# Complete Cayley Tables and Every Finite Group Obligation

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-20261013-advance-20261022-r05-finite-group-foundations · 2026-10-13 · 100 questions

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



## 1. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0001/)

Finite group · qf-str22-group-table-qualification · candidate





## 2. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0002/)

Finite group · qf-str22-group-table-qualification · candidate





## 3. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0003/)

Finite group · qf-str22-group-table-qualification · candidate





## 4. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0004/)

Finite group · qf-str22-group-table-qualification · candidate





## 5. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0005/)

Finite group · qf-str22-group-table-qualification · candidate





## 6. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0006/)

Finite group · qf-str22-group-table-qualification · candidate





## 7. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0007/)

Finite group · qf-str22-group-table-qualification · candidate





## 8. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0008/)

Finite group · qf-str22-group-table-qualification · candidate





## 9. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0009/)

Finite group · qf-str22-group-table-qualification · candidate





## 10. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0010/)

Finite group · qf-str22-group-table-qualification · candidate





## 11. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0011/)

Finite group · qf-str22-group-table-qualification · candidate





## 12. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0012/)

Finite group · qf-str22-group-table-qualification · candidate





## 13. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0013/)

Finite group · qf-str22-group-table-qualification · candidate





## 14. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0014/)

Finite group · qf-str22-group-table-qualification · candidate





## 15. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0015/)

Finite group · qf-str22-group-table-qualification · candidate





## 16. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0016/)

Finite group · qf-str22-group-table-qualification · candidate





## 17. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0017/)

Finite group · qf-str22-group-table-qualification · candidate





## 18. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0018/)

Finite group · qf-str22-group-table-qualification · candidate





## 19. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0019/)

Finite group · qf-str22-group-table-qualification · candidate





## 20. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0020/)

Finite group · qf-str22-group-table-qualification · candidate





## 21. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0021/)

Finite group · qf-str22-group-table-qualification · candidate





## 22. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0022/)

Finite group · qf-str22-group-table-qualification · candidate





## 23. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0023/)

Finite group · qf-str22-group-table-qualification · candidate





## 24. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0024/)

Finite group · qf-str22-group-table-qualification · candidate





## 25. Closure of Every Cell in a Complete Cayley Table

The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).

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

Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0025/)

Finite group · qf-str22-group-table-qualification · candidate





## 26. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0026/)

Finite group · qf-str22-group-associativity-audit · candidate





## 27. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0027/)

Finite group · qf-str22-group-associativity-audit · candidate





## 28. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0028/)

Finite group · qf-str22-group-associativity-audit · candidate





## 29. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0029/)

Finite group · qf-str22-group-associativity-audit · candidate





## 30. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0030/)

Finite group · qf-str22-group-associativity-audit · candidate





## 31. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0031/)

Finite group · qf-str22-group-associativity-audit · candidate





## 32. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0032/)

Finite group · qf-str22-group-associativity-audit · candidate





## 33. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0033/)

Finite group · qf-str22-group-associativity-audit · candidate





## 34. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0034/)

Finite group · qf-str22-group-associativity-audit · candidate





## 35. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0035/)

Finite group · qf-str22-group-associativity-audit · candidate





## 36. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0036/)

Finite group · qf-str22-group-associativity-audit · candidate





## 37. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0037/)

Finite group · qf-str22-group-associativity-audit · candidate





## 38. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0038/)

Finite group · qf-str22-group-associativity-audit · candidate





## 39. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0039/)

Finite group · qf-str22-group-associativity-audit · candidate





## 40. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0040/)

Finite group · qf-str22-group-associativity-audit · candidate





## 41. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0041/)

Finite group · qf-str22-group-associativity-audit · candidate





## 42. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0042/)

Finite group · qf-str22-group-associativity-audit · candidate





## 43. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0043/)

Finite group · qf-str22-group-associativity-audit · candidate





## 44. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0044/)

Finite group · qf-str22-group-associativity-audit · candidate





## 45. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0045/)

Finite group · qf-str22-group-associativity-audit · candidate





## 46. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0046/)

Finite group · qf-str22-group-associativity-audit · candidate





## 47. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0047/)

Finite group · qf-str22-group-associativity-audit · candidate





## 48. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0048/)

Finite group · qf-str22-group-associativity-audit · candidate





## 49. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0049/)

Finite group · qf-str22-group-associativity-audit · candidate





## 50. Exhaustive Associativity over All Ordered Triples

The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.

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

Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0050/)

Finite group · qf-str22-group-associativity-audit · candidate





## 51. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0051/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 52. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0052/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 53. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0053/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 54. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0054/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 55. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0055/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 56. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0056/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 57. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0057/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 58. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0058/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 59. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0059/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 60. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0060/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 61. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0061/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 62. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0062/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 63. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0063/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 64. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0064/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 65. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0065/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 66. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0066/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 67. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0067/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 68. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0068/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 69. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0069/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 70. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0070/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 71. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0071/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 72. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0072/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 73. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0073/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 74. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0074/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 75. Two-Sided Identity and Two-Sided Inverse Inventories

The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.

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

Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0075/)

Finite group · qf-str22-group-identity-inverse-inventory · candidate





## 76. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0076/)

Finite group · qf-str22-group-full-group-audit · candidate





## 77. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0077/)

Finite group · qf-str22-group-full-group-audit · candidate





## 78. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0078/)

Finite group · qf-str22-group-full-group-audit · candidate





## 79. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0079/)

Finite group · qf-str22-group-full-group-audit · candidate





## 80. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0080/)

Finite group · qf-str22-group-full-group-audit · candidate





## 81. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes, Yes, [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0081/)

Finite group · qf-str22-group-full-group-audit · candidate





## 82. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes, Yes, [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0082/)

Finite group · qf-str22-group-full-group-audit · candidate





## 83. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes, Yes, [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0083/)

Finite group · qf-str22-group-full-group-audit · candidate





## 84. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes, Yes, [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0084/)

Finite group · qf-str22-group-full-group-audit · candidate





## 85. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0085/)

Finite group · qf-str22-group-full-group-audit · candidate





## 86. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0086/)

Finite group · qf-str22-group-full-group-audit · candidate





## 87. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0087/)

Finite group · qf-str22-group-full-group-audit · candidate





## 88. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0088/)

Finite group · qf-str22-group-full-group-audit · candidate





## 89. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0089/)

Finite group · qf-str22-group-full-group-audit · candidate





## 90. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0090/)

Finite group · qf-str22-group-full-group-audit · candidate





## 91. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0091/)

Finite group · qf-str22-group-full-group-audit · candidate





## 92. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0092/)

Finite group · qf-str22-group-full-group-audit · candidate





## 93. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0093/)

Finite group · qf-str22-group-full-group-audit · candidate





## 94. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0094/)

Finite group · qf-str22-group-full-group-audit · candidate





## 95. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0095/)

Finite group · qf-str22-group-full-group-audit · candidate





## 96. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0096/)

Finite group · qf-str22-group-full-group-audit · candidate





## 97. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0097/)

Finite group · qf-str22-group-full-group-audit · candidate





## 98. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0098/)

Finite group · qf-str22-group-full-group-audit · candidate





## 99. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0099/)

Finite group · qf-str22-group-full-group-audit · candidate





## 100. Close Every Group Obligation on the Specified Finite Carrier

The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.

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

A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r05-finite-group-0100/)

Finite group · qf-str22-group-full-group-audit · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261022-r05-finite-group-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261013-advance-20261022-r05-finite-group-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261022-r05-finite-group-foundations/validation-report.json)