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Symmetric Groups and the Sign Homomorphism: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

From one-line notation to a disjoint-cycle decomposition, with the right-hand factor acting first

Example

In $S_{7}$ , let

$α = (0 1 2) (3 4), β = (0 3) (5 6), σ = α β .$

The right-hand factor acts first. The one-line form of $σ$ is $[4, 2, 0, 1, 3, 6, 5]$ , and its disjoint-cycle decomposition is

$σ = (0 4 3 1 2) (5 6) .$

Facts & Assumptions

Given: The displayed permutations in $S_{7}$ , with composition from right to left.

[L1]

A permutation of a finite set is recovered by following each unused point until its orbit closes, producing its disjoint-cycle decomposition (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation).

Verification

technique · direct

1.1

Applying $β$ and then $α$ gives $σ (0) = 4$ , $σ (1) = 2$ , $σ (2) = 0$ , $σ (3) = 1$ , $σ (4) = 3$ , $σ (5) = 6$ , and $σ (6) = 5$ , so the one-line form is $[4, 2, 0, 1, 3, 6, 5]$ .

givenL1

2.1

Starting at $0$ gives $0 \mapsto 4 \mapsto 3 \mapsto 1 \mapsto 2 \mapsto 0$ , and the only unused points satisfy $5 \mapsto 6 \mapsto 5$ . These cycles are disjoint and reproduce every value in step 1.1, so $σ = (0 4 3 1 2) (5 6)$ .

step 1.1L1∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

Overlapping cycles need not commute

Statement refuted

False claim. Any two cycles commute, even when their supports overlap.

Facts & Assumptions

Given: The cycles $α = (0 1)$ and $β = (1 2)$ in $S_{3}$ .

[L1]

Cycles with disjoint supports commute; disjointness is a hypothesis of that result (Cycles with disjoint supports commute).

Counterexample

technique · direct

1.1

With the right-hand factor acting first, $α β (0) = α (0) = 1$ , whereas $β α (0) = β (1) = 2$ .

givenL1

2.1

Therefore $α β \neq β α$ . Their supports overlap at $1$ , so this explicit pair refutes the claim and shows why the disjoint-support hypothesis in [L1] is necessary.

step 1.1L1∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

The six elements of $S_{3}$ : one-line form, cycle structure, inversions, and sign

Example

The elements of $S_{3}$ have the following one-line forms, cycle decompositions, inversion numbers, and signs:

permutation	one-line form	inversion number	sign
$id$	$[0, 1, 2]$	$0$	$+ 1$
$(0 1)$	$[1, 0, 2]$	$1$	$- 1$
$(0 2)$	$[2, 1, 0]$	$3$	$- 1$
$(1 2)$	$[0, 2, 1]$	$1$	$- 1$
$(0 1 2)$	$[1, 2, 0]$	$2$	$+ 1$
$(0 2 1)$	$[2, 0, 1]$	$2$	$+ 1$

Thus $A_{3} = {id, (0 1 2), (0 2 1)}$ .

Facts & Assumptions

Given: The symmetric group $S_{3}$ acting on $3 = {0, 1, 2}$ .

[L1]

The inversion number counts decreasing pairs in one-line notation, sign is $(- 1)$ to that number, and $A_{3}$ is the set of even permutations (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations, The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[L2]

A three-element set has $3! = 6$ permutations (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Verification

technique · direct

1.1

By [L2] there are $3! = 6$ permutations. Listing the three transpositions, the two three-cycles, and the identity gives six distinct maps. Counting decreasing pairs in each displayed one-line form gives respectively $0, 1, 3, 1, 2, 2$ , and [L1] gives the signs shown.

givenL1L2

2.1

The rows with sign $+ 1$ are exactly the identity and the two three-cycles, so the displayed set is precisely $A_{3}$ .

step 1.1L1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

Two different transposition factorisations of the same permutation have the same parity

Example

In $S_{3}$ , the three-cycle $(0 1 2)$ has the two factorizations

$(0 1 2) = (0 2) (0 1) = (0 2) (0 1) (1 2) (1 2) .$

Their lengths are $2$ and $4$ , so both are even, as the parity theorem predicts.

Facts & Assumptions

Given: Products in $S_{3}$ act from right to left.

[L1]

Every transposition factorisation of a permutation has parity equal to its inversion sign (Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$ ).

[L2]

Inversion sign is $(- 1)$ raised to the number of decreasing pairs in one-line notation (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

Verification

technique · direct

1.1

The product $(0 2) (0 1)$ sends $0 \mapsto 1 \mapsto 2 \mapsto 0$ , hence equals $(0 1 2)$ , and the final pair $(1 2) (1 2)$ is the identity, so the four-factor product is the same permutation.

givenL1

2.1

The one-line form of $(0 1 2)$ is $[1, 2, 0]$ , with two inversions and sign $+ 1$ by [L2]. The factor counts $2$ and $4$ are both even and therefore both have parity $+ 1$ , in agreement with [L1].

step 1.1L1L2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

A five-cycle is even and a six-cycle is odd

Example

The cycle $(0 1 2 3 4)$ is even, while $(0 1 2 3 4 5)$ is odd.

Facts & Assumptions

Given: The displayed cycles in symmetric groups on sets containing their entries.

[L1]

A cycle of length $k$ has sign $(- 1)^{k - 1}$ (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

Verification

technique · direct

1.1

The five-cycle is a product of $5 - 1 = 4$ transpositions and has sign $(- 1)^{4} = + 1$ ; the six-cycle is a product of $6 - 1 = 5$ transpositions and has sign $(- 1)^{5} = - 1$ .

givenL1

2.1

Thus the five-cycle is even and the six-cycle is odd. The shift by one comes from the number of transpositions in the standard cycle factorisation.

step 1.1L1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

$A_{4}$ consists of the identity, eight $3$ -cycles, and three products of disjoint transpositions

Example

The alternating group $A_{4}$ consists of

the identity;
the eight three-cycles $(0 1 2)$ , $(0 2 1)$ , $(0 1 3)$ , $(0 3 1)$ , $(0 2 3)$ , $(0 3 2)$ , $(1 2 3)$ , and $(1 3 2)$ ;
the three products $(0 1) (2 3)$ , $(0 2) (1 3)$ , and $(0 3) (1 2)$ .

Facts & Assumptions

Given: The symmetric group $S_{4}$ and its alternating subgroup $A_{4}$ .

[L1]

Every permutation has a unique disjoint-cycle type, and a cycle of length $k$ has sign $(- 1)^{k - 1}$ (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation, A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[L2]

The alternating group has $4! / 2 = 12$ elements in degree $4$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

Verification

technique · direct

1.1

The possible cycle types in $S_{4}$ are the identity; one transposition; two disjoint transpositions; a three-cycle with one fixed point; and a four-cycle. By [L1], exactly the identity, the three-cycles, and the products of two disjoint transpositions are even. There are two orientations on each of the four three-point supports and three partitions into two unordered pairs, giving exactly the displayed list.

givenL1

2.1

The list has $1 + 8 + 3 = 12$ elements, and [L2] gives $∣ A_{4} ∣ = 4! / 2 = 12$ , so it contains every even permutation and no other element.

step 1.1L1L2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

$A_{4}$ has no subgroup of order $6$

Example

The group $A_{4}$ has no subgroup of order $6$ .

Facts & Assumptions

Given: The explicit list of the elements of $A_{4}$ and a hypothetical subgroup $H \leq A_{4}$ with $∣ H ∣ = 6$ .

[L1]

$A_{4}$ consists of the identity, eight three-cycles, and three products of disjoint transpositions ( $A_{4}$ consists of the identity, eight $3$ -cycles, and three products of disjoint transpositions).

[L2]

Every subgroup of index $2$ is normal (Every subgroup of index two is normal).

[L3]

If $H$ is a subgroup of a finite group $G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Verification

technique · contradiction

1.1

Suppose, for contradiction, that $∣ H ∣ = 6$ . Since $∣ A_{4} ∣ = 12$ by [L1], [L3] gives $[A_{4} : H] = 2$ , and [L2] makes $H$ normal.

assume-contraL1L2L3

2.1

The complement of $H$ in $A_{4}$ has six elements, so it cannot contain all eight three-cycles from [L1]; hence $H$ contains a three-cycle $g$ .

step 1.1L1

3.1

Write $g = (a b c)$ and let $d$ be the fourth symbol. The three-cycle $u = (a b d)$ belongs to $A_{4}$ by [L1], so normality and direct evaluation give $u g u^{- 1} = (b d c) = : h \in H$ . Likewise $h \in A_{4}$ and $h g h^{- 1} = (a d b) = : k \in H$ . Subgroup closure also puts $g^{- 1}, h^{- 1}, k^{- 1}$ in $H$ .

step 2.1L1

4.1

The seven elements $e, g, g^{- 1}, h, h^{- 1}, k, k^{- 1}$ are distinct: their displayed supports or orientations differ. This contradicts $∣ H ∣ = 6$ , so no subgroup of order $6$ exists.

step 3.1L1discharge-contradiction∎

Symmetric Groups and the Sign Homomorphism: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

From one-line notation to a disjoint-cycle decomposition, with the right-hand factor acting first

Example

Facts & Assumptions

Verification

Overlapping cycles need not commute

Statement refuted

Facts & Assumptions

Counterexample

The six elements of S3S3​: one-line form, cycle structure, inversions, and sign

Example

Facts & Assumptions

Verification

Two different transposition factorisations of the same permutation have the same parity

Example

Facts & Assumptions

Verification

A five-cycle is even and a six-cycle is odd

Example

Facts & Assumptions

Verification

A4A4​ consists of the identity, eight 33-cycles, and three products of disjoint transpositions

Example

Facts & Assumptions

Verification

A4A4​ has no subgroup of order 66

Example

Facts & Assumptions

Verification

Sources

The six elements of $S_{3}$ : one-line form, cycle structure, inversions, and sign

$A_{4}$ consists of the identity, eight $3$ -cycles, and three products of disjoint transpositions

$A_{4}$ has no subgroup of order $6$