Normal subgroups of exponent p in p-groups

DefinitionBerkovich's question on normal subgroups of exponent p, for p > n#✓fully proved

For every prime \(p\) and every natural number \(n < p\), if a finite \(p\)-group \(G\) has a subgroup of order \(p^n\) all of whose elements \(x\) satisfy \(x^p = 1\), then \(G\) has a normal subgroup of order \(p^n\) all of whose elements \(x\) satisfy \(x^p = 1\).

Berkovich asked this for \(p\) sufficiently large relative to \(n\); the explicit bound \(p > n\) is the one of the resolution. For \(n \ge 1\) the condition \(x^p = 1\) on a group of order \(p^n\) says that it has exponent exactly \(p\).

def normalExponentPSubgroupOfLargePrime : Prop :=
  ∀ p n : ℕ, p.Prime → n < p → ∀ (G : Type) [Group G] [Finite G], IsPGroup p G →
    (∃ H : Subgroup G, Nat.card H = p ^ n ∧ ∀ x ∈ H, x ^ p = 1) →
      ∃ N : Subgroup G, N.Normal ∧ Nat.card N = p ^ n ∧ ∀ x ∈ N, x ^ p = 1

Used by 1

Used byNormal subgroups of exponent p in p-groups when p > n

TheoremNormal subgroups of exponent p in p-groups when p > n#●ready to prove

For every prime \(p\) and every natural number \(n < p\), if a finite \(p\)-group \(G\) has a subgroup \(H\) of order \(p^n\) with \(x^p = 1\) for all \(x \in H\), then \(G\) has a normal subgroup \(N\) of order \(p^n\) with \(x^p = 1\) for all \(x \in N\).

The Kourovka Archive answers "Yes, it is true if \(p > n\)" (A. Mann): by induction a subgroup \(H\) of exponent \(p\) and order \(p^n\) may be taken normal in a maximal subgroup \(M\); then \(H\) lies in the \((p-1)\)-st term \(\zeta_{p-1}(M)\) of the upper central series of \(M\), and the elements of order \(p\) of \(\zeta_{p-1}(M)\) form a normal subgroup of \(G\) containing \(H\).

public theorem_wanted normalExponentPSubgroupOfLargePrime_holds :
    normalExponentPSubgroupOfLargePrime

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 15.24 (Ya. G. Berkovich); resolution: A. Mann, letter of 1 October 2002, as recorded in the Kourovka Archive, Problem 15.24, Archive answer: Yes if p > n (H of exponent p and order p^n normal in a maximal subgroup M lies in ζ_{p-1}(M), whose elements of order p form a normal subgroup of G), https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesBerkovich's question on normal subgroups of exponent p, for p > n

1 fully proved1 ready to prove

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