{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-group-action-01",
  "mko_id": "mko-finite-group-action",
  "title": {
    "zh-Hant": "六元素群對四點的完整作用",
    "en": "A complete six-element action on four points"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-group-action-01",
    "source": {
      "mko_id": "mko-finite-group-action",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "2e31b8a871f6fd32317885aa68f63e2380a6c3a7d97c031b152b653c6a879de0",
      "english_sha256": "d2fb382c41e0c0388963c9f896b8d1cc26ad6fce451783c0262bb3ae5312cc51"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "明示24格表為有效群作用，但不是free或transitive。",
      "en": "The explicit 24-entry table is a valid group action, but it is neither free nor transitive."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "G={e,r,r²,s,rs,r²s}，依序編碼(0,0),(1,0),(2,0),(0,1),(1,1),(2,1)。乘法(i,j)(k,l)=(i+(−1)^j k mod3,j+l mod2)。點載體X={0,1,2,3}；(i,j)作用x<3為i+(−1)^j x mod3，並固定3。",
          "en": "G={e,r,r2,s,rs,r2s}, encoded respectively (0,0),(1,0),(2,0),(0,1),(1,1),(2,1), with (i,j)(k,l)=(i+(-1)^j*k mod3,j+l mod2). Points X={0,1,2,3}; (i,j) acts on x<3 by i+(-1)^j*x mod3 and fixes 3."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "G=\\{e,r,r^2,s,rs,r^2s\\}",
        "after_tex": "\\begin{array}{c|cccccc}*&e&r&r^2&s&rs&r^2s\\\\\\hline e&e&r&r^2&s&rs&r^2s\\\\r&r&r^2&e&rs&r^2s&s\\\\r^2&r^2&e&r&r^2s&s&rs\\\\s&s&r^2s&rs&e&r^2&r\\\\rs&rs&s&r^2s&r&e&r^2\\\\r^2s&r^2s&rs&s&r^2&r&e\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "將給定配對公式逐36對展開，全部輸出在同一六元素載體，沒有缺格。",
          "en": "Expand the pair formula for all 36 products; every output belongs to the same six-element carrier and no cell is missing."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "labels依序綁定六個pair，故同名r²不代表額外第七元素。",
            "en": "The six labels bind the six pairs; r^2 is not an additional seventh element."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "(i,j)(k,l)=(i+(-1)^jk,j+l)\\quad(\\bmod(3,2))",
        "after_tex": "((i,j)(k,l))(m,n)=(i+(-1)^jk+(-1)^{j+l}m,j+l+n)=(i,j)((k,l)(m,n));\\quad e=(0,0);\\quad(i,j)^{-1}=(-(-1)^ji,j)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩種括號的座標公式相同，證全部216 triples結合律；(0,0)兩側中性，列出的逆元兩側乘回(0,0)。",
          "en": "Both bracketings have identical coordinates, proving associativity for all 216 triples; (0,0) is a two-sided identity and the inverse formula multiplies back on both sides."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "指數j+l只依mod2，因(−1)²=1，公式對所有代表一致；逆元清單e,r²,r,s,rs,r²s完整。",
            "en": "The sign depends only on j+l modulo 2 since (-1)^2=1; the full inverse list is e,r^2,r,s,rs,r^2s."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "X=\\{0,1,2,3\\}",
        "after_tex": "\\begin{array}{c|cccc}\\cdot&0&1&2&3\\\\\\hline e&0&1&2&3\\\\r&1&2&0&3\\\\r^2&2&0&1&3\\\\s&0&2&1&3\\\\rs&1&0&2&3\\\\r^2s&2&1&0&3\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全部24個作用值由同一公式展開。對x<3，g·(h·x)=i+(−1)^j k+(−1)^{j+l}x=(gh)·x；3兩邊皆固定。e逐點固定。",
          "en": "Expand all 24 action values. For x<3, g*(h*x)=i+(-1)^j*k+(-1)^(j+l)*x=(gh)*x; both sides fix 3, and e fixes every point."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "表格對每個g與每個point恰有一值；相容式涵蓋全部6²·4=144 triples，不是只核每列像排列。",
            "en": "Every group-point pair has one value; the formula covers all 144 triples, not merely permutation rows."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S02",
          "S03"
        ],
        "before_tex": "g^{-1}(g\\cdot x)=(g^{-1}g)\\cdot x=e\\cdot x=x",
        "after_tex": "x\\mapsto g\\cdot x\\text{ has inverse }x\\mapsto g^{-1}\\cdot x",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "已證作用公理與群逆元提供每列的逆函數，所以每個g確實誘導排列。",
          "en": "The established action axioms and group inverse supply an inverse function for each row, so every g induces a permutation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "排列是作用公理的後果，不能只由每列看似排列反證公理。",
            "en": "Being permutations follows from the axioms; visually permutation-like rows do not prove those axioms."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S03"
        ],
        "before_tex": "s\\ne e,\\quad s\\cdot0=0",
        "after_tex": "\\text{the action is not free}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非單位s固定point0，直接否證free所需的無非單位固定元。",
          "en": "Nonidentity s fixes point 0, directly refuting freeness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "有效群作用不要求free；此反例不破壞S03。",
            "en": "A valid action need not be free, so this does not contradict S03."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "g\\cdot0\\in\\{0,1,2\\}\\quad\\forall g\\in G",
        "after_tex": "3\\notin G\\cdot0\\Rightarrow\\text{not transitive}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0-column完整只有0、1、2，無g能把0送到3，故非transitive。",
          "en": "The complete 0-column contains only 0,1,2; no g sends 0 to 3, so the action is not transitive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "3本身屬point載體且被全部g固定，不能從載體中刪去來改結論。",
            "en": "3 is part of the point carrier and is fixed by all g; it cannot be removed to alter the conclusion."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S03",
          "S04"
        ],
        "before_tex": "K=\\{e,u\\},\\ u^2=e,\\quad Y=\\{0,1\\}",
        "after_tex": "\\begin{array}{c|cc}*&e&u\\\\\\hline e&e&u\\\\u&u&e\\end{array}\\quad\\begin{array}{c|cc}\\cdot&0&1\\\\\\hline e&0&1\\\\u&0&1\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "另列C₂完整群表與trivial作用表：所有元素作用恒等，所以作用相容，但u≠e也誘導恒等，非faithful。",
          "en": "A complete C2 table and trivial action are shown: every element acts identically, so compatibility holds, but nonidentity u also acts identically and the action is not faithful."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "此小模型群律由mod2加法保留；恆等作用的兩邊總是x。",
            "en": "The small group's laws descend from addition modulo 2, and both sides of trivial-action compatibility are x."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S03",
          "S04",
          "S05",
          "S06",
          "S07"
        ],
        "before_tex": "e\\cdot x=x,\\quad(gh)\\cdot x=g\\cdot(h\\cdot x)\\text{ for all indices}",
        "after_tex": "G\\curvearrowright X\\text{ is qualified; free/faithful/transitive require separate certificates}",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一完整模型已核單位與相容兩責任；額外性質不可自動加上。",
          "en": "Identity and compatibility are fully checked in the same model; extra properties cannot be added automatically."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S04",
            "S05",
            "S06",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "S05、S06及S07分別示範三個不同額外責任，並不混成作用定義。",
            "en": "S05,S06,S07 illustrate three distinct extra obligations rather than folding them into the action definition."
          }
        },
        "leaf_basis": "mko_definition"
      }
    ],
    "result": {
      "zh-Hant": "明示24格表為有效群作用，但不是free或transitive。",
      "en": "The explicit 24-entry table is a valid group action, but it is neither free nor transitive."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "每列是排列就必為作用。",
          "en": "Permutation rows always give an action."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "C₃作用在兩點可設e=id,r=swap,r²=id；r²r=e但id∘swap≠id，故相容失敗。",
          "en": "On two points assign C3 rows e=id,r=swap,r2=id; r2*r=e but id composed with swap is not id, so compatibility fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "有效作用必free。",
          "en": "Every valid action is free."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "本表s≠e且s·0=0。",
          "en": "Here s is not e yet fixes 0."
        }
      },
      {
        "claim": {
          "zh-Hant": "有效作用必transitive或faithful。",
          "en": "Every valid action is transitive or faithful."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S07的兩點trivial C₂作用既非transitive也非faithful。",
          "en": "The two-point trivial C2 action in S07 is neither transitive nor faithful."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設精確模3、模2整數算術、有限集合外延相等與群／作用的定義；不預設此運算已為群。",
          "en": "Assume exact arithmetic modulo 3 and 2, extensional finite-set equality, and definitions of groups/actions; this operation is not pre-assumed to be a group."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "以同一完整有限群Cayley table、完整point carrier及total action table，全稱核e·x=x與(gh)·x=g·(h·x)。每個g因此誘導排列；有效作用本身不推出faithful、free或transitive，有限抽樣pairs不足。",
          "en": "Use one complete finite Cayley table, complete point carrier and total action table. Check e*x=x and (g*h)*x=g*(h*x) for all relevant indices. Each group element then induces a permutation, but validity alone does not certify faithful, free or transitive action. Sampled pairs are insufficient."
        },
        "source": {
          "mko_id": "mko-finite-group-action",
          "sha256": "2e31b8a871f6fd32317885aa68f63e2380a6c3a7d97c031b152b653c6a879de0",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": []
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整Cayley與作用表及全稱公式核對，不由排列rows單獨判作用。",
      "en": "Complete Cayley/action tables and universal formulas; permutation rows alone do not qualify an action."
    },
    "md_symbols": [
      [
        "G",
        "六元素群載體",
        "six-element group carrier"
      ],
      [
        "r,s",
        "r³=e,s²=e,sr=r²s的labels",
        "labels with r^3=e,s^2=e,sr=r^2s"
      ],
      [
        "X",
        "四個points",
        "four points"
      ]
    ],
    "md_error_controls": [
      {
        "title": "同一群表 / Same group table",
        "zh": "全部積、逆元與cosets綁定這個完整Cayley表。",
        "en": "Bind every product, inverse and coset to this complete Cayley table."
      },
      {
        "title": "完整枚舉 / Complete enumeration",
        "zh": "任何全稱有限性質都需全載體或全稱公式，不用proper sample。",
        "en": "Every universal finite property needs the full carrier or a universal formula, not a proper sample."
      }
    ],
    "md_open_items": []
  },
  "revision": {
    "id": "ocme-new142-reader-adoption-20261021",
    "collection": "coverage-20261021",
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    "original_source_sha256": "66876f6517e8e0f296b1b8edfb9302c71f46b693da283ac405ce212032fd0684",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
    "issue_ids": [],
    "parent_manifest_sha256": "d29f4cae8229ebd24f60d496cb1d3cd4ad8c0a4dca9d44daab0c3245c544283a",
    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
  }
}