# Complete Finite Subgroup and Homomorphism Obligations

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-20261023-r03-subgroup-homomorphism-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. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], nonempty candidate subset=[0, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0001/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 2. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 0, 3, 4, 5, 1], [0, 1, 2, 3, 4, 5], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4]], nonempty candidate subset=[1, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0002/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 3. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 3, 0, 4, 5, 2], [3, 4, 1, 5, 2, 0], [0, 1, 2, 3, 4, 5], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4]], nonempty candidate subset=[1, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0003/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 4. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [0, 1, 2, 3, 4, 5], [5, 3, 0, 4, 1, 2], [3, 0, 1, 5, 2, 4]], nonempty candidate subset=[1, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0004/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 5. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [5, 4, 0, 1, 3, 2], [0, 1, 2, 3, 4, 5], [4, 0, 1, 2, 5, 3]], nonempty candidate subset=[1, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0005/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 6. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty candidate subset=[1, 3, 5] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0006/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 7. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]], nonempty candidate subset=[0, 1, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0007/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 8. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 0, 3, 5, 1, 2], [0, 1, 2, 3, 4, 5], [3, 2, 0, 4, 5, 1], [5, 3, 4, 1, 2, 0], [1, 4, 5, 2, 0, 3], [2, 5, 1, 0, 3, 4]], nonempty candidate subset=[0, 1, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0008/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 9. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 3, 0, 5, 2, 1], [3, 0, 1, 4, 5, 2], [0, 1, 2, 3, 4, 5], [5, 4, 3, 2, 1, 0], [2, 5, 4, 1, 0, 3], [1, 2, 5, 0, 3, 4]], nonempty candidate subset=[0, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0009/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 10. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 2, 5, 0, 3, 1], [2, 0, 4, 1, 5, 3], [5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5], [3, 5, 1, 4, 0, 2], [1, 3, 0, 5, 2, 4]], nonempty candidate subset=[0, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0010/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 11. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4], [5, 3, 4, 1, 2, 0], [4, 5, 1, 0, 3, 2], [0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3]], nonempty candidate subset=[0, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0011/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 12. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 2, 4, 5, 1, 0], [2, 0, 3, 4, 5, 1], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty candidate subset=[0, 3, 5] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0012/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 13. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 0, 3], [2, 5, 0, 4, 3, 1], [3, 2, 4, 1, 5, 0], [4, 0, 3, 5, 1, 2], [5, 3, 1, 0, 2, 4]], nonempty candidate subset=[0, 1, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0013/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 14. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 0, 5, 2, 1, 3], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [2, 3, 4, 0, 5, 1], [1, 4, 3, 5, 0, 2], [3, 5, 0, 1, 2, 4]], nonempty candidate subset=[0, 1, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0014/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 15. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0], [0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4]], nonempty candidate subset=[0, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0015/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 16. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0], [1, 4, 0, 2, 5, 3], [0, 1, 2, 3, 4, 5], [3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4]], nonempty candidate subset=[0, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0016/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 17. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 5, 1, 4, 0, 2], [5, 4, 3, 2, 1, 0], [1, 3, 0, 5, 2, 4], [4, 2, 5, 0, 3, 1], [0, 1, 2, 3, 4, 5], [2, 0, 4, 1, 5, 3]], nonempty candidate subset=[0, 3, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0017/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 18. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [1, 3, 0, 4, 5, 2], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty candidate subset=[0, 3, 5] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0018/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 19. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 0, 2, 4], [2, 5, 0, 4, 3, 1], [3, 0, 4, 1, 5, 2], [4, 2, 3, 5, 1, 0], [5, 4, 1, 2, 0, 3]], nonempty candidate subset=[0, 1, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0019/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 20. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 0, 5, 1, 2, 4], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [1, 3, 4, 0, 5, 2], [2, 4, 3, 5, 0, 1], [4, 5, 0, 2, 1, 3]], nonempty candidate subset=[0, 1, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0020/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 21. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 5, 0, 2, 1, 4], [5, 2, 1, 4, 3, 0], [0, 1, 2, 3, 4, 5], [2, 4, 3, 0, 5, 1], [1, 3, 4, 5, 0, 2], [4, 0, 5, 1, 2, 3]], nonempty candidate subset=[0, 2, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0021/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 22. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], nonempty candidate subset=[0, 2, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0022/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 23. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 4, 1, 0, 3], [5, 4, 3, 2, 1, 0], [4, 3, 0, 5, 2, 1], [1, 2, 5, 0, 3, 4], [0, 1, 2, 3, 4, 5], [3, 0, 1, 4, 5, 2]], nonempty candidate subset=[0, 2, 4] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0023/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 24. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2], [1, 2, 4, 0, 5, 3], [3, 0, 1, 5, 2, 4], [0, 1, 2, 3, 4, 5]], nonempty candidate subset=[0, 2, 5] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0024/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 25. Product Closure of a Nonempty Subset in a Qualified Finite Group

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 0, 5, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 1, 2, 4], [4, 5, 1, 2, 3, 0], [5, 2, 3, 4, 0, 1]], nonempty candidate subset=[0, 1, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).

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

Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0025/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subset-closure · candidate





## 26. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], nonempty subset=[0, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], 0, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0026/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 27. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 0, 3, 4, 5, 1], [0, 1, 2, 3, 4, 5], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4]], nonempty subset=[1, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 2, 4], 1, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0027/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 28. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 3, 0, 4, 5, 2], [3, 4, 1, 5, 2, 0], [0, 1, 2, 3, 4, 5], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4]], nonempty subset=[1, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 2, 4], 2, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0028/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 29. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [0, 1, 2, 3, 4, 5], [5, 3, 0, 4, 1, 2], [3, 0, 1, 5, 2, 4]], nonempty subset=[1, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 4], 3, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0029/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 30. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [5, 4, 0, 1, 3, 2], [0, 1, 2, 3, 4, 5], [4, 0, 1, 2, 5, 3]], nonempty subset=[1, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 4], 4, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0030/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 31. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty subset=[1, 3, 5] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [1, 3, 5], 5, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0031/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 32. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]], nonempty subset=[0, 1, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], 0, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0032/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 33. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 0, 3, 5, 1, 2], [0, 1, 2, 3, 4, 5], [3, 2, 0, 4, 5, 1], [5, 3, 4, 1, 2, 0], [1, 4, 5, 2, 0, 3], [2, 5, 1, 0, 3, 4]], nonempty subset=[0, 1, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], 1, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0033/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 34. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 3, 0, 5, 2, 1], [3, 0, 1, 4, 5, 2], [0, 1, 2, 3, 4, 5], [5, 4, 3, 2, 1, 0], [2, 5, 4, 1, 0, 3], [1, 2, 5, 0, 3, 4]], nonempty subset=[0, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], 2, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0034/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 35. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 2, 5, 0, 3, 1], [2, 0, 4, 1, 5, 3], [5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5], [3, 5, 1, 4, 0, 2], [1, 3, 0, 5, 2, 4]], nonempty subset=[0, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], 3, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0035/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 36. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4], [5, 3, 4, 1, 2, 0], [4, 5, 1, 0, 3, 2], [0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3]], nonempty subset=[0, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], 4, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0036/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 37. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 2, 4, 5, 1, 0], [2, 0, 3, 4, 5, 1], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty subset=[0, 3, 5] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 5], 5, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0037/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 38. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 0, 3], [2, 5, 0, 4, 3, 1], [3, 2, 4, 1, 5, 0], [4, 0, 3, 5, 1, 2], [5, 3, 1, 0, 2, 4]], nonempty subset=[0, 1, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], 0, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0038/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 39. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 0, 5, 2, 1, 3], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [2, 3, 4, 0, 5, 1], [1, 4, 3, 5, 0, 2], [3, 5, 0, 1, 2, 4]], nonempty subset=[0, 1, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 4], 1, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0039/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 40. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0], [0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4]], nonempty subset=[0, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], 2, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0040/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 41. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0], [1, 4, 0, 2, 5, 3], [0, 1, 2, 3, 4, 5], [3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4]], nonempty subset=[0, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], 3, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0041/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 42. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 5, 1, 4, 0, 2], [5, 4, 3, 2, 1, 0], [1, 3, 0, 5, 2, 4], [4, 2, 5, 0, 3, 1], [0, 1, 2, 3, 4, 5], [2, 0, 4, 1, 5, 3]], nonempty subset=[0, 3, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 4], 4, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0042/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 43. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [1, 3, 0, 4, 5, 2], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4], [0, 1, 2, 3, 4, 5]], nonempty subset=[0, 3, 5] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 3, 5], 5, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0043/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 44. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 0, 2, 4], [2, 5, 0, 4, 3, 1], [3, 0, 4, 1, 5, 2], [4, 2, 3, 5, 1, 0], [5, 4, 1, 2, 0, 3]], nonempty subset=[0, 1, 3] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], 0, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0044/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 45. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 0, 5, 1, 2, 4], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [1, 3, 4, 0, 5, 2], [2, 4, 3, 5, 0, 1], [4, 5, 0, 2, 1, 3]], nonempty subset=[0, 1, 3] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], 1, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0045/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 46. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[3, 5, 0, 2, 1, 4], [5, 2, 1, 4, 3, 0], [0, 1, 2, 3, 4, 5], [2, 4, 3, 0, 5, 1], [1, 3, 4, 5, 0, 2], [4, 0, 5, 1, 2, 3]], nonempty subset=[0, 2, 3] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 3], 2, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0046/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 47. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], nonempty subset=[0, 2, 3] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 3], 3, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0047/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 48. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 4, 1, 0, 3], [5, 4, 3, 2, 1, 0], [4, 3, 0, 5, 2, 1], [1, 2, 5, 0, 3, 4], [0, 1, 2, 3, 4, 5], [3, 0, 1, 4, 5, 2]], nonempty subset=[0, 2, 4] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 4], 4, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0048/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 49. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2], [1, 2, 4, 0, 5, 3], [3, 0, 1, 5, 2, 4], [0, 1, 2, 3, 4, 5]], nonempty subset=[0, 2, 5] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 5], 5, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0049/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 50. Explicit Subgroup Closure, Identity and Inverse Obligations

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 0, 5, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 1, 2, 4], [4, 5, 1, 2, 3, 0], [5, 2, 3, 4, 0, 1]], nonempty subset=[0, 1, 3] normalized as a set. Give (size,subset,group identity,product closed?,contains identity?,inverse closed?,all subgroup checks?).

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

Use the already-qualified group’s two-sided identity and inverses, retaining each stated audit result separately. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 1, 3], 0, Yes, Yes, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0050/)

Subgroups and group homomorphisms · qf-mod23-subgroup-subgroup-audit · candidate





## 51. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[0, 3, 2, 1, 4] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 3, 2, 1, 4], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0051/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 52. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 0, 3, 4, 5, 1], [0, 1, 2, 3, 4, 5], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], total mapping=[0, 1, 2, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 1, 2, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0052/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 53. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[3, 1, 2, 0, 4] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 1, 2, 0, 4], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0053/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 54. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [0, 1, 2, 3, 4, 5], [5, 3, 0, 4, 1, 2], [3, 0, 1, 5, 2, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], total mapping=[2, 1, 0, 0, 2, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 0, 0, 2, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0054/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 55. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[2, 1, 0, 3, 4] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 1, 0, 3, 4], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0055/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 56. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], total mapping=[1, 0, 2, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0056/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 57. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[2, 1, 4, 0, 3] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 1, 4, 0, 3], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0057/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 58. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 0, 3, 5, 1, 2], [0, 1, 2, 3, 4, 5], [3, 2, 0, 4, 5, 1], [5, 3, 4, 1, 2, 0], [1, 4, 5, 2, 0, 3], [2, 5, 1, 0, 3, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], total mapping=[2, 1, 0, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 0, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0058/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 59. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[1, 4, 0, 3, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [1, 4, 0, 3, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0059/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 60. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 2, 5, 0, 3, 1], [2, 0, 4, 1, 5, 3], [5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5], [3, 5, 1, 4, 0, 2], [1, 3, 0, 5, 2, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], total mapping=[1, 2, 0, 0, 2, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 2, 0, 0, 2, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0060/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 61. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[0, 4, 2, 1, 3] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 4, 2, 1, 3], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0061/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 62. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 2, 4, 5, 1, 0], [2, 0, 3, 4, 5, 1], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], total mapping=[0, 1, 2, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 1, 2, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0062/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 63. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[4, 1, 2, 0, 3] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 1, 2, 0, 3], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0063/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 64. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 0, 5, 2, 1, 3], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [2, 3, 4, 0, 5, 1], [1, 4, 3, 5, 0, 2], [3, 5, 0, 1, 2, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], total mapping=[2, 1, 1, 0, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 1, 0, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0064/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 65. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[3, 1, 0, 2, 4] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 1, 0, 2, 4], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0065/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 66. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0], [1, 4, 0, 2, 5, 3], [0, 1, 2, 3, 4, 5], [3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], total mapping=[1, 0, 2, 0, 2, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 0, 2, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0066/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 67. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[4, 1, 0, 3, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 1, 0, 3, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0067/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 68. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [1, 3, 0, 4, 5, 2], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], total mapping=[0, 2, 1, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 2, 1, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0068/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 69. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[4, 0, 2, 3, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 0, 2, 3, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0069/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 70. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 0, 5, 1, 2, 4], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [1, 3, 4, 0, 5, 2], [2, 4, 3, 5, 0, 1], [4, 5, 0, 2, 1, 3]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], total mapping=[2, 1, 1, 0, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 1, 0, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0070/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 71. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[0, 3, 1, 4, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 3, 1, 4, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0071/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 72. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], total mapping=[1, 0, 2, 0, 2, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 0, 2, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0072/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 73. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[3, 0, 2, 4, 1] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 0, 2, 4, 1], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0073/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 74. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2], [1, 2, 4, 0, 5, 3], [3, 0, 1, 5, 2, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], total mapping=[0, 2, 1, 1, 0, 2] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 2, 1, 1, 0, 2], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0074/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 75. Operation Preservation for Every Source Pair

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], total mapping=[2, 0, 3, 1, 4] in source-label order. Give (source size,target size,mapping,every bad [a,b] pair,homomorphism?). A bad pair violates f(a*b)=f(a)*f(b); use nested increasing source-label order.

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

Compare both sides for every ordered pair. Passing operation preservation alone makes no injectivity or isomorphism claim. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 0, 3, 1, 4], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0075/)

Subgroups and group homomorphisms · qf-mod23-subgroup-homomorphism-audit · candidate





## 76. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[0, 3, 2, 1, 4]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 3, 2, 1, 4], [0], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0076/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 77. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 0, 3, 4, 5, 1], [0, 1, 2, 3, 4, 5], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], mapping=[0, 1, 2, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 1, 2, 1, 0, 2], [1, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0077/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 78. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[3, 1, 2, 0, 4]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 1, 2, 0, 4], [2], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0078/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 79. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [0, 1, 2, 3, 4, 5], [5, 3, 0, 4, 1, 2], [3, 0, 1, 5, 2, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], mapping=[2, 1, 0, 0, 2, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 0, 0, 2, 1], [2, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0079/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 80. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[2, 1, 0, 3, 4]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 1, 0, 3, 4], [4], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0080/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 81. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], mapping=[1, 0, 2, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 1, 0, 2], [2, 5], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0081/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 82. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[2, 1, 4, 0, 3]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 1, 4, 0, 3], [1], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0082/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 83. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 0, 3, 5, 1, 2], [0, 1, 2, 3, 4, 5], [3, 2, 0, 4, 5, 1], [5, 3, 4, 1, 2, 0], [1, 4, 5, 2, 0, 3], [2, 5, 1, 0, 3, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], mapping=[2, 1, 0, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 0, 1, 0, 2], [1, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0083/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 84. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[1, 4, 0, 3, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [1, 4, 0, 3, 2], [3], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0084/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 85. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 2, 5, 0, 3, 1], [2, 0, 4, 1, 5, 3], [5, 4, 3, 2, 1, 0], [0, 1, 2, 3, 4, 5], [3, 5, 1, 4, 0, 2], [1, 3, 0, 5, 2, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], mapping=[1, 2, 0, 0, 2, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 2, 0, 0, 2, 1], [2, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0085/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 86. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[0, 4, 2, 1, 3]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 4, 2, 1, 3], [0], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0086/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 87. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 2, 4, 5, 1, 0], [2, 0, 3, 4, 5, 1], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3], [1, 5, 0, 2, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], mapping=[0, 1, 2, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 1, 2, 1, 0, 2], [2, 5], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0087/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 88. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[4, 1, 2, 0, 3]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 1, 2, 0, 3], [2], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0088/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 89. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 0, 5, 2, 1, 3], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [2, 3, 4, 0, 5, 1], [1, 4, 3, 5, 0, 2], [3, 5, 0, 1, 2, 4]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], mapping=[2, 1, 1, 0, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 1, 0, 0, 2], [1, 2], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0089/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 90. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[3, 1, 0, 2, 4]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 1, 0, 2, 4], [4], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0090/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 91. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0], [1, 4, 0, 2, 5, 3], [0, 1, 2, 3, 4, 5], [3, 2, 5, 4, 0, 1], [2, 0, 3, 5, 1, 4]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], mapping=[1, 0, 2, 0, 2, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 0, 2, 1], [1, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0091/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 92. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[4, 1, 0, 3, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 1, 0, 3, 2], [1], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0092/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 93. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [1, 3, 0, 4, 5, 2], [5, 2, 4, 0, 1, 3], [2, 0, 5, 1, 3, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], mapping=[0, 2, 1, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 2, 1, 1, 0, 2], [1, 5], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0093/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 94. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[4, 0, 2, 3, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [4, 0, 2, 3, 1], [3], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0094/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 95. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[3, 0, 5, 1, 2, 4], [0, 1, 2, 3, 4, 5], [5, 2, 1, 4, 3, 0], [1, 3, 4, 0, 5, 2], [2, 4, 3, 5, 0, 1], [4, 5, 0, 2, 1, 3]], target size=3, target table=[[2, 0, 1], [0, 1, 2], [1, 2, 0]], mapping=[2, 1, 1, 0, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [2, 1, 1, 0, 0, 2], [1, 2], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0095/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 96. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[0, 3, 1, 4, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [0, 3, 1, 4, 2], [0], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0096/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 97. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], target size=3, target table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], mapping=[1, 0, 2, 0, 2, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [1, 0, 2, 0, 2, 1], [1, 3], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0097/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 98. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[3, 0, 2, 4, 1]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [3, 0, 2, 4, 1], [2], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0098/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 99. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=6, source table=[[2, 4, 5, 1, 3, 0], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2], [1, 2, 4, 0, 5, 3], [3, 0, 1, 5, 2, 4], [0, 1, 2, 3, 4, 5]], target size=3, target table=[[1, 2, 0], [2, 0, 1], [0, 1, 2]], mapping=[0, 2, 1, 1, 0, 2]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, 3, [0, 2, 1, 1, 0, 2], [1, 5], [0, 1, 2], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0099/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate





## 100. Kernel, Image and Injectivity of a Qualified Finite Homomorphism

Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. source size=5, source 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]], target size=5, target 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]], mapping=[2, 0, 3, 1, 4]. Require this map to preserve all source products. Give (source size,target size,mapping,source labels mapping to the target identity,sorted distinct image,injective?).

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

Find the kernel using the target identity, not an assumed label. For this complete finite total map, injectivity is equivalent to image size equaling source size; constant homomorphisms need not be injective. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (5, 5, [2, 0, 3, 1, 4], [4], [0, 1, 2, 3, 4], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261023-r03-subgroup-homomorphism-0100/)

Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations/validation-report.json)