Groups with complemented subgroup lattices

DefinitionKourovka Problem 11.128 on normal subgroups of \(K\)-groups#✓fully proved

Call a group \(G\) a \(K\)-group when its subgroup lattice is complemented: every subgroup \(A\) of \(G\) has a subgroup \(B\) with \(A \cap B = 1\) and \(\langle A, B \rangle = G\). This is the assertion (M. Emaldi) that every normal subgroup of a \(K\)-group, viewed as a group in its own right, is again a \(K\)-group.

def normalSubgroupOfKGroupIsKGroup : Prop :=
  ∀ (G : Type) [Group G], ComplementedLattice (Subgroup G) → ∀ N : Subgroup G, N.Normal →
    ComplementedLattice (Subgroup N)

Used by 1

Used byNormal subgroups of \(K\)-groups need not be \(K\)-groups

TheoremNormal subgroups of \(K\)-groups need not be \(K\)-groups#●ready to prove

There is a group \(G\) whose subgroup lattice is complemented (a \(K\)-group) with a normal subgroup \(N\) whose own subgroup lattice is not complemented; so Kourovka Problem 11.128 has a negative answer.

The Kourovka Archive answers 11.128 "No, it is not" (Obraztsov).

public theorem_wanted normalSubgroupOfKGroupIsKGroup_refuted : ¬ normalSubgroupOfKGroupIsKGroup

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 11.128 (M. Emaldi); resolution: V. N. Obraztsov, J. Austral. Math. Soc. (Ser. A) 61, no. 2 (1996), 267-288, Kourovka Archive, Problem 11.128: "No, it is not", https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesKourovka Problem 11.128 on normal subgroups of \(K\)-groups

1 fully proved1 ready to prove

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