{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r01-orbit-stabilizer-theorem-0001","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5abf4c324285a356906f8bfdbc27bb441652163e9c12228d785789581f91a2d1","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101001,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1],[1,0]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":0,\"point_count\":2}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(2, 0, [0, 1], 2, [0], 1, 2, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 0, [0, 1], 2, [0], 1, 2, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0002","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b464d6b21c4a82a79207a69a4d579fdcb75ce4122214544ed1441de9eeb4874","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101002,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2],[1,2,0],[2,0,1]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":1,\"point_count\":3}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(3, 1, [0, 1, 2], 3, [0], 1, 3, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 1, [0, 1, 2], 3, [0], 1, 3, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0003","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbabac4167b14268d09e521f7e1a042eea205fcfa8ba9ae65ef779a3d72bc294","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101003,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0004","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8d4a68eb4a6a1ae2ac44f200c482cc3632c24cc87ed8f9401b994a42a67d7ce","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101004,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":3,\"point_count\":5}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(5, 3, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 3, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0005","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04ec0de4b8031ef295cf8113d7ad466dfe551d36020e3cedda0236587cdf6c08","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101005,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_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]],\"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,\"point\":4,\"point_count\":6}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0006","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3a010b9eb10e7dda37054500bc8dc26ce49af56fe9302438248cdb0f67625da","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101006,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2],[1,0,2]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":3}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(2, 2, [2], 1, [0, 1], 2, 2, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0007","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9664ad7e218d2561b2700bde8936ba36d096f5d6ddd633aca78cb36a43df0220","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101007,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3],[1,2,0,3],[2,0,1,3]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(3, 2, [0, 1, 2], 3, [0], 1, 3, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 2, [0, 1, 2], 3, [0], 1, 3, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0008","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e711d85c4d6bd71cd57654933bac169a44a10e610850f339d2e94d0eb30c3ea2","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101008,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4],[1,2,3,0,4],[2,3,0,1,4],[3,0,1,2,4]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":2,\"point_count\":5}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 2, [0, 1, 2, 3], 4, [0], 1, 4, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0009","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1bea59d14867db6695fcae07961996c182a206401c395fbfe798a19189249051","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101009,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,4,0,5],[2,3,4,0,1,5],[3,4,0,1,2,5],[4,0,1,2,3,5]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":2,\"point_count\":6}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0010","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27bf9034eefea0a4f37ad21811875bb853686c30cb2ffee6c47ebb4d2368a596","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101010,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,5,0,6],[2,3,4,5,0,1,6],[3,4,5,0,1,2,6],[4,5,0,1,2,3,6],[5,0,1,2,3,4,6]],\"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,\"point\":2,\"point_count\":7}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(6, 2, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 2, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0011","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a600f5ffc7870282e195a79c9eacbe2259e40cc913ea844e0e908e772336ac7","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101011,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3],[1,0,2,3]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(2, 2, [2], 1, [0, 1], 2, 2, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0012","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c05e71a5b300f0eee4f8d9e1feb815801dcbbad86d8fd67baf264c2bf7744c4d","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101012,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4],[1,2,0,3,4],[2,0,1,3,4]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":1,\"point_count\":5}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(3, 1, [0, 1, 2], 3, [0], 1, 3, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 1, [0, 1, 2], 3, [0], 1, 3, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0013","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"670ffc66f02f6c5bb75a2c8b59499286de5bb39adaececcd5b98f88bf72cae38","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101013,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,0,4,5],[2,3,0,1,4,5],[3,0,1,2,4,5]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":0,\"point_count\":6}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(4, 0, [0, 1, 2, 3], 4, [0], 1, 4, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 0, [0, 1, 2, 3], 4, [0], 1, 4, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0014","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd5ec8a97bb75f668a2b3795f5403e05e519ce432459109bfacb6d3a3df1ce81","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101014,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,0,5,6],[2,3,4,0,1,5,6],[3,4,0,1,2,5,6],[4,0,1,2,3,5,6]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":6,\"point_count\":7}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(5, 6, [6], 1, [0, 1, 2, 3, 4], 5, 5, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 6, [6], 1, [0, 1, 2, 3, 4], 5, 5, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0015","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7983a38793c2fb277cc0856233114293f070cc3656a17e92a99cb8b4e8d3b090","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101015,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,0,6,7],[2,3,4,5,0,1,6,7],[3,4,5,0,1,2,6,7],[4,5,0,1,2,3,6,7],[5,0,1,2,3,4,6,7]],\"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,\"point\":6,\"point_count\":8}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(6, 6, [6], 1, [0, 1, 2, 3, 4, 5], 6, 6, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 6, [6], 1, [0, 1, 2, 3, 4, 5], 6, 6, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0016","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08f817b35ae349dbfb3dce28064ff6715b56aa3fff9d938daaf1d96914ee046d","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101016,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4],[1,0,2,3,4]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":0,\"point_count\":5}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(2, 0, [0, 1], 2, [0], 1, 2, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 0, [0, 1], 2, [0], 1, 2, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0017","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56fbbb1355f8c5b84484338f34f9499bc732aee7ea259476bcee849f1f68e3f1","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101017,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5],[1,2,0,3,4,5],[2,0,1,3,4,5]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":4,\"point_count\":6}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(3, 4, [4], 1, [0, 1, 2], 3, 3, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 4, [4], 1, [0, 1, 2], 3, 3, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0018","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe8b179cf19095a7d61ab00671e5388263ac8a871333e7164b1444af5d8495ad","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101018,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,0,4,5,6],[2,3,0,1,4,5,6],[3,0,1,2,4,5,6]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":3,\"point_count\":7}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(4, 3, [0, 1, 2, 3], 4, [0], 1, 4, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 3, [0, 1, 2, 3], 4, [0], 1, 4, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0019","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa2907e90d747544aae4ed649cf80610e01a5379c640256f4c9737eec2ca07da","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101019,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,0,5,6,7],[2,3,4,0,1,5,6,7],[3,4,0,1,2,5,6,7],[4,0,1,2,3,5,6,7]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":2,\"point_count\":8}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 2, [0, 1, 2, 3, 4], 5, [0], 1, 5, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0020","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eda9a85aad33afd86252aa609b33835d764062fd052afab23f412218105b2d1c","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101020,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,5,0,6,7,8],[2,3,4,5,0,1,6,7,8],[3,4,5,0,1,2,6,7,8],[4,5,0,1,2,3,6,7,8],[5,0,1,2,3,4,6,7,8]],\"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,\"point\":1,\"point_count\":9}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(6, 1, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 1, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0021","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c5de53c2aff4b2ad49b74a298e81660f4e25973f3503a6ef593973055b9e8e4","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101021,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5],[1,0,2,3,4,5]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":6}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(2, 2, [2], 1, [0, 1], 2, 2, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (2, 2, [2], 1, [0, 1], 2, 2, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0022","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8412322f899866b1ac75418adbc0e1b094ed78eafa59ac42b13b91c30731659f","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101022,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,0,3,4,5,6],[2,0,1,3,4,5,6]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":0,\"point_count\":7}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(3, 0, [0, 1, 2], 3, [0], 1, 3, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (3, 0, [0, 1, 2], 3, [0], 1, 3, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0023","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66d13bf89cc775003dd933d97411fc5648e1ec60ef15ce018d521d58ceb5094d","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101023,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,0,4,5,6,7],[2,3,0,1,4,5,6,7],[3,0,1,2,4,5,6,7]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":6,\"point_count\":8}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(4, 6, [6], 1, [0, 1, 2, 3], 4, 4, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (4, 6, [6], 1, [0, 1, 2, 3], 4, 4, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0024","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0818b1e9fe9a63afd4042e35e1905063a9318043b1735ec79b569a4df252c9e3","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101024,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,0,5,6,7,8],[2,3,4,0,1,5,6,7,8],[3,4,0,1,2,5,6,7,8],[4,0,1,2,3,5,6,7,8]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":5,\"point_count\":9}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(5, 5, [5], 1, [0, 1, 2, 3, 4], 5, 5, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (5, 5, [5], 1, [0, 1, 2, 3, 4], 5, 5, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0025","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af5a7deabb86754180c9f391710a3c96138a012aa47c138e3b735d1112ac36c2","family_id":"qf-mod41-orbitstab-counting-identity","seed":202611101025,"semantic_template":"qf-mod41-orbitstab-counting-identity/{\"action_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,0,6,7,8,9],[2,3,4,5,0,1,6,7,8,9],[3,4,5,0,1,2,6,7,8,9],[4,5,0,1,2,3,6,7,8,9],[5,0,1,2,3,4,6,7,8,9]],\"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,\"point\":4,\"point_count\":10}"},"fields":{"title":"The Same Point Binds Both Factors in Orbit-Stabilizer Counting","stem":"Use complete 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (group_order, point, orbit, orbit_size, stabilizer, stabilizer_size, product, equal).","answer":{"kind":"expression","value":"(6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)","explanation":"Compute the entire orbit and stabilizer of this same point. Their sizes multiply to the complete group order. A coincidental numerical product with a sample or another point is not this theorem certificate. The exact result, retaining the requested order, is (6, 4, [0, 1, 2, 3, 4, 5], 6, [0], 1, 6, Yes)."},"skills":["The Same Point Binds Both Factors in Orbit-Stabilizer Counting"],"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-20261110-r01-orbit-stabilizer-theorem-0026","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec46548a5d5ee5ee1592321e45f821c3f9efe6971e11ca377127009f288692a1","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101026,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1],[1,0]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":0,\"point_count\":2}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0027","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c8d3ee1a350af175ca818efa78551b508fe660d117baf4cf2806b758e47f925","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101027,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2],[1,2,0],[2,0,1]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":1,\"point_count\":3}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0028","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c9a622156be61dd1dafc4e8b07895d659d2c248c0dc4b458d0508bf11e84a10","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101028,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0029","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24b484d07391f650af95c71d32e0683e064652dbec3e40c22d31613f28b3700a","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101029,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":3,\"point_count\":5}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 3], [1, 4], [2, 0], [3, 1], [4, 2]], [[0, [2]], [1, [3]], [2, [4]], [3, [0]], [4, [1]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 3], [1, 4], [2, 0], [3, 1], [4, 2]], [[0, [2]], [1, [3]], [2, [4]], [3, [0]], [4, [1]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0030","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2421e7af68a4027725bd34c86e8682faa3c40788680d9e1fad3abbf3162a06b8","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101030,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_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]],\"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,\"point\":4,\"point_count\":6}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0031","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b7b74bd6fda6f1a2adeac6823f7830e16bc4286194d2dd4ac363b0fa55d6fe4","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101031,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2],[1,0,2]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":3}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0032","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"781afcafac26069a43b083d9de7f33321b27b60be8a455cfc01a614d843e24e2","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101032,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3],[1,2,0,3],[2,0,1,3]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 0], [2, 1]], [[0, [1]], [1, [2]], [2, [0]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 0], [2, 1]], [[0, [1]], [1, [2]], [2, [0]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0033","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2208675bb28a5c90e849ae6781ee6a0b97270654043684b19d1881087377a8b3","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101033,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4],[1,2,3,0,4],[2,3,0,1,4],[3,0,1,2,4]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":2,\"point_count\":5}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 0], [3, 1]], [[0, [2]], [1, [3]], [2, [0]], [3, [1]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0034","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0958ff6043f6cca47b8c3bf18fa748f85cd20643a60f1e6755dabf29fd39914c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101034,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,4,0,5],[2,3,4,0,1,5],[3,4,0,1,2,5],[4,0,1,2,3,5]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":2,\"point_count\":6}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0035","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dc0e41890e3013bedf2e5cf28e32984cbf43757e384ea72c70a973024b6282c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101035,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,5,0,6],[2,3,4,5,0,1,6],[3,4,5,0,1,2,6],[4,5,0,1,2,3,6],[5,0,1,2,3,4,6]],\"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,\"point\":2,\"point_count\":7}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 3], [2, 4], [3, 5], [4, 0], [5, 1]], [[0, [4]], [1, [5]], [2, [0]], [3, [1]], [4, [2]], [5, [3]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 5], [4, 0], [5, 1]], [[0, [4]], [1, [5]], [2, [0]], [3, [1]], [4, [2]], [5, [3]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0036","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a01c27ed3c45c18dd9b21addc2c37ec889f89cdc6392ab977d0e9f15e5fc9a0c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101036,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3],[1,0,2,3]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0037","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e869bfe3b49408323f6573ffbec14ac0ea320b01dd696dc12bc27851f84f30d2","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101037,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4],[1,2,0,3,4],[2,0,1,3,4]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":1,\"point_count\":5}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 0]], [[0, [2]], [1, [0]], [2, [1]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0038","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6155627cafdfd25cecd4b3ec7f79b5420ce786183c1981dea909ed0b9bf242c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101038,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,0,4,5],[2,3,0,1,4,5],[3,0,1,2,4,5]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":0,\"point_count\":6}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 0], [1, 1], [2, 2], [3, 3]], [[0, [0]], [1, [1]], [2, [2]], [3, [3]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [2, 2], [3, 3]], [[0, [0]], [1, [1]], [2, [2]], [3, [3]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0039","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"889a879dd256b5c2a9b06a58b1919b0a78a462d2b137a0dca302edf89dd751a6","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101039,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,0,5,6],[2,3,4,0,1,5,6],[3,4,0,1,2,5,6],[4,0,1,2,3,5,6]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":6,\"point_count\":7}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6]], [[6, [0, 1, 2, 3, 4]]], Yes, [6], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6]], [[6, [0, 1, 2, 3, 4]]], Yes, [6], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0040","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"652adc42d710bc4607c32b9dcd82254d007e6af46ce764f4b69d2758f014e12d","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101040,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,0,6,7],[2,3,4,5,0,1,6,7],[3,4,5,0,1,2,6,7],[4,5,0,1,2,3,6,7],[5,0,1,2,3,4,6,7]],\"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,\"point\":6,\"point_count\":8}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6]], [[6, [0, 1, 2, 3, 4, 5]]], Yes, [6], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6]], [[6, [0, 1, 2, 3, 4, 5]]], Yes, [6], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0041","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31f32910d27c38573a58186278c075c1f0350933e410fb391a093749799ea7d1","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101041,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4],[1,0,2,3,4]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":0,\"point_count\":5}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [[0, [0]], [1, [1]]], Yes, [0, 1], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0042","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"036d41fcad3230432ec6632c848544dd349ff85417ae8392dca7265f1f51bb26","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101042,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5],[1,2,0,3,4,5],[2,0,1,3,4,5]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":4,\"point_count\":6}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 4], [1, 4], [2, 4]], [[4, [0, 1, 2]]], Yes, [4], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 4], [2, 4]], [[4, [0, 1, 2]]], Yes, [4], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0043","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1de4c5fd169491bcacc8f3dd716bc67675496aaf4f4f840e8e31df3ffdb8c6a0","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101043,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,0,4,5,6],[2,3,0,1,4,5,6],[3,0,1,2,4,5,6]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":3,\"point_count\":7}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 3], [1, 0], [2, 1], [3, 2]], [[0, [1]], [1, [2]], [2, [3]], [3, [0]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 3], [1, 0], [2, 1], [3, 2]], [[0, [1]], [1, [2]], [2, [3]], [3, [0]]], Yes, [0, 1, 2, 3], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0044","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ae53f34e994cea406d79feffaf48420f7d56cea8e36b3f1f111b14ae1a24f22","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101044,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,0,5,6,7],[2,3,4,0,1,5,6,7],[3,4,0,1,2,5,6,7],[4,0,1,2,3,5,6,7]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":2,\"point_count\":8}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 3], [2, 4], [3, 0], [4, 1]], [[0, [3]], [1, [4]], [2, [0]], [3, [1]], [4, [2]]], Yes, [0, 1, 2, 3, 4], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0045","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9de53e9543a3b28781515f23426e266f737500c1b687d9880ed3923ab03018ae","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101045,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,5,0,6,7,8],[2,3,4,5,0,1,6,7,8],[3,4,5,0,1,2,6,7,8],[4,5,0,1,2,3,6,7,8],[5,0,1,2,3,4,6,7,8]],\"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,\"point\":1,\"point_count\":9}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], [[0, [5]], [1, [0]], [2, [1]], [3, [2]], [4, [3]], [5, [4]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], [[0, [5]], [1, [0]], [2, [1]], [3, [2]], [4, [3]], [5, [4]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0046","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4c01cf71fd2b7d576818db20554664ab2a823e9a87261ae6bd72cc6e3eff93f","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101046,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5],[1,0,2,3,4,5]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"point\":2,\"point_count\":6}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 2], [1, 2]], [[2, [0, 1]]], Yes, [2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0047","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e41ebedcccefc78c3157f21db774d15aa3c53bfb17170e68936543fc26971724","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101047,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,0,3,4,5,6],[2,0,1,3,4,5,6]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"point\":0,\"point_count\":7}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 0], [1, 1], [2, 2]], [[0, [0]], [1, [1]], [2, [2]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [2, 2]], [[0, [0]], [1, [1]], [2, [2]]], Yes, [0, 1, 2], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0048","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9ab85b9340bb5cb6e5eae4f30f3066bae693aeaec638248d0e95c3f1222c05a","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101048,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,0,4,5,6,7],[2,3,0,1,4,5,6,7],[3,0,1,2,4,5,6,7]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"point\":6,\"point_count\":8}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 6], [1, 6], [2, 6], [3, 6]], [[6, [0, 1, 2, 3]]], Yes, [6], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 6], [1, 6], [2, 6], [3, 6]], [[6, [0, 1, 2, 3]]], Yes, [6], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0049","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24889e4a806be034178a51712cb9b28842bb9a5505238b86c0a84f8534854d5c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101049,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,0,5,6,7,8],[2,3,4,0,1,5,6,7,8],[3,4,0,1,2,5,6,7,8],[4,0,1,2,3,5,6,7,8]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"point\":5,\"point_count\":9}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 5], [1, 5], [2, 5], [3, 5], [4, 5]], [[5, [0, 1, 2, 3, 4]]], Yes, [5], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 5], [1, 5], [2, 5], [3, 5], [4, 5]], [[5, [0, 1, 2, 3, 4]]], Yes, [5], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0050","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3f7c9df4b0f0b95343f863736f5a90462860a4ef3d05fb15751752fb8d8145c","family_id":"qf-mod41-orbitstab-action-fibers","seed":202611101050,"semantic_template":"qf-mod41-orbitstab-action-fibers/{\"action_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,0,6,7,8,9],[2,3,4,5,0,1,6,7,8,9],[3,4,5,0,1,2,6,7,8,9],[4,5,0,1,2,3,6,7,8,9],[5,0,1,2,3,4,6,7,8,9]],\"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,\"point\":4,\"point_count\":10}"},"fields":{"title":"All Action-Map Fibers Have the Same Stabilizer Cardinality","stem":"Use complete 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Give (action_map, fibers, equal_fiber_sizes, orbit, certificate).","answer":{"kind":"expression","value":"([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)","explanation":"List every [g,g*x] in group-index order and every [orbit point,fiber] in point order. The action map is surjective onto its orbit, not necessarily onto the whole point carrier. Each fiber has the same size as this point stabilizer. The exact result, retaining the requested order, is ([[0, 4], [1, 5], [2, 0], [3, 1], [4, 2], [5, 3]], [[0, [2]], [1, [3]], [2, [4]], [3, [5]], [4, [0]], [5, [1]]], Yes, [0, 1, 2, 3, 4, 5], action_map_surjective_onto_orbit)."},"skills":["All Action-Map Fibers Have the Same Stabilizer Cardinality"],"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-20261110-r01-orbit-stabilizer-theorem-0051","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5011794eeaada30ae186589f5d96ac65c289f7df932e8aa0c48336ef9463a3ab","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101051,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1],[1,0]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"orbit_point\":1,\"point\":0,\"point_count\":2}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0052","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13a2346e364d7b06aa247f2e8cd883ea8abfee511a11481e6dfee1a0775c1690","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101052,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2],[1,2,0],[2,0,1]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"orbit_point\":0,\"point\":1,\"point_count\":3}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(0, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0053","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0dcbb55b25fdcb4707bee7274d963cdf28b49530e736b4fabd64d39300ea3ce","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101053,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"orbit_point\":1,\"point\":2,\"point_count\":4}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 3, [3], [3], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 3, [3], [3], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0054","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59769f03edc8d34e0fefb8d1c65b3442b4b15e81d1dbf526e3776af100cc023a","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101054,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"orbit_point\":2,\"point\":3,\"point_count\":5}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(2, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0055","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"813c1dc5016f32cd917dbc7a3375b24dde3753a287b3445f38ef3f9db7a0c98a","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101055,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_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]],\"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,\"orbit_point\":3,\"point\":4,\"point_count\":6}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=3. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(3, 5, [5], [5], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (3, 5, [5], [5], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0056","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34c4e1c8d6ae61311f25d1d1c52482038f396615072348a93580391dd6fcb3b7","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101056,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2],[1,0,2]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"orbit_point\":2,\"point\":2,\"point_count\":3}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0057","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd8a2eb00bf06ee389a5d20b2fbca7a511011ce3478c1bb2d49dd8fdba8d51a5","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101057,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3],[1,2,0,3],[2,0,1,3]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"orbit_point\":0,\"point\":2,\"point_count\":4}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(0, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0058","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49e17701d8eb2d8328dc4c5d30bd3c8bd57ce45c1550c6accb64789b5ae54899","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101058,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4],[1,2,3,0,4],[2,3,0,1,4],[3,0,1,2,4]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"orbit_point\":2,\"point\":2,\"point_count\":5}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(2, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0059","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91b9204edbd832146f16f6493d0e9e98ad5c8b4da02b67bba1b23887641a1233","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101059,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,4,0,5],[2,3,4,0,1,5],[3,4,0,1,2,5],[4,0,1,2,3,5]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"orbit_point\":1,\"point\":2,\"point_count\":6}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0060","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4214a351e6f96ba3fd2bd437b3874b0c973a61cd35b48d2bbcf815933d638497","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101060,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,5,0,6],[2,3,4,5,0,1,6],[3,4,5,0,1,2,6],[4,5,0,1,2,3,6],[5,0,1,2,3,4,6]],\"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,\"orbit_point\":0,\"point\":2,\"point_count\":7}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(0, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0061","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c9afaf69bc4067a8628711504fb6535bc4441eb5de6e12ae9ce2e01d2208105","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101061,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3],[1,0,2,3]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"orbit_point\":2,\"point\":2,\"point_count\":4}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0062","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f4de6f9a37429cac36889e63a1a85bee6136eb4c89adc3099e8693f54bfff9c","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101062,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4],[1,2,0,3,4],[2,0,1,3,4]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"orbit_point\":1,\"point\":1,\"point_count\":5}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0063","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10ad4515ea06e69b61d424684f340c243e127fc7d84a2d8818c6e49c5249d495","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101063,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,0,4,5],[2,3,0,1,4,5],[3,0,1,2,4,5]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"orbit_point\":1,\"point\":0,\"point_count\":6}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0064","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8310ec84faf7bbb62867e95d9f65926749d12f2a34bfe79c28131027c216f0d4","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101064,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,0,5,6],[2,3,4,0,1,5,6],[3,4,0,1,2,5,6],[4,0,1,2,3,5,6]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"orbit_point\":6,\"point\":6,\"point_count\":7}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(6, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0065","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4b170dbcbe6cf62449d197f55d5e5678a0a0faa54978ce6379dd4957d34e095","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101065,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,0,6,7],[2,3,4,5,0,1,6,7],[3,4,5,0,1,2,6,7],[4,5,0,1,2,3,6,7],[5,0,1,2,3,4,6,7]],\"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,\"orbit_point\":6,\"point\":6,\"point_count\":8}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(6, 0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0066","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"099843a1e4ec60abeb00857c974ada8026e91bae3ccb9e47f3a9ae45b97544f0","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101066,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4],[1,0,2,3,4]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"orbit_point\":0,\"point\":0,\"point_count\":5}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=0. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(0, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (0, 0, [0], [0], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0067","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f650202d3d519b275199ea233760ad0fa41265952bfc78f98141a1176882da14","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101067,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5],[1,2,0,3,4,5],[2,0,1,3,4,5]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"orbit_point\":4,\"point\":4,\"point_count\":6}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=4. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(4, 0, [0, 1, 2], [0, 1, 2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (4, 0, [0, 1, 2], [0, 1, 2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0068","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95ccd0fe9d68404ae027bdb3a705ef2c6319a3c673c6a095ac7cda64beac1bad","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101068,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,0,4,5,6],[2,3,0,1,4,5,6],[3,0,1,2,4,5,6]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"orbit_point\":1,\"point\":3,\"point_count\":7}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0069","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"294131ad7045cb8b4db0e627cd7ea29cd544f3881a5b406501a17b93360e25f3","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101069,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,0,5,6,7],[2,3,4,0,1,5,6,7],[3,4,0,1,2,5,6,7],[4,0,1,2,3,5,6,7]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"orbit_point\":1,\"point\":2,\"point_count\":8}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 4, [4], [4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0070","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45159460862d4385de697ed4f48c964e0002effde8d36b9d1fc4c935646137df","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101070,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,5,0,6,7,8],[2,3,4,5,0,1,6,7,8],[3,4,5,0,1,2,6,7,8],[4,5,0,1,2,3,6,7,8],[5,0,1,2,3,4,6,7,8]],\"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,\"orbit_point\":3,\"point\":1,\"point_count\":9}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=3. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(3, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (3, 2, [2], [2], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0071","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e0e36d86b0d9433115470e5039463cd4f9ef99f54d7e972a3f81893a8f32bab","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101071,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5],[1,0,2,3,4,5]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"orbit_point\":2,\"point\":2,\"point_count\":6}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=2. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (2, 0, [0, 1], [0, 1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0072","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95f6dfb5e23f3c3438d80844bd55c8a7768a6a32522df07f8232c899b79a11ba","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101072,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,0,3,4,5,6],[2,0,1,3,4,5,6]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"orbit_point\":1,\"point\":0,\"point_count\":7}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=1. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (1, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0073","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbd3b45289c7589fce57850422d9556694538cfcb66bb62b78ef166e70b4bb59","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101073,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,0,4,5,6,7],[2,3,0,1,4,5,6,7],[3,0,1,2,4,5,6,7]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"orbit_point\":6,\"point\":6,\"point_count\":8}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=6. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(6, 0, [0, 1, 2, 3], [0, 1, 2, 3], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (6, 0, [0, 1, 2, 3], [0, 1, 2, 3], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0074","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0308b8f492f77b35eb64329139f29d8353c3df743d20f854fefc94ec4065284f","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101074,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,0,5,6,7,8],[2,3,4,0,1,5,6,7,8],[3,4,0,1,2,5,6,7,8],[4,0,1,2,3,5,6,7,8]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"orbit_point\":5,\"point\":5,\"point_count\":9}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=5. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(5, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (5, 0, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0075","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3036afe5774a3d5da2cdf92b1f3eca8a332a51a9e7d262ce292cfdd652f31082","family_id":"qf-mod41-orbitstab-coset-fiber-bijection","seed":202611101075,"semantic_template":"qf-mod41-orbitstab-coset-fiber-bijection/{\"action_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,0,6,7,8,9],[2,3,4,5,0,1,6,7,8,9],[3,4,5,0,1,2,6,7,8,9],[4,5,0,1,2,3,6,7,8,9],[5,0,1,2,3,4,6,7,8,9]],\"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,\"orbit_point\":5,\"point\":4,\"point_count\":10}"},"fields":{"title":"A Fiber Equals a Left Coset of the Same Point Stabilizer","stem":"Use complete 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Use orbit_point=5. Give (orbit_point, representative, fiber, left_coset, equal, certificate).","answer":{"kind":"expression","value":"(5, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)","explanation":"Require the target to lie in this orbit. Select the least group-index representative g0 sending x there, form the full fiber and left coset g0*Stab(x), sort their elements and compare. Do not substitute a right coset or another point stabilizer. The exact result, retaining the requested order, is (5, 1, [1], [1], Yes, fiber_equals_left_coset_of_same_point_stabilizer)."},"skills":["A Fiber Equals a Left Coset of the Same Point Stabilizer"],"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-20261110-r01-orbit-stabilizer-theorem-0076","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27f2988347f6dfcffa7b7ed89802ea34d7f49d503c084b3d41fd67fc1abed5f5","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101076,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1],[1,0]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"other_point\":1,\"point\":0,\"point_count\":2,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(2)={0,…,1}, total action_table=[[0, 1], [1, 0]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0077","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77dd447654862360443cc16d8c730327db63a089972dc9358c0fb9bcf89f9364","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101077,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2],[1,2,0],[2,0,1]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"other_point\":2,\"point\":1,\"point_count\":3,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0078","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b6d31191b0a052886d1fb2cad8a3fb7e99f138104daf045166e14076ed8cd21","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101078,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":4,\"sampled_elements\":[0,1,2]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2], [0, 2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [0, 2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0079","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c996aafa35738e2ed6f21120d655eb8e44fc348109a30b62802c884c39f57b07","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101079,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"other_point\":4,\"point\":3,\"point_count\":5,\"sampled_elements\":[0,1,2,3]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3], other_point=4. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4], [0], 4, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4], [0], 4, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0080","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcc39ea596fad200657bc9d3e6aed4e85d1a8fa486b0fca9a621b1c318c22354","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101080,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_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]],\"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,\"other_point\":5,\"point\":4,\"point_count\":6,\"sampled_elements\":[0,1,2,3,4]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete 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, point carrier Fin(6)={0,…,5}, total action_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]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0081","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17b212a1f72422218aab40253b481b537975b5bd8ff0338d734229e19ba271a9","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101081,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2],[1,0,2]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"other_point\":0,\"point\":2,\"point_count\":3,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(3)={0,…,2}, total action_table=[[0, 1, 2], [1, 0, 2]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=0. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [2], [2], [0, 1], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0082","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a42adf24408cd6decdd4fb85c68a3465c39d7876f332e8081fd62f3477bd585","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101082,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3],[1,2,0,3],[2,0,1,3]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":4,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 2, 0, 3], [2, 0, 1, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [2], [0, 1, 2], [0], 3, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [0, 1, 2], [0], 3, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0083","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd94c96a3e3fe89929ebf154b938eafe7b881befdf61344aa30ccf31aa07927b","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101083,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4],[1,2,3,0,4],[2,3,0,1,4],[3,0,1,2,4]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":5,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 3, 0, 4], [2, 3, 0, 1, 4], [3, 0, 1, 2, 4]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [2, 3], [0, 1, 2, 3], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0084","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de0791358f8c58638feaf70436538dab806f090daa15f3d240c018f48aac6c8f","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101084,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,4,0,5],[2,3,4,0,1,5],[3,4,0,1,2,5],[4,0,1,2,3,5]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":6,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 0, 5], [2, 3, 4, 0, 1, 5], [3, 4, 0, 1, 2, 5], [4, 0, 1, 2, 3, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [2], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0085","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65f163db694df0ca53750f84fe7a3192273c2790b520f3ab10134a34998e3139","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101085,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,5,0,6],[2,3,4,5,0,1,6],[3,4,5,0,1,2,6],[4,5,0,1,2,3,6],[5,0,1,2,3,4,6]],\"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,\"other_point\":3,\"point\":2,\"point_count\":7,\"sampled_elements\":[0,1,2,3,4]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete 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, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 0, 6], [2, 3, 4, 5, 0, 1, 6], [3, 4, 5, 0, 1, 2, 6], [4, 5, 0, 1, 2, 3, 6], [5, 0, 1, 2, 3, 4, 6]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4], [0, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0086","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c4cbbd659ddf0d6e7a302d28b306503ced07305322f98e2c0d848b26e5375c3","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101086,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3],[1,0,2,3]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":4,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(4)={0,…,3}, total action_table=[[0, 1, 2, 3], [1, 0, 2, 3]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0087","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b0973290a26a7c464695e5b2ece7dabac8b981d07a1c8e58f5c798d39843721","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101087,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4],[1,2,0,3,4],[2,0,1,3,4]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"other_point\":2,\"point\":1,\"point_count\":5,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 2, 0, 3, 4], [2, 0, 1, 3, 4]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [1, 2], [0, 1, 2], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0088","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56ded58bbeccd209f5310ab4a0e6307326b5e9cc6c2e1a95ea834ebca0724878","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101088,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5],[1,2,3,0,4,5],[2,3,0,1,4,5],[3,0,1,2,4,5]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"other_point\":1,\"point\":0,\"point_count\":6,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 0, 4, 5], [2, 3, 0, 1, 4, 5], [3, 0, 1, 2, 4, 5]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [0], [0, 1, 2, 3], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1, 2, 3], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0089","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d05d58a4a0e03a0d11f6a98f776d7b7e98879bead6424ae21d0b5580593343b1","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101089,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,4,0,5,6],[2,3,4,0,1,5,6],[3,4,0,1,2,5,6],[4,0,1,2,3,5,6]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"other_point\":0,\"point\":6,\"point_count\":7,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 0, 5, 6], [2, 3, 4, 0, 1, 5, 6], [3, 4, 0, 1, 2, 5, 6], [4, 0, 1, 2, 3, 5, 6]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=0. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [6], [6], [0, 1, 2, 3, 4], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [6], [6], [0, 1, 2, 3, 4], 0, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0090","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8500c0a3e833f62aae085da1dcd3956dbaecabb337718a49133f7a9368508a45","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101090,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,0,6,7],[2,3,4,5,0,1,6,7],[3,4,5,0,1,2,6,7],[4,5,0,1,2,3,6,7],[5,0,1,2,3,4,6,7]],\"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,\"other_point\":7,\"point\":6,\"point_count\":8,\"sampled_elements\":[0,1,2,3,4]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete 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, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 0, 6, 7], [2, 3, 4, 5, 0, 1, 6, 7], [3, 4, 5, 0, 1, 2, 6, 7], [4, 5, 0, 1, 2, 3, 6, 7], [5, 0, 1, 2, 3, 4, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=7. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4], [6], [6], [0, 1, 2, 3, 4, 5], 7, [0, 1, 2, 3, 4, 5], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [6], [6], [0, 1, 2, 3, 4, 5], 7, [0, 1, 2, 3, 4, 5], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0091","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b9a04df6b2817ff576c091786548f58dcd10103fa6f55d9228c680b7b58fa18","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101091,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4],[1,0,2,3,4]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"other_point\":1,\"point\":0,\"point_count\":5,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(5)={0,…,4}, total action_table=[[0, 1, 2, 3, 4], [1, 0, 2, 3, 4]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [0], [0, 1], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0092","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88af63daf8d0041d8cbfcc1de11426301a8c34353b7f323d00bcfd2bc38950ce","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101092,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5],[1,2,0,3,4,5],[2,0,1,3,4,5]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"other_point\":5,\"point\":4,\"point_count\":6,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 2, 0, 3, 4, 5], [2, 0, 1, 3, 4, 5]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [4], [4], [0, 1, 2], 5, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [4], [4], [0, 1, 2], 5, [0, 1, 2], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0093","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b322839e81274176810f7ee6733a8b83cbdd59867c73c559a4b49dd0c1ae59b2","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101093,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,3,0,4,5,6],[2,3,0,1,4,5,6],[3,0,1,2,4,5,6]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"other_point\":4,\"point\":3,\"point_count\":7,\"sampled_elements\":[0,1,2]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 3, 0, 4, 5, 6], [2, 3, 0, 1, 4, 5, 6], [3, 0, 1, 2, 4, 5, 6]] and specified point=3. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=4. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2], [0, 1, 3], [0, 1, 2, 3], [0], 4, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [0, 1, 3], [0, 1, 2, 3], [0], 4, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0094","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb57c8acc6715e556f97647ee21108b73627c6878212da45e9ab1c2f16f5391f","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101094,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,0,5,6,7],[2,3,4,0,1,5,6,7],[3,4,0,1,2,5,6,7],[4,0,1,2,3,5,6,7]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":8,\"sampled_elements\":[0,1,2]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 0, 5, 6, 7], [2, 3, 4, 0, 1, 5, 6, 7], [3, 4, 0, 1, 2, 5, 6, 7], [4, 0, 1, 2, 3, 5, 6, 7]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2], [2, 3, 4], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2], [2, 3, 4], [0, 1, 2, 3, 4], [0], 3, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0095","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03e3cb1948d5ac665ee94aba3c67b316785b75e69b00683e9c7e42255ae32fb0","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101095,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,5,0,6,7,8],[2,3,4,5,0,1,6,7,8],[3,4,5,0,1,2,6,7,8],[4,5,0,1,2,3,6,7,8],[5,0,1,2,3,4,6,7,8]],\"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,\"other_point\":2,\"point\":1,\"point_count\":9,\"sampled_elements\":[0,1,2,3,4]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete 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, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 0, 6, 7, 8], [2, 3, 4, 5, 0, 1, 6, 7, 8], [3, 4, 5, 0, 1, 2, 6, 7, 8], [4, 5, 0, 1, 2, 3, 6, 7, 8], [5, 0, 1, 2, 3, 4, 6, 7, 8]] and specified point=1. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=2. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4], [1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0], 2, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0096","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e6456e02752a9b59562204f43552343b84d4518edde55e5d4344ab9af740c81","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101096,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5],[1,0,2,3,4,5]],\"group_table\":[[0,1],[1,0]],\"identity\":0,\"other_point\":3,\"point\":2,\"point_count\":6,\"sampled_elements\":[0]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1], [1, 0]], identity=0, point carrier Fin(6)={0,…,5}, total action_table=[[0, 1, 2, 3, 4, 5], [1, 0, 2, 3, 4, 5]] and specified point=2. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0], other_point=3. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0], [2], [2], [0, 1], 3, [0, 1], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0097","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed2666d2787655ca21dfad4fda7a7402bd5bb0bbbb12215fa35efac535299426","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101097,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6],[1,2,0,3,4,5,6],[2,0,1,3,4,5,6]],\"group_table\":[[0,1,2],[1,2,0],[2,0,1]],\"identity\":0,\"other_point\":1,\"point\":0,\"point_count\":7,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2], [1, 2, 0], [2, 0, 1]], identity=0, point carrier Fin(7)={0,…,6}, total action_table=[[0, 1, 2, 3, 4, 5, 6], [1, 2, 0, 3, 4, 5, 6], [2, 0, 1, 3, 4, 5, 6]] and specified point=0. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=1. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [0, 1], [0, 1, 2], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [0, 1], [0, 1, 2], [0], 1, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0098","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5160719122a8baac3c9fdde7085e580a5d053ef134fdda2d0f677b4235e3f5a3","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101098,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7],[1,2,3,0,4,5,6,7],[2,3,0,1,4,5,6,7],[3,0,1,2,4,5,6,7]],\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"other_point\":7,\"point\":6,\"point_count\":8,\"sampled_elements\":[0,1]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0, point carrier Fin(8)={0,…,7}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 0, 4, 5, 6, 7], [2, 3, 0, 1, 4, 5, 6, 7], [3, 0, 1, 2, 4, 5, 6, 7]] and specified point=6. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1], other_point=7. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1], [6], [6], [0, 1, 2, 3], 7, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1], [6], [6], [0, 1, 2, 3], 7, [0, 1, 2, 3], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0099","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5104e4384751fcc9e49f35947c70417a1c2a3310ed5a7c4d19f9964db92a80fc","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101099,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7,8],[1,2,3,4,0,5,6,7,8],[2,3,4,0,1,5,6,7,8],[3,4,0,1,2,5,6,7,8],[4,0,1,2,3,5,6,7,8]],\"group_table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"identity\":0,\"other_point\":6,\"point\":5,\"point_count\":9,\"sampled_elements\":[0,1,2,3]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete group table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], identity=0, point carrier Fin(9)={0,…,8}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 0, 5, 6, 7, 8], [2, 3, 4, 0, 1, 5, 6, 7, 8], [3, 4, 0, 1, 2, 5, 6, 7, 8], [4, 0, 1, 2, 3, 5, 6, 7, 8]] and specified point=5. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3], other_point=6. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [5], [5], [0, 1, 2, 3, 4], 6, [0, 1, 2, 3, 4], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3], [5], [5], [0, 1, 2, 3, 4], 6, [0, 1, 2, 3, 4], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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-20261110-r01-orbit-stabilizer-theorem-0100","batch_id":"batch-20261014-advance-20261110-r01-orbit-stabilizer-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"465fb4e81e9e160880ad5dc61171b2f7182844a71151a35ec7ecc97205044e92","family_id":"qf-mod41-orbitstab-binding-boundary","seed":202611101100,"semantic_template":"qf-mod41-orbitstab-binding-boundary/{\"action_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,0,6,7,8,9],[2,3,4,5,0,1,6,7,8,9],[3,4,5,0,1,2,6,7,8,9],[4,5,0,1,2,3,6,7,8,9],[5,0,1,2,3,4,6,7,8,9]],\"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,\"other_point\":5,\"point\":4,\"point_count\":10,\"sampled_elements\":[0,1,2,3,4]}"},"fields":{"title":"A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate","stem":"Use complete 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, point carrier Fin(10)={0,…,9}, total action_table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 0, 6, 7, 8, 9], [2, 3, 4, 5, 0, 1, 6, 7, 8, 9], [3, 4, 5, 0, 1, 2, 6, 7, 8, 9], [4, 5, 0, 1, 2, 3, 6, 7, 8, 9], [5, 0, 1, 2, 3, 4, 6, 7, 8, 9]] and specified point=4. Row g/column h gives g*h, and action row g/column x gives g*x. Certify every group/action axiom, then enumerate all group elements with the same action and point binding. Free, faithful or transitive action is not required. Proper nonempty unique sampled_elements=[0, 1, 2, 3, 4], other_point=5. Give (sampled_elements, sample_image, full_orbit, point_stabilizer, other_point, other_stabilizer, sample_or_mixed_binding_certificate, boundary).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)","explanation":"Report the sample image, full orbit, this point stabilizer and other point stabilizer separately. The fixed No withholds a certificate from incomplete or mixed bindings; it does not negate special sample-image coverage or equality of two point stabilizers. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [0, 1, 2, 4, 5], [0, 1, 2, 3, 4, 5], [0], 5, [0], No, proper_sample_or_other_point_binding_cannot_certify_orbit_stabilizer_identity)."},"skills":["A Proper Sample or Mixed Point Binding Does Not Supply the Theorem Certificate"],"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."}}