LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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.

Every subgroup of a finite $p$ -group has order a power of $p$

If $P$ is a finite $p$ -group and $H \leq P$ , then $H$ is finite and

$∣ H ∣ = p^{k}$

for some $k \in N$ . If $∣ P ∣ = p^{n}$ , then $k \leq n$ .

Given: A finite $p$ -group $P$ with $∣ P ∣ = p^{n}$ and a subgroup $H \leq P$ .

[L1]

A finite $p$ -group has order $p^{n}$ for a prime $p$ and a natural $n$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

[L3]

technique · direct

1.1L1L2

By [L2], the finite set $H$ has positive order dividing $p^{n}$ .

2.1step 1.1L3

By uniqueness in [L3], no prime other than $p$ can divide $∣ H ∣$ , so $∣ H ∣ = p^{k}$ for some natural $k \leq n$ .

3.1step 2.1algebra∎

This includes the trivial subgroup, whose order is $1 = p^{0}$ , and proves that every subgroup of $P$ is a finite $p$ -group.

Depends on