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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)
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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.

The left cosets of a subgroup partition the group

Statement

For a subgroup H≤G, the set of distinct left cosets {gH:g∈G} is a partition of G: every element belongs to a left coset, every coset is nonempty, and two left cosets are either equal or disjoint.

Facts & Assumptions

Given: A group G and a subgroup H≤G.

[F1]

A relation is an equivalence relation when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set A/∼).

[L1]

The equivalence classes of an equivalence relation on a set are nonempty, cover the set and are pairwise equal or disjoint (The equivalence classes of an equivalence relation are nonempty, cover A, and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

[L2]

For a,b∈G, b∈aH if and only if a−1b∈H, and aH=bH if and only if a−1b∈H (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[F2]

Because H≤G, it contains the identity and is closed under inverses and products (Subgroup).

Proof

technique · direct
1.1

Define a∼b when a−1b∈H. Since the given H is a subgroup, a−1a=e∈H, so the relation is reflexive.

givenF1F2L2
1.2

If a∼b, then a−1b∈H, so its inverse b−1a belongs to H and b∼a; thus the relation is symmetric.

givenF1F2L2
1.3

If a∼b and b∼c, subgroup closure gives a−1c=(a−1b)(b−1c)∈H, so a∼c; thus the relation is transitive.

givenF1F2L2
2.1

By steps 1.1 to 1.3, ∼ is an equivalence relation. Its class at a is {b:a−1b∈H}=aH by [L2].

step 1.1step 1.2step 1.3F1L2
3.1

The conclusion follows from [L1] applied to these equivalence classes.

step 2.1L1∎

Depends on

Used by

Dependency tree · two levels

22 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