LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-13

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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.
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A normal subgroup of $A_{n}$ containing one $3$ -cycle equals $A_{n}$ for $n \geq 5$

Statement

Let $n \geq 5$ . If $N ⊴ A_{n}$ contains one $3$ -cycle, then $N = A_{n}$ .

Facts & Assumptions

Given: $n \geq 5$ and a normal subgroup $N ⊴ A_{n}$ containing a $3$ -cycle.

[F1]

Normality makes a subgroup contain every conjugate of each of its elements (Normal subgroup: invariance under conjugation).

[F2]

All $3$ -cycles form one conjugacy class in $A_{n}$ (All $3$ -cycles are conjugate in $A_{n}$ for $n \geq 5$ ).

[F3]

The $3$ -cycles generate $A_{n}$ ( $A_{n}$ is generated by $3$ -cycles for every $n \geq 3$ ).

Proof

technique · direct

1.1

By [F1], $N$ contains the entire $A_{n}$ -conjugacy class of its given $3$ -cycle.

2.1

By [F2], this means that $N$ contains every $3$ -cycle.

F2step 1.1

3.1

Since those cycles generate $A_{n}$ by [F3], one has $N = A_{n}$ .

F3step 2.1∎

Depends on

Used by

Aₙ is simple for every n≥5 Theorem

Dependency tree · two levels

12 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

T. Judson, Abstract Algebra: Theory and Applications, Simplicity of $A_n$ (standard reference, not scraped)

A normal subgroup of AnAn​ containing one 33-cycle equals AnAn​ for n≥5n≥5