{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-subgroup-homomorphism"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-subgroup-homomorphism-01","title":{"zh-Hant":"子群與群同態：在 ℤ/4 與 ℤ/2 的完整表上核對","en":"Subgroups and homomorphisms: auditing on the complete tables of ℤ/4 and ℤ/2"},"summary":{"zh-Hant":"G={0,1,2,3} 以 (a+b) mod 4 為運算，K={0,1} 以 (a+b) mod 2 為運算，兩者都給出完整 Cayley 表。（一）證明 H={0,2} 是 G 的子群，H'={0,1} 不是；（二）證明映射表 f(0)=0、f(1)=1、f(2)=0、f(3)=1 是同態 G→K（涵蓋全部 16 個輸入對）；（三）求 kernel 與 image，說明 f 滿射但不單射，因此不是同構；（四）說明常數映射 h≡1 不是同態。","en":"G={0,1,2,3} with operation (a+b) mod 4 and K={0,1} with operation (a+b) mod 2 are given by complete Cayley tables. (1) Prove that H={0,2} is a subgroup of G and H'={0,1} is not; (2) prove that the mapping table f(0)=0, f(1)=1, f(2)=0, f(3)=1 is a homomorphism G→K, covering all 16 input pairs; (3) find the kernel and the image, and show that f is surjective but not injective, hence not an isomorphism; (4) show that the constant map h≡1 is not a homomorphism."},"mko_ids":["mko-subgroup-homomorphism"]}],"title":"概念推導導讀"}