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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

$\alpha=(0\,1\,2)(3\,4),\qquad \beta=(0\,3)(5\,6),\qquad \sigma=\alpha\beta.$

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

$\sigma=(0\,4\,3\,1\,2)(5\,6).$

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 $\beta$ and then $\alpha$ gives $\sigma(0)=4$ , $\sigma(1)=2$ , $\sigma(2)=0$ , $\sigma(3)=1$ , $\sigma(4)=3$ , $\sigma(5)=6$ , and $\sigma(6)=5$ , so the one-line form is $[4,2,0,1,3,6,5]$ .

givenL1

2.1

Starting at $0$ gives $0\mapsto4\mapsto3\mapsto1\mapsto2\mapsto0$ , and the only unused points satisfy $5\mapsto6\mapsto5$ . These cycles are disjoint and reproduce every value in step 1.1, so $\sigma=(0\,4\,3\,1\,2)(5\,6)$ .

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 $\alpha=(0\,1)$ and $\beta=(1\,2)$ 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, $\alpha\beta(0)=\alpha(0)=1$ , whereas $\beta\alpha(0)=\beta(1)=2$ .

givenL1

2.1

Therefore $\alpha\beta\ne\beta\alpha$ . 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
$\operatorname{id}$	$[0,1,2]$	$0$	$+1$
$(0\,1)$	$[1,0,2]$	$1$	$-1$
$(0\,2)$	$[2,1,0]$	$3$	$-1$
$(1\,2)$	$[0,2,1]$	$1$	$-1$
$(0\,1\,2)$	$[1,2,0]$	$2$	$+1$
$(0\,2\,1)$	$[2,0,1]$	$2$	$+1$

Thus $A_3=\{\operatorname{id},(0\,1\,2),(0\,2\,1)\}$ .

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 $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$ , and even and odd permutations, The alternating group $A_n=\ker(\operatorname{sgn})$ of even permutations).

[L2]

A three-element set has $3!=6$ permutations (A finite set $A$ with $\lvert A\rvert = n$ 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 $\sigma$ has parity $(-1)^{\operatorname{inv}(\sigma)}$ ).

[L2]

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

Verification

technique · direct

1.1

The product $(0\,2)(0\,1)$ sends $0\mapsto1\mapsto2\mapsto0$ , 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 $\operatorname{sgn}(\sigma)=(-1)^{n-c(\sigma)}$ 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 $\operatorname{sgn}(\sigma)=(-1)^{n-c(\sigma)}$ 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 $n\ge2$ , $2\,|A_n|=n!$ , 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 $H\le A_4$ 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 $ugu^{-1}=(b\,d\,c)=:h\in H$ . Likewise $h\in A_4$ and $hgh^{-1}=(a\,d\,b)=:k\in H$ . 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∎

Sources

Standard references

Recommended treatments; not extraction sources.

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 S3S_3S3​: 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

A4A_4A4​ consists of the identity, eight 333-cycles, and three products of disjoint transpositions

Example

Facts & Assumptions

Verification

A4A_4A4​ has no subgroup of order 666

Example

Facts & Assumptions

Verification

Sources

Standard references

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 $6$