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On the power graph of a certain gyrogroup

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on-the-power-graph-of-a-certain-gyrogroup-3Number Theorymath.NTmath.RTposed by Yogendra Singh, Anand Kumar Tiwari, Fawad Ali, Mani Shankar Pandeyrecorded: open · 1 machine check, unexamined

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Statement

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

Context

Candidate 3 of the open problems stated in "On the power graph of a certain gyrogroup", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

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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 n1n\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 CpC_{p^\infty}.

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

    Assume P(G)P(G) is complete. For any a,bGa,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

    (ab)c=a(bc).(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 CpC_{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.

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    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-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.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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