LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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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.

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

Statement

Let $G$ be a group (Group and abelian group) and let $g \in G$ . Define the left translation $\lambda_g : G \to G$ by $\lambda_g(x) = gx$ and the right translation $\rho_g : G \to G$ by $\rho_g(x) = xg$ . Then:

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

Facts & Assumptions

Given: A group $G$ with identity $e$ , an element $g \in G$ with inverse $g^{-1}$ , and the maps $\lambda_g(x) = gx$ , $\rho_g(x) = xg$ (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

1.1

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

1.2

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

1.3

For every $x \in G$ , $\lambda_{g^{-1}}(\lambda_g(x)) = g^{-1}(gx) = (g^{-1}g)x = ex = x$ , and $\lambda_g(\lambda_{g^{-1}}(x)) = g(g^{-1}x) = (g g^{-1})x = ex = x$ .

1.4

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

1.5

Existence in claim 3: $g(g^{-1}h) = (g g^{-1})h = eh = h$ , so $x = g^{-1}h$ solves $gx = h$ ; and $(hg^{-1})g = h(g^{-1}g) = he = h$ , so $x = hg^{-1}$ solves $xg = h$ .

2.1

Claim 1 is steps 1.1 and 1.2.

step 1.1step 1.2

2.2

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

step 1.3step 1.4L3

2.3

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

step 1.1step 1.2

3.1

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

step 2.1step 2.2step 1.5step 2.3∎

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 $\lambda_g$ and $\rho_g$ permute $G$ . Inversion pairs $g$ with $g^{-1}$ (In a group $e^{-1} = e$ , $(g^{-1})^{-1} = g$ and $(gh)^{-1} = h^{-1}g^{-1}$ , 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; $(\mathbb{N},+)$ is such a monoid, recorded on the companion page.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Cancellation property (Wikipedia) (standard reference, not scraped)
Group (mathematics) (Wikipedia) (standard reference, not scraped)

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