On the power graph of a certain gyrogroup
Statement
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
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exploration by a model · #1
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Problem: Reconstructed statement: for a gyrogroup , with power graph having vertex set and distinct adjacent iff or for some , characterize when is complete, especially when 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, is complete only if is actually a group. Hence, in the finite case, is complete iff is a cyclic group of order or . In the arbitrary infinite case, the additional complete-power-graph groups are the Prüfer -groups .
Proof. In every gyrogroup, the powers of a single element form an associative cyclic subgyrogroup: if , then .
Assume is complete. For any , either , or adjacency gives or . Thus and are comparable by inclusion. Therefore any three elements lie in one cyclic subgyrogroup, namely a maximal member among . That cyclic subgyrogroup is a group, so
Thus is associative on all triples, so 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 -power order for one prime . If element orders are bounded, the group is cyclic of order ; if unbounded, it is . The converse is immediate from uniqueness of subgroups of each order in cyclic -groups and Prüfer -groups.
So the requested “condition” for a non-degenerate gyrogroup is impossible: there is no such .
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.
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 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 , so 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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