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For each odd prime p, an explicit group of order p³ᵖ⁺¹ and exponent p² has its set of p-th powers equal to a subgroup isomorphic to UT₃(F_p) × C_pᵖ⁻¹, which is nonabelian of exponent p. This gives a uniform negative answer to the odd-prime part of the problem. The Kourovka editors have accepted the case p = 3; the uniform write-up for all odd primes is team-checked and has been sent to the editors.

Review status: Editor-accepted. Model and setup attribution supplied by the project team.

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A counterexample of order 660,602,880 gives a negative answer, even when the nilpotent subgroup is normal and has nilpotency class two. An involution has a largest centralizer among nonidentity elements but does not centralize this subgroup. The write-up gives a self-contained proof without computer enumeration.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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A nonabelian group of order 512 has equally many endomorphisms and partial isomorphisms, giving a negative answer. The endomorphism count has a direct proof; the partial-isomorphism count uses exhaustive finite quadratic-map enumeration. The PDF includes a counting certificate and describes its verification; no claim of minimal order is made.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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If n left or right cosets do not cover a finite group G, their complement has at least |G|/2ⁿ elements. The proof establishes this sharp bound through matrix-rank estimates, with equality attained by coordinate index-two cosets. A note by another author deposited on Zenodo on 2 August 2026 gives the same rank argument, so the problem was solved before our write-up.

Review status: Solved elsewhere. Model and setup attribution supplied by the project team.

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The write-up describes N/[F, N] for a minimal free presentation of a finitely presented group. Its abstract isomorphism type is determined by the Schur multiplier and a free abelian summand, independently of the chosen minimal presentation. It also discusses the limits of computing these invariants from an arbitrary finite presentation.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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The answer depends on whether the rank is fixed. For each fixed rank, a computable bound on basis lengths follows from existing algorithms for fixed subgroups. A single bound for all ranks is impossible: an explicit family of automorphisms of norm 2 provides counterexamples. The Kourovka editors have since accepted an independent solution by Olga Kharlampovich; our write-up remains available for reference.

Review status: Solved elsewhere. Model and setup attribution supplied by the project team.

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The commutator width of Aut(Fₙ) is infinite for every n ≥ 3. The argument draws on published results about quasimorphisms and verbal width, connecting the Notebook question to existing theorems rather than claiming a new general theorem.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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In the group of increasing bijections of the rationals with bounded displacement, the right-relatively convex subgroups are exactly the isolated subgroups and exactly the convex ℓ-subgroups. Their lattice is distributive but not a chain, giving a negative answer. The main ingredient is an orbital sandwich theorem, proved with a coding construction adapted from Hyde, Jonušas, Mitchell and Péresse.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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The submitted argument constructs two good subnormal subgroups whose intersection is not good, in a countable locally finite 2-group of derived length at most three. It proves goodness using an explicit augmentation-ideal bound. This proposed counterexample is still pending team review.

Review status: Pending review. Model and setup attribution supplied by the project team.

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A finite group G = ⟨A, B⟩ with d(A) = d(B) = 2, coprime orders |A| and |B|, and d(G) = 4 gives a negative answer already for d = 2. The subgroups live in a direct power of PSL(2, 11); the proof combines exact finite counts in this group of order 660 with Lagrange’s theorem and the fact that a homomorphism is determined by its values on generators. Two archived checkers agree, but an independent GAP reproduction is still pending. This proposed counterexample is still pending team review.

Review status: Pending review. Model and setup attribution supplied by the project team.

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Over an explicit finitely presented noncommutative algebra R of characteristic two, a matrix σ in GL₃(R) has [[σ, T₁], T₂] of infinite order for every pair of nonidentity matrices with (Tᵢ − I)² = 0, so no such double commutator is unipotent. This refutes the statement for noncommutative R; it makes no claim about commutative rings or rings in which 2 is invertible. The proof rests on Gerasimov’s theorem on units of ring coproducts and a free-product tree argument, without finite enumeration. This proposed counterexample is still pending team review.

Review status: Pending review. Model and setup attribution supplied by the project team.

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The invertible monomial subgroup of GL₂ over the algebraic closure of F₂ meets every conjugacy class but is not parabolic. This refutes the general-linear assertion of part (b) as literally stated, in dimension two and characteristic two. The same monomial family appears in the Notebook archive for part (a), so no novelty is claimed, and other groups of Lie type are not addressed. This proposed counterexample is still pending team review.

Review status: Pending review. Model and setup attribution supplied by the project team.

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The proposed analogue fails for every rank n ≥ 2: explicit examples have n + 1 successive orbit elements generating the free group, while the first n generate a subgroup of infinite index. The write-up also gives counterexamples to a normal-closure variant at rank 2 and a subset variant at rank 3.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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For the explicit integer N = 2¹²⁴²¹⁰³⁵⁰, a finite 2-group of order 2¹¹⁰ᴺ⁺¹ has every abelian normal subgroup generated by at most 32N + 1 elements, while it contains an elementary abelian subgroup requiring 65N generators. This gives a negative answer to Pyber’s question. The group multiplication is given by finite deterministic formulas, and the proof is self-contained without enumeration of the group or its subgroups.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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A solvable group of order 46,875,000,000 has a faithful transitive action of degree 23,437,500 with distinct points α and β such that every element taking α to β fixes exactly one point. This refutes Müller’s conjecture already for solvable groups. The group is an extension of F₅¹² by a group of order 192, and the fixed-point calculation is carried out directly on twelve affine blocks.

Review status: Team-checked. Model and setup attribution supplied by the project team.

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