TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

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.

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

Statement

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. See Sylow I: every finite group has a Sylow $p$ -subgroup.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ , hence a Sylow $p$ -subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$ -subgroup).

[L2]

If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p) .$ . (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L3]

For a finite group $G$ and a prime $p$ , let ${Syl}_{p} (G)$ be the set of Sylow $p$ -subgroups (def-sylow-p-subgroup). Define $n_{p} (G) : = ∣ {Syl}_{p} (G) ∣ .$ This cardinal is defined even before existence is proved because ${Syl}_{p} (G)$ is a subset of the finite power set of $G$ ; thm-sylow-first-theorem later shows it is nonzero. (The number $n_{p} (G)$ of Sylow $p$ -subgroups).

[L4]

Let $H \leq G$ be a subgroup (def-subgroup). The normalizer of $H$ in $G$ is $N_{G} (H) : = {g \in G : g H g^{- 1} = H} .$ Thus $g \in N_{G} (H)$ exactly when the conjugation automorphism $c_{g}$ preserves $H$ setwise (thm-conjugation-is-an-automorphism). The subgroup property is proved in lem-centralizers-and-normalizers-are-subgroups. (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L5]

Let $G$ be a finite group and $H \leq G$ . Then $∣ G ∣ = [G : H] ∣ H ∣ .$ Consequently, under the canonical embedding $ι : N \to Z$ , $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Let a finite $p$ -subgroup $H$ act by left multiplication on $G / P$ .

L1L2L3L4L5givenalgebra

2.1

Its cardinality is prime to $p$ , so the fixed-point congruence gives a fixed coset $g P$ , and the fixed-coset condition is exactly $H \leq g P g^{- 1}$ .

step 1.1givenalgebra

3.1

Taking $H$ Sylow, Lagrange's theorem turns containment into equality; applying this to any Sylow $p$ -subgroup $Q$ gives $Q = g P g^{- 1}$ .

step 2.1givenalgebra

4.1

If $H = {1}$ , then $X^{H} = X$ and step 2.1 returns any coset, with ${1} \leq g P g^{- 1}$ for every $g$ . If $p ∤ ∣ G ∣$ , then $a = 0$ , so $P = {1}$ is the Sylow $p$ -subgroup and the only $p$ -subgroup of $G$ is ${1}$ itself, which is already equal to it. This proves the stated claim.

step 3.1givenalgebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 82 results over 15 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

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

Sylow II: in a finite group every pp-subgroup lies in a conjugate of any Sylow pp-subgroup, and the Sylow pp-subgroups form a single conjugacy class