ .

[L1]

A block is a nonempty subset $B$ such that for every $g \in G$ one has either $g \cdot B = B$ or $(g \cdot B) \cap B = \emptyset$ (Blocks and block systems for a group action).

[L2]

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

Proof

technique · direct

1.1L1choose

For the forward direction, let $H$ satisfy $G_{α} \leq H \leq G$ and put $B_{H} : = H \cdot α$ . If $(g \cdot B_{H}) \cap B_{H} \neq \emptyset$ , choose $g h_{1} \cdot α = h_{2} \cdot α$ with $h_{1}, h_{2} \in H$ . Then $h_{2}^{- 1} g h_{1} \in G_{α} \leq H$ , so $g \in H$ and therefore $g \cdot B_{H} = B_{H}$ . Thus $B_{H}$ is a block, and clearly $α \in B_{H}$ .

1.2L1

For the converse direction, let $B$ be a block containing $α$ , and let $G_{B} : = {g \in G : g \cdot B = B}$ . If $s \in G_{α}$ , then $s \cdot α = α \in B$ , so $(s \cdot B) \cap B \neq \emptyset$ ; [L1] gives $s \cdot B = B$ , hence $G_{α} \leq G_{B} \leq G$ .

1.3L1L2choose

For the converse direction, every $g \in G_{B}$ sends $α \in B$ back into $B$ , so $G_{B} \cdot α \subseteq B$ . Conversely, if $β \in B$ , choose $g \in G$ with $g \cdot α = β$ by [L2]. Then $β \in (g \cdot B) \cap B$ , so [L1] gives $g \cdot B = B$ and therefore $g \in G_{B}$ . Hence $β = g \cdot α \in G_{B} \cdot α$ , so $B = G_{B} \cdot α$ .

2.1step 1.1step 1.3∎

Step 1.3 shows that $B \mapsto G_{B} \mapsto G_{B} \cdot α$ returns $B$ . Step 1.1 gives $H \cdot α$ as a block containing $α$ , and if $g \in G_{H \cdot α}$ then $g \cdot α \in H \cdot α$ , so $g \cdot α = h \cdot α$ for some $h \in H$ ; thus $h^{- 1} g \in G_{α} \leq H$ , and hence $g \in H$ . Therefore $G_{H \cdot α} = H$ . The two assignments are mutually inverse.

Depends on

Used by

A transitive action on more than one point is primitive exactly when a point stabilizer is maximal Corollary

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory, Chapter 4 (standard reference, not scraped)
K. Conrad, Transitive Group Actions (standard reference, not scraped)