Answers are grounded in this paper and cite their sources.
Let πΊ be a finite group and let πΎ β‘ ( πΊ ) = β« 1 0 π β’ π β‘ ( πΊ , π₯ ) π π₯ = β π β π β‘ ( πΊ ) π π β‘ ( πΊ ) π π β‘ ( πΊ ) + 1 be the Sylow-integral invariant associated with the Sylow polynomial of πΊ . We show that small values of πΎ β‘ ( πΊ ) do not force supersolvability. More precisely, if π» is any finite group and π π‘ = β π β π β‘ ( π» ) πΆ π π‘ , then πΎ β’ ( π» Γ π π‘ ) β 0 as π‘ β β . Consequently, for every Ι > 0 there exists a finite solvable group πΊ which is not supersolvable and satisfies πΎ β‘ ( πΊ ) < Ι . We also compute πΎ β’ ( πΆ π Γ π΄ 4 ) explicitly and record infinite subfamilies with πΎ β‘ ( πΊ ) < 6 / 5 .
No takes yet. Share an insight, caveat, or question.
Liriano et al. (2026) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: