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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Inversion induces a bijection $g H \mapsto H g^{- 1}$ from left cosets to right cosets

Statement

For $H \leq G$ , the rule

$g H ⟼ H g^{- 1}$

is a well-defined bijection from the set of left cosets of $H$ to the set of right cosets of $H$ . Its inverse sends $H g$ to $g^{- 1} H$ .

Facts & Assumptions

Given: A group $G$ and a subgroup $H \leq G$ .

[L1]

For left cosets, $g H = k H$ if and only if $g^{- 1} k \in H$ ; for right cosets, $H g = H k$ if and only if $g k^{- 1} \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L2]

In a group, $(u v)^{- 1} = v^{- 1} u^{- 1}$ and $(u^{- 1})^{- 1} = u$ (In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

[L3]

A subgroup is closed under inverses (Subgroup).

[F1]

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

Proof

technique · direct

1.1givenL1L2L3

If $g H = k H$ , then $g^{- 1} k \in H$ by [L1], so $(g^{- 1} k)^{- 1} = k^{- 1} g \in H$ by subgroup inverse closure. The right-coset criterion gives $H g^{- 1} = H k^{- 1}$ , so the rule is well defined.

1.2L1L2

Define the reverse rule by $H g \mapsto g^{- 1} H$ . The same argument, with left and right interchanged, shows that it is well defined.

2.1step 1.1step 1.2L2F1∎

The two composites send $g H$ to $(g^{- 1})^{- 1} H = g H$ and $H g$ to $H (g^{- 1})^{- 1} = H g$ . Thus the rules are inverse bijections.

Inversion induces a bijection $g H \mapsto H g^{- 1}$ from left cosets to right cosets

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Inversion induces a bijection gH↦Hg−1gH↦Hg−1 from left cosets to right cosets

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Inversion induces a bijection $g H \mapsto H g^{- 1}$ from left cosets to right cosets