Endomorphisms and partial isomorphisms

DefinitionKourovka Problem 19.20 (Cameron)#✓fully proved

The statement that every finite group \(G\) with \(|\mathrm{End}(G)| = |\mathrm{PIso}(G)|\) is abelian. Here \(\mathrm{End}(G)\) is the set of endomorphisms of \(G\), and \(\mathrm{PIso}(G)\) is the set of partial isomorphisms of \(G\): the triples \((H, K, f)\) of subgroups \(H\), \(K\) of \(G\) and an isomorphism \(f \colon H \to K\).

def endPartialIsoCountImpliesAbelian : Prop :=
  ∀ (G : Type) [Group G] [Finite G],
    Nat.card (G →* G) = Nat.card (Σ H : Subgroup G, Σ K : Subgroup G, (H ≃* K)) →
      ∀ x y : G, x * y = y * x

Used by 1

Used byEqual endomorphism and partial isomorphism counts do not force commutativity

TheoremEqual endomorphism and partial isomorphism counts do not force commutativity#●ready to prove

There is a finite non-abelian group \(G\) with \(|\mathrm{End}(G)| = |\mathrm{PIso}(G)|\). Here \(\mathrm{End}(G)\) is the set of endomorphisms of \(G\), and \(\mathrm{PIso}(G)\) is the set of partial isomorphisms of \(G\): the triples \((H, K, f)\) of subgroups \(H\), \(K\) of \(G\) and an isomorphism \(f \colon H \to K\). This answers Kourovka Problem 19.20 (P. J. Cameron) in the negative.

K. Muliarchyk's Theorem 1.1 (2026) gives a non-abelian group of order \(512\) for which both counts equal \(129 \cdot 2^{24}\); the Kourovka Notebook (21st issue) answers 19.20 "No, it is not".

public theorem_wanted endPartialIsoCountImpliesAbelian_refuted : ¬ endPartialIsoCountImpliesAbelian

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 19.20 (P. J. Cameron); resolution: K. Muliarchyk, A nonabelian group with equally many endomorphisms and partial isomorphisms: a computer-assisted counterexample to Kourovka Problem 19.20, preprint, 2026, Theorem 1.1: a nonabelian group of order 512, nilpotency class two and exponent four, with |PIso(G)| = |End(G)| = 129 · 2^24 (the partial isomorphisms counted by exhaustive enumeration); the Kourovka Notebook (21st issue) marks 19.20 solved: No, it is not, https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/kourovka_19.20_solution.pdf.

Uses 1

Statement usesKourovka Problem 19.20 (Cameron)

1 fully proved1 ready to prove

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