Harmonic index tuples: complete classifications in A₅ and C₂ × S₄, and the group orders that admit a counterexample

Ginosar's Question 1 is false at length 5: (6,6,6,10,15) is A₅-harmonic but not Z-harmonic, so the Margolis-Schnabel bound 4 is sharp

Corollary 3.2: the five cosets extend to a coset partition of A₅ with index multiset 6,6,6,10,10,10,15,15,15

The four further counterexamples of Section 5: (6,6,6,15,20) and (6,6,10,12,15) in A₅, and (4,6,6,6,6) and (6,6,6,6,16) in C₂ x S₄ of order 48

Lemma 2.1 as a residue obstruction on integers with no tuple in it, and the definitions and machinery

New: the (6,6,6,15,20) and (6,6,10,12,15) families also extend to coset partitions of A₅, with profiles 6,6,6,15⁶,20² and 6,6,10³,12²,15³, both non-Z-harmonic

Controls: two too-large, two too-small, three non-vacuity, and Sun's Conjecture 1.2 surviving all five counterexamples

External: in A₅ exactly 3 index multisets of length at most 5 are harmonic-but-not-Z-harmonic, in C₂ x S₄ exactly 2, and none at length at most 4 in either – so the source's five rows are all of them

A₅ has exactly 59 subgroups: the 59 bitmasks of ClassifyData.lean are in bijection with Subgroup ↥(alternatingGroup (Fin 5)), by the generation chain (completeness) and an explicit subOfMask (soundness)

Margolis–Schnabel's Theorem B confirmed in A₅ at every length ≤ 4: every A₅-harmonic n-tuple with n ≤ 4 is Z-harmonic, proved by exhausting the subgroup lattice

The classification at length 5: an A₅-harmonic 5-tuple is Z-harmonic unless its multiset is 6,6,6,10,15, 6,6,6,15,20 or 6,6,10,12,15 — so the three A₅ counterexamples of arXiv:2608.15873 are all of them

Controls for the classification: two too-large, one too-small, three non-vacuity

Five index multisets A₅ cannot realise at all: (5,6), (5,12), (6,6,6,6,10), (6,6,6,6,20), (6,6,6,10,12) are not A₅-harmonic

C₂ × S₄ has exactly 98 subgroups: the 98 bitmasks of ClassifyData48.lean are in bijection with Subgroup (Multiplicative (ZMod 2) × Equiv.Perm (Fin 4)), together with G48L, the exhaustive 48-element list the family lacked

Margolis–Schnabel's Theorem B confirmed in C₂ × S₄ at every length ≤ 4: every C₂ × S₄-harmonic n-tuple with n ≤ 4 is Z-harmonic, proved by exhausting the subgroup lattice

The classification at length 5 in C₂ × S₄: a C₂ × S₄-harmonic 5-tuple is Z-harmonic unless its multiset is 4,6,6,6,6 or 6,6,6,6,16 — so the two order-48 counterexamples of arXiv:2608.15873 are all of them, and with HT10 all five of the source's rows are a complete list in both of its groups

Controls for the C₂ × S₄ classification: two too-large, two too-small, four non-vacuity

Seven index multisets of length 5 that C₂ × S₄ cannot realise at all — (6,6,6,6,8), (4,4,4,8,12), (4,4,4,12,16), (2,8,8,8,12), (2,8,8,12,16), (4,6,8,8,8), (4,6,8,8,16) — and these are exactly the minimal non-realisable multisets that no counting argument excludes

New: both order-48 families of arXiv:2608.15873 extend to coset partitions of C₂ × S₄, with profiles 4,6,6,6,6,12 (one extra coset) and 6,6,6,6,12,16,16,24,24,24 (five, the minimum), both still not Z-harmonic

The classification machinery parametrised by the multiplication table: for any finite group presented as a Model — a table, a numbering, and four finite checks — the listed masks are exactly its subgroups, the Bool search is complete on G-harmonic tuples, and a checked residue witness gives Z-harmonicity

The source's own index multiset (4,6,6,6,6) is already C₂ × S₃-harmonic — a counterexample to Ginosar's Question 1 in a group of order 12, four times smaller than the smallest order arXiv:2608.15873 reports

Three necessary conditions on a G-harmonic tuple, proved for any finite group: each entry divides |G|, ∑ᵢ |G|/aᵢ ≤ |G|, and distinct entries are never coprime

The chain lemma and its corollaries: a G-harmonic tuple whose entries are pairwise comparable under divisibility is Z-harmonic, so no group of prime-power order carries a counterexample, at any length

No group of order < 12 carries a counterexample of any length, and C₂ × S₃ does — so the smallest order of a group admitting a G-harmonic tuple that is not Z-harmonic is exactly 12

New: the order-12 counterexample extends to a coset partition of C₂ × S₃ with index profile 4,6,6,6,6,12, still not Z-harmonic — one extra coset, the minimum possible

Controls for the order-12 result: three too-large, two too-small, four non-vacuity

Counterexamples are not sporadic: for every odd k ≥ 3 the group C₂ × D₂ₖ of order 4k realises the (k+2)-tuple (4, 2k, …, 2k), which is not Z-harmonic — so Ginosar's question fails in infinitely many groups, starting at the smallest order that admits any failure

External: which orders below 31 admit a counterexample — 12, 20, 24, 28, 30 do, 1–11, 13–17, 19, 21, 22, 23, 25, 26, 27, 29 do not, and (6,6,6,6,8), which C₂ × S₄ cannot realise (HT17), is realised by a group of order 24

Refined counting for a G-harmonic configuration against a subgroup P of index 2 and odd order: a coset of a subgroup of even order meets P in exactly half its elements, a coset of a subgroup of odd order lies wholly inside P or wholly outside, so ∑ᵢ |Cᵢ ∩ P| ≤ |P| and ∑ᵢ |Cᵢ P| ≤ |P| — two inequalities where Bounds.lean has one

No group of order 18 carries a counterexample to Ginosar's Question 1, at any length — the last order below 48 that journal/2026-08-30-harmonic-tuples-r3.md left undetermined, settled in the negative and by proof rather than by search

Controls for the order-18 theorem: (3,3,6,9) passes every necessary condition Bounds.lean proves and is still not Z-harmonic, so the new inequality does the work; three too-large, two too-small, two non-vacuity

No group of order 2pᵏ (p an odd prime) carries a counterexample to Ginosar's Question 1, at any length — 18 = 2·3² is the smallest such order not already covered by zharmonicₒf_divisors_comparable, and the next are 50, 54, 98, 162

External: the map of the group orders below 48 for Ginosar's Question 1 — complete except for order 42. New here: 36, 40 and 44 do admit a counterexample and 45 does not; with HT29's order 18, every order below 48 except 42 is decided