PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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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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$Φ (P / N) = Φ (P) N / N$ for a normal subgroup of a finite $p$ -group

Statement

If $N ⊴ P$ and $P$ is a finite $p$ -group, then

$Φ (P / N) = Φ (P) N / N .$

Facts & Assumptions

Given: A finite $p$ -group $P$ , a normal subgroup $N$ , and the quotient map $π : P \to P / N$ .

[L1]

For every finite $p$ -group $P$ , $Φ (P) = P^{'} P^{p}$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group).

[L2]

A surjective homomorphism sends the derived subgroup onto the derived subgroup of the target (Homomorphisms respect commutator subgroups and derived series).

[L3]

In $P / N$ , coset multiplication satisfies $(x N) (y N) = x y N$ , and subgroups of $P / N$ correspond to subgroups $H$ of $P$ containing $N$ by $H \mapsto H / N$ and inverse image (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ , Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

Proof

technique · direct

1.1

By [L2], $π (P^{'}) = (P / N)^{'}$ . Also $π (g^{p}) = π (g)^{p}$ , so $π (P^{p}) = (P / N)^{p}$ .

givenL1L2algebra

2.1

Apply [L1] in $P / N$ and use step 1.1: $Φ (P / N) = π (P^{'}) π (P^{p}) = π (P^{'} P^{p}) = π (Φ (P))$ . Directly, $π^{- 1} (π (Φ (P))) = Φ (P) N$ , so the correspondence in [L3] gives $π (Φ (P)) = Φ (P) N / N$ .

step 1.1L1L3algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

M. van Beek, Topics in Finite p-Groups, Lemma 3.6(ii) (standard reference, not scraped)

Φ(P/N)=Φ(P)N/NΦ(P/N)=Φ(P)N/N for a normal subgroup of a finite pp-group