The Cayley–Hamilton analogue for free groups fails in ranks three and four, and only one power works in rank two
Abstract
Problem 17.32 of the Kourovka Notebook, proposed by O. V. Bogopolski, asks whether the free group Fₙ obeys the Cayley–Hamilton pattern: if w ∈ Fₙ and φ ∈ Aut Fₙ satisfy genw, φ(w), …, φⁿ(w) = Fₙ, must already genw, φ(w), …, φⁿ⁻¹(w) = Fₙ? Two independent solutions appeared in August 2026 and are recorded in the Notebook's answer line; both refute the assertion at n = 2 and say nothing about any larger rank. We give explicit counterexamples at n = 3 and n = 4. Both are built from the rank-two automorphism φ(x) = yx, φ(y) = x by adjoining generators that φ either fixes (rank three) or transposes (rank four), and each is certified on both sides: explicit words exhibiting every free generator, and an explicit homomorphism to a permutation group under which the initial segment of the orbit fixes a point that some free generator moves. We also settle, for the rank-two data of the two published solutions, every exponent at once: genw, φᵏ(w) = F₂ holds if and only if k = 2, the index of the abelianised pair being the Fibonacci number F(k); the two published solutions record only the cases k = 1 and k = 2. A complementary observation explains why the rank-two failure needs a non-abelian witness at all: no abelian quotient can detect it. Every theorem, proposition, corollary and lemma below is machine-checked in Lean 4 against Mathlib, by kernel evaluation only; Section [sec:verif] says exactly what is and what is not formalised. Whether Problem 17.32 fails for every n ≥ 2 remains open.
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Claim ledger
Stated results
K1known2026-09-03
Kourovka Problem 17.32 has a negative answer at rank two: there are w in F₂ and phi in Aut F₂ with <w, phi w, phi² w> = F₂ and <w, phi w>!= F₂; equivalently not CayleyHamiltonAnalogue 2 (FreeGroup (Fin 2))
K2known2026-09-03
The generation half at rank two: phi(w) = y x², phi²(w) = xyxyx = w² x, x = w⁻2 phi²(w), y = x⁻1 w, hence <w, phi(w), phi²(w)> = F₂
K3known2026-09-03
The properness half at rank two: pi: F₂ -> S₃ with pi(x) = (1 2 3), pi(y) = (2 3) has pi(w) = pi(phi(w)) = (1 2), so <w, phi(w)> is a proper subgroup of F₂
K4known2026-09-03
phi(x) = y x, phi(y) = x is an automorphism of F₂ with inverse phi⁻1(x) = y, phi⁻1(y) = x y⁻1, and phi is not the identity
K5known2026-09-03
The pair skipping phi(w) generates: <w, phi²(w)> = F₂
K6routine2026-09-03
Controls for the rank-two counterexample: <x, phi(x)> = F₂ (properness is a property of the word, not of phi); pi(phi²(w)) is not in the point stabiliser (the S₃ test does not overreach); and for every commutative group A and every homomorphism rho: F₂ -> A, <rho(w), rho(phi(w))> = rho(F₂), so no abelian quotient can detect the properness
K7routine2026-09-03
The witnessing quotient is not degenerate: pi: F₂ -> S₃ is surjective
K8candidate2026-09-03
Kourovka Problem 17.32 also fails at rank three: for F₃ = <a,b,c>, phi(a) = b a, phi(b) = a, phi(c) = c and w = a b c, one has <w, phi w, phi² w>!= F₃ – witnessed by rho: F₃ onto S₃ with a -> (2 3), b -> (1 2), c -> (1 2), whose values (2 3), 1, 1 all fix a letter that rho(b) moves – while <w, phi w, phi² w, phi³ w> = F₃ by explicit words in eᵢ = w (phiⁱ w)⁻1; equivalently not CayleyHamiltonAnalogue 3 (FreeGroup (Fin 3))
K9candidate2026-09-03
Kourovka Problem 17.32 also fails at rank four: for F₄ = <a,b,c,d>, phi(a) = b a, phi(b) = a, phi(c) = d, phi(d) = c and w = a b c, one has <w, phi w, phi² w, phi³ w>!= F₄ – witnessed by rho: F₄ -> S₄ with a, d -> (1 2)(3 4) and b, c -> (1 3 2), whose image lies in A₄ (equality of the image with A₄ checked outside Lean) – while <w,..., phi⁴ w> = F₄; and the failure is about the initial segment, since <w, phi w, phi² w, phi⁴ w> already generates F₄
K10candidate2026-09-03
For the sources' rank-two data (w = xy, phi(x) = y x, phi(y) = x), <w, phiᵏ(w)> = F₂ if and only if k = 2, for every k >= 1: phi acts on the abelianisation by the Fibonacci matrix, so w and phiᵏ(w) have exponent-sum vectors (1,1) and (F(k+2), F(k+1)) and span a subgroup of Z² of index F(k); for k >= 3 the Lean witness is the linear form chiₖ with chiₖ(x) = F(k+1), chiₖ(y) = -F(k+2), whose values on w and phiᵏ(w) are -F(k) and 0 while chiₖ(x) = F(k+1) is coprime to F(k) >= 2
Provenance
- Generated by
- Machina Mathematica
- Released by
- Korea Superintelligence Labs
- Source context
- Problem 17.32 of the *Kourovka Notebook* (No. 21, Novosibirsk 2026 = arXiv:1401.0300v46, source file 21tkt.tex line 9316), proposed by O. V. Bogopolski:
- Snapshot
- 2026-09-07 03:53 UTC
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