Powerful p-groups

DefinitionKourovka 21.137, main question#✓fully proved

The proposition that there is a prime \(p\) such that for every finite \(p\)-group \(G\) and every subgroup \(P\) of \(G\) whose underlying set is exactly the set \(\{x^p : x \in G\}\) of \(p\)-th powers in \(G\), the subgroup \(P\) is powerful: its commutator subgroup \([P, P]\) is contained in the subgroup \(P^m\) generated by the \(m\)-th powers of the elements of \(P\), where \(m = p\) for odd \(p\) and \(m = 4\) for \(p = 2\).

Kourovka Problem 21.137 (L. Wilson) asks this for each prime; it is stated here for some prime, so that its refutation covers all of them.

def pthPowersSubgroupPowerful : Prop :=
  ∃ p : ℕ, p.Prime ∧ ∀ (G : Type) [Group G] [Finite G], IsPGroup p G →
    ∀ P : Subgroup G, (P : Set G) = {g | ∃ x : G, x ^ p = g} →
      ⁅P, P⁆ ≤ Subgroup.closure ((fun x : G => x ^ (if p = 2 then 4 else p)) '' (P : Set G))

Used by 1

Used byKourovka 21.137 fails for every prime

DefinitionKourovka 21.137, odd-prime form#✓fully proved

The proposition that there is an odd prime \(p\) such that for every finite \(p\)-group \(G\) of exponent \(p^2\) and every subgroup \(P\) of \(G\) whose underlying set is exactly the set \(\{x^p : x \in G\}\) of \(p\)-th powers in \(G\), the subgroup \(P\) is abelian.

Kourovka Problem 21.137 (L. Wilson) asks this for each odd prime; it is stated here for some odd prime, so that its refutation covers all of them.

def pthPowersSubgroupAbelianExponentPSq : Prop :=
  ∃ p : ℕ, p.Prime ∧ p ≠ 2 ∧ ∀ (G : Type) [Group G] [Finite G], IsPGroup p G →
    Monoid.exponent G = p ^ 2 → ∀ P : Subgroup G, (P : Set G) = {g | ∃ x : G, x ^ p = g} →
      ∀ x ∈ P, ∀ y ∈ P, x * y = y * x

Used by 1

Used byKourovka 21.137 fails for every odd prime

TheoremKourovka 21.137 fails for every prime#●ready to prove

For every prime \(p\) there is a finite \(p\)-group \(G\) with a subgroup \(P\) whose underlying set is exactly the set \(\{x^p : x \in G\}\) of \(p\)-th powers in \(G\), such that \(P\) is not powerful: the commutator subgroup \([P, P]\) is not contained in the subgroup \(P^m\) generated by the \(m\)-th powers of the elements of \(P\), where \(m = p\) for odd \(p\) and \(m = 4\) for \(p = 2\).

For odd \(p\) a counterexample is Muliarchyk's group of order \(p^{3p+1}\) and exponent \(p^2\), whose \(p\)-th powers form a non-abelian subgroup of exponent \(p\); for \(p = 2\) it is the Sylow \(2\)-subgroup of \(S_8\), whose squares form \(C_2 \times D_8\) (letter of A. Chang). The 21st issue of the Kourovka Notebook marks Problem 21.137 solved ("No, it need not be").

public theorem_wanted pthPowersSubgroupPowerful_refuted : ¬ pthPowersSubgroupPowerful

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 21.137 (L. Wilson); resolution: K. Muliarchyk, A counterexample to Kourovka Problem 21.137 for p = 3, preprint (2026), and A non-powerful subgroup of p-th powers for every odd prime, preprint (2026), Kourovka Notebook site; letter of A. Chang of 1 August 2026, Muliarchyk (every odd prime), abstract and Section 1: for each odd prime p an affine group V ⋊ B_p (V = F_p[r]/(r^(2p))) of order p^(3p+1) and exponent p² whose set of p-th powers is a subgroup isomorphic to UT₃(F_p) × C_p^(p-1), non-abelian of exponent p, hence not powerful; Kourovka Notebook (21st issue), Problem 21.137: No, it need not be (p = 3: Muliarchyk, order 3^15; p = 2: the Sylow 2-subgroup of S₈ has exponent 8 and its squares form the non-abelian, non-powerful subgroup C₂ × D₈, letter of A. Chang), https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/kourovka_21.137_solution.pdf.

Uses 1

Statement usesKourovka 21.137, main question

TheoremKourovka 21.137 fails for every odd prime#●ready to prove

For every odd prime \(p\) there is a finite \(p\)-group \(G\) of exponent \(p^2\) with a subgroup \(P\) whose underlying set is exactly the set \(\{x^p : x \in G\}\) of \(p\)-th powers in \(G\), such that \(P\) is not abelian.

Muliarchyk's counterexample has order \(p^{3p+1}\), and its subgroup of \(p\)-th powers is isomorphic to \(\mathrm{UT}_3(\mathbb{F}_p) \times C_p^{p-1}\). The 21st issue of the Kourovka Notebook marks Problem 21.137 solved ("No, it need not be"), citing this preprint for all odd primes.

public theorem_wanted pthPowersSubgroupAbelianExponentPSq_refuted :
    ¬ pthPowersSubgroupAbelianExponentPSq

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 21.137 (L. Wilson); resolution: K. Muliarchyk, A counterexample to Kourovka Problem 21.137 for p = 3, preprint (2026), and A non-powerful subgroup of p-th powers for every odd prime, preprint (2026), Kourovka Notebook site; letter of A. Chang of 1 August 2026, Muliarchyk (every odd prime), abstract and Section 1: for each odd prime p an affine group V ⋊ B_p (V = F_p[r]/(r^(2p))) of order p^(3p+1) and exponent p² whose set of p-th powers is a subgroup isomorphic to UT₃(F_p) × C_p^(p-1), non-abelian of exponent p, hence not powerful; Kourovka Notebook (21st issue), Problem 21.137: No, it need not be (p = 3: Muliarchyk, order 3^15; p = 2: the Sylow 2-subgroup of S₈ has exponent 8 and its squares form the non-abelian, non-powerful subgroup C₂ × D₈, letter of A. Chang), https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/kourovka_21.137_solution.pdf.

Uses 1

Statement usesKourovka 21.137, odd-prime form

2 fully proved2 ready to prove

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