Complete Left-Coset Partitions, Divisibility and Precisely Scoped Converses

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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).

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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.

Reviewed step-by-step derivation

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