Complete Finite Subgroup and Homomorphism Obligations
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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Subgroups and group homomorphisms · qf-mod23-subgroup-kernel-image · candidate
Canonical question JSONL · English JSONL · Validation report