LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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.

Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution

Statement

Let $G$ be a group (Group and abelian group) and let $g \in G$ . Define the left translation $λ_{g} : G \to G$ by $λ_{g} (x) = g x$ and the right translation $ρ_{g} : G \to G$ by $ρ_{g} (x) = x g$ . Then:

Cancellation. For all $x, y \in G$ : if $g x = g y$ then $x = y$ , and if $x g = y g$ then $x = y$ .
$λ_{g}$ and $ρ_{g}$ are bijections of $G$ (Injection, surjection, bijection), with inverse maps $λ_{g^{- 1}}$ and $ρ_{g^{- 1}}$ respectively.
For every $h \in G$ the equation $g x = h$ has exactly one solution in $G$ , namely $x = g^{- 1} h$ ; and $x g = h$ has exactly one solution, namely $x = h g^{- 1}$ .

Facts & Assumptions

Given: A group $G$ with identity $e$ , an element $g \in G$ with inverse $g^{- 1}$ , and the maps $λ_{g} (x) = g x$ , $ρ_{g} (x) = x g$ (Group and abelian group).

[L1]

The group axioms: associativity, $e$ a two-sided identity, and $g^{- 1} g = e = g g^{- 1}$ (Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

[L3]

A map is bijective exactly when it has a two-sided inverse map; injective means $f (x) = f (y)$ forces $x = y$ (Injection, surjection, bijection).

Proof

technique · direct

1.1L1

Suppose $g x = g y$ . Multiplying on the left by $g^{- 1}$ and regrouping, $x = e x = (g^{- 1} g) x = g^{- 1} (g x) = g^{- 1} (g y) = (g^{- 1} g) y = e y = y$ .

1.2L1

Suppose $x g = y g$ . Multiplying on the right by $g^{- 1}$ and regrouping, $x = x e = x (g g^{- 1}) = (x g) g^{- 1} = (y g) g^{- 1} = y (g g^{- 1}) = y e = y$ .

1.3L1

For every $x \in G$ , $λ_{g^{- 1}} (λ_{g} (x)) = g^{- 1} (g x) = (g^{- 1} g) x = e x = x$ , and $λ_{g} (λ_{g^{- 1}} (x)) = g (g^{- 1} x) = (g g^{- 1}) x = e x = x$ .

1.4L1

For every $x \in G$ , $ρ_{g^{- 1}} (ρ_{g} (x)) = (x g) g^{- 1} = x (g g^{- 1}) = x$ , and $ρ_{g} (ρ_{g^{- 1}} (x)) = (x g^{- 1}) g = x (g^{- 1} g) = x$ .

1.5L1

Existence in claim 3: $g (g^{- 1} h) = (g g^{- 1}) h = e h = h$ , so $x = g^{- 1} h$ solves $g x = h$ ; and $(h g^{- 1}) g = h (g^{- 1} g) = h e = h$ , so $x = h g^{- 1}$ solves $x g = h$ .

2.1step 1.1step 1.2

Claim 1 is steps 1.1 and 1.2.

2.2step 1.3step 1.4L3

By step 1.3 the map $λ_{g^{- 1}}$ is a two-sided inverse of $λ_{g}$ , so $λ_{g}$ is a bijection of $G$ with inverse $λ_{g^{- 1}}$ ; by step 1.4 the same holds for $ρ_{g}$ with inverse $ρ_{g^{- 1}}$ . This is claim 2.

2.3step 1.1step 1.2

Uniqueness in claim 3: if $g x = h = g x^{'}$ then $x = x^{'}$ by step 1.1, and if $x g = h = x^{'} g$ then $x = x^{'}$ by step 1.2.

3.1step 2.1step 2.2step 1.5step 2.3∎

Cancellation holds on both sides, both translations are bijections, and each of the two equations has exactly one solution.

Remarks

Claim 2 is the form used most often: it says every row and every column of the multiplication table of a group is a rearrangement of $G$ , since $λ_{g}$ and $ρ_{g}$ permute $G$ . Inversion pairs $g$ with $g^{- 1}$ (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), so the two families of translations are indexed by the same elements.
Cancellation is strictly weaker than invertibility. A commutative monoid can satisfy both cancellation laws without being a group; $(N, +)$ is such a monoid, recorded on the companion page.

Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Cancellation in a group: gx=gygx=gy or xg=ygxg=yg forces x=yx=y; equivalently left and right translation by gg are bijections of GG, so gx=hgx=h and xg=hxg=h each have exactly one solution

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution