False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

False statement: all subgroups of the same $p$ -power order are conjugate

Statement

False claim: all subgroups of the same $p$ -power order are conjugate. See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

Let $n \in N$ , so that $n = {0, 1, \dots, n - 1}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_{n} : = Sym (n) = Sym ({0, 1, \dots, n - 1}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[L3]

For $σ, τ \in S_{n}$ , there is a $g \in S_{n}$ with $τ = g σ g^{- 1}$ if and only if $σ$ and $τ$ have the same cycle type, including their numbers of fixed points. (Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type).

Refutation

technique · direct

1.1L1L2L3givenalgebra

In the $S_{4} = Sym ({0, 1, 2, 3})$ of [L2], the order-two subgroups generated by $(0 1)$ and $(0 1) (2 3)$ cannot be conjugate because their nonidentity generators have different cycle types.

2.1step 1.1L1L3givenalgebra∎

Two subgroups of order $2$ are conjugate exactly when their nonidentity elements are, so step 1.1 exhibits two subgroups of the same $2$ -power order that are not conjugate, refuting the universal claim. No conflict with [L1] arises: both have order $2$ while the Sylow $2$ -subgroups of $S_{4}$ have order $8$ , and [L1] asserts conjugacy only among subgroups of that maximal $p$ -power order. This proves the stated claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Keith Conrad, Consequences of the Sylow Theorems, Sections 1-5 (standard reference, not scraped)

False statement: all subgroups of the same pp-power order are conjugate