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

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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.

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Normal p complements pass to subgroups and p local normalizers

Statement

Let $G$ be a finite group, $p$ a prime, and suppose $G$ has a normal $p$ -complement $K$ (Normal p complement and p nilpotent group). Then:

every subgroup $H \leq G$ has a normal $p$ -complement, namely $K \cap H$ ;
every quotient $G / L$ by a normal subgroup $L ⊴ G$ has a normal $p$ -complement, namely $K L / L$ ;
in particular $N_{G} (Q)$ has a normal $p$ -complement for every nontrivial $p$ -subgroup $Q \leq G$ (P local normalizer for normal complement theory).

Facts & Assumptions

Given: A finite group $G$ , a prime $p$ , a normal $p$ -complement $K ⊴ G$ , a subgroup $H \leq G$ , and a normal subgroup $L ⊴ G$ .

[F1]

$K ⊴ G$ , $p ∤ ∣ K ∣$ , and $[G : K]$ is a power of $p$ ; equivalently $G$ has a normal $p^{'}$ -subgroup of $p$ -power index, and a finite group has a normal $p$ -complement exactly when it has such a subgroup (Normal p complement and p nilpotent group, Equivalent forms of having a normal p complement).

[F2]

If $K ⊴ G$ and $H \leq G$ , then $H \cap K ⊴ H$ , $H K \leq G$ and $H / (H \cap K) ≅ H K / K$ , so $[H : H \cap K] = ∣ H K / K ∣$ (Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ , If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ , Normal subgroup: invariance under conjugation).

[F3]

Orders divide: if $X \leq Y$ then $∣ X ∣$ divides $∣ Y ∣$ , and $∣ Y ∣ = ∣ X ∣ [Y : X]$ ; a group whose order divides a power of $p$ is a finite $p$ -group (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 $n \in N$ , Every subgroup of a finite $p$ -group has order a power of $p$ ).

[F4]

If $L ⊴ G$ with $L \leq K L$ , then $K L ⊴ G$ and $(G / L) / (K L / L) ≅ G / K L$ ; the quotient $K L / L$ is the image of $K$ under the natural map $G \to G / L$ , and $∣ K L / L ∣ = ∣ K ∣ / ∣ K \cap L ∣$ (Third isomorphism theorem for groups: $(G / K) / (N / K) ≅ G / N$ , The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ ).

[F5]

The natural map $G / K \to G / K L$ is an epimorphism with kernel $K L / K$ ; hence $G / K L ≅ (G / K) / (K L / K)$ and $∣ G / K L ∣$ divides $∣ G / K ∣$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ , The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , Normal subgroup: invariance under conjugation).

[F6]

A normalizer $N_{G} (Q)$ of a nontrivial $p$ -subgroup $Q$ is a subgroup of $G$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup, Subgroup).

Proof

technique · direct

1.1

$K \cap H ⊴ H$ by [F2], and $∣ K \cap H ∣$ divides $∣ K ∣$ by [F3], so $p ∤ ∣ K \cap H ∣$ .

F2F3

1.2

Moreover $[H : H \cap K] = ∣ H K / K ∣$ by [F2], and $H K / K \leq G / K$ , so $[H : H \cap K]$ divides $[G : K]$ by [F3] and is a power of $p$ . Hence $K \cap H$ is a normal $p^{'}$ -subgroup of $H$ of $p$ -power index, i.e. a normal $p$ -complement of $H$ by [F1].

F1F2F3

1.3

$K L ⊴ G$ and $K L / L ⊴ G / L$ by [F4]; and $∣ K L / L ∣ = ∣ K ∣ / ∣ K \cap L ∣$ divides $∣ K ∣$ , so $p ∤ ∣ K L / L ∣$ .

F3F4

1.4

By [F4] and [F5], $∣ (G / L) : (K L / L) ∣ = ∣ G / K L ∣$ divides $∣ G / K ∣$ , a power of $p$ ; hence $(G / L) : (K L / L)$ is a power of $p$ and $K L / L$ is a normal $p$ -complement of $G / L$ by [F1].

F1F4F5

2.1

For a nontrivial $p$ -subgroup $Q \leq G$ the normalizer $H : = N_{G} (Q)$ is a subgroup of $G$ by [F6], so step 1.2 gives that $K \cap N_{G} (Q)$ is a normal $p$ -complement of $N_{G} (Q)$ ; with steps 1.1 and 1.4 this establishes all three assertions. ∎

F6step 1.1step 1.2step 1.4

Depends on

Used by

Frobenius normal p complement theorem Theorem

Dependency tree · two levels

61 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)