Group_has_Subgroups_of_All_Prime_Power_Factors

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

Let $p$ be a prime. Let $G$ be a finite group of order $n$. If $p^k \divides n$ then $G$ has at least one subgroup of order $p^k$.

Why this node was retired

The posted statement is

theorem Group_has_Subgroups_of_All_Prime_Power_Factors {G : Type _} [Group G] [Fintype G] {p : ℕ} [Fact (Nat.Prime p)] : ∃ (P : Sylow p G), True := by sorry

The target only asks for one Sylow p-subgroup; it omits the exponent k and the source's claim for every prime-power divisor.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

For finite G, prime p and k∈ℕ with p^k dividing |G|, prove there exists H≤G with |H|=p^k.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.