{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0001","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f5b62315228a26ae5ee69245619d1c39fc9356946c180d0ef8845584047fb27","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233001,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0002","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0bbd97cb5e395b3dee11c058e44f1ccece3c9caa6c0da019646ab50e4f0ff0b","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233002,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[1,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0003","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aaa7e8cc6d622a97111bab2637e8fd0b832be99601827e920b881541d0ae11b2","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233003,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[1,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0004","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e871c7a3f3468ad7452779606f1fcdae1b1c71b80defa3e539301c9be10ae5d3","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233004,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[1,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0005","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0569539695c41d5fa7d14e9354702b28cf7ff3234b7ba13c99e176c7e707e35d","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233005,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[1,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0006","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bd6dd700d40ceae11821b85b29fa73c8f5a53ad37cfccf3d6eaa2151e663b03","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233006,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[1,3,5],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 5], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0007","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6e8a02deccbf859a2a639ba8f006226794cca697286b165f21659e852887490","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233007,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0008","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34ea2b48cdb589ed8e0fb509160efc555745a8dbe3f4e6a9db6583a39f302afa","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233008,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0009","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b0026d43996cafaf974470a28e4f86ceeabc916d13162f3cbfd9edfcd95c957","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233009,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0010","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1774c4b4c13e4f37c52e8067d6397398ccc9dc92c98e0c4e7af95e6d916af97","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233010,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0011","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e19f3bd719c2f5c9b401b59e159bf6c54907d3501c3a94da19af108541cef23","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233011,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0012","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e29ed9ce4ff1c06357601f832462a2202d726cfe2ddff3e11c0b0716e88627d","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233012,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,5],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 5], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0013","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ea07971b98bb96d25a2d6f158c1c3941321b33f4f4c9045474cebc4e5bb9257","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233013,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0014","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5e2a96e140f55f9369acf6a6213bdb30c9ce0cc0937a4f740c3823286b0f450","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233014,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0015","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e76399d3631a5ae3a501adb0507156934b6bc48392555f22e52313ae43ae4629","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233015,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0016","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a10b4b08b697f113eb5413a4381e85ef42aa1922cb2ac161607463684bbb3322","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233016,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0017","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c344c436b92976cb7b5f484e475e94b8e7271a3decc82c926b29d45623b8f16d","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233017,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0018","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"169a56d50b8fdbb8222cbb2686bc32aea09f5532a5f5bbd0b0d5e6235f2578c0","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233018,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,3,5],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 5], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0019","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"752d0dcc41fdb4e5ae92d12f1317b0a32d27ae760c3378f6c954be5079c3e8ee","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233019,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0020","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71057b586f3aefe786c783a89e0d428e422ff2589e89f73e47002de8fb277fde","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233020,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0021","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0b069886b5050f6bae95ac2a90a6e48e543808e2c6746d2187266b886376e3e","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233021,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,3],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 3], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0022","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e241d2d5dd52e27dbacaeda3927e5d79279e5cb89a845ba155540b6307ed39c8","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233022,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,3],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 3], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0023","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca7dc6f41cbd189f68834c09fddff5146762f208d4f42868d2d5a1dc2213158f","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233023,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0024","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06ce2a1ce5a6ad99ee5796b4c89a952eeabd952463a9741eced9e232da4bc69e","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233024,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,5],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 5], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0025","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"caeba7ae12bccb191234631c1d3f68abef65a223da3617f99cf7c7f584c96584","family_id":"qf-mod23-subgroup-subset-closure","seed":202610233025,"semantic_template":"qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Product Closure of a Nonempty Subset in a Qualified Finite Group","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], Yes)","explanation":"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)."},"skills":["Product Closure of a Nonempty Subset in a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0026","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7cdfb50a0282fbfe39d35cd2f96f132441552369f930cab125ced80b0d686dfd","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233026,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], 0, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0027","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3e7578016c9eb844e977248945a314e28bcc0f322697c061499aa7351ca8c98","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233027,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[1,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 2, 4], 1, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0028","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2970a38ff20c9a8a7896c23e76c3d6929fee52f5fc2402f144023cf66e609ad8","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233028,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[1,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 2, 4], 2, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0029","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"deb5e300443852ac9e925bd3a9c80b14bbe7f7b113c51daa1f6822bbef0fef2f","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233029,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[1,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 4], 3, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0030","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a13f54b52ca5888a64d216c9b2036977bae980e9e89702de8c4788775457df4f","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233030,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[1,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 4], 4, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0031","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24b8be4a1273bb55b2cc63c8f916f8a30f8bccbddda2f94bb4b60c0d945204ae","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233031,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[1,3,5],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [1, 3, 5], 5, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0032","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"551a443b0d009f3a2a4222ee22af41dbbcb55cca49c999a6156379069af5b100","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233032,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], 0, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0033","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e59983c79a116b09435c8ea51a298ef85158711fb4e31f08bafdfdc4a143d140","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233033,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], 1, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0034","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"508f630e96e20de565274a0d221a84361552b69b0d47141ab598c481ff8c7562","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233034,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], 2, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0035","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cc03e140eeff2e6beb28064b1e795b5f066a30d5d9bfbcd636a87aa74bf6eae","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233035,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], 3, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0036","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"281c4907c940c921d33ead196ea532f7a94a97b5f16f2e858410221e988379b8","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233036,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], 4, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0037","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6e8ff3f42817f7cd4aaffff5caff784457473230fe413d68d2a390a0a84f951","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233037,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,5],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 5], 5, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0038","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"887f4b02657168ab6365712d7124be299831d21cfd7d8a56a5bccb4e109c41fb","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233038,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], 0, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0039","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06976acc971c5e43dc50e9aaf357824358b07c82ec640ffe39797d85f34b1a40","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233039,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 4], 1, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0040","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"492c6b6885dd04e070420ec1917d0833a6ae950448b5d809886ff8c5be55691a","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233040,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], 2, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0041","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"819636ebc3b35532d0f1b427c9a7414888d980d6e0a5ffc80d0367f295756ee1","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233041,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], 3, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0042","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ad60361b586788d415e0a20a00e85731fc3faf6fe6b121c9c3a9f82abf16cfc","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233042,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 4], 4, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0043","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"320ed6c3e497a90ce2e9b5993b9e2252810dbcf1d2d987bc7efcdeea8df49faf","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233043,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,3,5],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 3, 5], 5, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0044","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4298741861593f944ee641c7a9342d021d171dc1af49a479ff148e24a4bfbdad","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233044,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], 0, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0045","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0feb380a764cad9796a3bb14f00e790c51939e3d872739a96bbc8eb07766085a","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233045,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], 1, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0046","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09832a1beb42ae02c9bc21a73c988e77848ff468a14737fcc208489cf3fa692d","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233046,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,3],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 3], 2, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0047","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77726f7762041e415b8c8d2c50468c1c9de804d1622762fa090650deb688afc4","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233047,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,3],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 3], 3, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0048","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a36cc6ebd6c83d99418e680996105e671cc66bd3673ef264d82ccc55d29da4d","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233048,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,4],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 4], 4, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0049","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9293bba306bf3b933f2d3275a202287263989b3b6d479f2077398300f00e5657","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233049,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,2,5],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 2, 5], 5, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0050","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4a2053aac02887388e706fdbe49752e0f8c012d0effdeaef4acced8c7278836","family_id":"qf-mod23-subgroup-subgroup-audit","seed":202610233050,"semantic_template":"qf-mod23-subgroup-subgroup-audit/{\"n\":6,\"subset\":[0,1,3],\"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]]}"},"fields":{"title":"Explicit Subgroup Closure, Identity and Inverse Obligations","stem":"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?).","answer":{"kind":"expression","value":"(6, [0, 1, 3], 0, Yes, Yes, Yes, Yes)","explanation":"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)."},"skills":["Explicit Subgroup Closure, Identity and Inverse Obligations"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0051","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b845d8ece65fc45ad041a2814451f14245cbceeb1d46491c3cc95b5a18229ac","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233051,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,3,2,1,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [0, 3, 2, 1, 4], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0052","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"501f57137e30713a66422c9c725f7490834e3a3203d8f3df48fef4aa2fdae106","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233052,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,1,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [0, 1, 2, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0053","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30008825d2693de4b9e701619f40c4f34338e80cf7e18575e34fcf149b756bdd","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233053,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[3,1,2,0,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [3, 1, 2, 0, 4], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0054","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36ad6aa8f77b99944ea23bf45f09184434ad895f283f382c49bd52c83ce56852","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233054,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,0,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [2, 1, 0, 0, 2, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0055","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26657079a7b43feeffffe1b67fefacdcf4f29d8fe9b7502eac750ced619a7ae4","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233055,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,0,3,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [2, 1, 0, 3, 4], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0056","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f8ae55c8fc141b4bd94513b396b4ea3da037a133ff541c80acc7b6184e68f6b","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233056,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[1,0,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0057","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a68cb4ee2c9b0084602f996a3f4ef959507b1d7f10c3d7ca8bcd9ed4393ad4f6","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233057,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,4,0,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [2, 1, 4, 0, 3], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0058","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34f4e5180268e30f58c6da50d693ad8140466e46adf9a3c779a8864d2b66a0e7","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233058,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,0,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [2, 1, 0, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0059","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1e99a4101e10a7dced93e65bcf4a316c42bcd7aa7b76ef6e94d7e7d1b8df9e0","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233059,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[1,4,0,3,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [1, 4, 0, 3, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0060","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f2471b15b65acb7453ae9228d30aeff9a21a51069d1db3060be72fd0a190b55","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233060,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[1,2,0,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [1, 2, 0, 0, 2, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0061","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4bd73f1526f0ee157f3cbb32dbdcf019cf6ffaf3a7008e5482a336082fa72a4","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233061,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,4,2,1,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [0, 4, 2, 1, 3], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0062","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e74b916103b0940762d58230dbe9102fb5593b1cfcba3de028daf1d5614712d6","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233062,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,1,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [0, 1, 2, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0063","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44d0e50e25ffc0162453961e2991c81c39e4be93916c3199e8a283675d7858b6","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233063,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[4,1,2,0,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [4, 1, 2, 0, 3], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0064","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e188716cc2813fb200e41d91f96092d1cb7923156fe351adbef14963fb400534","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233064,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,1,0,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [2, 1, 1, 0, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0065","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aaa6071221b619824657e25c8982ba308af56aec5f7f5c1b4c46c5d6fcef9aaa","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233065,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[3,1,0,2,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [3, 1, 0, 2, 4], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0066","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b167df98f94e903d91c938222994fab11b63ce95e9a9cb2df512029f423c7219","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233066,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[1,0,2,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 0, 2, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0067","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a621aae5dfaf33c0a975df051747eea88dd417318c76aa1712e546b90c012d2","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233067,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[4,1,0,3,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [4, 1, 0, 3, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0068","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e432b3d23203aedc7c230b865f885e57ea4b995426f50183f1a474d80bd89a0","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233068,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,2,1,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [0, 2, 1, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0069","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51f495f818e87e4790980139458736031a4f9368491a9577175a51dac360f30b","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233069,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[4,0,2,3,1],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [4, 0, 2, 3, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0070","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"414ec40fcb27aac1a42780f21b14ae38cc68d11b92dc05b762c5b92f004e1496","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233070,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,1,1,0,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [2, 1, 1, 0, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0071","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f66db799773d2d687b4a9d4be4a667d065e85b7d5bbe68e86a8d04318e4ab382","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233071,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,3,1,4,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [0, 3, 1, 4, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0072","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c475294f9a10943275610eb5817cb697a010516f81c5857f0e06a4b04649b63","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233072,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[1,0,2,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 0, 2, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0073","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50e87ae6b6897304abd87bbe5f2046c7bc477eec9e5112e428c47c17cb113c81","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233073,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[3,0,2,4,1],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [3, 0, 2, 4, 1], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0074","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66b2760acd74b8aa9c0fff8e1962cb52ab70865e58c7fb7e3620d17bdbb76277","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233074,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[0,2,1,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(6, 3, [0, 2, 1, 1, 0, 2], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0075","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32b442c86872297982ea7396589b400e5b6d5e020135db4b22fede22d05c25ac","family_id":"qf-mod23-subgroup-homomorphism-audit","seed":202610233075,"semantic_template":"qf-mod23-subgroup-homomorphism-audit/{\"mapping\":[2,0,3,1,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Operation Preservation for Every Source Pair","stem":"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.","answer":{"kind":"expression","value":"(5, 5, [2, 0, 3, 1, 4], [], Yes)","explanation":"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)."},"skills":["Operation Preservation for Every Source Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0076","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a111e9c1be6850cfea4d01281b4d41fb1eb59131de1c83f262f9ecfe2f430af7","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233076,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,3,2,1,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [0, 3, 2, 1, 4], [0], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0077","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c8949158b22ee05d25074e3870ea66927b810895e9dc0a73e3cf97fef16db98","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233077,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,1,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [0, 1, 2, 1, 0, 2], [1, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0078","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c19a9d4d41c08d67db76b458c59aa8ed3a251917f4aa6c7367d162f732e84a47","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233078,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[3,1,2,0,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [3, 1, 2, 0, 4], [2], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0079","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cedd7b9d1796d220dc5ecae189494545dffe32f6b2fb356665b755b742fd202b","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233079,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,0,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [2, 1, 0, 0, 2, 1], [2, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0080","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f973c5d8f78a511f5b68fcee422618b31b0865e19538baeb5316b558ec7e5db0","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233080,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,0,3,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [2, 1, 0, 3, 4], [4], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0081","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2867fab383a012edbe3aedb6ed3eb2a3f0d3fdcdc752634249b6de3116bc62b6","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233081,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[1,0,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 1, 0, 2], [2, 5], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0082","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ee93de45dbff2c8cd019ea0f83bd725b2bc993e72238b626e02bbaf2362a033","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233082,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,4,0,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [2, 1, 4, 0, 3], [1], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0083","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4a0b2fb7a5592a47b2eb0ddea45f950fe710f3626c2c93bd85645ae7fa1429f","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233083,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,0,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [2, 1, 0, 1, 0, 2], [1, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0084","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b243bcdc039747fa1d3b35e0dfbd8ef63cc0a63624189a8fd9f58143b40e468","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233084,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[1,4,0,3,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [1, 4, 0, 3, 2], [3], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0085","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"901c6e4b999e2ea377ac22c5b446b5e4d9e87f3ed0837bef0c1fe3d7b7cafe45","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233085,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[1,2,0,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [1, 2, 0, 0, 2, 1], [2, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0086","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1025381669b4ace63ecd752d53c6aa45efb6082ad83c907ccd7eb980a2af9b0","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233086,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,4,2,1,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [0, 4, 2, 1, 3], [0], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0087","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afc30bebff70c53759d73e4025c1f38381d411c576a404ac698f122b9519d72d","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233087,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,1,2,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [0, 1, 2, 1, 0, 2], [2, 5], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0088","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e09c1e9c23d2ca0b7e748d2022f425efd80d72f5ec8a802b7976657b35fe811d","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233088,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[4,1,2,0,3],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [4, 1, 2, 0, 3], [2], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0089","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a261e1aae6e0607d859bf9921628ff0239082ef9eaadeba4afd6d320ca10c9a8","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233089,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,1,0,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [2, 1, 1, 0, 0, 2], [1, 2], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0090","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf48d235dc2ac5bc4cf3523b40b43caa7f6d34acc4736d67d75f6e9f30dab7c2","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233090,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[3,1,0,2,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [3, 1, 0, 2, 4], [4], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0091","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd6bbd8e39a4bb4aaae745d0363aa09e80103bbf284eb8f4b509ba5f13c955a1","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233091,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[1,0,2,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 0, 2, 1], [1, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0092","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4680f1646beee6aec4b94c7bae196ad5fe969d007d1f9b93204c8cec1a349495","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233092,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[4,1,0,3,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [4, 1, 0, 3, 2], [1], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0093","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2fc4fe16b59952389790f494fd65d58ed399e73ae504703f48abed56627fec4e","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233093,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,2,1,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [0, 2, 1, 1, 0, 2], [1, 5], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0094","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d239f6959af82e6e4ce7942f1973ae1013385c2bab4f8c6d121b07853137729","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233094,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[4,0,2,3,1],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [4, 0, 2, 3, 1], [3], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0095","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"756e059a58ee1a7bfae9b96ab89c6c170801e5fdd8026ddaaa4df4d87d05cd6f","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233095,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,1,1,0,0,2],\"source_n\":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_n\":3,\"target_table\":[[2,0,1],[0,1,2],[1,2,0]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [2, 1, 1, 0, 0, 2], [1, 2], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0096","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46d21d928458d9dba18f9c55215aa59d5d4d8c00236dee070821618f846659f0","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233096,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,3,1,4,2],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [0, 3, 1, 4, 2], [0], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0097","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"047bd33dae63b914465a164a91ac0d701d5d6a95fdebb717e801cea0116ae948","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233097,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[1,0,2,0,2,1],\"source_n\":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_n\":3,\"target_table\":[[0,1,2],[1,2,0],[2,0,1]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [1, 0, 2, 0, 2, 1], [1, 3], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0098","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"115a11d017b455fcea31cde0d0c1e9f765815506c47bde293122e2e296691ed3","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233098,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[3,0,2,4,1],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [3, 0, 2, 4, 1], [2], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0099","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64c7443ae70eb834858bc68fab7486eea309caf65d4db14a1b0f51beda961b48","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233099,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[0,2,1,1,0,2],\"source_n\":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_n\":3,\"target_table\":[[1,2,0],[2,0,1],[0,1,2]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(6, 3, [0, 2, 1, 1, 0, 2], [1, 5], [0, 1, 2], No)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261023-r03-subgroup-homomorphism-0100","batch_id":"batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c7c0ce41de2c86f6bce497b78da9f1d46a2d5ea8d5b0f24e0729446d8c5328c","family_id":"qf-mod23-subgroup-kernel-image","seed":202610233100,"semantic_template":"qf-mod23-subgroup-kernel-image/{\"mapping\":[2,0,3,1,4],\"source_n\":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_n\":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]]}"},"fields":{"title":"Kernel, Image and Injectivity of a Qualified Finite Homomorphism","stem":"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?).","answer":{"kind":"expression","value":"(5, 5, [2, 0, 3, 1, 4], [4], [0, 1, 2, 3, 4], Yes)","explanation":"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)."},"skills":["Kernel, Image and Injectivity of a Qualified Finite Homomorphism"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}