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

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Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$

Statement

Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then the number of Sylow $p$ -subgroups satisfies $n_{p} (G) \equiv 1 (m o d p), n_{p} (G) ∣ m .$ 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]

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

[L3]

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 $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L4]

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).

[L5]

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

Proof

technique · direct

1.1L4L5givenalgebra

By [L5] the set ${Syl}_{p} (G)$ of [L4] is nonempty, so fix $P \in {Syl}_{p} (G)$ and let it act by conjugation on that set; the action is well defined because a conjugate of a Sylow $p$ -subgroup again has order $p^{a}$ .

2.1step 1.1L1L3L4givenalgebra

Suppose $P$ fixes $Q \in {Syl}_{p} (G)$ , so $P \leq N_{G} (Q)$ , and also $Q \leq N_{G} (Q)$ . Both have order $p^{a}$ , the full power of $p$ dividing $∣ G ∣$ and hence dividing $∣ N_{G} (Q) ∣$ , so both are Sylow $p$ -subgroups of $N_{G} (Q)$ ; by [L1] applied inside $N_{G} (Q)$ they are conjugate there, and $Q ⊴ N_{G} (Q)$ makes every such conjugate equal to $Q$ , so $P = Q$ . Thus $P$ itself is the only fixed point, and since $P$ is a finite $p$ -group acting on the finite set ${Syl}_{p} (G)$ , [L3] gives $n_{p} (G) \equiv 1 (m o d p)$ .

3.1step 1.1step 2.1L2givenalgebra∎

By [L2], $n_{p} (G) = [G : N_{G} (P)]$ . Since $P \leq N_{G} (P) \leq G$ , the index tower gives $[G : P] = [G : N_{G} (P)] [N_{G} (P) : P]$ , and $[G : P] = m$ because $∣ P ∣ = p^{a}$ ; hence $n_{p} (G)$ divides $m$ . This proves the stated claim.

Depends on

Used by

Cyclic sylow does not alone imply a normal p complement Counterexample
A5 as the smallest nonabelian simple group Example
Sylow data for finite groups of order at most 15 Example
Sylow subgroups of Aff(ℤ/5): n₂=5 and n₅=1 Example
The Sylow subgroups of A₅ Example
The Sylow subgroups of S₄ Example
The unique Sylow p-subgroup of Aff(ℤ/p²) Example
False statement: every group of order 42 has a normal Sylow 2-subgroup False statement
Every group of order 105 has normal Sylow 5- and 7-subgroups and is not simple Theorem
Every group of order 45 is abelian Theorem
Every group of order p²q for distinct primes has a normal Sylow subgroup Theorem

Dependency tree · two levels

20 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≡1(modp)np​≡1(modp) and np∣mnp​∣m when ∣G∣=pam∣G∣=pam with p∤mp∤m