# Complete Left-Coset Partitions, Divisibility and Precisely Scoped Converses

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0001/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 2. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0002/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 3. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0003/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 4. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0004/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 5. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0005/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 6. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0006/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 7. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0007/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 8. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0008/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 9. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0009/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 10. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0010/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 11. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0011/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 12. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0012/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 13. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0013/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 14. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0014/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 15. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0015/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 16. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0016/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 17. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0017/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 18. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0018/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 19. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0019/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 20. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0020/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 21. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0021/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 22. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0022/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 23. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0023/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 24. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0024/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 25. Divisor-Sized Candidates Still Need Complete Subgroup Closure

Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).

View answer and explanation([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)

Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0025/)

Lagrange theorem for finite groups · qf-mod42-lagrange-subgroup-qualification · candidate





## 26. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0026/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 27. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0027/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 28. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0028/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 29. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0029/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 30. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0030/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 31. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0031/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 32. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0032/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 33. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0033/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 34. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0034/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 35. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0035/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 36. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0036/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 37. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0037/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 38. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0038/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 39. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0039/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 40. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0040/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 41. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0041/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 42. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0042/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 43. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0043/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 44. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0044/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 45. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0045/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 46. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0046/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 47. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0047/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 48. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0048/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 49. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0049/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 50. Complete Left Cosets Are Equal or Disjoint and Cover the Group

Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).

View answer and explanation([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)

Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0050/)

Lagrange theorem for finite groups · qf-mod42-lagrange-left-coset-partition · candidate





## 51. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(6, 3, 2, 6, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0051/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 52. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(8, 4, 2, 8, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0052/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 53. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(10, 5, 2, 10, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0053/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 54. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(12, 6, 2, 12, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0054/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 55. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(14, 7, 2, 14, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0055/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 56. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(6, 3, 2, 6, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0056/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 57. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(8, 4, 2, 8, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0057/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 58. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(10, 5, 2, 10, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0058/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 59. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(12, 6, 2, 12, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0059/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 60. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(14, 7, 2, 14, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0060/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 61. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(6, 3, 2, 6, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0061/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 62. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(8, 4, 2, 8, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0062/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 63. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(10, 5, 2, 10, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0063/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 64. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(12, 6, 2, 12, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0064/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 65. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(14, 7, 2, 14, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0065/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 66. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(6, 3, 2, 6, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0066/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 67. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(8, 4, 2, 8, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0067/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 68. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(10, 5, 2, 10, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0068/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 69. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(12, 6, 2, 12, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0069/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 70. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(14, 7, 2, 14, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0070/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 71. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(6, 3, 2, 6, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0071/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 72. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(8, 4, 2, 8, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0072/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 73. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(10, 5, 2, 10, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0073/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 74. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(12, 6, 2, 12, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0074/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 75. The Index Times Subgroup Order Equals Group Order

Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).

View answer and explanation(14, 7, 2, 14, 0, Yes)

Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0075/)

Lagrange theorem for finite groups · qf-mod42-lagrange-divisibility-identity · candidate





## 76. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(6, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0076/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 77. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(8, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0077/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 78. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(10, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0078/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 79. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(12, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0079/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 80. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(14, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0080/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 81. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(6, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0081/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 82. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(8, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0082/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 83. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(10, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0083/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 84. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(12, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0084/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 85. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(14, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0085/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 86. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(6, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0086/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 87. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(8, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0087/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 88. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(10, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0088/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 89. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(12, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0089/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 90. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(14, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0090/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 91. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(6, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0091/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 92. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(8, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0092/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 93. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(10, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0093/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 94. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(12, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0094/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 95. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(14, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0095/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 96. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(6, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0096/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 97. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(8, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0097/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 98. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(10, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0098/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 99. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(12, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0099/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate





## 100. Subset Qualification and Subgroup Existence Are Different Converse Questions

Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).

size_divides_order, candidate_is_subgroup and the general converse claim are separate. The fixed No does not negate a possibly true subgroup flag or settle existence of a subgroup of that order.

View answer and explanation(14, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)

Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0100/)

Lagrange theorem for finite groups · qf-mod42-lagrange-converse-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations/validation-report.json)