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✓ 8 results · all verified · 0 also independently AI-judged

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Conjugacy in $S_{n}$ , Generation, and the Simplicity of $A_{n}$ — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Conjugating $(1 4) (2 5 3)$ by an explicit permutation in $S_{5}$

Example

Let $σ = (1 4) (2 5 3)$ and $g = (1 2 3 4 5)$ in $S_{5}$ . Then $g σ g^{- 1} = (2 5) (3 1 4) .$

Facts & Assumptions

Given: $σ = (1 4) (2 5 3)$ and $g = (1 2 3 4 5)$ in $S_{5}$ .

[F1]

Conjugating a cycle relabels each entry by the conjugating permutation (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F2]

Permutations in $S_{n}$ are conjugate exactly when they have the same cycle type (Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type).

Verification

technique · direct

1.1

The relabellings are $(g (1) g (4)) = (2 5)$ and $(g (2) g (5) g (3)) = (3 1 4)$ .

algebra

2.1

Apply [F1] to the two disjoint factors to obtain the displayed conjugate.

F1step 1.1

3.1

The result has one $2$ -cycle and one $3$ -cycle, hence the same cycle type as $σ$ , as [F2] requires; conjugating back by $g^{- 1}$ recovers $σ$ .

F2step 2.1algebra∎

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-13Open item page →

The five conjugacy classes of $S_{4}$ and the class equation $24 = 1 + 6 + 3 + 8 + 6$

Example

The conjugacy data for $S_{4}$ are

cycle type	representative	centralizer size	class size	parity
$1^{4}$	$1$	$24$	$1$	even
$2, 1^{2}$	$(12)$	$4$	$6$	odd
$2^{2}$	$(12) (34)$	$8$	$3$	even
$3, 1$	$(123)$	$3$	$8$	even
$4$	$(1234)$	$4$	$6$	odd

Facts & Assumptions

Given: The symmetric group $S_{4}$ .

[F1]

A cycle type $(c_{k})$ has centralizer size $\prod k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F2]

$S_{n}$ -classes are indexed by the tuples with $\sum k c_{k} = n$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ ), and $n! = \sum n! / \prod_{k} k^{c_{k}} c_{k}!$ over those tuples, the summand indexed by $(c_{k})$ being the size of the corresponding class (The class equation of $S_{n}$ is $n! = \sum_{\sum k c_{k} = n} n! / \prod_{k} k^{c_{k}} c_{k}!$ ).

[F3]

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

Verification

technique · counting

1.1

The five partitions of $4$ give exactly the five cycle types in the table.

F2algebra

2.1

Applying [F1] gives centralizer sizes $24, 4, 8, 3, 4$ ; dividing $24$ by them gives class sizes $1, 6, 3, 8, 6$ .

F1F2step 1.1algebra

3.1

Their sum is $24 = 1 + 6 + 3 + 8 + 6$ , which verifies [F2].

F2step 2.1algebra

4.1

Applying [F3] to the representatives gives the parity column.

F3step 1.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The seven conjugacy classes of $S_{5}$ and their centralizer and class sizes

Example

The conjugacy data for $S_{5}$ are

cycle type	centralizer size	class size	parity	splits in $A_{5}$ ?
$1^{5}$	$120$	$1$	even	no
$2, 1^{3}$	$12$	$10$	odd	--
$2^{2}, 1$	$8$	$15$	even	no
$3, 1^{2}$	$6$	$20$	even	no
$3, 2$	$6$	$20$	odd	--
$4, 1$	$4$	$30$	odd	--
$5$	$5$	$24$	even	yes

Facts & Assumptions

Given: The symmetric group $S_{5}$ .

[F1]

A cycle type $(c_{k})$ has centralizer size $\prod k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F2]

[F3]

[F4]

For $n \geq 2$ and $σ \in A_{n}$ , the $S_{n}$ -class of $σ$ splits into two $A_{n}$ -classes of equal size exactly when all cycle lengths in its decomposition, including $1$ -cycles for fixed points, are odd and no two are equal (For $n \geq 2$ , an $S_{n}$ -class of an even permutation splits in $A_{n}$ exactly when all cycle lengths, including $1$ -cycles, are odd and distinct).

Verification

technique · counting

1.1

The seven partitions of $5$ give the seven rows. Applying [F1] gives the centralizer column, and [F2] gives the class-size column.

F1F2algebra

2.1

The sizes sum to $1 + 10 + 15 + 20 + 20 + 30 + 24 = 120$ , verifying the class equation.

F2step 1.1algebra

3.1

Formula [F3] gives the parity column. Among the even rows, [F4] applies only to the single cycle of length $5$ , giving the final column.

F3F4step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The conjugacy classes of $A_{5}$ : sizes $1, 20, 15, 12, 12$ and the split $5$ -cycles

Example

The five conjugacy classes of $A_{5}$ have representatives and sizes

1 : 1, (123) : 20, (12) (34) : 15,

(12345) : 12, (13524) : 12.

The last two classes are the two halves of the $S_{5}$ class of $5$ -cycles.

Facts & Assumptions

Given: The alternating group $A_{5}$ .

[F1]

$S_{n}$ -classes are indexed by the tuples with $\sum k c_{k} = n$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ ); a permutation of type $(c_{k})$ has centralizer cardinality $\prod_{k} k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ); and $n! = \sum n! / \prod_{k} k^{c_{k}} c_{k}!$ over those tuples, the summand indexed by $(c_{k})$ being the size of the corresponding class (The class equation of $S_{n}$ is $n! = \sum_{\sum k c_{k} = n} n! / \prod_{k} k^{c_{k}} c_{k}!$ ).

[F2]

[F3]

[F4]

$∣ A_{5} ∣ = 5! / 2 = 60$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

Verification

technique · counting

1.1

By [F1] and [F2], the even $S_{5}$ types are $1^{5}$ , $3, 1^{2}$ , $2^{2}, 1$ , and $5$ , with symmetric class sizes $1, 20, 15, 24$ .

F1F2algebra

2.1

By [F3], the first three stay single classes, while the $5$ -cycle class splits into two equal classes of size $12$ .

F3step 1.1

3.1

Put $s = (12345)$ . The permutation $q = (2 3 5 4)$ satisfies $q s q^{- 1} = s^{2} = (13524)$ and is odd by [F2]. Every other conjugator from $s$ to $s^{2}$ differs from $q$ by an element centralizing $s$ ; such a centralizer element is determined by the image of $1$ and is therefore a power of the even $5$ -cycle $s$ . Thus every conjugator is odd, so $s$ and $s^{2}$ lie in the two different halves from step 2.1.

F2step 2.1algebra

4.1

The total $1 + 20 + 15 + 12 + 12 = 60$ agrees with [F4].

F4step 1.1step 2.1∎

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-13Open item page →

$V_{4} = {1, (12) (34), (13) (24), (14) (23)}$ is a proper nontrivial normal subgroup of $A_{4}$

Example

In $A_{4}$ , set $V_{4} = {1, (12) (34), (13) (24), (14) (23)} .$ Then $V_{4}$ is a proper nontrivial normal subgroup.

Facts & Assumptions

Given: The displayed subset $V_{4} \subseteq A_{4}$ .

[F1]

$A_{4}$ is the kernel of sign (The alternating group $A_{n} = \ker (sgn)$ of even permutations), and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are included as $1$ -cycles (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[F2]

Conjugation relabels every cycle entry (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F3]

A subgroup is normal when it is invariant under conjugation (Normal subgroup: invariance under conjugation).

[F4]

$∣ A_{4} ∣ = 4! / 2 = 12$ ( $A_{n}$ 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

Every displayed double transposition has two cycles and hence sign $(- 1)^{4 - 2} = + 1$ by [F1]. The product of two distinct nonidentity displayed elements is the third, and each is its own inverse; hence $V_{4}$ is a subgroup of $A_{4}$ .

F1algebra

1.2

By [F2], conjugation by any permutation relabels a double transposition to another double transposition. Thus $V_{4}$ is invariant under $A_{4}$ -conjugation, and [F3] makes it normal.

F2F3

2.1

Its order is $4$ , strictly between $1$ and $12 = ∣ A_{4} ∣$ from [F4], so it is nontrivial and proper.

F4algebra∎

False statementConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-13Open item page →

FALSE: two even permutations of the same cycle type are always conjugate in $A_{n}$

Statement refuted

Two elements of $A_{n}$ with the same cycle type must be conjugate in $A_{n}$ .

Facts & Assumptions

Given: The cycles $(123)$ and $(132)$ in $A_{4}$ .

[F1]

A $k$ -cycle 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), and $A_{n}$ is the kernel of sign (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[F2]

Conjugation relabels cycle entries (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F3]

Counterexample

technique · counterexample

1.1

In $A_{4}$ , the cycles $(123)$ and $(132)$ have sign $(- 1)^{2} = + 1$ by [F1] and have the same cycle type $(3, 1)$ .

F1algebra

2.1

The lengths $3$ and $1$ are odd and distinct, so [F3] says that the $S_{4}$ -class of $3$ -cycles splits into two $A_{4}$ -classes.

F3step 1.1

2.2

More explicitly, [F2] shows that conjugating $(123)$ to $(132)$ reverses the cyclic order on ${1, 2, 3}$ while fixing the remaining point; every such relabeling is odd. Hence no element of $A_{4}$ performs it.

F2step 1.1algebra

3.1

Thus the two displayed even permutations have the same cycle type but are not conjugate in $A_{4}$ .

step 1.1step 2.2∎

False statementConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-13Open item page →

FALSE: $A_{n}$ is simple for every $n \geq 4$

Statement refuted

$A_{n}$ is simple for every $n \geq 4$ .

Facts & Assumptions

Given: The boundary case $A_{4}$ .

[F1]

$A_{4}$ consists of the even permutations and has order $4! / 2 = 12$ (The alternating group $A_{n} = \ker (sgn)$ of even permutations, $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ); moreover $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are included as $1$ -cycles (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[F2]

Conjugation relabels cycle entries (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F3]

Conjugation-invariant subgroups are normal (Normal subgroup: invariance under conjugation), while a simple group has no proper nontrivial normal subgroup (Simple groups).

[F4]

The valid boundary is that $A_{n}$ is simple for every $n \geq 5$ ( $A_{n}$ is simple for every $n \geq 5$ ).

Counterexample

technique · counterexample

1.1

Each double transposition has two cycles and hence sign $(- 1)^{4 - 2} = + 1$ by [F1]. Thus $V = {1, (12) (34), (13) (24), (14) (23)}$ lies in $A_{4}$ , and direct multiplication shows it is a subgroup of order $4$ .

F1algebra

2.1

By [F2], conjugation relabels a double transposition to another member of $V$ . Thus [F3] makes $V ⊴ A_{4}$ .

F2F3step 1.1

2.2

Since $1 < 4 < 12$ , $V$ is proper and nontrivial, so [F3] shows that $A_{4}$ is not simple.

F1F3step 1.1

3.1

This refutes the proposed lower bound $4$ ; [F4] records that $5$ is the correct one.

F4step 2.2∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

FALSE: any transposition together with any $n$ -cycle generates $S_{n}$

Statement refuted

Every $n$ -cycle and every transposition together generate $S_{n}$ .

Facts & Assumptions

Given: The elements $c = (1234)$ and $t = (13)$ of $S_{4}$ .

[F1]

Relative to $(1 2 \dots n)$ , the neighboring transposition $(1 2)$ does generate $S_{n}$ with that cycle (For $n \geq 2$ , $(1 2 \dots n)$ and $(1 2)$ generate $S_{n}$ ).

[F2]

A generated subgroup is the smallest subgroup containing its generators (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F3]

Permutations act on the underlying set by composition (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Counterexample

technique · counterexample

1.1

In $S_{4}$ , let $c = (1234)$ and $t = (13)$ , and partition the symbols into $B_{1} = {1, 3}$ and $B_{2} = {2, 4}$ .

2.1

The cycle $c$ swaps $B_{1}$ and $B_{2}$ , while $t$ preserves each block. Therefore every word in $c, t$ preserves the two-block system setwise.

F2F3step 1.1

3.1

The permutation $(12)$ does not preserve that block system, so it is not in $⟨ c, t ⟩$ ; hence this generated subgroup is proper in $S_{4}$ .

F2step 2.1

4.1

Thus an arbitrary transposition need not work. The positive theorem [F1] requires a neighboring transposition relative to the chosen cycle.

F1step 3.1∎

Sources

Standard references

Recommended treatments; not extraction sources.

Conjugacy in SnSn​, Generation, and the Simplicity of AnAn​ — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Conjugating (1 4)(2 5 3)(14)(253) by an explicit permutation in S5S5​

Example

Facts & Assumptions

Verification

The five conjugacy classes of S4S4​ and the class equation 24=1+6+3+8+624=1+6+3+8+6

Example

Facts & Assumptions

Verification

The seven conjugacy classes of S5S5​ and their centralizer and class sizes

Example

Facts & Assumptions

Verification

The conjugacy classes of A5A5​: sizes 1,20,15,12,121,20,15,12,12 and the split 55-cycles

Example

Facts & Assumptions

Verification

V4={1,(12)(34),(13)(24),(14)(23)}V4​={1,(12)(34),(13)(24),(14)(23)} is a proper nontrivial normal subgroup of A4A4​

Example

Facts & Assumptions

Verification

FALSE: two even permutations of the same cycle type are always conjugate in AnAn​

Statement refuted

Facts & Assumptions

Counterexample

FALSE: AnAn​ is simple for every n≥4n≥4

Statement refuted

Facts & Assumptions

Counterexample

FALSE: any transposition together with any nn-cycle generates SnSn​

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard references

Conjugacy in $S_{n}$ , Generation, and the Simplicity of $A_{n}$ — Examples

Conjugating $(1 4) (2 5 3)$ by an explicit permutation in $S_{5}$

The five conjugacy classes of $S_{4}$ and the class equation $24 = 1 + 6 + 3 + 8 + 6$

The seven conjugacy classes of $S_{5}$ and their centralizer and class sizes

The conjugacy classes of $A_{5}$ : sizes $1, 20, 15, 12, 12$ and the split $5$ -cycles

$V_{4} = {1, (12) (34), (13) (24), (14) (23)}$ is a proper nontrivial normal subgroup of $A_{4}$

FALSE: two even permutations of the same cycle type are always conjugate in $A_{n}$

FALSE: $A_{n}$ is simple for every $n \geq 4$

FALSE: any transposition together with any $n$ -cycle generates $S_{n}$