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

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.

A finite product of normal $p$ -subgroups is a normal $p$ -subgroup

Statement

A finite product of normal $p$ -subgroups of a group is a normal $p$ -subgroup. The empty product is the trivial subgroup. See A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $p$ be a prime natural number (def-prime). A finite $p$ -group is a finite group $P$ (def-group, def-finite-cardinality) whose order has the form $∣ P ∣ = p^{n}$ for some $n \in N$ , with natural exponentiation as in def-nat-power. The case $n = 0$ permits the trivial group. A finite $p$ -group is nontrivial exactly when $n \geq 1$ . (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ . Here $H N : = {h n : h \in H, n \in N}$ . (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[L3]

If $H$ is a subgroup of a finite group $G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ ; in particular $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1L1L2givenalgebra

Induct on the number of factors; the empty product is the trivial normal $p$ -subgroup.

2.1step 1.1L2L3givenalgebra

Let $A$ and $B$ be normal $p$ -subgroups. By [L2], $A B$ is a subgroup, and $g A B g^{- 1} = A B$ for every ambient-group element $g$ , so $A B$ is normal. The multiplication map $A \times B \to A B$ is surjective. For a fixed factorization $a b$ , all its preimages are exactly $(a x, x^{- 1} b)$ with $x \in A \cap B$ . Thus every fibre has $∣ A \cap B ∣$ elements and $∣ A B ∣ = ∣ A ∣ ∣ B ∣ / ∣ A \cap B ∣$ . By [L3], $∣ A \cap B ∣$ is a power of $p$ , so $∣ A B ∣$ is a power of $p$ .

3.1step 1.1step 2.1givenalgebra∎

Applying step 2.1 repeatedly proves the result for every finite product, including one factor and repeated factors.

Depends on

Used by

Nothing in the library uses this result yet.

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, The Sylow Theorems, Sections 1-2 (standard reference, not scraped)

A finite product of normal pp-subgroups is a normal pp-subgroup