# Complete Subgroup Closure, Conjugation and All-Element Coset Equality

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

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

batch-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](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0001/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0002/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0003/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0004/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0005/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0006/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0007/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0008/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0009/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0010/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0011/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0012/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0013/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0014/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0015/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0016/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0017/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0018/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0019/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0020/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0021/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0022/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0023/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0024/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0025/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0026/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0027/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0028/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0029/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0030/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0031/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0032/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0033/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0034/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0035/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0036/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0037/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0038/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0039/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0040/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0041/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0042/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0043/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0044/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0045/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0046/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0047/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0048/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0049/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0050/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0051/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0052/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0053/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0054/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0055/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0056/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0057/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0058/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0059/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0060/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0061/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0062/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0063/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0064/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0065/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0066/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0067/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0068/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0069/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0070/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0071/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0072/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0073/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0074/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0075/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0076/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0077/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0078/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0079/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0080/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0081/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0082/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0083/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0084/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0085/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0086/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0087/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0088/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0089/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0090/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0091/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0092/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0093/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0094/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0095/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0096/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0097/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0098/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0099/)

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](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r01-finite-normal-subgroup-0100/)

Finite normal subgroup · qf-mod44-normal-nonnormal-witness · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations/validation-report.json)