Complete Subgroup Closure, Conjugation and All-Element Coset Equality
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations · 2026-10-15 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.
Field-order and source-binding data
1. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
2. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
3. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
4. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
5. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1], Yes, No, No, No, No, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1], Yes, No, No, No, No, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
6. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([1], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
7. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
8. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[1, 3, 5, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
9. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[1, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([1, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
10. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[1, 2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([1, 2], Yes, No, No, No, No, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 2], Yes, No, No, No, No, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
11. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([2], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
12. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
13. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
14. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[2, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([2, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
15. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([2, 3], Yes, No, No, No, No, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2, 3], Yes, No, No, No, No, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
16. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([3], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
17. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
18. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[1, 3, 5, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
19. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[3, 8]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([3, 8], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3, 8], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
20. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[3, 4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([3, 4], Yes, No, No, No, No, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3, 4], Yes, No, No, No, No, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
21. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
22. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
23. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
24. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[4, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([4, 9], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([4, 9], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
25. Subgroup Closure Precedes a Normal-Subgroup Certificate
Use 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 H=[4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).
View answer and explanation
([4, 5], Yes, No, No, No, No, subgroup_obligations_precede_normality)
Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([4, 5], Yes, No, No, No, No, subgroup_obligations_precede_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-subgroup-qualification · candidate
26. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
27. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
28. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
29. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
30. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes], [8, [0], Yes], [9, [0], Yes], [10, [0], Yes], [11, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes], [8, [0], Yes], [9, [0], Yes], [10, [0], Yes], [11, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
31. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
32. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
33. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
34. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes], [6, [1], Yes], [7, [1], Yes], [8, [1], Yes], [9, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes], [6, [1], Yes], [7, [1], Yes], [8, [1], Yes], [9, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
35. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
36. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
37. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
38. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
39. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
40. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes], [8, [2], Yes], [9, [2], Yes], [10, [2], Yes], [11, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes], [8, [2], Yes], [9, [2], Yes], [10, [2], Yes], [11, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
41. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
42. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
43. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
44. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes], [6, [3], Yes], [7, [3], Yes], [8, [3], Yes], [9, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes], [6, [3], Yes], [7, [3], Yes], [8, [3], Yes], [9, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
45. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
46. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
47. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
48. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
49. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
50. Every Group Element Must Conjugate the Entire Subgroup to Itself
Use 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 H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).
View answer and explanation
([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes], [8, [4], Yes], [9, [4], Yes], [10, [4], Yes], [11, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)
Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes], [8, [4], Yes], [9, [4], Yes], [10, [4], Yes], [11, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-conjugate-set-audit · candidate
51. Left and Right Cosets Must Agree for Every g
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
52. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
53. Left and Right Cosets Must Agree for Every g
Use 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 H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
54. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
55. Left and Right Cosets Must Agree for Every g
Use 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 H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
56. Left and Right Cosets Must Agree for Every g
Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
57. Left and Right Cosets Must Agree for Every g
Use 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 H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
58. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
59. Left and Right Cosets Must Agree for Every g
Use 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 H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
60. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
61. Left and Right Cosets Must Agree for Every g
Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
62. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
63. Left and Right Cosets Must Agree for Every g
Use 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 H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
64. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
65. Left and Right Cosets Must Agree for Every g
Use 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 H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
66. Left and Right Cosets Must Agree for Every g
Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
67. Left and Right Cosets Must Agree for Every g
Use 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 H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
68. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
69. Left and Right Cosets Must Agree for Every g
Use 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 H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
70. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
71. Left and Right Cosets Must Agree for Every g
Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
72. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
73. Left and Right Cosets Must Agree for Every g
Use 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 H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
74. Left and Right Cosets Must Agree for Every g
Use 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 H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
75. Left and Right Cosets Must Agree for Every g
Use 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 H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).
View answer and explanation
([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)
Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-left-right-coset-audit · candidate
76. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 0, 5, 1, 3], [3, 5, 4, 0, 2, 1], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
77. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
78. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 0, 1, 5, 4], [3, 2, 4, 5, 1, 0], [4, 5, 3, 2, 0, 1], [5, 4, 1, 0, 2, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
79. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 0, 4, 3, 1], [3, 4, 1, 5, 2, 0], [4, 3, 5, 1, 0, 2], [5, 2, 4, 0, 1, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
80. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 0, 1, 5, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
81. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 0, 5, 1, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
82. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
83. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 0, 5, 1, 3], [3, 5, 4, 0, 2, 1], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
84. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 5, 4, 1, 0], [3, 2, 1, 0, 5, 4], [4, 5, 3, 2, 0, 1], [5, 4, 0, 1, 3, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
85. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 5, 1, 3, 0], [3, 5, 4, 0, 2, 1], [4, 2, 1, 5, 0, 3], [5, 3, 0, 4, 1, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
86. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 4, 5, 0, 1], [3, 2, 1, 0, 5, 4], [4, 5, 0, 1, 2, 3], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
87. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 4, 1, 0, 3], [3, 4, 5, 0, 1, 2], [4, 3, 0, 5, 2, 1], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
88. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 0, 4, 3, 1], [3, 4, 1, 5, 2, 0], [4, 3, 5, 1, 0, 2], [5, 2, 4, 0, 1, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
89. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 0, 1, 5, 4], [3, 2, 4, 5, 1, 0], [4, 5, 3, 2, 0, 1], [5, 4, 1, 0, 2, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
90. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 5, 1, 3, 0], [3, 5, 4, 0, 2, 1], [4, 2, 1, 5, 0, 3], [5, 3, 0, 4, 1, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
91. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 5, 4, 1, 0], [3, 2, 1, 0, 5, 4], [4, 5, 3, 2, 0, 1], [5, 4, 0, 1, 3, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 5, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
92. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 2, 1, 5, 0, 3], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
93. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 3, 5, 1, 0, 2], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
94. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 0, 5, 1, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
95. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 0, 1, 5, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
96. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 4, 1, 0, 3], [3, 4, 5, 0, 1, 2], [4, 3, 0, 5, 2, 1], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
97. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 4, 5, 0, 1], [3, 2, 1, 0, 5, 4], [4, 5, 0, 1, 2, 3], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 4, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
98. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 3, 5, 1, 0, 2], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 2, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
99. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 2, 1, 5, 0, 3], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 1], 3, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
100. One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup
Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=2. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).
View answer and explanation
([0, 2], 4, 2, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)
Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 2], 4, 2, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality).
Reviewed step-by-step derivation
Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate
Canonical question JSONL · English JSONL · Validation report