LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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 times normal subgroup covers when the index is a p-power

Statement

Let $N$ be a finite group, $p$ a prime, $C ⊴ N$ a normal subgroup such that $N / C$ is a $p$ -group (Normal subgroup: invariance under conjugation, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ), and let $T \in {Syl}_{p} (N)$ be a Sylow $p$ -subgroup of $N$ . Then $N = T C$ .

Facts & Assumptions

Given: A finite group $N$ , a prime $p$ , a normal subgroup $C ⊴ N$ with $N / C$ a $p$ -group, and a Sylow $p$ -subgroup $T \leq N$ .

[F1]

$∣ N / C ∣ = [N : C] = p^{k}$ for some $k \geq 0$ ; write $∣ N ∣ = p^{a} m$ with $p ∤ m$ , so that $∣ T ∣ = p^{a}$ and $∣ C ∣ = ∣ N ∣ / [N : C] = p^{a - k} m$ (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , Sylow $p$ -subgroups of a finite group, Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[F2]

$T \cap C \leq T$ , so $T \cap C$ is a finite $p$ -group and $∣ T \cap C ∣$ divides $∣ C ∣$ (Every subgroup of a finite $p$ -group has order a power of $p$ , Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[F3]

$T C$ is a subgroup of $N$ with $∣ T C ∣ = ∣ T ∣ ∣ C ∣ / ∣ T \cap C ∣$ and $T C / C \leq N / C$ ; in particular $∣ T C / C ∣ = ∣ T ∣ / ∣ T \cap C ∣$ (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ , Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ , If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

Proof

technique · direct

1.1

By [F2] the order $∣ T \cap C ∣$ is a $p$ -power dividing $∣ C ∣ = p^{a - k} m$ ; since $m$ is prime to $p$ , $∣ T \cap C ∣$ divides $p^{a - k}$ , so $∣ T \cap C ∣ \leq p^{a - k}$ .

F1F2algebra

2.1

Hence $∣ T C / C ∣ = ∣ T ∣ / ∣ T \cap C ∣ \geq p^{a} / p^{a - k} = p^{k} = ∣ N / C ∣$ by [F1] and [F3].

F1F3step 1.1

3.1

Since $T C / C \leq N / C$ by [F3], the inequality of step 2.1 forces $T C / C = N / C$ , and then $∣ T C ∣ = ∣ N ∣$ because $∣ T C ∣ = ∣ T C / C ∣ \cdot ∣ C ∣$ by [F1] and [F3]; a subgroup of $N$ with as many elements as $N$ equals $N$ , so $N = T C$ . ∎

F1F3step 2.1

Depends on

Used by

Dependency tree · two levels

50 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

Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)