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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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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.

In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided

Statement

Let (M,∗,e) be a monoid (Semigroup and monoid) and let g∈M. If h is a left inverse of g and h′ is a right inverse of g (Left inverse, right inverse, and invertible element of a monoid), then h=h′. Consequently:

  1. an invertible element g has exactly one two-sided inverse, written g−1;
  2. for an invertible g, every left inverse of g and every right inverse of g equals g−1, so a single one-sided equation h∗g=e or g∗h=e already forces h=g−1;
  3. if g has both a left inverse and a right inverse then g is invertible.

Associativity is used, and is essential.

Facts & Assumptions

Given: A monoid (M,∗,e), an element g∈M, an element h∈M with h∗g=e, and an element h′∈M with g∗h′=e (Left inverse, right inverse, and invertible element of a monoid).

[A1]

h∗g=e.

[A2]

g∗h′=e.

[L1]

∗ is associative: (x∗y)∗z=x∗(y∗z) for all x,y,z∈M (Semigroup and monoid, Binary operation on a set; associativity, commutativity, and a subset closed under the operation).

[L2]

e is a two-sided identity: e∗x=x=x∗e for every x∈M (Semigroup and monoid, Left identity, right identity, and two-sided identity for a binary operation).

Proof

technique · direct
1.1A2L2

Since e is a right identity and g∗h′=e, we may write h=h∗e=h∗(g∗h′).

1.2A1L2

Since e is a left identity and h∗g=e, we may write h′=e∗h′=(h∗g)∗h′.

1.3L1

Associativity identifies the two right-hand sides: h∗(g∗h′)=(h∗g)∗h′.

2.1step 1.1step 1.2step 1.3

Combining, h=h∗(g∗h′)=(h∗g)∗h′=h′, which is the stated equality of a left inverse and a right inverse.

3.1step 2.1A1A2

Claim 3: if h is a left inverse and h′ a right inverse of g, then h=h′ by step 2.1, so this common element satisfies both h∗g=e and g∗h=e and is a two-sided inverse; hence g is invertible.

3.2step 2.1given

Claim 1: let u and v be two-sided inverses of g. Then u is a left inverse and v is a right inverse of g, so u=v by step 2.1; the two-sided inverse of an invertible element is therefore unique, and the notation g−1 is legitimate.

3.3step 2.1given

Claim 2: let g be invertible with two-sided inverse g−1. If h∗g=e then h is a left inverse and g−1 is a right inverse of g, so h=g−1 by step 2.1; if g∗h=e then g−1 is a left inverse and h a right inverse, so g−1=h by step 2.1.

4.1step 2.1step 3.1step 3.2step 3.3∎

A left inverse and a right inverse of the same element coincide, an invertible element has exactly one inverse, and either one-sided equation already identifies it.

Remarks

  • Where associativity enters. Step 1.3 is the only use of it, and it is indispensable: for a non-associative operation with a two-sided identity an element can have several distinct left inverses. This is why the lemma is stated for monoids and not for arbitrary operations with an identity, in contrast with A left identity and a right identity for the same binary operation are equal; hence there is at most one two-sided identity, which needs no associativity at all.

  • What this licenses. Group and abelian group defines a group as a monoid all of whose elements are invertible, and then writes g−1 and treats it as two-sided. Claim 1 is what makes that notation well defined, and claim 2 is the working form used throughout: to identify an inverse it suffices to verify one of the two equations, provided the element is already known to be invertible.

Depends on

Dependency tree · 2 levels

Sources