Question 24600: Subgroup Closure Precedes a Normal-Subgroup Certificate

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Original question and givens
Definition and scope
Step-by-step derivation
Reverse checks
Result
This question review and source binding

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R0048-24600 · q-20261015-advance-20261113-r01-finite-normal-subgroup-0025

Original question and givens

Use complete finite Cayley table=[[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, 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]], identity=4 and candidate subset H=[4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).

G001 — Use complete finite Cayley table=[[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, 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]], identity=4 and candidate subset H=[4, 5]. Row g/column h gives g*h. Validate all group axioms, then identity/product/inverse closure of the whole subset before treating it as a subgroup. Normality has a separate all-g obligation. Give (subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate).

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-normal-subgroup

Step-by-step derivation

S001 · implies

Before

source roles = [normal-subgroup-qualification]

After

derived = [[subset, contains_identity, product_closed, inverse_closed, subgroup, normal_subgroup, certificate]]

Conditions: none

List the full source field order without JavaScript,keeping original data,Boolean flags,mathematical objects and nonclaims distinct.

Step identifiers and dependencies

Rule: source_ordered_fields_without_JS · Uses: G001

S002 · implies

Before

source roles = [Normality ORIGINAL group, 32, [[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, 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]], 4]

After

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]], 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 dependencies

Rule: Group_complete_Cayley_signature_actual_identity · Uses: G001, S001

S003 · implies

Before

source roles = [Normality ORIGINAL group, 12, 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, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes], [Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes]], 1728, 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 dependencies

Rule: Group_ALL_associativity_triples_compact_complete_pair_matrix · Uses: G001, S002

S004 · implies

Before

source roles = [Normality ORIGINAL group, 4]

After

derived = [[8, 7, 6, 5, 4, 3, 2, 1, 0, 11, 10, 9], [[0, 8, 4, 4], [1, 7, 4, 4], [2, 6, 4, 4], [3, 5, 4, 4], [4, 4, 4, 4], [5, 3, 4, 4], [6, 2, 4, 4], [7, 1, 4, 4], [8, 0, 4, 4], [9, 11, 4, 4], [10, 10, 4, 4], [11, 9, 4, 4]], 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 dependencies

Rule: Group_ALL_two_sided_inverses · Uses: G001, S003

S005 · implies

Before

source roles = [Normality ORIGINAL H, [4, 5], 4]

After

derived = [Yes, [[4, 4, 4, Yes], [4, 5, 5, Yes], [5, 4, 5, Yes], [5, 5, 6, No]], No, [[4, 4, Yes], [5, 3, No]], No, No]

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 dependencies

Rule: Subgroup_identity_ALL_products_ALL_inverses_separate · Uses: G001, S004

S006 · implies

Before

source roles = [Normality ORIGINAL H, 0, [4, 5], 8]

After

derived = [[[4, 0, 8, 4], [5, 1, 8, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S005

S007 · implies

Before

source roles = [Normality ORIGINAL H, 1, [4, 5], 7]

After

derived = [[[4, 1, 7, 4], [5, 2, 7, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S006

S008 · implies

Before

source roles = [Normality ORIGINAL H, 2, [4, 5], 6]

After

derived = [[[4, 2, 6, 4], [5, 3, 6, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S007

S009 · implies

Before

source roles = [Normality ORIGINAL H, 3, [4, 5], 5]

After

derived = [[[4, 3, 5, 4], [5, 4, 5, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S008

S010 · implies

Before

source roles = [Normality ORIGINAL H, 4, [4, 5], 4]

After

derived = [[[4, 4, 4, 4], [5, 5, 4, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S009

S011 · implies

Before

source roles = [Normality ORIGINAL H, 5, [4, 5], 3]

After

derived = [[[4, 5, 3, 4], [5, 6, 3, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S010

S012 · implies

Before

source roles = [Normality ORIGINAL H, 6, [4, 5], 2]

After

derived = [[[4, 6, 2, 4], [5, 7, 2, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S011

S013 · implies

Before

source roles = [Normality ORIGINAL H, 7, [4, 5], 1]

After

derived = [[[4, 7, 1, 4], [5, 8, 1, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S012

S014 · implies

Before

source roles = [Normality ORIGINAL H, 8, [4, 5], 0]

After

derived = [[[4, 8, 0, 4], [5, 9, 0, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S013

S015 · implies

Before

source roles = [Normality ORIGINAL H, 9, [4, 5], 11]

After

derived = [[[4, 9, 11, 4], [5, 10, 11, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S014

S016 · implies

Before

source roles = [Normality ORIGINAL H, 10, [4, 5], 10]

After

derived = [[[4, 10, 10, 4], [5, 11, 10, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S015

S017 · implies

Before

source roles = [Normality ORIGINAL H, 11, [4, 5], 9]

After

derived = [[[4, 11, 9, 4], [5, 0, 9, 5]], [4, 5], 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 dependencies

Rule: Subgroup_EVERY_g_entire_conjugate_set · Uses: G001, S016

Reverse checks

S017 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
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

([4, 5], Yes, No, No, No, No, subgroup_obligations_precede_normality)

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: f35f8ee22f2439738679fddb35f496653d010cb75dc1351e27dfcbafe57453c5
candidate record SHA-256: 0d57dd2b31d7b5d2d77e5b2e125de7d02f82487cd7321b143b0302e4b2bd8f46
semantic review receipt SHA-256: 049a9e07e69fead88d93b34f0937b9470a06d71c1a8d097058fc83554529bdd0
question-first witness SHA-256: 0c71ebb418f3de347d232d159fc25e93036f4ab9166bcfb5ee5b4be86d5c55b9
Lean: NOT_RUN · human review: NOT_PERFORMED

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