CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Squaring is not a homomorphism on a nonabelian group

Statement refuted

The squaring map $s (g) = g^{2}$ need not be a group homomorphism. It fails on the nonabelian symmetric group $S_{3}$ .

Facts & Assumptions

Given: The permutations $τ = (1 2)$ and $ρ = (2 3)$ of ${1, 2, 3}$ , composed from right to left.

[L1]

The symmetric group on a set consists of its permutations under composition (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L2]

The symmetric group on every set is a group, and on a set with three distinct elements it is nonabelian ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L3]

Natural powers in a group satisfy $g^{0} = e$ and $g^{n + 1} = g^{n} g$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

Group powers satisfy $(g^{m})^{n} = g^{m n}$ , and if $g h = h g$ then $(g h)^{n} = g^{n} h^{n}$ ; that last claim is false in general without its hypothesis, and the equation can fail already at $n = 2$ (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).

[L5]

A group homomorphism $f$ satisfies $f (x y) = f (x) f (y)$ for all $x, y$ (Monoid homomorphism and group homomorphism).

Counterexample

technique · direct

1.1

By [L1] and [L2], $τ, ρ \in S_{3}$ , and their product is $τ ρ = (1 2 3)$ under the stated composition convention.

L1L2given

2.1

The two transpositions satisfy $τ^{2} = ρ^{2} = e$ , while [L3] gives $(τ ρ)^{2} = (1 3 2) \neq e$ . Here $ρ τ = (1 3 2) \neq (1 2 3) = τ ρ$ , so $τ$ and $ρ$ do not commute and the hypothesis of the power law in [L4] is not met; [L4] records that the equation can fail at $n = 2$ in exactly that case.

L3L4step 1.1

3.1

Therefore $s (τ ρ) = (τ ρ)^{2} \neq e = τ^{2} ρ^{2} = s (τ) s (ρ)$ , so $s$ violates the homomorphism identity [L5].

L5step 2.1∎

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: 52 results over 18 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

W. Stein, Elementary Number Theory, Section 4.1 (standard reference, not scraped)