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 $f : G \to H$ and $g, h \in G$ ,

$f (g) = f (h) ⟺ g \ker f = h \ker f .$

Facts & Assumptions

Given: A group homomorphism $f : G \to H$ and elements $g, h \in G$ .

[L1]

$\ker f$ is the set of elements sent to $e_{H}$ (The kernel and image of a group homomorphism).

[L2]

$f (h^{- 1} g) = f (h)^{- 1} f (g)$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

[L3]

For $K \leq G$ , $g K = h K$ if and only if $h^{- 1} g \in K$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

Proof

technique · direct

1.1

If $f (g) = f (h)$ , then [L2] gives $f (h^{- 1} g) = e_{H}$ , so $h^{- 1} g \in \ker f$ and [L3] gives $g \ker f = h \ker f$ .

L1L2L3givenalgebra

2.1

If $g \ker f = h \ker f$ , then [L3] gives $h^{- 1} g \in \ker f$ , so $f (h)^{- 1} f (g) = e_{H}$ and $f (g) = f (h)$ .

step 1.1L1L2L3givenalgebra

3.1

This proves the stated equivalence.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

21 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

Ernst, An Inquiry-Based Approach to Abstract Algebra, Homomorphisms (standard reference, not scraped)