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

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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

1.1L2given

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.

1.2L1L2given

${i d}_{X}$ 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 $f \circ {i d}_{X} = f = {i d}_{X} \circ f$ .

1.3L1L2

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 $f^{- 1} \circ f = {i d}_{X} = f \circ f^{- 1}$ .

1.4givenL1

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.

2.1step 1.1step 1.2step 1.3L3

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 $f^{- 1}$ .

2.2step 1.4given

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 $(τ \circ ρ) (c) = τ (ρ (c)) = τ (b) = a$ .

3.1step 2.2L2given

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.

4.1step 2.1step 3.1∎

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

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.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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