False statementConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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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∎

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: 69 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

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

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