CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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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.
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The Fitting subgroup of $A_{5}$ does not contain its centralizer

Statement refuted

For every finite group $G$ , one has $C_{G} (F (G)) \leq F (G)$ . For $A_{5}$ , one has $F (A_{5}) = 1$ and $C_{A_{5}} (F (A_{5})) = A_{5}$ , so solvability cannot be omitted.

Facts & Assumptions

Given: The alternating group $A_{5}$ and its Fitting subgroup (The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group).

[L1]

For every finite group $G$ , $F (G)$ is nilpotent and normal, and every normal nilpotent subgroup is contained in it (The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group).

[L2]

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

[L3]

The group $A_{5}$ is not solvable ( $A_{5}$ and $S_{n}$ for $n \geq 5$ are not solvable).

[L4]

Every nilpotent group is solvable (Nilpotent groups, and in particular finite $p$ -groups, are solvable).

[F1]

The centralizer $C_{G} (H)$ consists of the elements of $G$ that commute with every element of $H$ (The centralizer $C_{G} (H)$ of a subgroup).

Counterexample

technique · direct

1.1

By [L1], $F (A_{5})$ is normal and nilpotent. Simplicity [L2] leaves $F (A_{5}) = 1$ or $F (A_{5}) = A_{5}$ ; the second would make $A_{5}$ nilpotent and hence solvable by [L4], contradicting [L3]. Thus $F (A_{5}) = 1$ .

givenL1L2L3L4algebra

2.1

For $A_{5}$ , one has $F (A_{5}) = 1$ and $C_{A_{5}} (F (A_{5})) = A_{5}$ , since every element centralizes the trivial subgroup by [F1]. Therefore $C_{A_{5}} (F (A_{5})) ≰ F (A_{5})$ .

step 1.1F1givenalgebra∎

Depends on

Used by

FALSE: the Fitting subgroup always contains its centralizer False statement

Dependency tree · two levels

27 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

D. A. Craven, The Theory of p-Groups, §1.1 (standard reference, not scraped)
T. Judson, Abstract Algebra: Theory and Applications, Simplicity of $A_n$ (standard reference, not scraped)

The Fitting subgroup of A5A5​ does not contain its centralizer