第 25712 題：有限群第一同構定理

原題與已知條件

完整有限群、total homomorphism、kernel、image與quotient cosets的第一同構責任。操作=iso1-complete-qualification；資料={"source_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]],"source_identity":0,"target_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]],"target_identity":0,"mapping":[0,5,4,3,2,1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 完整有限群、total homomorphism、kernel、image與quotient cosets的第一同構責任。操作=iso1-complete-qualification；資料={"source_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]],"source_identity":0,"target_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]],"target_identity":0,"mapping":[0,5,4,3,2,1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

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。

• mko-first-isomorphism-theorem-finite-groups

逐步推導

source roles = [iso1-complete-qualification]

derived = [[source_order, target_order, mapping, homomorphism, first_failed_pair_or_null, kernel, image, normal_kernel, image_subgroup, certificate]]

source roles = [Iso1 ORIGINAL SOURCE, [[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]], 0]

derived = [6, [[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5]], 是]

source roles = [Iso1 ORIGINAL SOURCE, 6, 0]

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]], 216, [0, 5, 4, 3, 2, 1]]

source roles = [Iso1 ORIGINAL TARGET, [[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]], 0]

derived = [6, [[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5]], 是]

source roles = [Iso1 ORIGINAL TARGET, 6, 0]

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]], 216, [0, 5, 4, 3, 2, 1]]

source roles = [Iso1 ACTUAL kernel, [0], 0]

derived = [是, [[0, 0, 0, 是]], 是, [[0, 0, 是]], 是, 是, [[0, [0], 是], [1, [0], 是], [2, [0], 是], [3, [0], 是], [4, [0], 是], [5, [0], 是]], 是]

source roles = [Iso1 ACTUAL image, [0, 1, 2, 3, 4, 5], 0]

derived = [是, [[0, 0, 0, 是], [0, 1, 1, 是], [0, 2, 2, 是], [0, 3, 3, 是], [0, 4, 4, 是], [0, 5, 5, 是], [1, 0, 1, 是], [1, 1, 2, 是], [1, 2, 3, 是], [1, 3, 4, 是], [1, 4, 5, 是], [1, 5, 0, 是], [2, 0, 2, 是], [2, 1, 3, 是], [2, 2, 4, 是], [2, 3, 5, 是], [2, 4, 0, 是], [2, 5, 1, 是], [3, 0, 3, 是], [3, 1, 4, 是], [3, 2, 5, 是], [3, 3, 0, 是], [3, 4, 1, 是], [3, 5, 2, 是], [4, 0, 4, 是], [4, 1, 5, 是], [4, 2, 0, 是], [4, 3, 1, 是], [4, 4, 2, 是], [4, 5, 3, 是], [5, 0, 5, 是], [5, 1, 0, 是], [5, 2, 1, 是], [5, 3, 2, 是], [5, 4, 3, 是], [5, 5, 4, 是]], 是, [[0, 0, 是], [1, 5, 是], [2, 4, 是], [3, 3, 是], [4, 2, 是], [5, 1, 是]], 是, 是, [[0, [0, 1, 2, 3, 4, 5], 是], [1, [0, 1, 2, 3, 4, 5], 是], [2, [0, 1, 2, 3, 4, 5], 是], [3, [0, 1, 2, 3, 4, 5], 是], [4, [0, 1, 2, 3, 4, 5], 是], [5, [0, 1, 2, 3, 4, 5], 是]], 是]

source roles = [[0, 5, 4, 3, 2, 1], 0, 0]

derived = [[[0, 0, 0, 0, 是], [0, 1, 5, 5, 是], [0, 2, 4, 4, 是], [0, 3, 3, 3, 是], [0, 4, 2, 2, 是], [0, 5, 1, 1, 是], [1, 0, 5, 5, 是], [1, 1, 4, 4, 是], [1, 2, 3, 3, 是], [1, 3, 2, 2, 是], [1, 4, 1, 1, 是], [1, 5, 0, 0, 是], [2, 0, 4, 4, 是], [2, 1, 3, 3, 是], [2, 2, 2, 2, 是], [2, 3, 1, 1, 是], [2, 4, 0, 0, 是], [2, 5, 5, 5, 是], [3, 0, 3, 3, 是], [3, 1, 2, 2, 是], [3, 2, 1, 1, 是], [3, 3, 0, 0, 是], [3, 4, 5, 5, 是], [3, 5, 4, 4, 是], [4, 0, 2, 2, 是], [4, 1, 1, 1, 是], [4, 2, 0, 0, 是], [4, 3, 5, 5, 是], [4, 4, 4, 4, 是], [4, 5, 3, 3, 是], [5, 0, 1, 1, 是], [5, 1, 0, 0, 是], [5, 2, 5, 5, 是], [5, 3, 4, 4, 是], [5, 4, 3, 3, 是], [5, 5, 2, 2, 是]], 是, null, [0], [0, 1, 2, 3, 4, 5], 是, 是]

逐步反向核對

• S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S007 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S005 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(6, 6, [0, 5, 4, 3, 2, 1], 是, null, [0], [0, 1, 2, 3, 4, 5], 是, 是, homomorphism_kernel_image_and_quotient_qualifications_precede_the_theorem)

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。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。