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Wilbrink (1984): a (99,14,1,2)(99,14,1,2)(99,14,1,2) graph is not vertex-transitive

Proved
Conway99.conway_99_not_vertex_transitive

by Gabewhigham · Sep 6, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsgraph-theorystrongly-regular-graphs

No strongly regular graph with parameters (99,14,1,2)(99,14,1,2)(99,14,1,2) is vertex-transitive: if ggg is such a graph, then it is not the case that for every pair of vertices v,wv, wv,w there is a graph automorphism of ggg carrying vvv to www.

This is Wilbrink's theorem. In particular a hypothetical 999999-graph cannot be a Cayley graph, which rules out the group-theoretic constructions that produce most known strongly regular graphs — including the two realised members of the family λ=1\lambda = 1λ=1, μ=2\mu = 2μ=2, namely the Paley graph on 999 vertices and the Berlekamp–van Lint–Seidel graph. The statement is a genuine restriction rather than a vacuous one only in the sense that existence is open; proving it does not require knowing whether such a graph exists.

Preamble
import Mathlib.Combinatorics.SimpleGraph.StronglyRegular
import Mathlib.Combinatorics.SimpleGraph.Maps

open SimpleGraph
Formal statement
namespace Conway99

theorem conway_99_not_vertex_transitive {V : Type*} [Fintype V] {g : SimpleGraph V}
    [DecidableRel g.Adj] (h : g.IsSRGWith 99 14 1 2) :
    ¬ ∀ v w : V, ∃ f : g ≃g g, f v = w := by sorry

end Conway99
Source
H. A. Wilbrink, 'On the (99,14,1,2) strongly regular graph', in: Papers dedicated to J. J. Seidel, EUT Report 84-WSK-03, Eindhoven University of Technology, 1984, pp. 342-355, https://research.tue.nl/files/2449333/256699.pdf ; as summarised in https://en.wikipedia.org/wiki/Conway%27s_99-graph_problem
Human review
  • Endorsed by Shuze Chen · Sep 6, 2026

  • Endorsed by Gabewhigham · Sep 6, 2026

    Confirmed by the mission captain (proposal self-audit).

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