TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Sylow III*: $n_{p} (G) = [G : N_{G} (P)]$

Statement

If $P$ is a Sylow $p$ -subgroup of a finite group $G$ , then $n_{p} (G) = [G : N_{G} (P)] .$ 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 Statement.

[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]

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).

[L3]

Let $H \leq G$ . The rule $G / N_{G} (H) ⟶ {g H g^{- 1} : g \in G}, g N_{G} (H) ⟼ g H g^{- 1},$ is a well-defined bijection. If $G$ is finite, the number of distinct conjugates of $H$ is $[G : N_{G} (H)]$ . (The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$ ).

Proof

technique · direct

1.1L1L2L3givenalgebra

Conjugation is transitive on the Sylow $p$ -subgroups by Sylow II, and the stabilizer of $P$ is exactly $N_{G} (P)$ .

2.1step 1.1L2L3givenalgebra∎

By step 1.1 the set of Sylow $p$ -subgroups is exactly the conjugacy class of $P$ , so [L3] counts it as $[G : N_{G} (P)]$ , and [L2] identifies that count with $n_{p} (G)$ . If $P$ is the unique Sylow $p$ -subgroup, then $N_{G} (P) = G$ and both sides are $1$ ; if $p ∤ ∣ G ∣$ , then $P = {1}$ , again with $N_{G} (P) = G$ and both sides $1$ . This proves the stated claim.

Depends on

Used by

Cyclic sylow does not alone imply a normal p complement Counterexample
Sylow p-subgroups of Aut((ℤ/p)²): nₚ=p+1 Example
Sylow III: nₚ≡1 (mod p) and nₚ∣ m when |G|=pᵃ m with p∤ m 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

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

Sylow III*: np(G)=[G:NG(P)]np​(G)=[G:NG​(P)]