# Same-Point Orbit-Stabilizer Counts, Fibers and Left-Coset Certificates

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

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

batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations · 2026-10-14 · 100 questions

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

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

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



## 1. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

View answer and explanation(2, 0, [0, 1], 2, [0], 1, 2, Yes)

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 0, [0, 1], 2, [0], 1, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0001/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 2. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 1, [0, 1, 2], 3, [0], 1, 3, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0002/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 3. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0003/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 4. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 3, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0004/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 5. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0005/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 6. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

View answer and explanation(2, 2, [2], 1, [0, 1], 2, 2, Yes)

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0006/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 7. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 2, [0, 1, 2], 3, [0], 1, 3, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0007/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 8. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0008/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 9. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0009/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 10. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 2, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0010/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 11. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

View answer and explanation(2, 2, [2], 1, [0, 1], 2, 2, Yes)

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0011/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 12. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 1, [0, 1, 2], 3, [0], 1, 3, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0012/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 13. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 0, [0, 1, 2, 3], 4, [0], 1, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0013/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 14. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 6, [6], 1, [0, 1, 2, 3, 4], 5, 5, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0014/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 15. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 6, [6], 1, [0, 1, 2, 3, 4, 5], 6, 6, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0015/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 16. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

View answer and explanation(2, 0, [0, 1], 2, [0], 1, 2, Yes)

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 0, [0, 1], 2, [0], 1, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0016/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 17. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 4, [4], 1, [0, 1, 2], 3, 3, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0017/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 18. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 3, [0, 1, 2, 3], 4, [0], 1, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0018/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 19. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0019/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 20. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 1, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0020/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 21. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

View answer and explanation(2, 2, [2], 1, [0, 1], 2, 2, Yes)

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0021/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 22. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 0, [0, 1, 2], 3, [0], 1, 3, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0022/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 23. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 6, [6], 1, [0, 1, 2, 3], 4, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0023/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 24. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 5, [5], 1, [0, 1, 2, 3, 4], 5, 5, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0024/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 25. The Same Point Binds Both Factors in Orbit-Stabilizer Counting

Use complete group 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).

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

Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0025/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-counting-identity · candidate





## 26. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0026/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 27. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0027/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 28. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0028/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 29. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 3], [1, 4], [2, 0], [3, 1], [4, 2]], [[0, [2]], [1, [3]], [2, [4]], [3, [0]], [4, [1]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0029/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 30. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0030/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 31. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0031/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 32. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 0], [2, 1]], [[0, [1]], [1, [2]], [2, [0]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 0], [2, 1]], [[0, [1]], [1, [2]], [2, [0]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0032/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 33. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0033/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 34. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0034/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 35. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 5], [4, 0], [5, 1]], [[0, [4]], [1, [5]], [2, [0]], [3, [1]], [4, [2]], [5, [3]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0035/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 36. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0036/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 37. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0037/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 38. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 0], [1, 1], [2, 2], [3, 3]], [[0, [0]], [1, [1]], [2, [2]], [3, [3]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [2, 2], [3, 3]], [[0, [0]], [1, [1]], [2, [2]], [3, [3]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0038/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 39. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6]], [[6, [0, 1, 2, 3, 4]]], Yes, [6], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0039/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 40. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6]], [[6, [0, 1, 2, 3, 4, 5]]], Yes, [6], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0040/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 41. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0041/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 42. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 4], [1, 4], [2, 4]], [[4, [0, 1, 2]]], Yes, [4], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 4], [2, 4]], [[4, [0, 1, 2]]], Yes, [4], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0042/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 43. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 3], [1, 0], [2, 1], [3, 2]], [[0, [1]], [1, [2]], [2, [3]], [3, [0]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 3], [1, 0], [2, 1], [3, 2]], [[0, [1]], [1, [2]], [2, [3]], [3, [0]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0043/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 44. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0044/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 45. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], [[0, [5]], [1, [0]], [2, [1]], [3, [2]], [4, [3]], [5, [4]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0045/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 46. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0046/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 47. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 0], [1, 1], [2, 2]], [[0, [0]], [1, [1]], [2, [2]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [2, 2]], [[0, [0]], [1, [1]], [2, [2]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0047/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 48. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

View answer and explanation([[0, 6], [1, 6], [2, 6], [3, 6]], [[6, [0, 1, 2, 3]]], Yes, [6], action_map_surjective_onto_orbit)

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6]], [[6, [0, 1, 2, 3]]], Yes, [6], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0048/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 49. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 5], [1, 5], [2, 5], [3, 5], [4, 5]], [[5, [0, 1, 2, 3, 4]]], Yes, [5], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0049/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 50. All Action-Map Fibers Have the Same Stabilizer Cardinality

Use complete group 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).

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

List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0050/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-action-fibers · candidate





## 51. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0051/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 52. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0052/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 53. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 3, [3], [3], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0053/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 54. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0054/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 55. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=3. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (3, 5, [5], [5], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0055/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 56. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0056/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 57. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(0, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0057/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 58. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0058/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 59. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0059/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 60. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0060/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 61. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0061/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 62. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0062/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 63. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0063/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 64. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0064/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 65. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0065/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 66. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(0, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0066/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 67. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=4. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(4, 0, [0, 1, 2], [0, 1, 2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (4, 0, [0, 1, 2], [0, 1, 2], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0067/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 68. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0068/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 69. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0069/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 70. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=3. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(3, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (3, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0070/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 71. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0071/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 72. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0072/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 73. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

View answer and explanation(6, 0, [0, 1, 2, 3], [0, 1, 2, 3], Yes, fiber_equals_left_coset_of_same_point_stabilizer)

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3], [0, 1, 2, 3], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0073/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 74. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=5. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (5, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0074/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 75. A Fiber Equals a Left Coset of the Same Point Stabilizer

Use complete group 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=5. Give (orbit_point, representative, fiber, left_coset, equal, certificate).

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

Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (5, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0075/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-coset-fiber-bijection · candidate





## 76. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0076/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 77. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0077/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 78. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2], [0, 2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [0, 2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0078/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 79. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3], other_point=4. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4], [0], 4, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4], [0], 4, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0079/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 80. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0080/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 81. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=0. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [2], [2], [0, 1], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0081/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 82. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [2], [0, 1, 2], [0], 3, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [0, 1, 2], [0], 3, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0082/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 83. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0083/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 84. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [2], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0084/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 85. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3, 4], [0, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0085/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 86. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0086/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 87. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0087/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 88. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [0], [0, 1, 2, 3], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1, 2, 3], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0088/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 89. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=0. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [6], [6], [0, 1, 2, 3, 4], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [6], [6], [0, 1, 2, 3, 4], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0089/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 90. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=7. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3, 4], [6], [6], [0, 1, 2, 3, 4, 5], 7, [0, 1, 2, 3, 4, 5], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [6], [6], [0, 1, 2, 3, 4, 5], 7, [0, 1, 2, 3, 4, 5], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0090/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 91. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0091/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 92. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [4], [4], [0, 1, 2], 5, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [4], [4], [0, 1, 2], 5, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0092/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 93. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=4. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2], [0, 1, 3], [0, 1, 2, 3], [0], 4, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [0, 1, 3], [0, 1, 2, 3], [0], 4, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0093/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 94. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2], [2, 3, 4], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [2, 3, 4], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0094/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 95. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3, 4], [1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0095/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 96. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0096/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 97. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [0, 1], [0, 1, 2], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [0, 1], [0, 1, 2], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0097/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 98. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=7. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1], [6], [6], [0, 1, 2, 3], 7, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [6], [6], [0, 1, 2, 3], 7, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0098/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 99. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3], other_point=6. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3], [5], [5], [0, 1, 2, 3, 4], 6, [0, 1, 2, 3, 4], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3], [5], [5], [0, 1, 2, 3, 4], 6, [0, 1, 2, 3, 4], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0099/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate





## 100. A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate

Use complete group 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).

The fixed No withholds a theorem certificate from a proper sample or mixed point binding. It does not deny that the sample image may equal the full orbit or that two point stabilizers may coincide.

View answer and explanation([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)

Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0100/)

Orbit-stabilizer theorem · qf-mod41-orbitstab-binding-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations/validation-report.json)