# Prime Divisors, Exact Element Orders and Composite-Order Nonconverses

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-20261112-r01-cauchy-theorem-finite-groups-foundations · 2026-10-14 · 100 questions

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

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

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



## 1. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (6, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0001/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 2. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (8, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0002/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 3. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (10, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0003/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 4. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (12, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0004/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 5. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (14, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0005/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 6. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (6, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0006/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 7. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (8, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0007/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 8. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (10, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0008/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 9. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (12, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0009/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 10. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (14, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0010/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 11. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (6, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0011/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 12. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (8, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0012/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 13. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (10, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0013/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 14. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (12, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0014/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 15. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (14, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0015/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 16. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (6, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0016/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 17. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (8, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0017/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 18. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (10, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0018/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 19. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (12, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0019/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 20. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (14, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0020/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 21. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (6, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0021/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 22. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (8, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0022/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 23. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (10, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0023/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 24. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (12, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0024/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 25. Cauchy Requires a Prime Divisor of a Fully Certified Finite Group

Use complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Declared prime=2. Give (group_order, prime, qualified, divisibility_remainder, certificate).

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

Require primality and prime divisibility of the complete group order. Qualification is separate from producing an exact-order element witness. Finite model enumeration is not a proof of the general Cauchy theorem. The exact result, retaining the requested order, is (14, 2, Yes, 0, complete_group_and_prime_divisibility_hypotheses_hold).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0025/)

Cauchy theorem for finite groups · qf-mod43-cauchy-prime-divisor-qualification · candidate





## 26. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (6, [[0, 1], [1, 6], [2, 3], [3, 2], [4, 3], [5, 6]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0026/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 27. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (8, [[0, 1], [1, 8], [2, 4], [3, 8], [4, 2], [5, 8], [6, 4], [7, 8]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0027/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 28. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (10, [[0, 1], [1, 10], [2, 5], [3, 10], [4, 5], [5, 2], [6, 5], [7, 10], [8, 5], [9, 10]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0028/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 29. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (12, [[0, 1], [1, 12], [2, 6], [3, 4], [4, 3], [5, 12], [6, 2], [7, 12], [8, 3], [9, 4], [10, 6], [11, 12]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0029/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 30. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

View answer and explanation(14, [[0, 1], [1, 14], [2, 7], [3, 14], [4, 7], [5, 14], [6, 7], [7, 2], [8, 7], [9, 14], [10, 7], [11, 14], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group)

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (14, [[0, 1], [1, 14], [2, 7], [3, 14], [4, 7], [5, 14], [6, 7], [7, 2], [8, 7], [9, 14], [10, 7], [11, 14], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0030/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 31. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (6, [[0, 6], [1, 1], [2, 6], [3, 3], [4, 2], [5, 3]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0031/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 32. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (8, [[0, 8], [1, 1], [2, 8], [3, 4], [4, 8], [5, 2], [6, 8], [7, 4]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0032/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 33. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (10, [[0, 10], [1, 1], [2, 10], [3, 5], [4, 10], [5, 5], [6, 2], [7, 5], [8, 10], [9, 5]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0033/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 34. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (12, [[0, 12], [1, 1], [2, 12], [3, 6], [4, 4], [5, 3], [6, 12], [7, 2], [8, 12], [9, 3], [10, 4], [11, 6]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0034/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 35. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

View answer and explanation(14, [[0, 14], [1, 1], [2, 14], [3, 7], [4, 14], [5, 7], [6, 14], [7, 7], [8, 2], [9, 7], [10, 14], [11, 7], [12, 14], [13, 7]], Yes, complete_element_order_table_for_declared_finite_group)

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (14, [[0, 14], [1, 1], [2, 14], [3, 7], [4, 14], [5, 7], [6, 14], [7, 7], [8, 2], [9, 7], [10, 14], [11, 7], [12, 14], [13, 7]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0035/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 36. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (6, [[0, 3], [1, 6], [2, 1], [3, 6], [4, 3], [5, 2]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0036/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 37. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (8, [[0, 4], [1, 8], [2, 1], [3, 8], [4, 4], [5, 8], [6, 2], [7, 8]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0037/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 38. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (10, [[0, 5], [1, 10], [2, 1], [3, 10], [4, 5], [5, 10], [6, 5], [7, 2], [8, 5], [9, 10]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0038/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 39. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (12, [[0, 6], [1, 12], [2, 1], [3, 12], [4, 6], [5, 4], [6, 3], [7, 12], [8, 2], [9, 12], [10, 3], [11, 4]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0039/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 40. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

View answer and explanation(14, [[0, 7], [1, 14], [2, 1], [3, 14], [4, 7], [5, 14], [6, 7], [7, 14], [8, 7], [9, 2], [10, 7], [11, 14], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group)

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (14, [[0, 7], [1, 14], [2, 1], [3, 14], [4, 7], [5, 14], [6, 7], [7, 14], [8, 7], [9, 2], [10, 7], [11, 14], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0040/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 41. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (6, [[0, 2], [1, 3], [2, 6], [3, 1], [4, 6], [5, 3]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0041/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 42. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (8, [[0, 8], [1, 4], [2, 8], [3, 1], [4, 8], [5, 4], [6, 8], [7, 2]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0042/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 43. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (10, [[0, 10], [1, 5], [2, 10], [3, 1], [4, 10], [5, 5], [6, 10], [7, 5], [8, 2], [9, 5]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0043/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 44. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (12, [[0, 4], [1, 6], [2, 12], [3, 1], [4, 12], [5, 6], [6, 4], [7, 3], [8, 12], [9, 2], [10, 12], [11, 3]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0044/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 45. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

View answer and explanation(14, [[0, 14], [1, 7], [2, 14], [3, 1], [4, 14], [5, 7], [6, 14], [7, 7], [8, 14], [9, 7], [10, 2], [11, 7], [12, 14], [13, 7]], Yes, complete_element_order_table_for_declared_finite_group)

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (14, [[0, 14], [1, 7], [2, 14], [3, 1], [4, 14], [5, 7], [6, 14], [7, 7], [8, 14], [9, 7], [10, 2], [11, 7], [12, 14], [13, 7]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0045/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 46. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (6, [[0, 3], [1, 2], [2, 3], [3, 6], [4, 1], [5, 6]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0046/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 47. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (8, [[0, 2], [1, 8], [2, 4], [3, 8], [4, 1], [5, 8], [6, 4], [7, 8]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0047/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 48. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (10, [[0, 5], [1, 10], [2, 5], [3, 10], [4, 1], [5, 10], [6, 5], [7, 10], [8, 5], [9, 2]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0048/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 49. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

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

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (12, [[0, 3], [1, 4], [2, 6], [3, 12], [4, 1], [5, 12], [6, 6], [7, 4], [8, 3], [9, 12], [10, 2], [11, 12]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0049/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 50. Every Element Order Is the First Positive Identity Return

Use complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Give (group_order, element_order_rows, all_orders_divide_group_order, certificate).

View answer and explanation(14, [[0, 7], [1, 14], [2, 7], [3, 14], [4, 1], [5, 14], [6, 7], [7, 14], [8, 7], [9, 14], [10, 7], [11, 2], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group)

Enumerate every element and its least positive identity-return exponent, checking through the full finite group order. Report whether every computed order divides |G|. Do not count exponent0 or infer exact order from only g^p=e. The exact result, retaining the requested order, is (14, [[0, 7], [1, 14], [2, 7], [3, 14], [4, 1], [5, 14], [6, 7], [7, 14], [8, 7], [9, 14], [10, 7], [11, 2], [12, 7], [13, 14]], Yes, complete_element_order_table_for_declared_finite_group).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0050/)

Cauchy theorem for finite groups · qf-mod43-cauchy-element-order-table · candidate





## 51. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=3. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

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

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (6, 2, 3, [[1, 3], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0051/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 52. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=4. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(8, 2, 4, [[1, 4], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (8, 2, 4, [[1, 4], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0052/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 53. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=5. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(10, 2, 5, [[1, 5], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (10, 2, 5, [[1, 5], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0053/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 54. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=6. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(12, 2, 6, [[1, 6], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (12, 2, 6, [[1, 6], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0054/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 55. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=7. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(14, 2, 7, [[1, 7], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (14, 2, 7, [[1, 7], [2, 0]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0055/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 56. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=4. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

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

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (6, 2, 4, [[1, 4], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0056/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 57. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=5. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(8, 2, 5, [[1, 5], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (8, 2, 5, [[1, 5], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0057/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 58. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=6. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(10, 2, 6, [[1, 6], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (10, 2, 6, [[1, 6], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0058/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 59. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=7. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(12, 2, 7, [[1, 7], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (12, 2, 7, [[1, 7], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0059/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 60. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=8. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(14, 2, 8, [[1, 8], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (14, 2, 8, [[1, 8], [2, 1]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0060/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 61. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=5. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(6, 2, 5, [[1, 5], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (6, 2, 5, [[1, 5], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0061/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 62. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=6. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(8, 2, 6, [[1, 6], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (8, 2, 6, [[1, 6], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0062/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 63. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=7. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(10, 2, 7, [[1, 7], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (10, 2, 7, [[1, 7], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0063/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 64. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=8. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(12, 2, 8, [[1, 8], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (12, 2, 8, [[1, 8], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0064/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 65. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=9. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(14, 2, 9, [[1, 9], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (14, 2, 9, [[1, 9], [2, 2]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0065/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 66. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=0. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

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

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (6, 2, 0, [[1, 0], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0066/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 67. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=7. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(8, 2, 7, [[1, 7], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (8, 2, 7, [[1, 7], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0067/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 68. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=8. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(10, 2, 8, [[1, 8], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (10, 2, 8, [[1, 8], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0068/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 69. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=9. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(12, 2, 9, [[1, 9], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (12, 2, 9, [[1, 9], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0069/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 70. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=10. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(14, 2, 10, [[1, 10], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (14, 2, 10, [[1, 10], [2, 3]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0070/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 71. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=1. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

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

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (6, 2, 1, [[1, 1], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0071/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 72. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=0. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(8, 2, 0, [[1, 0], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (8, 2, 0, [[1, 0], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0072/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 73. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=9. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(10, 2, 9, [[1, 9], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (10, 2, 9, [[1, 9], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0073/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 74. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=10. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(12, 2, 10, [[1, 10], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (12, 2, 10, [[1, 10], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0074/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 75. An Order-p Witness Must Exclude Every Smaller Positive Return

Use complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Prime=2, supplied element=11. Give (group_order, prime, element, positive_power_rows, p_power_is_identity, no_smaller_positive_identity_power, certificate).

Check g^p=e separately from every smaller positive power being nonidentity. The identity satisfies g^p=e but is not an order-p witness.

View answer and explanation(14, 2, 11, [[1, 11], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is)

Require prime divisibility and exact element order p. List powers1 through p, verify the p-th is identity and every smaller positive one is not. The identity is not an order-p witness even though its p-th power is identity. The exact result, retaining the requested order, is (14, 2, 11, [[1, 11], [2, 4]], Yes, Yes, g_to_p_is_identity_and_no_smaller_positive_power_is).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0075/)

Cauchy theorem for finite groups · qf-mod43-cauchy-order-p-witness · candidate





## 76. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 0, 1], [3, 2, 1, 0]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=4. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

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

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (4, 4, [[0, 1], [1, 2], [2, 2], [3, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0076/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 77. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[1, 0, 3, 2], [0, 1, 2, 3], [3, 2, 1, 0], [2, 3, 0, 1]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=4. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

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

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (4, 4, [[0, 2], [1, 1], [2, 2], [3, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0077/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 78. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[2, 3, 0, 1], [3, 2, 1, 0], [0, 1, 2, 3], [1, 0, 3, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=4. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

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

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (4, 4, [[0, 2], [1, 2], [2, 1], [3, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0078/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 79. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[3, 2, 1, 0], [2, 3, 0, 1], [1, 0, 3, 2], [0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=4. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

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

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (4, 4, [[0, 2], [1, 2], [2, 2], [3, 1]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0079/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 80. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4], [4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 1], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 1], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0080/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 81. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[1, 0, 3, 2, 5, 4, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7], [3, 2, 1, 0, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5], [5, 4, 7, 6, 1, 0, 3, 2], [4, 5, 6, 7, 0, 1, 2, 3], [7, 6, 5, 4, 3, 2, 1, 0], [6, 7, 4, 5, 2, 3, 0, 1]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 1], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 1], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0081/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 82. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4], [0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0], [4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 1], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 1], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0082/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 83. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[3, 2, 1, 0, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5], [1, 0, 3, 2, 5, 4, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7], [7, 6, 5, 4, 3, 2, 1, 0], [6, 7, 4, 5, 2, 3, 0, 1], [5, 4, 7, 6, 1, 0, 3, 2], [4, 5, 6, 7, 0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 2], [3, 1], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 2], [3, 1], [4, 2], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0083/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 84. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0084/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 85. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[5, 4, 7, 6, 1, 0, 3, 2], [4, 5, 6, 7, 0, 1, 2, 3], [7, 6, 5, 4, 3, 2, 1, 0], [6, 7, 4, 5, 2, 3, 0, 1], [1, 0, 3, 2, 5, 4, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7], [3, 2, 1, 0, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5]] with identity=5, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 1], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 1], [6, 2], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0085/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 86. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0], [4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4], [0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6]] with identity=6, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 1], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 1], [7, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0086/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 87. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[7, 6, 5, 4, 3, 2, 1, 0], [6, 7, 4, 5, 2, 3, 0, 1], [5, 4, 7, 6, 1, 0, 3, 2], [4, 5, 6, 7, 0, 1, 2, 3], [3, 2, 1, 0, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5], [1, 0, 3, 2, 5, 4, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7]] with identity=7, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=8. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 1]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (8, 8, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 1]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0087/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 88. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]] with identity=0, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 1], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 1], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0088/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 89. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1]] with identity=1, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 1], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 1], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0089/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 90. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2]] with identity=2, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 1], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 1], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0090/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 91. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3]] with identity=3, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 1], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 1], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0091/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 92. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4]] with identity=4, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0092/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 93. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5]] with identity=5, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 1], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 1], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0093/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 94. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6]] with identity=6, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 1], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 1], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0094/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 95. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7]] with identity=7, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 1], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 1], [8, 2], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0095/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 96. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8]] with identity=8, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 1], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 1], [9, 2], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0096/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 97. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9]] with identity=9, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 1], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 1], [10, 2], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0097/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 98. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10]] with identity=10, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 1], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 1], [11, 2], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0098/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 99. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11]] with identity=11, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 1], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 1], [12, 2], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0099/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate





## 100. A Composite Divisor Need Not Be an Element Order

Use complete finite Cayley table=[[12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3], [13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2], [14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1], [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7], [9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6], [10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5], [11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4], [4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11], [5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10], [6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9], [7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14], [2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13], [3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12]] with identity=12, row g/column h meaning g*h. Certify closure, associativity, identity and inverses. Element order is the least positive exponent returning to this same identity, not just one exponent that does so. Requested composite divisor=16. Give (group_order, composite_divisor, element_order_rows, divisor_divides_order, exists_element_of_requested_order, boundary).

View answer and explanation(16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 1], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order)

Require actual divisibility and exhaust the entire element-order table with no element of that requested order. The final No is absence in this complete model, not a claim that composite element orders never occur in other groups. The exact result, retaining the requested order, is (16, 16, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2], [10, 2], [11, 2], [12, 1], [13, 2], [14, 2], [15, 2]], Yes, No, composite_divisor_need_not_be_an_element_order).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-0100/)

Cauchy theorem for finite groups · qf-mod43-cauchy-composite-divisor-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261112-r01-cauchy-theorem-finite-groups-foundations/validation-report.json)