{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0001","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3076f3930d151aca891d5693ac69a36a09386e3911e21ea76124d62c8d3f6aaf","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131001,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0002","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a58e56db3792ce890afbf066543bc42618f20e54da4e3e2570a5e28f0f76572d","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131002,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],\"identity\":0,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0003","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bd4f51b42f11cc5e0f083205623713bcf14f70abf33256ace5fde8dcf755ac3","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131003,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0,2,4,6]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and candidate subset H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0004","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46558470c1108f42870b0b77b2a8daab7640db074970422001f5968362576743","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131004,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and candidate subset H=[0, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0005","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45f7cb6ff80dd5bdf99d8fea8f915bbdebd30e7393fbf706b955bb0786b8644d","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131005,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0,1]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1], Yes, No, No, No, No, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1], Yes, No, No, No, No, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0006","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c146be2cb6fa200b8fa747244bb7f667fb7c9faed2cc0efa86fc36b5c53b78f","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131006,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[3,0,1,2],[0,1,2,3],[1,2,3,0],[2,3,0,1]],\"identity\":1,\"subset\":[1]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([1], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0007","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91169d0f79dc258fa3846a8e54f30bc75916b0178eead4c702c169bc6f69c8e0","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131007,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0008","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d75f968bb8c54b80daa76a4a62aace80845c1d605ec955386474b8651fbbc9f8","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131008,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[1,3,5,7]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and candidate subset H=[1, 3, 5, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0009","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52a33f5e7bae1291f82bcfe195b5e86686e557f2f86c27ad318d84e54cd49251","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131009,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1,6]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and candidate subset H=[1, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([1, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0010","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd7981c2a3fc8747d3a0e75cdf76fe8bf9e96360bcc598120d6d18808ef66a27","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131010,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[1,2]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and candidate subset H=[1, 2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([1, 2], Yes, No, No, No, No, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 2], Yes, No, No, No, No, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0011","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b98ea883dfd8ad1ae04f94785a251af2358e8d4f42d5eba374db25c03b01012e","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131011,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[2,3,0,1],[3,0,1,2],[0,1,2,3],[1,2,3,0]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([2], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0012","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dfbcd5c8641af543548559f18e4307c16a31ae798f2666659104a6ef2eff4d33","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131012,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0013","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b84eeb86b3e9c2c1ea865601081312b74842d2cacbae440f65b0c9c2b68ea5e0","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131013,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[0,2,4,6]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and candidate subset H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0014","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cc19f3a572aaf429de010606e4f903847a6b2a4f4cd7cb88a9eae4ac4f69437","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131014,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[2,7]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and candidate subset H=[2, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([2, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0015","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8fadb8e41c02bf79b12f9af5ee1ff7896930f3b2a2a8a3600febe1754c68fde","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131015,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[2,3]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and candidate subset H=[2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([2, 3], Yes, No, No, No, No, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([2, 3], Yes, No, No, No, No, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0016","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c615510ab52543db3f5c5d0c68a0e00de41e504e644c85f61abbf0e5896807f5","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131016,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[1,2,3,0],[2,3,0,1],[3,0,1,2],[0,1,2,3]],\"identity\":3,\"subset\":[3]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([3], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0017","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95ddf3e0c708e6f2d65f9d6a5354ccd4773c32ca1cdea5afcdc8fbabb253d17b","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131017,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0018","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cf7031f1ad3442084c2ec7189d65586945f578df7638ce629b993d1d9548872","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131018,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[1,3,5,7]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and candidate subset H=[1, 3, 5, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([1, 3, 5, 7], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0019","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ad698719b2fcc639059b31e6d9f8c91099c818cc0bf86c789f4c343fe873bcc","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131019,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[3,8]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and candidate subset H=[3, 8]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([3, 8], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3, 8], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0020","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a4033abb3cd15cb1422b5dbeb53bfff0bfa883aa7df1df36b161eb62f0e9b15","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131020,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[3,4]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and candidate subset H=[3, 4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([3, 4], Yes, No, No, No, No, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([3, 4], Yes, No, No, No, No, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0021","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7600b245e66e1f9658b18f7f3220be5050f25481ec1474f0aec7640ffaf838d8","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131021,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[0,1,2,3],[1,0,3,2],[2,3,1,0],[3,2,0,1]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0022","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a9b4d0102c8d414921abf939791ee93c074ae82e0a31be71d329ab9635eefbd","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131022,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0023","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3322864855210b5a17efd5355b9a4e1eae998ad97f3e1dc9ca2f462229f15630","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131023,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[0,2,4,6]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and candidate subset H=[0, 2, 4, 6]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([0, 2, 4, 6], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0024","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42c6f446e184e7d0d603ab4ebefe0ee536d685693b37da54a6e72720b03b2701","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131024,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[4,9]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and candidate subset H=[4, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([4, 9], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([4, 9], Yes, Yes, Yes, Yes, Yes, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0025","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f35f8ee22f2439738679fddb35f496653d010cb75dc1351e27dfcbafe57453c5","family_id":"qf-mod44-normal-subgroup-qualification","seed":202611131025,"semantic_template":"qf-mod44-normal-subgroup-qualification/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[4,5]}"},"fields":{"title":"Subgroup Closure Precedes a Normal-Subgroup Certificate","stem":"Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and candidate subset H=[4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).","answer":{"kind":"expression","value":"([4, 5], Yes, No, No, No, No, subgroup_obligations_precede_normality)","explanation":"Report identity, product and inverse closure, joint subgroup qualification and the normal-subgroup predicate separately. Finite size or order divisibility alone establishes neither layer. Trivial and whole subgroups remain valid normal cases. The exact result, retaining the requested order, is ([4, 5], Yes, No, No, No, No, subgroup_obligations_precede_normality)."},"skills":["Subgroup Closure Precedes a Normal-Subgroup 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-20261015-advance-20261113-r01-finite-normal-subgroup-0026","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"196328e9310c79687c1312d76e4ead1552c87dca7213059230486b22c24cd018","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131026,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0027","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0ad055bdb54d3b6cad3b494b071c348848a7c4c9b9973b3bfb3074234294dcb","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131027,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],\"identity\":0,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0028","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"716dc4918dae016674974ea609f31e32cb2d22b9d04e4e07f4f3d903414088aa","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131028,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0029","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a05d385066749adc1c3382f9035055cdba8b9f23e9fc6828c3567f3ecc1eb6cc","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131029,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0030","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5ffec5e77b9d754c4c8b6499dd7f99bdf037ba60ca9f09dccb24a3fecd00c75","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131030,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes], [8, [0], Yes], [9, [0], Yes], [10, [0], Yes], [11, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes], [4, [0], Yes], [5, [0], Yes], [6, [0], Yes], [7, [0], Yes], [8, [0], Yes], [9, [0], Yes], [10, [0], Yes], [11, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0031","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7671b4822d8a6d1b3ce503a1bd4995d2627ac02cd5d57bf87d2fecec25195d19","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131031,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[3,0,1,2],[0,1,2,3],[1,2,3,0],[2,3,0,1]],\"identity\":1,\"subset\":[0,1,2,3]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0032","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d99199989a4355f73ce7829806b8b4f35fca8dd75d0da631129dcec8c56258e","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131032,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[1]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0033","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d421291ebd16c9350e4ca572dbc0d28fa3f67087e268239bcdca3881aaf1a35","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131033,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[0,1,2,3,4,5,6,7]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0034","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b41bd3d04769b60cbb8b200f21b9da093b247b7565366feca27fc70a5dee43c0","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131034,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes], [6, [1], Yes], [7, [1], Yes], [8, [1], Yes], [9, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([1], [[0, [1], Yes], [1, [1], Yes], [2, [1], Yes], [3, [1], Yes], [4, [1], Yes], [5, [1], Yes], [6, [1], Yes], [7, [1], Yes], [8, [1], Yes], [9, [1], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0035","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"755d7f9932f33360ac8b9f62b6b918fb98f8adac9d82788619132b6e4cabad36","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131035,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[0,1,10,11,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0036","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c50b0d0bcf36323e9b780e5959cc01d9aeb417b0a35e6afd1c103bcdd45f326","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131036,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[2,3,0,1],[3,0,1,2],[0,1,2,3],[1,2,3,0]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0037","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f30517f43282fb531d8c514ecb43ca68f77c9974a2b2d2ac70dbad2cfa0c0250","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131037,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0038","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7903a90aa1020f89795a0832cc112a2f5fb44bbfb5b620479298ea02a9a8fa81","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131038,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0039","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1d5da20ba4cce0c3f917c0e862dceb5f265b777f75e3f11371e0574a33d5cbc","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131039,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0040","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b59556dccfd4dfe0c9924582e88b2dea0725336f90653b52dfca78f20a23aa3","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131040,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes], [8, [2], Yes], [9, [2], Yes], [10, [2], Yes], [11, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([2], [[0, [2], Yes], [1, [2], Yes], [2, [2], Yes], [3, [2], Yes], [4, [2], Yes], [5, [2], Yes], [6, [2], Yes], [7, [2], Yes], [8, [2], Yes], [9, [2], Yes], [10, [2], Yes], [11, [2], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0041","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a82ad88185fcdecfc03b6b6349f4e8a0999996340fc217291d77330dbc8c698f","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131041,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[1,2,3,0],[2,3,0,1],[3,0,1,2],[0,1,2,3]],\"identity\":3,\"subset\":[0,1,2,3]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0042","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eef8c13de6bc28978d52acf75a3a7269c2e039c972e5fed7db4f52e9a2824da7","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131042,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[3]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0043","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17e1708cbff99f9e2e516df8b5f6b3bd40a0a48567c34687e6036a5c5138b0e0","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131043,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[0,1,2,3,4,5,6,7]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0044","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4feedcf9ac4329e09758e16c05aadb6c82d492cb3e258e0773ee694e480a6ea3","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131044,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[3]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes], [6, [3], Yes], [7, [3], Yes], [8, [3], Yes], [9, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([3], [[0, [3], Yes], [1, [3], Yes], [2, [3], Yes], [3, [3], Yes], [4, [3], Yes], [5, [3], Yes], [6, [3], Yes], [7, [3], Yes], [8, [3], Yes], [9, [3], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0045","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90cef61ea1b3f281d350ed0645958a091e97d25f499c0f57ebc17f3f6699e223","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131045,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[0,1,10,11,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0046","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc66252c5103e93f0111fe8cd9d7f5b021dc476d571eadfcbc62515a1e86f99c","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131046,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[0,1,2,3],[1,0,3,2],[2,3,1,0],[3,2,0,1]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0], [[0, [0], Yes], [1, [0], Yes], [2, [0], Yes], [3, [0], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0047","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b71260058482c5af071e602598d550a41d2594833ae192ebcd8eff892262908","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131047,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0048","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba95579c1ec8acc6df7318d512b49c9238119a92b5e5900858483b2cb498f155","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131048,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[4]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and candidate subset H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0049","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d53610b228106903ff04e2b8d133feff0e398d7aaed8737966553d497e817e68","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131049,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0050","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"974aa4406eefd7a60cee8bf83ecb0a222a807f65babfb9c9bd793f05cb35d827","family_id":"qf-mod44-normal-conjugate-set-audit","seed":202611131050,"semantic_template":"qf-mod44-normal-conjugate-set-audit/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[4]}"},"fields":{"title":"Every Group Element Must Conjugate the Entire Subgroup to Itself","stem":"Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and candidate subset H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, conjugate_rows, normal, certificate).","answer":{"kind":"expression","value":"([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes], [8, [4], Yes], [9, [4], Yes], [10, [4], Yes], [11, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)","explanation":"Require a genuine subgroup. Enumerate [g,gHg^(-1),equals H] for every g, using sorted element sets. One escaping conjugate refutes normality; sampled successes or generators without a proved generation obligation are insufficient. The exact result, retaining the requested order, is ([4], [[0, [4], Yes], [1, [4], Yes], [2, [4], Yes], [3, [4], Yes], [4, [4], Yes], [5, [4], Yes], [6, [4], Yes], [7, [4], Yes], [8, [4], Yes], [9, [4], Yes], [10, [4], Yes], [11, [4], Yes]], Yes, every_group_element_conjugates_H_to_itself)."},"skills":["Every Group Element Must Conjugate the Entire Subgroup to Itself"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0051","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8c250040aa183d2b5a4e8859cfd338dfdaf0bec339979acaf0386db5c495c33","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131051,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0052","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4740da6446be2f8736770894bc6b18673bdd8f1163630a4bd087a6fbd3e96d4","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131052,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],\"identity\":0,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]], identity=0 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0053","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54313915af045a5da52d6b02e66d800d4d1399ea351904fd39ef4283fa193c38","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131053,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0054","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2893a74d5241b548ef86a7625f5604c538402d20ac4c9cb171a0544fa41a9e93","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131054,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8]],\"identity\":0,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=0 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0055","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9aee3d55b26b269dc5c8479bec0ea9234a586f681f6cdb54a98946adc9a57134","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131055,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0056","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aed49f17d0253adc3bd234bfd180d6e502167bba27f28f8c4a936dea9e7b2c8c","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131056,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[3,0,1,2],[0,1,2,3],[1,2,3,0],[2,3,0,1]],\"identity\":1,\"subset\":[0,1,2,3]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1]], identity=1 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0057","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f83bf871a9869fdb5d3b94484ffd52bacad53331c621642fd0292d00b040d9cc","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131057,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3]],\"identity\":1,\"subset\":[1]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0058","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0729dbfb9e8122258264736d7df5f0609b6ce3678e16bca258209107de7f1bfb","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131058,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5]],\"identity\":1,\"subset\":[0,1,2,3,4,5,6,7]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5]], identity=1 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0059","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72f9f0f6301a50782e773556f8932aa8ec487c85f2e63f477bd05096c8177bef","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131059,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7]],\"identity\":1,\"subset\":[1]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]], identity=1 and candidate subset H=[1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([1], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0060","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"858bb1bf49f9e8804676bceac8b7f0d7376a9ef6f66f70b41e630e54ef267e03","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131060,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9]],\"identity\":1,\"subset\":[0,1,10,11,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]], identity=1 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0061","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cc9545dea1a580b0ed2b1b2f6b4a81d19bd10cd930d610cc42689f335929889","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131061,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[2,3,0,1],[3,0,1,2],[0,1,2,3],[1,2,3,0]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3], [1, 2, 3, 0]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0062","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4af3474ad7c312abb08bf2f5a9e35234ffacec684976f4ae9e426b7a71fc91ff","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131062,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2]],\"identity\":2,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2]], identity=2 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0063","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b55f72c54679a4365f2ea72d43d5feda9cb4721ba1a5224f56ff3165602036e4","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131063,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0064","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7374bdeb60ffe42eeca2f99c50d494ee2e7c1dc5ad83377ec0ea44401cef51f","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131064,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6]],\"identity\":2,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]], identity=2 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0065","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"001f6a73ba7a91670b61018d72f16a0d4b0daac58bfa8a658befbe4863f71d17","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131065,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8]],\"identity\":2,\"subset\":[2]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8]], identity=2 and candidate subset H=[2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([2], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0066","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97bdb97c9f7d6fdd46c2ffab400e52d8934177c7a3cd620ec5ed533158deedab","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131066,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[1,2,3,0],[2,3,0,1],[3,0,1,2],[0,1,2,3]],\"identity\":3,\"subset\":[0,1,2,3]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2], [0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [1, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [2, [0, 1, 2, 3], [0, 1, 2, 3], Yes], [3, [0, 1, 2, 3], [0, 1, 2, 3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0067","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b362a9587f83794d0e8ee9e7735038f9f135cbbfac66ca3e7c6817146789ee28","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131067,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1]],\"identity\":3,\"subset\":[3]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0068","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7d578be62eb6fabd111abedea97d7b806d6c04ed77ea040db210d94ba89f115","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131068,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3]],\"identity\":3,\"subset\":[0,1,2,3,4,5,6,7]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3]], identity=3 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0069","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4967f171513391569e09eb77ed94cfd7117029e4b0d96afc542bb53aa2530f99","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131069,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4],[6,7,8,9,0,1,2,3,4,5]],\"identity\":3,\"subset\":[3]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4], [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]], identity=3 and candidate subset H=[3]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([3], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0070","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38b82ad7bf31a55b9700c674bb322c1db80ba6f63a92f06e3ef713038a6d738b","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131070,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6],[8,9,10,11,0,1,2,3,4,5,6,7]],\"identity\":3,\"subset\":[0,1,10,11,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6], [8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7]], identity=3 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [10, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes], [11, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0071","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9aaa184d6b9e4ca922b5b3702f5d50217aafcfa76f3c492c1e5b78f9d0220a89","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131071,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[0,1,2,3],[1,0,3,2],[2,3,1,0],[3,2,0,1]],\"identity\":0,\"subset\":[0]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[0, 1, 2, 3], [1, 0, 3, 2], [2, 3, 1, 0], [3, 2, 0, 1]], identity=0 and candidate subset H=[0]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0072","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a092e40c2df071b005ac8a7293161c78da0a4632f1912658edba1330fb05a92f","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131072,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4],[0,1,2,3,4,5],[1,2,3,4,5,0]],\"identity\":4,\"subset\":[0,1,2,3,4,5]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0]], identity=4 and candidate subset H=[0, 1, 2, 3, 4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [2, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [3, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [4, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes], [5, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0073","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a9071a095e1c1d7398710fc53213eeca48a50f14d18e03c1438cc917866b510","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131073,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6],[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2]],\"identity\":4,\"subset\":[4]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2]], identity=4 and candidate subset H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0074","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e2dfcb21f87e70b56543cca90f8304f08ee4a95b29d11536cff9cfe99651fdf","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131074,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[6,7,8,9,0,1,2,3,4,5],[7,8,9,0,1,2,3,4,5,6],[8,9,0,1,2,3,4,5,6,7],[9,0,1,2,3,4,5,6,7,8],[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,0],[2,3,4,5,6,7,8,9,0,1],[3,4,5,6,7,8,9,0,1,2],[4,5,6,7,8,9,0,1,2,3],[5,6,7,8,9,0,1,2,3,4]],\"identity\":4,\"subset\":[0,1,2,3,4,5,6,7,8,9]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[6, 7, 8, 9, 0, 1, 2, 3, 4, 5], [7, 8, 9, 0, 1, 2, 3, 4, 5, 6], [8, 9, 0, 1, 2, 3, 4, 5, 6, 7], [9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [2, 3, 4, 5, 6, 7, 8, 9, 0, 1], [3, 4, 5, 6, 7, 8, 9, 0, 1, 2], [4, 5, 6, 7, 8, 9, 0, 1, 2, 3], [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]], identity=4 and candidate subset H=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [2, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [4, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [5, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [6, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [7, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [8, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes], [9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0075","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"097de9acd3c0f045b65280c0b4e6e3a316827331c1b1f0f057d1023df07c982d","family_id":"qf-mod44-normal-left-right-coset-audit","seed":202611131075,"semantic_template":"qf-mod44-normal-left-right-coset-audit/{\"group_table\":[[8,9,10,11,0,1,2,3,4,5,6,7],[9,10,11,0,1,2,3,4,5,6,7,8],[10,11,0,1,2,3,4,5,6,7,8,9],[11,0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,7,8,9,10,11],[1,2,3,4,5,6,7,8,9,10,11,0],[2,3,4,5,6,7,8,9,10,11,0,1],[3,4,5,6,7,8,9,10,11,0,1,2],[4,5,6,7,8,9,10,11,0,1,2,3],[5,6,7,8,9,10,11,0,1,2,3,4],[6,7,8,9,10,11,0,1,2,3,4,5],[7,8,9,10,11,0,1,2,3,4,5,6]],\"identity\":4,\"subset\":[4]}"},"fields":{"title":"Left and Right Cosets Must Agree for Every g","stem":"Use complete finite Cayley table=[[8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7], [9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1], [3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2], [4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3], [5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4], [6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5], [7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6]], identity=4 and candidate subset H=[4]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subgroup, coset_rows, all_left_right_equal, certificate).","answer":{"kind":"expression","value":"([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)","explanation":"Require subgroup qualification and compare the entire gH and Hg for each g in group-index order. Their all-g equality is equivalent to normality. Do not infer well-defined quotient multiplication here or conflate nonabelian left/right cosets. The exact result, retaining the requested order, is ([4], [[0, [0], [0], Yes], [1, [1], [1], Yes], [2, [2], [2], Yes], [3, [3], [3], Yes], [4, [4], [4], Yes], [5, [5], [5], Yes], [6, [6], [6], Yes], [7, [7], [7], Yes], [8, [8], [8], Yes], [9, [9], [9], Yes], [10, [10], [10], Yes], [11, [11], [11], Yes]], Yes, left_and_right_cosets_agree_for_every_g_exactly_when_H_is_normal)."},"skills":["Left and Right Cosets Must Agree for Every g"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0076","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cda514f1eb3b5330388546bbeefa7a9923645c6f38bfc3f4af684e24c76f00ef","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131076,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,4,0,5,1,3],[3,5,4,0,2,1],[4,2,3,1,5,0],[5,3,1,2,0,4]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 0, 5, 1, 3], [3, 5, 4, 0, 2, 1], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0077","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ef15633c404f40d44b5d87e62435e6429bf401add7c22f844a9c8e0559917c8","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131077,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,5,0,4,3,1],[3,4,5,0,1,2],[4,3,1,2,5,0],[5,2,3,1,0,4]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0078","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b932659d5614394c94ac2762d05ef059f9ad4726d8388ae02f1e75bf6905169","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131078,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,3,0,1,5,4],[3,2,4,5,1,0],[4,5,3,2,0,1],[5,4,1,0,2,3]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 0, 1, 5, 4], [3, 2, 4, 5, 1, 0], [4, 5, 3, 2, 0, 1], [5, 4, 1, 0, 2, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0079","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b084bdd05ae9b1f254b490becbb7d12b1ccfa194e5a79c8729988a3d7f6a4d1","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131079,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,5,0,4,3,1],[3,4,1,5,2,0],[4,3,5,1,0,2],[5,2,4,0,1,3]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 0, 4, 3, 1], [3, 4, 1, 5, 2, 0], [4, 3, 5, 1, 0, 2], [5, 2, 4, 0, 1, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0080","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3f3df6cb3ef3f8dc387aeed893e7dab00d7a47dc7123871b6bc14c9604d71ff","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131080,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,3,0,1,5,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,4,3,2,1,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 0, 1, 5, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0081","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f35ce9e12091b57862764555901823e995788fcb71a1438901767f9d87728688","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131081,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,4,0,5,1,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,3,4,1,2,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 0, 5, 1, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0082","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04075f003ca30640b7eeb032c99e152425e2cf6fbba66e213e50b7a3411f01a3","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131082,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,5,0,4,3,1],[3,4,5,0,1,2],[4,3,1,2,5,0],[5,2,3,1,0,4]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0083","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4442ff3a75f4dea3e70d2bb4a0dc44a8101d2e5d53fb3ea05fd0f9e2eca06f8","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131083,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,4,0,5,1,3],[3,5,4,0,2,1],[4,2,3,1,5,0],[5,3,1,2,0,4]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 0, 5, 1, 3], [3, 5, 4, 0, 2, 1], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0084","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"595960aaf21d9f84730fc0b102408c58cf547c8de9aa37e7267e504b5139c291","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131084,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,3,5,4,1,0],[3,2,1,0,5,4],[4,5,3,2,0,1],[5,4,0,1,3,2]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 5, 4, 1, 0], [3, 2, 1, 0, 5, 4], [4, 5, 3, 2, 0, 1], [5, 4, 0, 1, 3, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 5, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0085","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c38e4d9aa9424b4a19350a9aef876427ce668223f723c6d8278c012c106a9df6","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131085,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,4,5,1,3,0],[3,5,4,0,2,1],[4,2,1,5,0,3],[5,3,0,4,1,2]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 5, 1, 3, 0], [3, 5, 4, 0, 2, 1], [4, 2, 1, 5, 0, 3], [5, 3, 0, 4, 1, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0086","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eff9a573f170b047c97b20c91d1a3c79608a9ff4c9744afb8f4ba34b7bd805a2","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131086,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,3,4,5,0,1],[3,2,1,0,5,4],[4,5,0,1,2,3],[5,4,3,2,1,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 4, 5, 0, 1], [3, 2, 1, 0, 5, 4], [4, 5, 0, 1, 2, 3], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 4, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0087","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9d80d39f39dd731da7b7466110945e34d367cee9a4cec4d6c125528f5755434","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131087,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,5,4,1,0,3],[3,4,5,0,1,2],[4,3,0,5,2,1],[5,2,1,4,3,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 4, 1, 0, 3], [3, 4, 5, 0, 1, 2], [4, 3, 0, 5, 2, 1], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0088","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56cf2c204c3eb732172d33feb92954ef9ed5a79cc82763c61cf343117dc01ded","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131088,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,5,0,4,3,1],[3,4,1,5,2,0],[4,3,5,1,0,2],[5,2,4,0,1,3]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 0, 4, 3, 1], [3, 4, 1, 5, 2, 0], [4, 3, 5, 1, 0, 2], [5, 2, 4, 0, 1, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 5, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0089","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f78cd4fd6d11a71d192fa808881f44c8b05818e18779b48f57c96b8a9a52387","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131089,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,3,0,1,5,4],[3,2,4,5,1,0],[4,5,3,2,0,1],[5,4,1,0,2,3]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 0, 1, 5, 4], [3, 2, 4, 5, 1, 0], [4, 5, 3, 2, 0, 1], [5, 4, 1, 0, 2, 3]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0090","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"18c3cd81168ec5988cadd44bb3e4e1a4c99b7748f206514f3167167a115e7193","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131090,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,4,5,1,3,0],[3,5,4,0,2,1],[4,2,1,5,0,3],[5,3,0,4,1,2]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 5, 1, 3, 0], [3, 5, 4, 0, 2, 1], [4, 2, 1, 5, 0, 3], [5, 3, 0, 4, 1, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0091","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"680ea0ebd03932d7aa90b85f261359febc3bfc3f1672669133442a2585a43bb1","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131091,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,3,5,4,1,0],[3,2,1,0,5,4],[4,5,3,2,0,1],[5,4,0,1,3,2]],\"identity\":0,\"subset\":[0,1],\"witness_g\":5,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 5, 4, 1, 0], [3, 2, 1, 0, 5, 4], [4, 5, 3, 2, 0, 1], [5, 4, 0, 1, 3, 2]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=5, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 5, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 5, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0092","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4700fbf0e4bbf4591fdfa493986523e57bdccc1fd4a3f794d386f75a2800d90e","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131092,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,2,1,5,0,3],[5,3,4,1,2,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 2, 1, 5, 0, 3], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 3, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0093","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e42745c6babf913fc46831294f6b09099e93546770949dde5c25e4b6a74874a","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131093,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,3,5,1,0,2],[5,2,1,4,3,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 3, 5, 1, 0, 2], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 2, 5, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0094","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6794eab2efdaaf2f99ba9ae8b3d4c9d2489b44341fed79156acf43eaf9079995","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131094,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,4,0,5,1,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,3,4,1,2,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 4, 0, 5, 1, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 4, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0095","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d99f066e8e5e999f46496df6ed34884b1853cf4b02bfee1f4052f7f09a2f9553","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131095,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,3,0,1,5,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,4,3,2,1,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 3, 0, 1, 5, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 3, 2, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0096","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"481732d49d3ee60a3fb9148a07eef069fb8c1833d46fb8c13108ce57ec939288","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131096,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,3,2,5,4],[2,5,4,1,0,3],[3,4,5,0,1,2],[4,3,0,5,2,1],[5,2,1,4,3,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 3, 2, 5, 4], [2, 5, 4, 1, 0, 3], [3, 4, 5, 0, 1, 2], [4, 3, 0, 5, 2, 1], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 4, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0097","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1dbef3a0558303855cbccfca9965c41b6c03032186066f021f16dc4f7a7b7b90","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131097,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,3,4,5,0,1],[3,2,1,0,5,4],[4,5,0,1,2,3],[5,4,3,2,1,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":4,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 3, 4, 5, 0, 1], [3, 2, 1, 0, 5, 4], [4, 5, 0, 1, 2, 3], [5, 4, 3, 2, 1, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 4, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 4, 1, 2, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0098","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f30e06dfd4bc3d5506e1ff8d1739819a1f76ebab36793d3cc0d6cf195a52d385","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131098,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,3,5,1,0,2],[5,2,1,4,3,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":2,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 3, 5, 1, 0, 2], [5, 2, 1, 4, 3, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=2, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 2, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 2, 1, 3, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0099","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd7998b8dc737d9b61dac5b003d8c1b9b7cef7e2ee9e489e672843fed06b6ad8","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131099,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,5,4,3,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,2,1,5,0,3],[5,3,4,1,2,0]],\"identity\":0,\"subset\":[0,1],\"witness_g\":3,\"witness_h\":1}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 5, 4, 3, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 2, 1, 5, 0, 3], [5, 3, 4, 1, 2, 0]], identity=0 and candidate subset H=[0, 1]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=3, h=1. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 1], 3, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 1], 3, 1, 2, 4, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r01-finite-normal-subgroup-0100","batch_id":"batch-20261015-advance-20261113-r01-finite-normal-subgroup-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba34fe6f4817bdf8f2ec2119d16d0e01843d8da6a8e9262b4ecfbcd1707d2d63","family_id":"qf-mod44-normal-nonnormal-witness","seed":202611131100,"semantic_template":"qf-mod44-normal-nonnormal-witness/{\"group_table\":[[0,1,2,3,4,5],[1,0,4,5,2,3],[2,5,0,4,3,1],[3,4,5,0,1,2],[4,3,1,2,5,0],[5,2,3,1,0,4]],\"identity\":0,\"subset\":[0,2],\"witness_g\":4,\"witness_h\":2}"},"fields":{"title":"One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup","stem":"Use complete finite Cayley table=[[0, 1, 2, 3, 4, 5], [1, 0, 4, 5, 2, 3], [2, 5, 0, 4, 3, 1], [3, 4, 5, 0, 1, 2], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]], identity=0 and candidate subset H=[0, 2]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Supplied g=4, h=2. Give (subgroup, g, h, inverse_g, escaping_conjugate, normal, certificate).","answer":{"kind":"expression","value":"([0, 2], 4, 2, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)","explanation":"Require h∈H and g*h*g^(-1) outside H, with the inverse from the same complete table. Report the exact escaping element and No normality; this branch does not substitute a failed subgroup test for a nonnormal-subgroup witness. The exact result, retaining the requested order, is ([0, 2], 4, 2, 5, 3, No, one_h_in_H_with_g_h_g_inverse_outside_H_disproves_normality)."},"skills":["One Explicit Escaping Conjugate Refutes Normality of a Genuine Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}