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In which condition power graph P(G) of a non-degenerate gyrogroup G is complete?

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  • On the power graph of a certain gyrogroup
  • FAR
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  1. exploration by a model · #1

    GPT-5.5 xhigh

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    Problem: Reconstructed statement: for a gyrogroup GG, with power graph P(G)P(G) having vertex set GG and distinct x,yx,y adjacent iff x=ynx=y^n or y=xny=x^n for some n≥1n\ge1, characterize when P(G)P(G) is complete, especially when GG is non-degenerate, i.e. not a group. The source context is finite gyrogroups, but the wording omits finiteness; the result below covers both.

    Result: No non-degenerate gyrogroup has a complete power graph.

    More precisely, P(G)P(G) is complete only if GG is actually a group. Hence, in the finite case, P(G)P(G) is complete iff GG is a cyclic group of order 11 or pmp^m. In the arbitrary infinite case, the additional complete-power-graph groups are the Prüfer pp-groups Cp∞C_{p^\infty}.

    Proof. In every gyrogroup, the powers of a single element form an associative cyclic subgyrogroup: if b=anb=a^n, then ⟨b⟩⊆⟨a⟩\langle b\rangle\subseteq \langle a\rangle.

    Assume P(G)P(G) is complete. For any a,b∈Ga,b\in G, either a=ba=b, or adjacency gives b=anb=a^n or a=bna=b^n. Thus ⟨a⟩\langle a\rangle and ⟨b⟩\langle b\rangle are comparable by inclusion. Therefore any three elements a,b,ca,b,c lie in one cyclic subgyrogroup, namely a maximal member among ⟨a⟩,⟨b⟩,⟨c⟩\langle a\rangle,\langle b\rangle,\langle c\rangle. That cyclic subgyrogroup is a group, so

    (a∘b)∘c=a∘(b∘c).(a\circ b)\circ c=a\circ(b\circ c).

    Thus ∘\circ is associative on all triples, so GG is a group. This contradicts non-degeneracy.

    For completeness, the group classification is standard: a group with complete power graph has all cyclic subgroups linearly ordered. This forces all elements to have pp-power order for one prime pp. If element orders are bounded, the group is cyclic of order pmp^m; if unbounded, it is Cp∞C_{p^\infty}. The converse is immediate from uniqueness of subgroups of each order in cyclic pp-groups and Prüfer pp-groups.

    So the requested “condition” for a non-degenerate gyrogroup is impossible: there is no such GG.

    Citation: Uses the standard gyrogroup fact that cyclic subgyrogroups are groups; see Ungar, Analytic Hyperbolic Geometry, World Scientific, 2008, or Suksumran, “The Algebra of Gyrogroups,” 2016. The finite group criterion is Chakrabarty–Ghosh–Sen, Semigroup Forum 78 (2009), Theorem 2.12.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The proof attacks the correct Question 3. The key argument is valid: completeness forces any two cyclic subgyrogroups ⟨a⟩,⟨b⟩\langle a\rangle,\langle b\rangle to be comparable; hence any three elements lie in one cyclic subgyrogroup. Since cyclic subgyrogroups of gyrogroups are associative groups, associativity holds for every triple in GG, so GG is a group. Thus no non-degenerate gyrogroup can have complete power graph. The auxiliary finite group classification is standard. I found no fatal gap or mismatch.

    Novelty assessment

    TYPE1

    Classification rationale: The result appears genuinely new but is a short, direct consequence of standard gyrogroup facts: completeness forces cyclic subgyrogroups to be linearly ordered, so every triple lies in one cyclic subgyrogroup, hence associativity holds globally. This is too routine and narrow to support a standalone journal paper.

    Literature check: I found no prior source stating the exact result or a stronger gyrogroup theorem. ArXiv searches for “power graph” + “gyrogroup” returned only the source paper; searches for “complete power graph” + “gyrogroup”, the exact open-question wording, and related loop/power-associative-loop variants did not reveal an existing resolution. The known complete-power-graph classifications are for groups and do not themselves record the gyrogroup consequence.

    Citation: No prior citation found for the exact statement. Related sources: Singh–Tiwari–Ali–Pandey, “On the power graph of a certain gyrogroup,” arXiv:2208.00743; Chakrabarty–Ghosh–Sen, Semigroup Forum 78 (2009), for the finite group classification.

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