CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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$[S_{n}, S_{n}] = A_{n}$ for $n \geq 2$ , and $[A_{n}, A_{n}] = A_{n}$ for $n \geq 5$

Statement

For $n \geq 2$ , $[S_{n}, S_{n}] = A_{n}$ . For $n \geq 5$ , $[A_{n}, A_{n}] = A_{n}$ .

Facts & Assumptions

Given: The groups $S_{n}$ and $A_{n}$ in the stated ranges of $n$ .

[F1]

$[G, G]$ is generated by the commutators $[g, h] = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F2]

The commutator subgroup is normal (The commutator subgroup is normal).

[F3]

For $N ⊴ G$ , the quotient $G / N$ is abelian exactly when $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[F4]

$A_{n}$ is the kernel of sign (The alternating group $A_{n} = \ker (sgn)$ of even permutations) and is generated by $3$ -cycles for $n \geq 3$ ( $A_{n}$ is generated by $3$ -cycles for every $n \geq 3$ ).

[F5]

$A_{n}$ is simple for $n \geq 5$ ( $A_{n}$ is simple for every $n \geq 5$ ).

Proof

technique · direct

1.1

Since $S_{n} / A_{n}$ is the abelian sign image, [F3] gives $[S_{n}, S_{n}] \subseteq A_{n}$ .

F3F4algebra

1.2

Every $3$ -cycle satisfies $(a b c) = [(b c), (a b)]$ under the convention in [F1], so [F4] gives $A_{n} \subseteq [S_{n}, S_{n}]$ for $n \geq 3$ .

F1F4algebra

1.3

For $n \geq 5$ , the $3$ -cycles $(1 2 3)$ and $(3 4 5)$ do not commute, so $[A_{n}, A_{n}]$ is nontrivial by [F1].

F1algebra

2.1

At $n = 2$ , both $A_{2}$ and the commutator subgroup of the abelian group $S_{2}$ are trivial. Thus steps 1.1--1.2 prove the first formula for all $n \geq 2$ .

F1F4step 1.1step 1.2

3.1

This subgroup is normal by [F2], and simplicity [F5] therefore forces $[A_{n}, A_{n}] = A_{n}$ .

F2F5step 1.3∎

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: 44 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

J. S. Milne, Group Theory (standard reference, not scraped)

[Sn,Sn]=An[Sn​,Sn​]=An​ for n≥2n≥2, and [An,An]=An[An​,An​]=An​ for n≥5n≥5