# Question 24977: A Trivial Kernel Is Equivalent to Injectivity, Not Yet the First Isomorphism Theorem [Return to catalogue page 125](https://ocme.evemisslab.com/en/questions/page/125/) [← Question derivation index](https://ocme.evemisslab.com/en/questions/) · [Original question batch](https://ocme.evemisslab.com/en/practice/batch-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-foundations/#q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0077) This teaching derivation has second-context AI mathematical and bilingual semantic review, limited to the model and assumptions stated in this question. It has no human review or new Lean proof and does not promote the original question or MKO to Canonical truth. R0048-24977 · q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0077 ## Original question and givens Use complete source Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], source identity=0, target table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], target identity=0, and total mapping=[0, 1, 2, 3]. Row x/column y gives x*y in the respective table. Both groups qualify first, then every pair must satisfy f(x*y)=f(x)*f(y). A total map alone is not a homomorphism. Give (kernel, injective, trivial_kernel, equivalence_holds, quotient_or_first_isomorphism_claim, boundary). - G001 — Use complete source Cayley table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], source identity=0, target table=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], target identity=0, and total mapping=[0, 1, 2, 3]. Row x/column y gives x*y in the respective table. Both groups qualify first, then every pair must satisfy f(x*y)=f(x)*f(y). A total map alone is not a homomorphism. Give (kernel, injective, trivial_kernel, equivalence_holds, quotient_or_first_isomorphism_claim, boundary). ## Definition and scope Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate. - [mko-finite-group-homomorphism-kernel](https://ocme.evemisslab.com/en/mko/mko-finite-group-homomorphism-kernel/) ## Step-by-step derivation S001 · impliesBefore source roles = [kernel-trivial-injective-boundary] After derived = [[kernel, injective, trivial_kernel, equivalence_holds, quotient_or_first_isomorphism_claim, boundary]] Conditions: none List the full source field order without JavaScript,keeping original data,Boolean flags,mathematical objects and nonclaims distinct. Step identifiers and dependenciesRule: source_ordered_fields_without_JS · Uses: G001 S002 · impliesBefore source roles = [Kernel ORIGINAL SOURCE group, 64, [[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], 0] After derived = [4, [[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3]], Yes] Conditions: none Validate the complete closed n-by-n table and actual identity label,not an assumed0.Mod44 allows32 and mod45 allows64;do not mix carriers or tables. Step identifiers and dependenciesRule: Group_complete_Cayley_signature_actual_identity · Uses: G001, S001 S003 · impliesBefore source roles = [Kernel ORIGINAL SOURCE group, 4, FOR EVERY x,y,z in0..n-1] After derived = [[[Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes]], 64, Yes] Conditions: none Entry(x,y) checks (xy)z=x(yz) for every z,covering all n^3 triples,not samples.The complete table is already displayed,so do not print millions of redundant lookups. Step identifiers and dependenciesRule: Group_ALL_associativity_triples_compact_complete_pair_matrix · Uses: G001, S002 S004 · impliesBefore source roles = [Kernel ORIGINAL SOURCE group, 0] After derived = [[0, 3, 2, 1], [[0, 0, 0, 0], [1, 3, 0, 0], [2, 2, 0, 0], [3, 1, 0, 0]], Yes] Conditions: none Every left and right inverse must return the same actual identity;associativity gives uniqueness.This completes the group premises. Step identifiers and dependenciesRule: Group_ALL_two_sided_inverses · Uses: G001, S003 S005 · impliesBefore source roles = [Kernel ORIGINAL TARGET group, 64, [[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], 0] After derived = [4, [[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3]], Yes] Conditions: none Validate the complete closed n-by-n table and actual identity label,not an assumed0.Mod44 allows32 and mod45 allows64;do not mix carriers or tables. Step identifiers and dependenciesRule: Group_complete_Cayley_signature_actual_identity · Uses: G001, S004 S006 · impliesBefore source roles = [Kernel ORIGINAL TARGET group, 4, FOR EVERY x,y,z in0..n-1] After derived = [[[Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes]], 64, Yes] Conditions: none Entry(x,y) checks (xy)z=x(yz) for every z,covering all n^3 triples,not samples.The complete table is already displayed,so do not print millions of redundant lookups. Step identifiers and dependenciesRule: Group_ALL_associativity_triples_compact_complete_pair_matrix · Uses: G001, S005 S007 · impliesBefore source roles = [Kernel ORIGINAL TARGET group, 0] After derived = [[0, 3, 2, 1], [[0, 0, 0, 0], [1, 3, 0, 0], [2, 2, 0, 0], [3, 1, 0, 0]], Yes] Conditions: none Every left and right inverse must return the same actual identity;associativity gives uniqueness.This completes the group premises. Step identifiers and dependenciesRule: Group_ALL_two_sided_inverses · Uses: G001, S006 S008 · impliesBefore source roles = [[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], [[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]], [0, 1, 2, 3]] After derived = [[[0, 0, 0, 0, Yes], [0, 1, 1, 1, Yes], [0, 2, 2, 2, Yes], [0, 3, 3, 3, Yes], [1, 0, 1, 1, Yes], [1, 1, 2, 2, Yes], [1, 2, 3, 3, Yes], [1, 3, 0, 0, Yes], [2, 0, 2, 2, Yes], [2, 1, 3, 3, Yes], [2, 2, 0, 0, Yes], [2, 3, 1, 1, Yes], [3, 0, 3, 3, Yes], [3, 1, 0, 0, Yes], [3, 2, 1, 1, Yes], [3, 3, 2, 2, Yes]], Yes, null] Conditions: none For these same source and target groups,check f(xy)=f(x)f(y) for every pair and retain the first failed row/null.A total map need not be a homomorphism.Failure blocks kernel theorems;an internal fallback is not the true identity preimage. Step identifiers and dependenciesRule: Kernel_complete_TOTAL_map_ALL_product_pairs_before_theorems · Uses: G001, S007 S009 · impliesBefore source roles = [[0, 1, 2, 3], 0, 0] After derived = [[0], 0, 0] Conditions: none The kernel is the complete preimage of the actual target identity.Cancel f(e)=f(e)^2 in the target group to obtain f(e)=e_target,so the true kernel is nonempty;do not assume identity0. Step identifiers and dependenciesRule: Kernel_actual_TARGET_identity_complete_preimage · Uses: G001, S008 S010 · impliesBefore source roles = [Kernel ACTUAL identity preimage, [0], 0] After derived = [Yes, [[0, 0, 0, Yes]], Yes, [[0, 0, Yes]], Yes, Yes] Conditions: none H is nonempty and unique.Check identity membership,all products and every inverse separately;their conjunction qualifies a subgroup.Size divisibility or finiteness alone is insufficient. Step identifiers and dependenciesRule: Subgroup_identity_ALL_products_ALL_inverses_separate · Uses: G001, S009 S011 · impliesBefore source roles = [Kernel ACTUAL identity preimage, 0, [0], 0] After derived = [[[0, 0, 0, 0]], [0], Yes] Conditions: none In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators. Step identifiers and dependenciesRule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S010 S012 · impliesBefore source roles = [Kernel ACTUAL identity preimage, 1, [0], 3] After derived = [[[0, 1, 3, 0]], [0], Yes] Conditions: none In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators. Step identifiers and dependenciesRule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S011 S013 · impliesBefore source roles = [Kernel ACTUAL identity preimage, 2, [0], 2] After derived = [[[0, 2, 2, 0]], [0], Yes] Conditions: none In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators. Step identifiers and dependenciesRule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S012 S014 · impliesBefore source roles = [Kernel ACTUAL identity preimage, 3, [0], 1] After derived = [[[0, 3, 1, 0]], [0], Yes] Conditions: none In the same table compute g*h then multiply by g inverse for every h.Check set equality for every g,not sampled elements or unproved generators. Step identifiers and dependenciesRule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S013 S015 · impliesBefore source roles = [[0], Yes, 0] After derived = [f(kl)=f(k)f(l)=e, f(k^-1)=f(k)^-1=e, f(gkg^-1)=f(g)e*f(g)^-1=e] Conditions: none Products,inverses and every conjugate map to target identity,giving the general closure/normality argument alongside this complete finite audit.No quotient/isomorphism is claimed. Step identifiers and dependenciesRule: Kernel_GENERAL_closure_normality_from_homomorphism · Uses: G001, S014 S016 · impliesBefore source roles = [0, [0], [0, 1, 2, 3]] After derived = [[[0, [0]], [1, [1]], [2, [2]], [3, [3]]], Yes, Yes, Yes, No] Conditions: none If f is injective,f(k)=f(e) forces k=e.If K={e},image equality gives a^-1b=e and a=b.The fixed No withholds quotient/first-isomorphism claims,not this iff or injectivity. Step identifiers and dependenciesRule: Kernel_trivial_actual_identity_iff_injective_NO_quotient_claim · Uses: G001, S015 ## Reverse checks - S016 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S015 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S014 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S013 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S012 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S011 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S010 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S009 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S008 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. - S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence. ## Result ([0], Yes, Yes, Yes, No, trivial_kernel_iff_injective_without_quotient_or_first_isomorphism_claim) The full substitutions and premise checks give the original answer; source identity,format and status are unchanged. ## This question review and source binding Review date: 2026-10-26 Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof. Source hashes and review receipts - input commit: 0f7be61e2c22317a4075561dc822c87d881a50db - original question SHA-256: e5a3f08ecd3feae56921b79170fb1e2c65811e6f4a991f988ce71ad906b2474a - candidate record SHA-256: 6910251a89de3f00b696854d759932788c1abbe2964bdaab9695ed2ac4b0dfd3 - semantic review receipt SHA-256: 049a9e07e69fead88d93b34f0937b9470a06d71c1a8d097058fc83554529bdd0 - question-first witness SHA-256: 0c71ebb418f3de347d232d159fc25e93036f4ab9166bcfb5ee5b4be86d5c55b9 - Lean: NOT_RUN · human review: NOT_PERFORMED [Same-source full JSON](https://ocme.evemisslab.com/data/question-derivations/questions/q-20261015-advance-20261114-r01-finite-group-homomorphism-kernel-0077.json)