Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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.

Every group of order p2q for distinct primes has a normal Sylow subgroup

Statement

Every group of order p2q for distinct primes has a normal Sylow subgroup. See Sylow III: np≡1(modp) and np∣m when ∣G∣=pam with p∤m.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let ∣G∣=pam with p∤m. Then the number of Sylow p-subgroups satisfies np(G)≡1(modp),np(G)∣m.. (Sylow III: np≡1(modp) and np∣m when ∣G∣=pam with p∤m).

[L2]

A Sylow p-subgroup of a finite group is normal if and only if it is the unique Sylow p-subgroup. (A Sylow p-subgroup is normal if and only if it is unique).

Proof

technique · direct
1.1L1L2givenalgebra

If p>q, the restrictions np∣q and np≡1(modp) force np=1.

2.1step 1.1givenalgebra

If q>p, then nq∈{1,p2}; the second value forces q∣(p−1)(p+1), hence the sole exceptional pair (p,q)=(2,3).

3.1step 2.1givenalgebra

For order 12, if the four Sylow 3-subgroups are nonnormal, their eight nonidentity elements leave exactly four elements, so every Sylow 2-subgroup is that same four-element complement and is normal.

4.1step 1.1step 2.1step 3.1L2givenalgebra∎

The two orderings exhaust the hypothesis, since p and q are distinct: step 1.1 settles p>q with a normal Sylow p-subgroup, and step 2.1 settles q>p with a normal Sylow q-subgroup except at (p,q)=(2,3), which step 3.1 settles with a normal Sylow 2-subgroup. Every case therefore produces a normal Sylow subgroup, and [L2] turns each uniqueness count n=1 into normality. This proves the stated claim.

Depends on

Used by

Dependency tree · two levels

8 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