{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-first-isomorphism-theorem-finite-groups-01",
  "mko_id": "mko-first-isomorphism-theorem-finite-groups",
  "title": {
    "zh-Hant": "非滿射同態的正確image同構",
    "en": "The correct image isomorphism for a nonsurjection"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-first-isomorphism-theorem-finite-groups-01",
    "source": {
      "mko_id": "mko-first-isomorphism-theorem-finite-groups",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "5586ce5ed3d47467ffd6c485dfcb495844143a17783d770ce2572420bfa28065",
      "english_sha256": "f9e622d65b015b588d1e29eb89655426b79abaf193e2f4763bec7b9e8c6aeadb"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "f:C₄→C₄,f(a)=2a mod4給C₄/{0,2}≅{0,2}，不是整個codomain C₄。",
      "en": "For f:C4 to C4,f(a)=2a mod 4, C4/{0,2} is isomorphic to {0,2}, not the full codomain C4."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "G=Q=C₄，兩者皆採下列完整模4加法表；f(0)=0,f(1)=2,f(2)=0,f(3)=2。",
          "en": "G=Q=C4, both using the full addition-modulo-4 table below; f(0)=0,f(1)=2,f(2)=0,f(3)=2."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "G=Q=C_4",
        "after_tex": "\\begin{array}{c|cccc}+&0&1&2&3\\\\\\hline 0&0&1&2&3\\\\1&1&2&3&0\\\\2&2&3&0&1\\\\3&3&0&1&2\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "來源與目標是同一完整四元素群模型，各有此16格表；結合律由整數下降，identity0、逆元0,3,2,1全部核。",
          "en": "Source and target are the same complete four-element group model, each with this 16-entry table; associativity descends from integers, identity is 0 and inverse list 0,3,2,1 is complete."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "相同模型不等於f恒等，f仍需獨立檢查。",
            "en": "Identical group models do not make f the identity; the map still needs checking."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "f:G\\to Q",
        "after_tex": "(f(0),f(1),f(2),f(3))=(0,2,0,2)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整total map的四個值都屬Q。",
          "en": "All four values of the complete total map lie in Q."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "Q含1、3但f不一定命中，保留image與codomain界線。",
            "en": "Q contains 1,3, which f need not hit; retain the image/codomain distinction."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01",
          "S02"
        ],
        "before_tex": "f((a+b)\\bmod4)\\equiv2(a+b)\\pmod4",
        "after_tex": "\\begin{array}{c|cccc}f(a+b)&0&1&2&3\\\\\\hline 0&0&2&0&2\\\\1&2&0&2&0\\\\2&0&2&0&2\\\\3&2&0&2&0\\\\\\end{array}\\quad=f(a)+f(b)\\ (\\bmod4)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全部16pairs由分配律與取餘保持同值，完整驗同態。",
          "en": "Distributivity and reduction give equality for all 16 pairs, fully verifying the homomorphism."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "不能只看cardinality或f(0)=0便稱同態。",
            "en": "Neither cardinality nor f(0)=0 alone qualifies a homomorphism."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S01",
          "S02",
          "S03"
        ],
        "before_tex": "K=f^{-1}(0),\\quad I=f(G)",
        "after_tex": "K=\\{0,2\\},\\quad I=\\{0,2\\};\\quad0+0=0,0+2=2,2+0=2,2+2=0\\ (\\bmod4)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整preimage與image皆明列；四乘積與逆元0,2給子群，g+k−g=k對所有g證K正規。",
          "en": "The full preimage and image are explicit; four products and inverses 0,2 qualify subgroups, and g+k-g=k for all g proves K normal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "K位於來源，I位於目標，即使數字集合相同仍有不同角色。",
            "en": "K belongs to the source and I to the target, with distinct roles despite identical numeral sets."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S04"
        ],
        "before_tex": "A=0+K=2+K,\\quad B=1+K=3+K",
        "after_tex": "A=\\{0,2\\},\\quad B=\\{1,3\\},\\quad A\\sqcup B=G",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全部四cosets去重成完整兩類partition。",
          "en": "All four cosets deduplicate to a complete two-class partition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "商元素是sets A、B，非某一唯一代表0、1。",
            "en": "Quotient elements are sets A,B rather than unique representatives 0,1."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S01",
          "S04",
          "S05"
        ],
        "before_tex": "a,b=0,1,2,3",
        "after_tex": "\\begin{array}{c|cccc}(a+b)+K&0&1&2&3\\\\\\hline 0&A&B&A&B\\\\1&B&A&B&A\\\\2&A&B&A&B\\\\3&B&A&B&A\\\\\\end{array}\\qquad\\begin{array}{c|cc}+&A&B\\\\\\hline A&A&B\\\\B&B&A\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整16格代表pair類指標在四個2×2區塊各常值，證商加法良定並得到全部四格商表。",
          "en": "All 16 representative-pair class indices are constant on each 2-by-2 block, proving well-defined quotient addition and yielding its full four-entry table."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "由父結合律下降，加上identity A、inverse A與B各自身，商群資格完整。",
            "en": "Associativity descends, identity is A, and A,B are each self-inverse, fully qualifying the quotient."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S02",
          "S05",
          "S06"
        ],
        "before_tex": "\\bar f(a+K)=f(a)",
        "after_tex": "f(0)=f(2)=0\\Rightarrow\\bar f(A)=0;\\quad f(1)=f(3)=2\\Rightarrow\\bar f(B)=2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩cosets全部代表值一致，先證induced map代表獨立。",
          "en": "Every representative within each coset has the same value, first proving representative independence of the induced map."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S05",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "不能只選0、1然後忽略2、3，四個代表皆已核。",
            "en": "One cannot check only 0,1 and ignore 2,3; all four representatives are checked."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S04",
          "S06",
          "S07"
        ],
        "before_tex": "A+A=A,A+B=B,B+A=B,B+B=A",
        "after_tex": "0+0=0,0+2=2,2+0=2,2+2=0\\ (\\bmod4)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個商product的像與I中product逐對相符，所以bar f是homomorphism。",
          "en": "The images of all four quotient products match products in I, so bar f is a homomorphism."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "S06",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "兩邊使用不同載體但對應相同四格，不能只以大小2推同構。",
            "en": "The carriers differ but all four products correspond; size two alone is not an isomorphism proof."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "S04",
          "S07"
        ],
        "before_tex": "\\bar f(A)=0\\ne2=\\bar f(B)",
        "after_tex": "\\bar f\\text{ injective and onto }I=\\{0,2\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個來源像不同證injective，全部I元素皆命中證surjective-to-image。",
          "en": "Distinct images prove injectivity, and both I elements are hit, proving surjectivity onto the image."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "未命中codomain的1、3，因此不能寫surjective onto Q。",
            "en": "Codomain values 1,3 are missed, so onto Q cannot be claimed."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "S06",
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "\\bar f:G/K\\to I\\text{ well-defined bijective homomorphism}",
        "after_tex": "G/K\\cong I\\subsetneq Q",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整良定、保群運算與雙射三責任成立，認證指定模型第一同構結論。",
          "en": "Well-definedness, operation preservation and bijectivity establish the first-isomorphism conclusion for this model."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06",
            "S07",
            "S08",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "此finite certificate不取代一般抽象證明；也不把proper image偷換整個Q。",
            "en": "This finite certificate does not replace the general abstract proof or substitute Q for the proper image."
          }
        },
        "leaf_basis": "mko_definition"
      }
    ],
    "result": {
      "zh-Hant": "f:C₄→C₄,f(a)=2a mod4給C₄/{0,2}≅{0,2}，不是整個codomain C₄。",
      "en": "For f:C4 to C4,f(a)=2a mod 4, C4/{0,2} is isomorphic to {0,2}, not the full codomain C4."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "G/K≅Q因f是同態。",
          "en": "G/K is isomorphic to Q because f is a homomorphism."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "G/K有2元素而Q有4，f未滿射。",
          "en": "G/K has 2 elements but Q has 4, and f is not onto."
        }
      },
      {
        "claim": {
          "zh-Hant": "只看|G/K|=|I|便證同構。",
          "en": "Equal sizes |G/K|=|I| prove isomorphism."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "需實際bar f並驗全部product與代表獨立；數目不提供運算證書。",
          "en": "An actual induced map must respect all products and representatives; sizes give no operation certificate."
        }
      },
      {
        "claim": {
          "zh-Hant": "bar f(B)可依代表1或3取不同值。",
          "en": "bar f(B) may differ for representatives 1 and 3."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(1)=f(3)=2，良定性要求同值。",
          "en": "f(1)=f(3)=2, and well-definedness requires agreement."
        }
      }
    ],
    "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4算術與群、同態、商coset、同構的定義；加法記號。",
          "en": "Assume exact modulo-4 arithmetic and definitions of groups, homomorphisms, quotient cosets and isomorphism, in additive notation."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "來源與目標均須以完整Cayley table認證，f為complete total homomorphism；kernel、image與quotient cosets先完整建立。誘導映射bar f(g ker f)=f(g)必須對每個coset的所有representatives給同值，並完整驗乘法保存、injective與surjective-to-image。若f不surjective，結論仍僅為G/ker f≅im f，不得把image換成整個codomain。有限枚舉只證指定模型；非同態不得套用定理。",
          "en": "Qualify both complete Cayley tables and the complete total homomorphism f first. Construct the full kernel, image and kernel-coset quotient. The induced map bar_f(g*kernel)=f(g) must agree on every representative of every coset, preserve every product and be injective and surjective onto the image. Equal cardinalities alone do not prove an isomorphism. If f is not surjective, do not replace the image by the entire codomain. Finite enumeration certifies the specified model, not the general abstract theorem; a nonhomomorphism cannot invoke the theorem."
        },
        "source": {
          "mko_id": "mko-first-isomorphism-theorem-finite-groups",
          "sha256": "5586ce5ed3d47467ffd6c485dfcb495844143a17783d770ce2572420bfa28065",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "cardinality_quotient",
        "lhs": "4/2",
        "rhs": "2",
        "step": "S10"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整finite homomorphism、kernel、image、quotient及induced map，全代表核對。",
      "en": "Complete finite homomorphism, kernel, image, quotient and induced map with every representative checked."
    },
    "md_symbols": [
      [
        "K,I",
        "來源kernel與目標image",
        "source kernel and target image"
      ],
      [
        "\\bar f",
        "由coset到image的映射",
        "map from cosets to the image"
      ]
    ],
    "md_error_controls": [
      {
        "title": "image界線 / Image boundary",
        "zh": "結論只到I={0,2}，f不滿射時不可換Q。",
        "en": "The conclusion targets I={0,2}; without surjectivity it cannot be replaced by Q."
      },
      {
        "title": "代表先核 / Representatives first",
        "zh": "先查A的0、2及B的1、3，再談induced map。",
        "en": "Check A's representatives 0,2 and B's 1,3 before the induced map."
      }
    ],
    "md_open_items": []
  },
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    "bilingual_sync": "complete_source_bound",
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    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
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}