CounterexampleConstruction: AI-adaptedVerification: 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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$(g h)^{n} = g^{n} h^{n}$ fails without commutativity: two transpositions in $Sym ({1, 2, 3})$ with $(g h)^{2} \neq g^{2} h^{2}$

Statement refuted

False claim: in every group $G$ , $(g h)^{n} = g^{n} h^{n}$ for all $g, h \in G$ and all $n$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

The claim is true when $g$ and $h$ commute, which is exactly what Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute asserts and exactly why that lemma carries the hypothesis in its statement and in its title. Without the hypothesis it fails, already at $n = 2$ : in $Sym ({1, 2, 3})$ take the transpositions $g = (1 2)$ and $h = (2 3)$ . Then $g^{2} h^{2} = i d$ while $(g h)^{2} \neq i d$ .

Facts & Assumptions

Given: $X = {1, 2, 3}$ with $1$ , $2$ , $3$ pairwise distinct; $Sym (X)$ with composition $(f \circ k) (x) = f (k (x))$ and identity $i d$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition); and $g = (1 2)$ , $h = (2 3)$ , so $g$ exchanges $1$ and $2$ and fixes $3$ , while $h$ exchanges $2$ and $3$ and fixes $1$ .

[L1]

$Sym (X)$ is a group under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

[L2]

$f^{2} = f^{σ (1)} = f^{1} \circ f = f \circ f$ , and $f^{0} = i d$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

Two permutations are equal exactly when they agree at every point of $X$ (Injection, surjection, bijection, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L4]

The refuted claim: $(g h)^{n} = g^{n} h^{n}$ in every group, for all elements and all $n$ .

Counterexample

technique · direct

1.1

$g \circ g = i d$ : $g$ sends $1 \mapsto 2 \mapsto 1$ and $2 \mapsto 1 \mapsto 2$ and fixes $3$ twice. Hence $g^{2} = i d$ .

L2L3given

1.2

$h \circ h = i d$ for the same reason, with the pair $2, 3$ in place of $1, 2$ . Hence $h^{2} = i d$ .

L2L3given

1.3

The composite $g \circ h$ sends $1 \mapsto h (1) = 1 \mapsto g (1) = 2$ , sends $2 \mapsto h (2) = 3 \mapsto g (3) = 3$ , and sends $3 \mapsto h (3) = 2 \mapsto g (2) = 1$ .

L3given

2.1

$g^{2} h^{2} = i d \circ i d = i d$ .

step 1.1step 1.2L1

2.2

$(g h)^{2} = (g \circ h) \circ (g \circ h)$ sends $1 \mapsto 2 \mapsto 3$ by step 1.3, so it does not fix $1$ , since $3 \neq 1$ .

step 1.3L2L3given

3.1

Therefore $(g h)^{2} \neq i d = g^{2} h^{2}$ , and the claim of [L4] is false; the elements $g$ and $h$ do not commute, since $g \circ h$ sends $2$ to $3$ while $h \circ g$ sends $2 \mapsto g (2) = 1 \mapsto h (1) = 1$ .

step 2.1step 2.2L3L4∎

Remarks

What survives without commutativity is the expansion $(g h)^{2} = g h g h$ , which is all that associativity gives. Turning it into $g^{2} h^{2}$ requires exchanging the middle $h$ and $g$ , and that exchange is the commuting hypothesis in Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, nothing weaker.
Why this matters for the library. The published Laws of integer exponents states $(a b)^{n} = a^{n} b^{n}$ with no commuting hypothesis and is correct, because it is a statement about a field, where multiplication is commutative by definition. Carrying that form over to groups would be a false generalisation, and this item is the witness.
The same pair $g, h$ witnesses that $Sym ({1, 2, 3})$ is not abelian ( $Sym ({1, 2, 3})$ has exactly six elements, is non-abelian, and its elements have orders $1$ , $2$ and $3$ ), which is no accident. At $n = 2$ the equation $(g h)^{2} = g^{2} h^{2}$ reads $g h g h = g g h h$ , and cancelling $g$ on the left and $h$ on the right (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution) turns it into $h g = g h$ ; so at this exponent the law holds for a pair exactly when the pair commutes, and every non-abelian group supplies a counterexample.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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)
Abelian group (Wikipedia) (standard reference, not scraped)
Exponentiation (Wikipedia) (standard reference, not scraped)

(gh)n=gnhn(gh)n=gnhn fails without commutativity: two transpositions in Sym⁡({1,2,3})Sym({1,2,3}) with (gh)2≠g2h2(gh)2=g2h2