CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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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$(gh)^{n} = g^{n}h^{n}$ fails without commutativity: two transpositions in $\operatorname{Sym}(\{1,2,3\})$ with $(gh)^{2} \ne g^{2}h^{2}$

Statement refuted

False claim: in every group $G$ , $(gh)^{n} = g^{n} h^{n}$ 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 $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ 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 $\operatorname{Sym}(\{1,2,3\})$ take the transpositions $g = (1\,2)$ and $h = (2\,3)$ . Then $g^{2} h^{2} = \mathrm{id}$ while $(gh)^{2} \ne \mathrm{id}$ .

Facts & Assumptions

Given: $X = \{1,2,3\}$ with $1$ , $2$ , $3$ pairwise distinct; $\operatorname{Sym}(X)$ with composition $(f \circ k)(x) = f(k(x))$ and identity $\mathrm{id}$ (The symmetric group $\operatorname{Sym}(X)$ : 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]

$\operatorname{Sym}(X)$ is a group under composition ( $\operatorname{Sym}(X)$ 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]

$f^{2} = f^{\sigma(1)} = f^{1} \circ f = f \circ f$ , and $f^{0} = \mathrm{id}$ (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 $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition).

[L4]

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

Counterexample

technique · direct

1.1

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

L2L3given

1.2

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

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

$g^{2} h^{2} = \mathrm{id} \circ \mathrm{id} = \mathrm{id}$ .

step 1.1step 1.2L1

2.2

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

step 1.3L2L3given

3.1

Therefore $(gh)^{2} \ne \mathrm{id} = g^{2} h^{2}$ , 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 $(gh)^{2} = ghgh$ , 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 $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute, nothing weaker.
Why this matters for the library. The published Laws of integer exponents states $(ab)^{n} = a^{n}b^{n}$ 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 $\operatorname{Sym}(\{1,2,3\})$ is not abelian ( $\operatorname{Sym}(\{1,2,3\})$ 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 $(gh)^{2} = g^{2}h^{2}$ reads $ghgh = gghh$ , and cancelling $g$ on the left and $h$ on the right (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $gx = h$ and $xg = h$ each have exactly one solution) turns it into $hg = gh$ ; 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 56 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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} = g^{n}h^{n}(gh)n=gnhn fails without commutativity: two transpositions in Sym⁡({1,2,3})\operatorname{Sym}(\{1,2,3\})Sym({1,2,3}) with (gh)2≠g2h2(gh)^{2} \ne g^{2}h^{2}(gh)2=g2h2