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

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.

$Φ (P \times Q) = Φ (P) \times Φ (Q)$ for finite $p$ -groups

Statement

For finite $p$ -groups $P$ and $Q$ ,

$Φ (P \times Q) = Φ (P) \times Φ (Q) .$

Facts & Assumptions

Given: Finite $p$ -groups $P, Q$ .

[F1]

The external direct product has componentwise multiplication, and this operation makes $P \times Q$ a group (The external direct product $G \times H$ with componentwise multiplication, $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[F2]

The commutator subgroup is generated by the elements $[g, h] = g h g^{- 1} h^{- 1}$ , and natural powers are defined recursively from the group operation (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L1]

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

Proof

technique · direct

1.1givenF1F2algebra

Componentwise commutators and powers give $(P \times Q)^{'} = P^{'} \times Q^{'}$ and $(P \times Q)^{p} = P^{p} \times Q^{p}$ .

2.1step 1.1L1algebra∎

By [L1] and step 1.1, $Φ (P \times Q) = (P^{'} \times Q^{'}) (P^{p} \times Q^{p}) = P^{'} P^{p} \times Q^{'} Q^{p} = Φ (P) \times Φ (Q)$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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(iii) (standard reference, not scraped)

Φ(P×Q)=Φ(P)×Φ(Q)Φ(P×Q)=Φ(P)×Φ(Q) for finite pp-groups