LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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.

Dedekind's modular law for subgroup products

Statement

Let $A, B, C \leq G$ with $A \leq C$ . If $A B$ is a subgroup of $G$ , then $A (B \cap C) = A B \cap C .$ The equality also holds as an equality of subsets whenever the displayed products are formed; the subgroup hypothesis ensures that both sides are subgroups in later applications.

Facts & Assumptions

Given: Subgroups $A, B, C \leq G$ with $A \leq C$ , and with $A B$ a subgroup.

[F1]

A subgroup contains the identity and inverses and is closed under products (Subgroup).

Proof

technique · direct

1.1

If $x \in A (B \cap C)$ , write $x = a b$ with $a \in A$ and $b \in B \cap C$ ; then $x \in A B$ , and $a, b \in C$ gives $x \in C$ , so $x \in A B \cap C$ .

givenF1

1.2

If $x \in A B \cap C$ , write $x = a b$ with $a \in A$ and $b \in B$ ; since $a, x \in C$ , one has $b = a^{- 1} x \in C$ , hence $b \in B \cap C$ and $x \in A (B \cap C)$ .

givenF1

2.1

The two inclusions prove $A (B \cap C) = A B \cap C$ .

step 1.1step 1.2∎

Depends on

Subgroup

Used by

Dependency tree · two levels

5 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 6 (standard reference, not scraped)
K. Conrad, Subgroup Series I (standard reference, not scraped)
K. Igusa, Notes on Jordan-Hölder, section 5 (standard reference, not scraped)