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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements

Statement

For every set $X$ , the triple $(Sym (X), \circ, {i d}_{X})$ of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition is a group (Group and abelian group); the inverse of a permutation $f$ is its inverse function $f^{- 1}$ .

If $X$ contains three distinct elements $a$ , $b$ , $c$ , then $Sym (X)$ is not abelian: the transpositions $τ = (a b)$ and $ρ = (b c)$ satisfy $τ \circ ρ \neq ρ \circ τ$ .

Facts & Assumptions

Given: A set $X$ ; $Sym (X)$ the set of bijections $X \to X$ with the operation $f \circ g$ defined by $(f \circ g) (x) = f (g (x))$ and the element ${i d}_{X}$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition); and, for the second claim, three distinct elements $a, b, c \in X$ together with the transpositions $τ = (a b)$ and $ρ = (b c)$ of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition.

[L1]

A composite of two bijections $X \to X$ is a bijection $X \to X$ , so $\circ$ is a binary operation on $Sym (X)$ ; ${i d}_{X}$ is a bijection; and a bijection $f$ has a two-sided inverse function $f^{- 1}$ , which is itself a bijection (Injection, surjection, bijection, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L2]

Two functions $X \to X$ are equal exactly when they agree at every point of $X$ .

[L3]

A group is a monoid in which every element is invertible; a monoid is an associative operation with a two-sided identity (Group and abelian group, Semigroup and monoid, Left inverse, right inverse, and invertible element of a monoid).

Proof

technique · direct

1.1

Composition is associative: for $f, g, h \in Sym (X)$ and $x \in X$ , both $((f \circ g) \circ h) (x)$ and $(f \circ (g \circ h)) (x)$ evaluate to $f (g (h (x)))$ , so the two composites agree at every point and are equal.

L2given

1.2

${i d}_{X}$ is a two-sided identity: for $f \in Sym (X)$ and $x \in X$ , $(f \circ {i d}_{X}) (x) = f (x)$ and $({i d}_{X} \circ f) (x) = f (x)$ , so $f \circ {i d}_{X} = f = {i d}_{X} \circ f$ .

L1L2given

1.3

Every $f \in Sym (X)$ is invertible in $Sym (X)$ : the inverse function $f^{- 1}$ is again a bijection $X \to X$ , hence lies in $Sym (X)$ , and it satisfies $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ for every $x$ , that is $f^{- 1} \circ f = {i d}_{X} = f \circ f^{- 1}$ .

L1L2

1.4

The transposition $τ = (a b)$ satisfies $τ (a) = b$ , $τ (b) = a$ and $τ (x) = x$ for $x \notin {a, b}$ ; the transposition $ρ = (b c)$ satisfies $ρ (b) = c$ , $ρ (c) = b$ and $ρ (x) = x$ for $x \notin {b, c}$ . Both are bijections of $X$ , being their own inverses.

givenL1

2.1

By steps 1.1 and 1.2 the pair $(Sym (X), \circ)$ with the element ${i d}_{X}$ is a monoid; by step 1.3 every element of it is invertible; hence it is a group, and the inverse of $f$ is the inverse function $f^{- 1}$ .

step 1.1step 1.2step 1.3L3

2.2

Evaluate the two composites at $c$ . Since $c \neq a$ and $c \neq b$ , $τ (c) = c$ , so $(ρ \circ τ) (c) = ρ (τ (c)) = ρ (c) = b$ . And $ρ (c) = b$ , so $(τ \circ ρ) (c) = τ (ρ (c)) = τ (b) = a$ .

step 1.4given

3.1

The two composites take different values at $c$ , because $a \neq b$ ; hence $τ \circ ρ \neq ρ \circ τ$ and $Sym (X)$ is not abelian.

step 2.2L2given

4.1

$Sym (X)$ is a group under composition, and it fails to be abelian as soon as $X$ has three distinct elements.

step 2.1step 3.1∎

Remarks

"Three distinct elements", not a cardinality hypothesis. The second claim is stated and proved with three named, pairwise distinct points of $X$ . No notion of the size of $X$ is used, so the statement is available for any $X$ whatever, finite or not, and needs nothing about counting.
For $X$ with at most two elements $Sym (X)$ is abelian: it has at most two elements itself, and any group with at most two elements is abelian, since one of any two of its elements is then the identity.
The witness pair $τ, ρ$ is reused on the companion page to show that the exponent law $(g h)^{n} = g^{n} h^{n}$ genuinely needs its commuting hypothesis.

Depends on

Used by

Dependency tree · two levels

11 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

Symmetric group (Wikipedia) (standard reference, not scraped)
Function composition (Wikipedia) (standard reference, not scraped)

Sym⁡(X)Sym(X) is a group under composition, and it is non-abelian whenever XX has at least three distinct elements