原題與已知條件
完整有限群、subgroup與全部coset partition及代表元責任。操作=quotient-normal-coset-partition;資料={"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":[0,2,4,6,8,10]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。
G001 — 完整有限群、subgroup與全部coset partition及代表元責任。操作=quotient-normal-coset-partition;資料={"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":[0,2,4,6,8,10]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。
逐步推導
S001 · implies
操作前
source roles = [quotient-normal-coset-partition]
操作後
derived = [[subgroup, normal, cosets, partition_certificate, coset_count, certificate]]
條件:無
本頁列完整來源欄序,不需JavaScript;LP三種mode雖同為6欄,角色仍由原mode選,不由答案形狀猜。
步驟識別與依賴
規則:source_ordered_fields_without_JS · 依賴:G001
S002 · implies
操作前
source roles = [[[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]], 0]
操作後
derived = [12, [[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], [8, 8, 8, 8], [9, 9, 9, 9], [10, 10, 10, 10], [11, 11, 11, 11]], 是]
條件:無
先核整個closed Cayley table與實際identity標籤,不假設0或套舊上限32。
步驟識別與依賴
規則:Quotient_complete_group_table_actual_identity · 依賴:G001, S001
S003 · implies
操作前
source roles = [12, 0]
操作後
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, 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, 是, 是, 是, 是], [是, 是, 是, 是, 是, 是, 是, 是, 是, 是, 是, 是]], 1728, [0, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]]
條件:無
每個(x,y)矩陣格核全部z的(xy)z=x(yz),共n³責任;每個inverse同時左右乘回同一identity。完整查核不靠抽樣。
步驟識別與依賴
規則:Quotient_ALL_group_associativity_triples_and_two_sided_inverses · 依賴:G001, S002
S004 · implies
操作前
source roles = [0, [0, 2, 4, 6, 8, 10]]
操作後
derived = [[[0, 0, 0, 是], [0, 2, 2, 是], [0, 4, 4, 是], [0, 6, 6, 是], [0, 8, 8, 是], [0, 10, 10, 是], [2, 0, 2, 是], [2, 2, 4, 是], [2, 4, 6, 是], [2, 6, 8, 是], [2, 8, 10, 是], [2, 10, 0, 是], [4, 0, 4, 是], [4, 2, 6, 是], [4, 4, 8, 是], [4, 6, 10, 是], [4, 8, 0, 是], [4, 10, 2, 是], [6, 0, 6, 是], [6, 2, 8, 是], [6, 4, 10, 是], [6, 6, 0, 是], [6, 8, 2, 是], [6, 10, 4, 是], [8, 0, 8, 是], [8, 2, 10, 是], [8, 4, 0, 是], [8, 6, 2, 是], [8, 8, 4, 是], [8, 10, 6, 是], [10, 0, 10, 是], [10, 2, 0, 是], [10, 4, 2, 是], [10, 6, 4, 是], [10, 8, 6, 是], [10, 10, 8, 是]], [[0, 0, 是], [2, 10, 是], [4, 8, 是], [6, 6, 是], [8, 4, 是], [10, 2, 是]], 是, [[0, [0, 2, 4, 6, 8, 10]], [1, [0, 2, 4, 6, 8, 10]], [2, [0, 2, 4, 6, 8, 10]], [3, [0, 2, 4, 6, 8, 10]], [4, [0, 2, 4, 6, 8, 10]], [5, [0, 2, 4, 6, 8, 10]], [6, [0, 2, 4, 6, 8, 10]], [7, [0, 2, 4, 6, 8, 10]], [8, [0, 2, 4, 6, 8, 10]], [9, [0, 2, 4, 6, 8, 10]], [10, [0, 2, 4, 6, 8, 10]], [11, [0, 2, 4, 6, 8, 10]]], 是]
條件:無
先含identity、全product與inverse closure成真subgroup;再對所有g核共軛H。positive branches須normal,boundary須真正nonnormal,不混同未成subgroup。
步驟識別與依賴
規則:Quotient_genuine_subgroup_before_actual_normality · 依賴:G001, S003
S005 · implies
操作前
source roles = [[0, 2, 4, 6, 8, 10], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]
操作後
derived = [[[0, 2, 4, 6, 8, 10], [1, 3, 5, 7, 9, 11]], [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], 是]
條件:無
每個元素g在gH中;若aH交bH,ah1=bh2給b^-1a∈H,從而兩完整coset相同。列全g後去重且依來源排序,元素是coset SETS,不是最小代表元本身。
步驟識別與依賴
規則:Quotient_ALL_LEFT_coset_SETS_complete_disjoint_partition · 依賴:G001, S004
S006 · implies
操作前
source roles = [0, 0, [0, 2, 4, 6, 8, 10], [0, 2, 4, 6, 8, 10]]
操作後
derived = [[[0, 0, 0, 0], [0, 2, 2, 0], [0, 4, 4, 0], [0, 6, 6, 0], [0, 8, 8, 0], [0, 10, 10, 0], [2, 0, 2, 0], [2, 2, 4, 0], [2, 4, 6, 0], [2, 6, 8, 0], [2, 8, 10, 0], [2, 10, 0, 0], [4, 0, 4, 0], [4, 2, 6, 0], [4, 4, 8, 0], [4, 6, 10, 0], [4, 8, 0, 0], [4, 10, 2, 0], [6, 0, 6, 0], [6, 2, 8, 0], [6, 4, 10, 0], [6, 6, 0, 0], [6, 8, 2, 0], [6, 10, 4, 0], [8, 0, 8, 0], [8, 2, 10, 0], [8, 4, 0, 0], [8, 6, 2, 0], [8, 8, 4, 0], [8, 10, 6, 0], [10, 0, 10, 0], [10, 2, 0, 0], [10, 4, 2, 0], [10, 6, 4, 0], [10, 8, 6, 0], [10, 10, 8, 0]], [0], 是]
條件:無
每組coset的每對代表元都要落在同一product coset,僅用首代表元造表還不足以證良定;保完整可能index集合。
步驟識別與依賴
規則:Quotient_EVERY_coset_pair_ALL_representative_products · 依賴:G001, S005
S007 · implies
操作前
source roles = [0, 1, [0, 2, 4, 6, 8, 10], [1, 3, 5, 7, 9, 11]]
操作後
derived = [[[0, 1, 1, 1], [0, 3, 3, 1], [0, 5, 5, 1], [0, 7, 7, 1], [0, 9, 9, 1], [0, 11, 11, 1], [2, 1, 3, 1], [2, 3, 5, 1], [2, 5, 7, 1], [2, 7, 9, 1], [2, 9, 11, 1], [2, 11, 1, 1], [4, 1, 5, 1], [4, 3, 7, 1], [4, 5, 9, 1], [4, 7, 11, 1], [4, 9, 1, 1], [4, 11, 3, 1], [6, 1, 7, 1], [6, 3, 9, 1], [6, 5, 11, 1], [6, 7, 1, 1], [6, 9, 3, 1], [6, 11, 5, 1], [8, 1, 9, 1], [8, 3, 11, 1], [8, 5, 1, 1], [8, 7, 3, 1], [8, 9, 5, 1], [8, 11, 7, 1], [10, 1, 11, 1], [10, 3, 1, 1], [10, 5, 3, 1], [10, 7, 5, 1], [10, 9, 7, 1], [10, 11, 9, 1]], [1], 是]
條件:無
每組coset的每對代表元都要落在同一product coset,僅用首代表元造表還不足以證良定;保完整可能index集合。
步驟識別與依賴
規則:Quotient_EVERY_coset_pair_ALL_representative_products · 依賴:G001, S006
S008 · implies
操作前
source roles = [1, 0, [1, 3, 5, 7, 9, 11], [0, 2, 4, 6, 8, 10]]
操作後
derived = [[[1, 0, 1, 1], [1, 2, 3, 1], [1, 4, 5, 1], [1, 6, 7, 1], [1, 8, 9, 1], [1, 10, 11, 1], [3, 0, 3, 1], [3, 2, 5, 1], [3, 4, 7, 1], [3, 6, 9, 1], [3, 8, 11, 1], [3, 10, 1, 1], [5, 0, 5, 1], [5, 2, 7, 1], [5, 4, 9, 1], [5, 6, 11, 1], [5, 8, 1, 1], [5, 10, 3, 1], [7, 0, 7, 1], [7, 2, 9, 1], [7, 4, 11, 1], [7, 6, 1, 1], [7, 8, 3, 1], [7, 10, 5, 1], [9, 0, 9, 1], [9, 2, 11, 1], [9, 4, 1, 1], [9, 6, 3, 1], [9, 8, 5, 1], [9, 10, 7, 1], [11, 0, 11, 1], [11, 2, 1, 1], [11, 4, 3, 1], [11, 6, 5, 1], [11, 8, 7, 1], [11, 10, 9, 1]], [1], 是]
條件:無
每組coset的每對代表元都要落在同一product coset,僅用首代表元造表還不足以證良定;保完整可能index集合。
步驟識別與依賴
規則:Quotient_EVERY_coset_pair_ALL_representative_products · 依賴:G001, S007
S009 · implies
操作前
source roles = [1, 1, [1, 3, 5, 7, 9, 11], [1, 3, 5, 7, 9, 11]]
操作後
derived = [[[1, 1, 2, 0], [1, 3, 4, 0], [1, 5, 6, 0], [1, 7, 8, 0], [1, 9, 10, 0], [1, 11, 0, 0], [3, 1, 4, 0], [3, 3, 6, 0], [3, 5, 8, 0], [3, 7, 10, 0], [3, 9, 0, 0], [3, 11, 2, 0], [5, 1, 6, 0], [5, 3, 8, 0], [5, 5, 10, 0], [5, 7, 0, 0], [5, 9, 2, 0], [5, 11, 4, 0], [7, 1, 8, 0], [7, 3, 10, 0], [7, 5, 0, 0], [7, 7, 2, 0], [7, 9, 4, 0], [7, 11, 6, 0], [9, 1, 10, 0], [9, 3, 0, 0], [9, 5, 2, 0], [9, 7, 4, 0], [9, 9, 6, 0], [9, 11, 8, 0], [11, 1, 0, 0], [11, 3, 2, 0], [11, 5, 4, 0], [11, 7, 6, 0], [11, 9, 8, 0], [11, 11, 10, 0]], [0], 是]
條件:無
每組coset的每對代表元都要落在同一product coset,僅用首代表元造表還不足以證良定;保完整可能index集合。
步驟識別與依賴
規則:Quotient_EVERY_coset_pair_ALL_representative_products · 依賴:G001, S008
S010 · implies
操作前
source roles = [[0, 2, 4, 6, 8, 10], 是, [[0, 2, 4, 6, 8, 10], [1, 3, 5, 7, 9, 11]]]
操作後
derived = [[[0, 1], [1, 0]], 0, [[0, 0], [1, 1]], 是, 否]
條件:無
若a換ah、b換bk,(ah)(bk)=ab(b^-1hb)k∈abH,normality才保代表元替換不變。associativity由G傳下,eH為identity、a^-1H為inverse。固定No只不自動提供commutativity證書,不說具體cyclic quotient非交換,也不引第一同構定理。
步驟識別與依賴
規則:Quotient_GENERAL_normality_representative_replacement_and_group_laws · 依賴:G001, S009