Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Two elements have the same image under a homomorphism if and only if they lie in the same coset of its kernel

Statement

Two elements have the same image under a homomorphism if and only if they lie in the same coset of its kernel.

For a homomorphism f:G→H and g,h∈G,

f(g)=f(h)⟺gker⁡f=hker⁡f.

Facts & Assumptions

Given: A group homomorphism f:G→H and elements g,h∈G.

[L1]

ker⁡f is the set of elements sent to eH (The kernel and image of a group homomorphism).

[L2]

f(h−1g)=f(h)−1f(g) (A group homomorphism automatically satisfies f(e)=e′ and f(g−1)=f(g)−1, and f(gn)=f(g)n for every n∈Z; for monoid homomorphisms preservation of the identity must be assumed).

[L3]

For K≤G, gK=hK if and only if h−1g∈K (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

Proof

technique · direct
1.1L1L2L3givenalgebra

If f(g)=f(h), then [L2] gives f(h−1g)=eH, so h−1g∈ker⁡f and [L3] gives gker⁡f=hker⁡f.

2.1step 1.1L1L2L3givenalgebra

If gker⁡f=hker⁡f, then [L3] gives h−1g∈ker⁡f, so f(h)−1f(g)=eH and f(g)=f(h).

3.1step 1.1step 2.1∎

This proves the stated equivalence.

Depends on

Dependency tree · 2 levels

Sources