CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

1.1F3F4algebra

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

1.2F1F4algebra

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 $n \geq 3$ .

1.3F1F4algebra

For $n \geq 5$ , put $α = (0 1 2)$ and $β = (2 3 4)$ in $A_{n}$ . These do not commute, since $(α β) (2) = 3$ whereas $(β α) (2) = 0$ . Hence $[A_{n}, A_{n}]$ is nontrivial by [F1].

2.1F1F4step 1.1step 1.2

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 $n \geq 2$ .

3.1F2F5step 1.3∎

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

Depends on

Used by

A composition series and the derived series of S₃ Example
Composition and derived series of S₄ Example
Every solvable group is abelian False statement
A₅ and Sₙ for n≥5 are not solvable Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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