TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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.

For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$

Statement

Let $N ⊴ G$ . The left cosets form a group $G / N$ under

$(g N) (h N) = g h N .$

Its identity is $N = e N$ , and the inverse of $g N$ is $g^{- 1} N$ .

Facts & Assumptions

Given: A group $G$ and a normal subgroup $N ⊴ G$ .

[L1]

Coset multiplication $(g N) (h N) = g h N$ is well defined when $N$ is normal (Coset multiplication $(g H) (h H) = g h H$ is well defined if and only if $H$ is normal).

[F1]

The quotient set $G / N$ consists of the left cosets of $N$ , with the proposed product $(g N) (h N) = g h N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[F2]

A group operation is associative, has a two-sided identity, and gives every element a two-sided inverse (Group and abelian group).

Proof

technique · direct

1.1L1F1

By [L1], the formula in [F1] is a binary operation on the coset set, independent of representatives.

1.2F1F2algebra

For $g, h, k \in G$ , one has $((g N) (h N)) (k N) = (g h) k N = g (h k) N = (g N) ((h N) (k N))$ . Also $(e N) (g N) = g N = (g N) (e N)$ , so $N = e N$ is the identity.

1.3F1F2algebra

The products $(g N) (g^{- 1} N)$ and $(g^{- 1} N) (g N)$ both equal $e N = N$ , so $g^{- 1} N$ is the inverse of $g N$ .

2.1step 1.1step 1.2step 1.3F2∎

Steps 1.1 through 1.3 verify the binary operation, associativity, identity, and inverse axioms; therefore $G / N$ is a group with the stated identity and inverses.

Depends on

Dependency tree · 2 levels

Sources

T. W. Judson, Abstract Algebra: Theory and Applications, Factor Groups and Normal Subgroups