Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-01
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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.
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Inversion induces a bijection gH↦Hg−1 from left cosets to right cosets

Statement

For H≤G, the rule

gH⟼Hg−1

is a well-defined bijection from the set of left cosets of H to the set of right cosets of H. Its inverse sends Hg to g−1H.

Facts & Assumptions

Given: A group G and a subgroup H≤G.

[L1]

For left cosets, gH=kH if and only if g−1k∈H; for right cosets, Hg=Hk if and only if gk−1∈H (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[L3]

A subgroup is closed under inverses (Subgroup).

[F1]

A map with a two-sided inverse is a bijection (Injection, surjection, bijection, Equinumerous sets, A≈B and A⪯B).

Proof

technique · direct
1.1

If gH=kH, then g−1k∈H by [L1], so (g−1k)−1=k−1g∈H by subgroup inverse closure. The right-coset criterion gives Hg−1=Hk−1, so the rule is well defined.

givenL1L2L3
1.2

Define the reverse rule by Hg↦g−1H. The same argument, with left and right interchanged, shows that it is well defined.

L1L2
2.1

The two composites send gH to (g−1)−1H=gH and Hg to H(g−1)−1=Hg. Thus the rules are inverse bijections.

step 1.1step 1.2L2F1∎

Depends on

Used by

Dependency tree · two levels

14 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