LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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The symmetric groups $S_{n}$ are solvable for $n \leq 4$

Statement

The symmetric groups $S_{n}$ are solvable for every $n \leq 4$ .

Facts & Assumptions

Given: The symmetric groups on at most four letters.

[L1]

A group is solvable exactly when it has a finite subnormal series with abelian factors (A group is solvable if and only if it has a subnormal series with abelian factors).

[L2]

The alternating group is the sign kernel, and sign is a homomorphism (The alternating group $A_{n} = \ker (sgn)$ of even permutations, The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ).

[L3]

Conjugation relabels cycles and therefore relabels a double transposition to another double transposition (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

Proof

technique · direct

1.1

The groups $S_{0}$ and $S_{1}$ are trivial, and $S_{2}$ is cyclic of order $2$ , so all three are solvable by [L1].

L1algebra

1.2

For $S_{3}$ , the subgroup $A_{3}$ is normal as the sign kernel by [L2], has order $3$ , and the quotient $S_{3} / A_{3}$ has order $2$ . Thus $1 ◃ A_{3} ◃ S_{3}$ is a subnormal series with abelian factors, so $S_{3}$ is solvable by [L1].

L1L2algebra

1.3

In $A_{4}$ , let $V_{4} = {1, (12) (34), (13) (24), (14) (23)} .$ Each listed double transposition is even, and direct multiplication shows that these four elements form a subgroup. By [L3], conjugation by any element of $A_{4}$ permutes the three nonidentity double transpositions, so $V_{4} ⊴ A_{4}$ . The quotient $A_{4} / V_{4}$ has order $3$ , hence is abelian, and $V_{4}$ is abelian of order $4$ . Since $A_{4}$ is the sign kernel in $S_{4}$ , the chain $1 ◃ V_{4} ◃ A_{4} ◃ S_{4}$ has abelian factors $V_{4}$ , $A_{4} / V_{4}$ , and $S_{4} / A_{4}$ . Therefore $S_{4}$ is solvable by [L1].

L1L2L3algebra

2.1

Steps 1.1, 1.2, and 1.3 cover every $n \leq 4$ .

step 1.1step 1.2step 1.3∎

Depends on

Used by

Every polynomial of degree at most four is solvable by radicals Corollary

Dependency tree · two levels

17 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, Chapter 6 (standard reference, not scraped)
J. Ash, Basic Abstract Algebra, Chapter 4 (standard reference, not scraped)

The symmetric groups SnSn​ are solvable for n≤4n≤4