LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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.
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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

Statement

Let $G$ be a group (Group and abelian group) with identity $e$ . For all $g, h \in G$ :

$e^{- 1} = e$ ;
$(g^{- 1})^{- 1} = g$ ; in particular inversion $g \mapsto g^{- 1}$ is its own two-sided inverse as a map $G \to G$ , hence a bijection of $G$ ;
$(g h)^{- 1} = h^{- 1} g^{- 1}$ ;
$(g h)^{- 1} = g^{- 1} h^{- 1}$ holds if and only if $g$ and $h$ commute. So the reversal of order in claim 3 cannot be dropped in general, and in an abelian group it may be.

Facts & Assumptions

Given: A group $G$ with identity $e$ , and elements $g, h \in G$ ; $g^{- 1}$ denotes the unique two-sided inverse of $g$ (Group and abelian group, Left inverse, right inverse, and invertible element of a monoid).

[L1]

Uniqueness of inverses in the sharp form: if $x$ is invertible and $y * x = e$ or $x * y = e$ , then $y = x^{- 1}$ ; and an element with a left and a right inverse is invertible with that common element as inverse (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).

[L2]

The group axioms: $*$ is associative, $e$ is a two-sided identity, and every element has a two-sided inverse (Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

Proof

technique · direct

1.1

$e e = e$ by the identity law, so $e$ is a two-sided inverse of $e$ ; since inverses are unique, $e^{- 1} = e$ , which is claim 1.

L1L2

1.2

The defining equations of $g^{- 1}$ are $g^{- 1} g = e$ and $g g^{- 1} = e$ ; read with $g^{- 1}$ in the role of the element being inverted, they say that $g$ is a two-sided inverse of $g^{- 1}$ . Uniqueness gives $(g^{- 1})^{- 1} = g$ , which is the equation of claim 2.

L1L2

1.3

Compute $(g h) (h^{- 1} g^{- 1}) = g (h h^{- 1}) g^{- 1} = g e g^{- 1} = g g^{- 1} = e$ , using associativity to regroup and the identity law twice.

1.4

Compute likewise $(h^{- 1} g^{- 1}) (g h) = h^{- 1} (g^{- 1} g) h = h^{- 1} e h = h^{- 1} h = e$ .

2.1

By steps 1.3 and 1.4 the element $h^{- 1} g^{- 1}$ is a two-sided inverse of $g h$ , so $g h$ is invertible and $(g h)^{- 1} = h^{- 1} g^{- 1}$ by uniqueness; this is claim 3.

step 1.3step 1.4L1

2.2

Inversion is a map $G \to G$ by claim (G3) and uniqueness, and step 1.2 says it composed with itself is the identity map of $G$ , so it is a bijection of $G$ onto itself; this completes claim 2.

step 1.2L1L2

3.1

Suppose $g h = h g$ . Applying step 2.1 to the pair $(h, g)$ gives $(h g)^{- 1} = g^{- 1} h^{- 1}$ , and $g h = h g$ gives $(g h)^{- 1} = (h g)^{- 1}$ ; hence $(g h)^{- 1} = g^{- 1} h^{- 1}$ .

step 2.1given

3.2

Conversely suppose $(g h)^{- 1} = g^{- 1} h^{- 1}$ . Taking inverses of both sides and using step 2.1 on the right and step 1.2 on the left gives $g h = ((g h)^{- 1})^{- 1} = (g^{- 1} h^{- 1})^{- 1} = (h^{- 1})^{- 1} (g^{- 1})^{- 1} = h g$ .

step 1.2step 2.1

4.1

Steps 3.1 and 3.2 give claim 4: $(g h)^{- 1} = g^{- 1} h^{- 1}$ holds exactly when $g$ and $h$ commute, so the reversal in claim 3 is essential precisely for non-commuting pairs, and is harmless in an abelian group.

step 3.1step 3.2

5.1

Claims 1, 2, 3 and 4 are established in steps 1.1, 2.2, 2.1 and 4.1.

step 1.1step 2.1step 2.2step 4.1∎

Remarks

Claim 4 is what makes the wording of claim 3 more than a stylistic preference: a pair with $(g h)^{- 1} \neq g^{- 1} h^{- 1}$ exists in a group exactly when some two of its elements fail to commute. That non-abelian groups exist is settled below by $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, which shows $Sym (X)$ is non-abelian whenever $X$ has three distinct elements.
Claim 2 is used constantly in the form "inversion is a bijection": a statement quantified over all $g$ may be re-read as a statement quantified over all $g^{- 1}$ without loss.

Depends on

Used by

Dependency tree · two levels

6 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

Group (mathematics) (Wikipedia) (standard reference, not scraped)
Inverse element (Wikipedia) (standard reference, not scraped)

In a group e−1=ee−1=e, (g−1)−1=g(g−1)−1=g and (gh)−1=h−1g−1(gh)−1=h−1g−1, the order of the last product being essential