{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0053",
  "pilot_number": "R0048-24903",
  "number": 24903,
  "family_id": "qf-mod45-kernel-fiber-coset-equivalence",
  "title": {
    "zh-Hant": "第 24903 題：有限群同態的核",
    "en": "Question 24903: Image Equality Agrees with a Inverse Times b Membership in the Kernel"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0053",
    "batch_id": "batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-finite-group-homomorphism-kernel"
    ],
    "title_zh": "有限群同態的核",
    "stem_zh": "完整來源／目標Cayley table與complete total map逐對同態核對。操作=kernel-fiber-coset-equivalence；資料={\"source_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]],\"source_identity\":0,\"target_table\":[[0,1],[1,0]],\"target_identity\":0,\"mapping\":[0,1,0,1,0,1,0,1],\"a\":2,\"b\":4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(2, 4, 0, 0, 是, 2, 是, [0, 2, 4, 6], [0, 2, 4, 6], 是, 是, f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset)",
      "explanation_zh": "Wave6第八層核對=(2, 4, 0, 0, 是, 2, 是, [0, 2, 4, 6], [0, 2, 4, 6], 是, 是, f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset)；依完整明示資料獨立重算；kernel是目標identity的完整preimage，subgroup、forall g,k的normality、fibers與injectivity均完整枚舉；非同態不得套kernel-normality；本節點不宣稱quotient或第一同構定理。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "有限群同態的核"
      ]
    },
    "variant": {
      "family_id": "qf-mod45-kernel-fiber-coset-equivalence",
      "seed": 202611141053,
      "parameters": {
        "source_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
          ]
        ],
        "source_identity": 0,
        "target_table": [
          [
            0,
            1
          ],
          [
            1,
            0
          ]
        ],
        "target_identity": 0,
        "mapping": [
          0,
          1,
          0,
          1,
          0,
          1,
          0,
          1
        ],
        "a": 2,
        "b": 4,
        "domain": "Complete_finite_group_homomorphism_kernel",
        "operation": "kernel-fiber-coset-equivalence",
        "semantic_template": "qf-mod45-kernel-fiber-coset-equivalence/{\"a\":2,\"b\":4,\"mapping\":[0,1,0,1,0,1,0,1],\"source_identity\":0,\"source_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]],\"target_identity\":0,\"target_table\":[[0,1],[1,0]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、非同態kernel、未驗normal的quotient、class-size公式、finite-quadrature Gaussian定理、bias-variance decomposition、LP最優性、minimum vertex cover、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-15",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0053",
    "batch_id": "batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "3c6d56c5ec3a2441535ea994ff8134c568d6f71855fb0f519f3144b60c0da24e",
      "family_id": "qf-mod45-kernel-fiber-coset-equivalence",
      "seed": 202611141053,
      "semantic_template": "qf-mod45-kernel-fiber-coset-equivalence/{\"a\":2,\"b\":4,\"mapping\":[0,1,0,1,0,1,0,1],\"source_identity\":0,\"source_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]],\"target_identity\":0,\"target_table\":[[0,1],[1,0]]}"
    },
    "fields": {
      "title": "Image Equality Agrees with a Inverse Times b Membership in the Kernel",
      "stem": "Use complete source 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]], source identity=0, target table=[[0, 1], [1, 0]], target identity=0, and total mapping=[0, 1, 0, 1, 0, 1, 0, 1]. Row x/column y gives x*y in the respective table. Both groups qualify first, then every pair must satisfy f(x*y)=f(x)*f(y). A total map alone is not a homomorphism. Given source a=2, b=4. Give (a, b, image_a, image_b, equal_images, left_division_a_inverse_b, difference_in_kernel, fiber, left_coset, fiber_equals_coset, equivalence_holds, certificate).",
      "answer": {
        "kind": "expression",
        "value": "(2, 4, 0, 0, Yes, 2, Yes, [0, 2, 4, 6], [0, 2, 4, 6], Yes, Yes, f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset)",
        "explanation": "Require homomorphism qualification. Compute a^(-1)*b, its kernel membership, the full fiber of f(a) and left coset a*K. This fiber is nonempty because it contains a; no surjectivity onto the target is assumed. equal_images and difference_in_kernel may both be No while equivalence_holds is Yes. The exact result, retaining the requested order, is (2, 4, 0, 0, Yes, 2, Yes, [0, 2, 4, 6], [0, 2, 4, 6], Yes, Yes, f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset)."
      },
      "skills": [
        "Image Equality Agrees with a Inverse Times b Membership in the Kernel"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "完整來源／目標Cayley table與complete total map逐對同態核對。操作=kernel-fiber-coset-equivalence；資料={\"source_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]],\"source_identity\":0,\"target_table\":[[0,1],[1,0]],\"target_identity\":0,\"mapping\":[0,1,0,1,0,1,0,1],\"a\":2,\"b\":4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use complete source 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]], source identity=0, target table=[[0, 1], [1, 0]], target identity=0, and total mapping=[0, 1, 0, 1, 0, 1, 0, 1]. Row x/column y gives x*y in the respective table. Both groups qualify first, then every pair must satisfy f(x*y)=f(x)*f(y). A total map alone is not a homomorphism. Given source a=2, b=4. Give (a, b, image_a, image_b, equal_images, left_division_a_inverse_b, difference_in_kernel, fiber, left_coset, fiber_equals_coset, equivalence_holds, certificate)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source roles = [kernel-fiber-coset-equivalence]",
      "after": "derived = [[a, b, image_a, image_b, equal_images, left_division_a_inverse_b, difference_in_kernel, fiber, left_coset, fiber_equals_coset, equivalence_holds, certificate]]",
      "relation": "implies",
      "rule_id": "source_ordered_fields_without_JS",
      "conditions": [],
      "explanation": {
        "zh-Hant": "直接列來源完整欄序，不需JavaScript；原題、布林旗標、數學物件與非聲稱保持分層。",
        "en": "List the full source field order without JavaScript,keeping original data,Boolean flags,mathematical objects and nonclaims distinct."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "source roles = [Kernel ORIGINAL SOURCE group, 64, [[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]], 0]",
      "after": "derived = [8, [[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5], [6, 6, 6, 6], [7, 7, 7, 7]], 是]",
      "relation": "implies",
      "rule_id": "Group_complete_Cayley_signature_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗完整n×n閉合表與實際identity標籤，非預設0。mod44上限32，mod45上限64；同一carrier不能混入另一表。",
        "en": "Validate the complete closed n-by-n table and actual identity label,not an assumed0.Mod44 allows32 and mod45 allows64;do not mix carriers or tables."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "source roles = [Kernel ORIGINAL SOURCE group, 8, FOR EVERY x,y,z in0..n-1]",
      "after": "derived = [[[是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是]], 512, 是]",
      "relation": "implies",
      "rule_id": "Group_ALL_associativity_triples_compact_complete_pair_matrix",
      "conditions": [],
      "explanation": {
        "zh-Hant": "矩陣(x,y)逐全部z核(xy)z=x(yz)，故涵蓋全部n³三元組，不是抽樣。表已在前步完整列出，不重印數百萬同義查表項。",
        "en": "Entry(x,y) checks (xy)z=x(yz) for every z,covering all n^3 triples,not samples.The complete table is already displayed,so do not print millions of redundant lookups."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "source roles = [Kernel ORIGINAL SOURCE group, 0]",
      "after": "derived = [[0, 7, 6, 5, 4, 3, 2, 1], [[0, 0, 0, 0], [1, 7, 0, 0], [2, 6, 0, 0], [3, 5, 0, 0], [4, 4, 0, 0], [5, 3, 0, 0], [6, 2, 0, 0], [7, 1, 0, 0]], 是]",
      "relation": "implies",
      "rule_id": "Group_ALL_two_sided_inverses",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個元素的左、右逆都要回到同一實際identity；結合律使兩側逆唯一。這才完成群前提。",
        "en": "Every left and right inverse must return the same actual identity;associativity gives uniqueness.This completes the group premises."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "source roles = [Kernel ORIGINAL TARGET group, 64, [[0, 1], [1, 0]], 0]",
      "after": "derived = [2, [[0, 0, 0, 0], [1, 1, 1, 1]], 是]",
      "relation": "implies",
      "rule_id": "Group_complete_Cayley_signature_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗完整n×n閉合表與實際identity標籤，非預設0。mod44上限32，mod45上限64；同一carrier不能混入另一表。",
        "en": "Validate the complete closed n-by-n table and actual identity label,not an assumed0.Mod44 allows32 and mod45 allows64;do not mix carriers or tables."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "source roles = [Kernel ORIGINAL TARGET group, 2, FOR EVERY x,y,z in0..n-1]",
      "after": "derived = [[[是, 是], [是, 是]], 8, 是]",
      "relation": "implies",
      "rule_id": "Group_ALL_associativity_triples_compact_complete_pair_matrix",
      "conditions": [],
      "explanation": {
        "zh-Hant": "矩陣(x,y)逐全部z核(xy)z=x(yz)，故涵蓋全部n³三元組，不是抽樣。表已在前步完整列出，不重印數百萬同義查表項。",
        "en": "Entry(x,y) checks (xy)z=x(yz) for every z,covering all n^3 triples,not samples.The complete table is already displayed,so do not print millions of redundant lookups."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "source roles = [Kernel ORIGINAL TARGET group, 0]",
      "after": "derived = [[0, 1], [[0, 0, 0, 0], [1, 1, 0, 0]], 是]",
      "relation": "implies",
      "rule_id": "Group_ALL_two_sided_inverses",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個元素的左、右逆都要回到同一實際identity；結合律使兩側逆唯一。這才完成群前提。",
        "en": "Every left and right inverse must return the same actual identity;associativity gives uniqueness.This completes the group premises."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "source roles = [[[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]], [[0, 1], [1, 0]], [0, 1, 0, 1, 0, 1, 0, 1]]",
      "after": "derived = [[[0, 0, 0, 0, 是], [0, 1, 1, 1, 是], [0, 2, 0, 0, 是], [0, 3, 1, 1, 是], [0, 4, 0, 0, 是], [0, 5, 1, 1, 是], [0, 6, 0, 0, 是], [0, 7, 1, 1, 是], [1, 0, 1, 1, 是], [1, 1, 0, 0, 是], [1, 2, 1, 1, 是], [1, 3, 0, 0, 是], [1, 4, 1, 1, 是], [1, 5, 0, 0, 是], [1, 6, 1, 1, 是], [1, 7, 0, 0, 是], [2, 0, 0, 0, 是], [2, 1, 1, 1, 是], [2, 2, 0, 0, 是], [2, 3, 1, 1, 是], [2, 4, 0, 0, 是], [2, 5, 1, 1, 是], [2, 6, 0, 0, 是], [2, 7, 1, 1, 是], [3, 0, 1, 1, 是], [3, 1, 0, 0, 是], [3, 2, 1, 1, 是], [3, 3, 0, 0, 是], [3, 4, 1, 1, 是], [3, 5, 0, 0, 是], [3, 6, 1, 1, 是], [3, 7, 0, 0, 是], [4, 0, 0, 0, 是], [4, 1, 1, 1, 是], [4, 2, 0, 0, 是], [4, 3, 1, 1, 是], [4, 4, 0, 0, 是], [4, 5, 1, 1, 是], [4, 6, 0, 0, 是], [4, 7, 1, 1, 是], [5, 0, 1, 1, 是], [5, 1, 0, 0, 是], [5, 2, 1, 1, 是], [5, 3, 0, 0, 是], [5, 4, 1, 1, 是], [5, 5, 0, 0, 是], [5, 6, 1, 1, 是], [5, 7, 0, 0, 是], [6, 0, 0, 0, 是], [6, 1, 1, 1, 是], [6, 2, 0, 0, 是], [6, 3, 1, 1, 是], [6, 4, 0, 0, 是], [6, 5, 1, 1, 是], [6, 6, 0, 0, 是], [6, 7, 1, 1, 是], [7, 0, 1, 1, 是], [7, 1, 0, 0, 是], [7, 2, 1, 1, 是], [7, 3, 0, 0, 是], [7, 4, 1, 1, 是], [7, 5, 0, 0, 是], [7, 6, 1, 1, 是], [7, 7, 0, 0, 是]], 是, null]",
      "relation": "implies",
      "rule_id": "Kernel_complete_TOTAL_map_ALL_product_pairs_before_theorems",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先對同一source/target每對(x,y)驗f(xy)=f(x)f(y)，保首failed row/null。total map不是homomorphism；失敗即擋所有kernel定理，不把內部fallback當真正identity preimage。",
        "en": "For these same source and target groups,check f(xy)=f(x)f(y) for every pair and retain the first failed row/null.A total map need not be a homomorphism.Failure blocks kernel theorems;an internal fallback is not the true identity preimage."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "source roles = [[0, 1, 0, 1, 0, 1, 0, 1], 0, 0]",
      "after": "derived = [[0, 2, 4, 6], 0, 0]",
      "relation": "implies",
      "rule_id": "Kernel_actual_TARGET_identity_complete_preimage",
      "conditions": [],
      "explanation": {
        "zh-Hant": "kernel逐全部source元素取目標實際identity的preimage。f(e)=f(e)^2在target群消去得f(e)=e_target，故真kernel非空；不能假設target identity=0。",
        "en": "The kernel is the complete preimage of the actual target identity.Cancel f(e)=f(e)^2 in the target group to obtain f(e)=e_target,so the true kernel is nonempty;do not assume identity0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, [0, 2, 4, 6], 0]",
      "after": "derived = [是, [[0, 0, 0, 是], [0, 2, 2, 是], [0, 4, 4, 是], [0, 6, 6, 是], [2, 0, 2, 是], [2, 2, 4, 是], [2, 4, 6, 是], [2, 6, 0, 是], [4, 0, 4, 是], [4, 2, 6, 是], [4, 4, 0, 是], [4, 6, 2, 是], [6, 0, 6, 是], [6, 2, 0, 是], [6, 4, 2, 是], [6, 6, 4, 是]], 是, [[0, 0, 是], [2, 6, 是], [4, 4, 是], [6, 2, 是]], 是, 是]",
      "relation": "implies",
      "rule_id": "Subgroup_identity_ALL_products_ALL_inverses_separate",
      "conditions": [],
      "explanation": {
        "zh-Hant": "H非空且無重複；含identity、任兩元素乘積、每個逆元分別核全域，三者合取才可稱subgroup。order divisibility或有限性不足。",
        "en": "H is nonempty and unique.Check identity membership,all products and every inverse separately;their conjunction qualifies a subgroup.Size divisibility or finiteness alone is insufficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 0, [0, 2, 4, 6], 0]",
      "after": "derived = [[[0, 0, 0, 0], [2, 2, 0, 2], [4, 4, 0, 4], [6, 6, 0, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 1, [0, 2, 4, 6], 7]",
      "after": "derived = [[[0, 1, 7, 0], [2, 3, 7, 2], [4, 5, 7, 4], [6, 7, 7, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 2, [0, 2, 4, 6], 6]",
      "after": "derived = [[[0, 2, 6, 0], [2, 4, 6, 2], [4, 6, 6, 4], [6, 0, 6, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 3, [0, 2, 4, 6], 5]",
      "after": "derived = [[[0, 3, 5, 0], [2, 5, 5, 2], [4, 7, 5, 4], [6, 1, 5, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 4, [0, 2, 4, 6], 4]",
      "after": "derived = [[[0, 4, 4, 0], [2, 6, 4, 2], [4, 0, 4, 4], [6, 2, 4, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 5, [0, 2, 4, 6], 3]",
      "after": "derived = [[[0, 5, 3, 0], [2, 7, 3, 2], [4, 1, 3, 4], [6, 3, 3, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 6, [0, 2, 4, 6], 2]",
      "after": "derived = [[[0, 6, 2, 0], [2, 0, 2, 2], [4, 2, 2, 4], [6, 4, 2, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "source roles = [Kernel ACTUAL identity preimage, 7, [0, 2, 4, 6], 1]",
      "after": "derived = [[[0, 7, 1, 0], [2, 1, 1, 2], [4, 3, 1, 4], [6, 5, 1, 6]], [0, 2, 4, 6], 是]",
      "relation": "implies",
      "rule_id": "Subgroup_EVERY_g_entire_conjugate_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按同一表先乘g*h，再乘g的右逆，遍歷全部h；再對全部g核集合等式，不由取樣或生成元猜normal。",
        "en": "In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "source roles = [[0, 2, 4, 6], 是, 0]",
      "after": "derived = [f(kl)=f(k)f(l)=e, f(k^-1)=f(k)^-1=e, f(gkg^-1)=f(g)e*f(g)^-1=e]",
      "relation": "implies",
      "rule_id": "Kernel_GENERAL_closure_normality_from_homomorphism",
      "conditions": [],
      "explanation": {
        "zh-Hant": "乘積、逆元與全g共軛的像都回到target identity，這給一般closure/normality理由，和本題完整枚舉互相核對；不宣稱quotient/isomorphism。",
        "en": "Products,inverses and every conjugate map to target identity,giving the general closure/normality argument alongside this complete finite audit.No quotient/isomorphism is claimed."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "source roles = [2, 4, 6, [0, 1, 0, 1, 0, 1, 0, 1], [0, 2, 4, 6]]",
      "after": "derived = [2, 是, 是, [0, 2, 4, 6], [0, 2, 4, 6], 是, 是]",
      "relation": "implies",
      "rule_id": "Kernel_actual_LEFT_division_fiber_LEFT_coset_two_directions",
      "conditions": [],
      "explanation": {
        "zh-Hant": "f(a)=f(b)等價f(a)^-1f(b)=e，故用a^-1*b而非b*a^-1。任意y在fiber iff a^-1y∈K；y↦a^-1y與k↦ak互逆，給完整fiber=aK。fiber含a，不假設target滿射。False/False的iff仍True。",
        "en": "Image equality is equivalent to f(a)^-1f(b)=e,using a^-1*b,not b*a^-1.For any y,fiber membership is equivalent to a^-1y in K;the inverse maps y->a^-1y and k->ak give fiber=aK.The fiber contains a without target surjectivity.False/False still makes the iff true."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      2,
      4,
      0,
      0,
      true,
      2,
      true,
      [
        0,
        2,
        4,
        6
      ],
      [
        0,
        2,
        4,
        6
      ],
      true,
      true,
      "f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset"
    ]
  },
  "conclusion": {
    "canonical_value": "(2, 4, 0, 0, 是, 2, 是, [0, 2, 4, 6], [0, 2, 4, 6], 是, 是, f(a)=f(b)_iff_a_inverse_b_in_kernel_and_nonempty_fiber_is_a_left_coset)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations",
      "batch_path": "public/data/questions/batches/batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations/questions.jsonl",
      "batch_sha256": "3f7d62a92a3b7dc4115370810169c8da49b4e2bf4b448b54efee63c57f32120d",
      "original_line": 53,
      "original_line_sha256": "53977d04df24a47ee0aec8d5c13331a1963fca876c4c19539668e722aac2e064",
      "question_compact_sha256": "3c6d56c5ec3a2441535ea994ff8134c568d6f71855fb0f519f3144b60c0da24e",
      "english_row_compact_sha256": "11b0265a52f92a3c5ed19870c5585aeb70b38a46404a18f40f18dcf360ae9c3f",
      "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-finite-group-homomorphism-kernel": {
        "sha256": "2d8504a53dc49db696cced0fde35020459fee128373b3d2efc52f01168374d11",
        "english_sha256": "4c8c7e9116861cd389dc48d302f74f3de3e520a7450771096acb9a4ec434b806",
        "original_path": "public/data/mko/finite-group-homomorphism-kernel.json",
        "english_path": "public/data/i18n/editions/mko-finite-group-homomorphism-kernel.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "7f1649723787c6d7cf1abe85636fd3740f2c3cc1784071db7e38069361c81fd4",
    "original_candidate_sha256": "7866a514e766f037e17063fd8a32045bc59e242a120a22ea0c9a9d664ab668af",
    "original_candidate_line_sha256": "7866a514e766f037e17063fd8a32045bc59e242a120a22ea0c9a9d664ab668af"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-009-p49",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "049a9e07e69fead88d93b34f0937b9470a06d71c1a8d097058fc83554529bdd0",
      "question_first_witness_sha256": "0c71ebb418f3de347d232d159fc25e93036f4ab9166bcfb5ee5b4be86d5c55b9"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 20,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "7b257f58862273ed7e08282005bca518dca70fc3ca122d815d3b3b5bf0bd6d99",
      "arithmetic_steps": 0
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}