{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0001","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38d01829aca3a9b691944e4f7a53241719ca502d3660ac08e9fa95c2b772dbc6","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111001,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_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]],\"identity\":0,\"subset\":[0,2,4]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0002","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9c46f602e5fd43f67c6158148814a846cacf9b0a4fb32eecec8e53d9fbcaef7","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111002,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0,2,4,6]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0003","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b6ed91dbb58803747cd9dc23edafa90a8027b35cd95acb29b82f1ca40def802","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111003,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0004","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cab77c0a9eb47cd493414f1d7b87f57493e1e096f5be8009fdaf52d6460a96da","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111004,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0005","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d5fce702b93d2eea102f99ef600e5841ccce9acbb1d479a3075dabeb7cd3ca4","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111005,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12]],\"identity\":0,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0006","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c04486f943e4747a3e4b2f8e2ea2137b4d953a47b2a57c8301b54bd77cbb7b7","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111006,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[1,3,5]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0007","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9b285005200e477da4e5c20f5293809dcabd2dcdf0598be243c43d6b30aa926","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111007,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[1,3,5,7]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0008","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe29c8114c4e38eb867e315ada7783902c92f6e243ada98578bb7f006534267c","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111008,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0009","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1191c8e3eb61218b0402b5b3f7e65a0c306a49fe483f3e5048304dafedf43b1e","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111009,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0010","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17f85229197a3b43307d562cd5affc4c93cff4b205d8733a9ea152d612499f11","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111010,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11]],\"identity\":1,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0011","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7492c2a7a517db5941fdf5308d49473844e5fd9031a73f636ba09b9993aa098c","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111011,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,2,4]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0012","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"328eef4a786499aaf7d216da07dac5bee0e2a4c41a6848adcf660da993089d6b","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111012,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[0,2,4,6]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0013","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2dbc4a33d2b697116b90e2e3263464a14091bb64269145e98de9d175cf57a7c9","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111013,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0014","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6599a6a6b0a77ae2c4576648fa4746864791afaae40d709da57a12a783f81d2c","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111014,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0015","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"910ee1559dd461798a882f496437040aede1a70918b8b3a7b45f6120bd3ba333","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111015,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":2,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0016","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6cef94d63fa595087bb244fc3c92626b6511a8c790120d38ae9677526a4ccf9","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111016,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[1,3,5]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0017","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce3c9b8bae7614fad59b61aceda2d3e5e888ed7f131fa96b521d454e764b42d7","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111017,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[1,3,5,7]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0018","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19f321dc33f1a757ed236c3f3b7a6dd58d0072c4982472131d6f49c9762bbfbb","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111018,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0019","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0806862ced43185842790e65daef1c58366b9c410d8d818aecd01bd2ac8776ad","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111019,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0020","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ed15ab71d4cbf58d623ac31205609d8ce84956c45e5f1ba4f3b21cca18e346d","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111020,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9]],\"identity\":3,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([1, 3, 5, 7, 9, 11, 13], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0021","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aba7952cdef08c5ffafd728f8e5a9dbec65ce2549a28d6f824b253ef09fbd473","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111021,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,2,4]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0022","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d2871aa642db8393b1ff9aa59aae5cf2d2f620a2e759d1af07e4320a001d79b","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111022,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[0,2,4,6]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0023","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"093a41b709068edbbb0de835c6a49a4bb7fea9669161c55b336dfdfdfc72f673","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111023,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0024","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f1123d8d9ebfc937c163df4e6e6fef9c9c9fc3fd9cbfd0d37e991c7148ded12","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111024,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0025","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cdaa7cdb4871efbe42b5d2f51c6c9d488341e96a64ca6f7b73fe640db21bf46","family_id":"qf-mod42-lagrange-subgroup-qualification","seed":202611111025,"semantic_template":"qf-mod42-lagrange-subgroup-qualification/{\"group_table\":[[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8]],\"identity\":4,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Divisor-Sized Candidates Still Need Complete Subgroup Closure","stem":"Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)","explanation":"Report identity, product and inverse closure separately, then their joint subgroup flag. No closure obligation is replaced by a sampled product or the subset size. The exact result, retaining the requested order, is ([0, 2, 4, 6, 8, 10, 12], Yes, Yes, Yes, Yes, complete_Cayley_table_subgroup_certificate)."},"skills":["Divisor-Sized Candidates Still Need Complete Subgroup Closure"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0026","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0248e6aeffcf2e70d005624f18bc787f8f480cf10808a3f2c791b58c8c447137","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111026,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_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]],\"identity\":0,\"subset\":[0,2,4]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0027","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8f7c71f30a48422c563044dfc34fc89535cdfa95fc34588555bd33ce0cccbd5","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111027,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0,2,4,6]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0028","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"963850377ba350f3199c5c231bfe6635b6fcaec4a66b7a6080c8f4ed7e564fe7","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111028,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0029","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8402b224e6f2f2ebfd8e4723fc51b27d46067978c19b9ac3d5f0d2b328923050","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111029,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0030","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8c9faad655b15472216b42b59366d05743f4723efb6f00815070739d0a38a0d","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111030,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12]],\"identity\":0,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0031","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc65d1680813670336671977f150a2742c5e8a39ea95a77074bb4a4f1a2f5a58","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111031,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[1,3,5]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0032","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"116d4fffb2e4a35466376d08c6d8ee2bcc68c6a4b7ff243327efb855a36caa2c","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111032,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[1,3,5,7]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0033","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30c1ada495f35fb66c0aae9254e1e736a54a62d11fae30efe21cc9dafa5b415c","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111033,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0034","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9bdfbbe3368636541625690d837fd6f8add16826f2c4f48d2d19b41e2fcb6d9","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111034,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0035","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28bb74f18ee78d5192b9c76e26bcbc77174f84c6dd428615571c93e018fa538d","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111035,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11]],\"identity\":1,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0036","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa8e7d41e46d1d4fea0bc66943931beff367f9110969ef97c1ea8e69026382d5","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111036,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,2,4]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0037","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"823b14918cf34f17f535c77f13e6f6a4d0911ddaa2d42489e4fc344e74450914","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111037,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[0,2,4,6]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0038","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e583586644f96671f1c0b7cffa219ff81d64b226b3d76a76f5004ec780c36f9","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111038,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0039","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d99ad189455dd82d23bd06660a8f508aaf6eb57c2b9298e168791d8415cce12b","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111039,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0040","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15889c9fdf259a9e6f929136ae71bf95d0796c00346bbc178387957d7ce4891d","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111040,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":2,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0041","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cee17421d7f29ebad68cca195ca2779826483c30768cb6eb5daa91244c8eebbb","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111041,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[1,3,5]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0042","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d4d016c139e0b8ea1c1885b998670b1c7a7a2422adbfdd9fd42bbfd02b5c682","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111042,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[1,3,5,7]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0043","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf021bdc651096deddc63815063f3c2efd2b2d65217302b3254f21fa0360732a","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111043,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0044","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af451fe52e4b8bcd7e1fa8ab95bdc0eefcf60bf8eb5930ed0e16f9b566e78c8b","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111044,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0045","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa6e907e2acd1bbd859620ad722a292c5d4b16ffba423dfdd4fdeea606a6e552","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111045,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9]],\"identity\":3,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0046","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0d3685e9efc36c55607673e6e0c9a130f7278a72aaede5c023152266576b342","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111046,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,2,4]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4]], [1, [1, 3, 5]]], Yes, [0, 1, 2, 3, 4, 5], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0047","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"070ffdfcbba05eb7b43a8392ae44be57f5b5fe287d0616d418c256fcfc4fa708","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111047,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[0,2,4,6]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6]], [1, [1, 3, 5, 7]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0048","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e626098086d67b39c10e8f0a395056a8d648e14637197652edc9516556138233","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111048,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8]], [1, [1, 3, 5, 7, 9]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0049","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9a9fe8d91faaf7cf9bc33ad4513a7a5335415100a8bb9fefe37613adddf5ce9","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111049,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10]], [1, [1, 3, 5, 7, 9, 11]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0050","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e341b239acee5778cf49e4bc603e2e94acec87edb562fbc152d0cbbd347f614e","family_id":"qf-mod42-lagrange-left-coset-partition","seed":202611111050,"semantic_template":"qf-mod42-lagrange-left-coset-partition/{\"group_table\":[[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8]],\"identity\":4,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"Complete Left Cosets Are Equal or Disjoint and Cover the Group","stem":"Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (representative_coset_rows, equal_sizes, flattened_elements, partition_certificate, certificate).","answer":{"kind":"expression","value":"([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)","explanation":"Require a genuine subgroup H. Enumerate every left coset gH, using the least original group-index representative for each distinct coset; order cosets by the source canonical element-list order. Left translation is a bijection from H to gH. Report equal sizes and exact flattened whole-group coverage. Do not replace gH by Hg for a nonabelian group. The exact result, retaining the requested order, is ([[0, [0, 2, 4, 6, 8, 10, 12]], [1, [1, 3, 5, 7, 9, 11, 13]]], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], Yes, left_cosets_equal_or_disjoint_and_cover_G)."},"skills":["Complete Left Cosets Are Equal or Disjoint and Cover the 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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0051","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dd81a3baf074fe686b9948b0572ec7029067c40de869a21e806ba4bfbcf1639","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111051,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_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]],\"identity\":0,\"subset\":[0,2,4]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(6, 3, 2, 6, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0052","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7cd2913c65355e1c6b20e750065f2e1817be915760c493d6b0db9c8426927f8c","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111052,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0,2,4,6]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(8, 4, 2, 8, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0053","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7575226e20cc10584ff206b7b16233ec3fac233959b80ba741343a36d1fd13b4","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111053,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(10, 5, 2, 10, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0054","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26a40f257c641a4f8c472266472f5b18ae50d5e545982e70b435867577c48b81","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111054,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(12, 6, 2, 12, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0055","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"200cdfb5698cc2d72fdc8f3f323c59f98347dfb42386f9096a63c9e8df539365","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111055,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12]],\"identity\":0,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(14, 7, 2, 14, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0056","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d7c0aae633fbae3754c251537d5dfeb12563804e2123582f930ea3f8e8ea2bc","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111056,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[1,3,5]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(6, 3, 2, 6, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0057","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49a47d21b763f5e563288fa8841eda5e64f269da9beb808ac3569fb5ae15ad52","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111057,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[1,3,5,7]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(8, 4, 2, 8, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0058","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4ecc2d12694038edfd2a34f3223eb4b13c6bf4f32749ec72c346d6e5885fe46","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111058,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(10, 5, 2, 10, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0059","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6b7ca8f553ab6af31c10e0226cf7076ed498017574b5c36ada2cf321451130c","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111059,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(12, 6, 2, 12, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0060","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1531e60d5797e3bf4d4799ba439987cd522386ac3d3ca89e0f43fa8a600517ad","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111060,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11]],\"identity\":1,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(14, 7, 2, 14, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0061","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b8cd0082e562a4919158d9880f07b2659504ff12f1c5e185d5630b8851a6eaa","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111061,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,2,4]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(6, 3, 2, 6, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0062","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"401370ab3d401016a6eceab7b527920846775f63a32d3a3d2cb0e65f1a6132f8","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111062,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[0,2,4,6]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(8, 4, 2, 8, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0063","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5c913199fafb30ea3e91681776d1e75866dd254068a099e686afe4fb6300a9f","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111063,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(10, 5, 2, 10, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0064","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c8412e9cda88c0f95ec21d9e9de279f17c1b2a0b47877a65046a41556714865","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111064,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(12, 6, 2, 12, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0065","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ce7e67a087da03621a1491d185698bb0471ff16733250bfce3207537397597b","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111065,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":2,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(14, 7, 2, 14, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0066","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90e66077e19153439b5ba8df2b7fd83bf31f1901f51858b210c55fc259c34de2","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111066,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[1,3,5]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and subset=[1, 3, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(6, 3, 2, 6, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0067","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d98f52581c46f80ac8020ac153744533e11d6ab0f8b4b5a7a38a4471b6acc6d","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111067,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[1,3,5,7]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and subset=[1, 3, 5, 7]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(8, 4, 2, 8, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0068","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2de0a81f84bfcfc4d51bb46918302d47a269f7a0a5fb64adee382371bd6db4db","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111068,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[1,3,5,7,9]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and subset=[1, 3, 5, 7, 9]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(10, 5, 2, 10, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0069","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"edcc190f18c603e46e1d04186df122f55e7c53516fbe2e8b316674a691429078","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111069,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[1,3,5,7,9,11]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and subset=[1, 3, 5, 7, 9, 11]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(12, 6, 2, 12, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0070","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b9a7eec243f379ec83fabd5938fec5349fe8c6248b28380eb00e7f2729eb62b","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111070,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9]],\"identity\":3,\"subset\":[1,3,5,7,9,11,13]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and subset=[1, 3, 5, 7, 9, 11, 13]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(14, 7, 2, 14, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0071","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d1e343915ef37211cca05af3688cbb461ea94d4fb3df4390d22febc61ce08fb","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111071,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,2,4]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and subset=[0, 2, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(6, 3, 2, 6, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (6, 3, 2, 6, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0072","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0691693dd226a948de502f7d30ac37edf56cbbafb7bb99cc3c290e08a5b2291","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111072,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[0,2,4,6]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and subset=[0, 2, 4, 6]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(8, 4, 2, 8, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (8, 4, 2, 8, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0073","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97523e3e7ec2086e59738fa1a858160101c1e9cec3712cc00349fe6c30d349c9","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111073,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[0,2,4,6,8]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and subset=[0, 2, 4, 6, 8]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(10, 5, 2, 10, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (10, 5, 2, 10, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0074","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"628e1551a65051ffb76fd9bd3f67b85db9cf4cf1b3eab067c8729341405ba5d6","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111074,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[0,2,4,6,8,10]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and subset=[0, 2, 4, 6, 8, 10]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(12, 6, 2, 12, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (12, 6, 2, 12, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0075","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c62ee72e8a8fa2775f50553a79f00b781fc5fed59719c31da8ab229f60fb6c04","family_id":"qf-mod42-lagrange-divisibility-identity","seed":202611111075,"semantic_template":"qf-mod42-lagrange-divisibility-identity/{\"group_table\":[[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8]],\"identity\":4,\"subset\":[0,2,4,6,8,10,12]}"},"fields":{"title":"The Index Times Subgroup Order Equals Group Order","stem":"Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and subset=[0, 2, 4, 6, 8, 10, 12]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, subgroup_order, index, product, remainder, equal).","answer":{"kind":"expression","value":"(14, 7, 2, 14, 0, Yes)","explanation":"Require complete subgroup qualification and every left coset. Count |G|, |H| and [G:H], then compare the product and retain the exact divisibility remainder0. This is a finite equal-size partition certificate. The exact result, retaining the requested order, is (14, 7, 2, 14, 0, Yes)."},"skills":["The Index Times Subgroup Order Equals Group Order"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0076","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f3b671358be432f250db8ed0aa0ea4d401a326d159e7d5f1e479f6b76d56f35","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111076,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[0,1],\"group_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]],\"identity\":0}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(6, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0077","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cacf488263ce2389fb035644a5f3f133a80bc011fd934f6366bb7c4e7c16b5dc","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111077,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[0,1],\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(8, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0078","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0fff661c0e3052ff8581305d20aae5466868ffd19648b5c560249711c18ca5a","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111078,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[0,1],\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(10, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0079","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b0181ae52079d4f2728aaf1ef9385454636e09a1eee4dbe37d820a918b70415","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111079,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[0,1],\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(12, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0080","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"894fd1f313ee3de5ac2831ae3f0bcb4f4c4f2add8ae1af4387a47f333d32f6e1","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111080,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[0,1],\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12]],\"identity\":0}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]], identity=0 and candidate_subset=[0, 1]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(14, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [0, 1], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0081","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98cd652df1b3beee1f4014f05cbf627e52a789d420b63800fa6c1ed6d6e37d4b","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111081,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[1,2],\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(6, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0082","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8f6b113f4b5b9151bdd0f201ca862907df5841a2242769717f31d4be710dc84","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111082,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[1,2],\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(8, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0083","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4071301bc571beed5b5eec832c941236119a43823cccdd4fbdb0a862b4963da","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111083,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[1,2],\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(10, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0084","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e90a3faa170a0196e2eadab97898a54ce0703d7a51040505ac9bd102cfbd839e","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111084,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[1,2],\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(12, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0085","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45cbca399a31619d5f6eefdca2d94d55423c0e503563fa20ec829b6505f9973a","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111085,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[1,2],\"group_table\":[[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11]],\"identity\":1}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]], identity=1 and candidate_subset=[1, 2]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(14, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [1, 2], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0086","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4a252e09bfb8942696e1b2e309c9523578743ba7295917128d577a2b7b4415f","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111086,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[2,3],\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(6, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0087","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14ded4e88c627aa7142ae28b966c73adb0fdfca941c581b63de68681310d2c3c","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111087,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[2,3],\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(8, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0088","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7c4da95f7e1a02c91a36a7ca6589756ff91edb1412393afd54815b867f2e5cb","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111088,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[2,3],\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(10, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0089","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36119680db28cd11c381a792fd0968f2945c6432d8ed97aa39edc9cd0ad5c52c","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111089,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[2,3],\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(12, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0090","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c35a5d15842da2c1e871e9d88754b0d006606585ea7a7f3a3c411b9ed8f0249","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111090,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[2,3],\"group_table\":[[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":2}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=2 and candidate_subset=[2, 3]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(14, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [2, 3], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0091","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e885ab87c932a08cdf67684937ff48a41a3f21861b36993493579016f3088370","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111091,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[3,4],\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(6, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0092","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82c466af4041c9529ee4d68f2c71323abaf7d4d479c728a8e64fb4db4042fc0d","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111092,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[3,4],\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(8, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0093","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52adfe9e83f68ffdffb8463d4194fec38660073dde775305dc577a7aaad9502c","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111093,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[3,4],\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(10, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0094","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8410d5c66dd74573c7388a1abe832ead06867bae853398dff3b5b71dd8b7f48","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111094,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[3,4],\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(12, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0095","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c29693a4f1dac344484d5b9a408a3308f85e6f14a348273aaad6902514ffd6b","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111095,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[3,4],\"group_table\":[[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8],[10,11,12,13,0,1,2,3,4,5,6,7,8,9]],\"identity\":3}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=3 and candidate_subset=[3, 4]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(14, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [3, 4], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0096","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e26dcbb825899f0c2c6865d0ad193093919592379da4633c9d64583b36c65a9b","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111096,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[4,5],\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(6, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (6, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0097","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"507a1864cd5df0ccc5188da3522700c06649783abb9fcad1e6ccd41371bf222e","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111097,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[4,5],\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(8, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (8, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0098","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9748873428a7ad2cbcdccfece9d7ad12ddde01a99d4b16d2691192dfd59948b0","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111098,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[4,5],\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(10, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (10, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0099","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d38242d5c5b8ca44a5e98d09b6718a129f0567327c40fa76b706927fb68461e5","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111099,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[4,5],\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(12, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (12, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-0100","batch_id":"batch-20261014-advance-20261111-r01-lagrange-theorem-finite-groups-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a5a06b31b39a95f8ae4d9cd8f7e8d5e6b04fee5095a2fa62ed23fd9700a9a21","family_id":"qf-mod42-lagrange-converse-boundary","seed":202611111100,"semantic_template":"qf-mod42-lagrange-converse-boundary/{\"candidate_subset\":[4,5],\"group_table\":[[10,11,12,13,0,1,2,3,4,5,6,7,8,9],[11,12,13,0,1,2,3,4,5,6,7,8,9,10],[12,13,0,1,2,3,4,5,6,7,8,9,10,11],[13,0,1,2,3,4,5,6,7,8,9,10,11,12],[0,1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3,4,5,6,7,8,9,10,11,12,13,0],[2,3,4,5,6,7,8,9,10,11,12,13,0,1],[3,4,5,6,7,8,9,10,11,12,13,0,1,2],[4,5,6,7,8,9,10,11,12,13,0,1,2,3],[5,6,7,8,9,10,11,12,13,0,1,2,3,4],[6,7,8,9,10,11,12,13,0,1,2,3,4,5],[7,8,9,10,11,12,13,0,1,2,3,4,5,6],[8,9,10,11,12,13,0,1,2,3,4,5,6,7],[9,10,11,12,13,0,1,2,3,4,5,6,7,8]],\"identity\":4}"},"fields":{"title":"Subset Qualification and Subgroup Existence Are Different Converse Questions","stem":"Use the complete finite Cayley table=[[10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [13, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 12, 13, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=4 and candidate_subset=[4, 5]. Row g/column h gives g*h. First certify every group axiom, then audit the entire candidate for identity, product and inverse closure; cardinality alone is not subgroup qualification. Give (group_order, candidate_subset, candidate_size, size_divides_order, candidate_is_subgroup, general_converse_claim, boundary).","answer":{"kind":"expression","value":"(14, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)","explanation":"Report whether this candidate size divides |G| and whether this particular candidate is actually a subgroup. The final fixed No withholds a general converse from cardinality alone; it does not negate a true candidate flag or prove/deny existence of some other subgroup of that order. The exact result, retaining the requested order, is (14, [4, 5], 2, Yes, No, No, divisor_sized_subset_need_not_be_a_subgroup_no_existence_converse_claim)."},"skills":["Subset Qualification and Subgroup Existence Are Different Converse Questions"],"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."}}