{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0001","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e6113836d2b8cc08300006fc995d0d53be52d476e8d233d2ace05893485d590","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261001,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"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]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley 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]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0002","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c1657bf683eac7318d5e5cdca09709297c28a1bfbceec8dccf8bb5540a07463","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261002,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0003","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d89e1beb996e9b3d4b62c0631eca5ba19fd3d522d6d9b1f2202a58f4d818e26","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261003,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0004","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df9957186f6f162c8523c4c0a8187117a671994507b7debc79ffd646b21131ec","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261004,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0005","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7e6332884c1f7e2266fdea7f10f63f3002c733fb6e373d1fcec0d091b49d350","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261005,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0006","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1d51e5a359c125afe498ef06fb9b4931203238795e5a814e29f2b05c88dcab9","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261006,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0007","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43e456c02fa6e84651abfa9c2a20296fc829b0e8a57a1649537c0ffe096e5bf2","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261007,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0008","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"844478419e89187c2ab14023b4f6b495e194c8179fcbfd3f9f7a4eaf4904923d","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261008,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0009","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1859d755ca4726fedff869bfbdc8930add86a6af9c77cc8db1e600ebe9c12c69","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261009,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0010","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6690d29e8486f02fdd420c468bb75c581d020132b83cfce15828681bb824ce53","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261010,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0011","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a018427ff345045aa809202dde72cbcbc88dbe01bbc16b26cff4ea77a9a2a186","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261011,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0012","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9ca28a0db6f44abb7eecb0dcd0af108c1d03db85c80f24a2603b2c3d56fbb54","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261012,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0013","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ced005ed4fbe0e61efe040b9f914a58267ffc969876a22412e4ed8648f715e0","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261013,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0014","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0577bc14c9091222eaaf1e2879b339257f007ece6c0f18286b180ca54b931192","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261014,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0015","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4eccf491dd72b2c8cc5d74e102c5c3803ea2af73ca68258a90e878bf54fed6a9","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261015,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0016","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4905d9ecc8669f83c272a08612c45c3f642c5d48778f93f30f07a8a0ca63c6ed","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261016,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0017","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5640685ae1594645f3a8093718804fa6f028fc72a5bbb456bde02fad0ce5cd41","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261017,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0018","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"024edda7600385c4d28df0470398f6dab40c8a325e44392e931bf30fcfad28de","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261018,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0019","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1ff48cbc49119a9a3a471c1f7cdae2b824c1129e8cd2882d640bb04ca864261","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261019,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0020","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b9cd469cc5227d6417e0c40ee428837b58cbeb43669d9f38734d1f3bedce719","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261020,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0021","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdb026ba76555f180872266d42d3b4af09168864d54af84c1cdf64bd49bed75b","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261021,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0022","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8529dce1a8aab69063b4577b2a767b4d002e6dcdb9498842f99ccef2f0511cbe","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261022,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0023","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f2672dee6c1dc5a60d6e5aebc70e261e2ef8d9ab389950bdbcb803c7681c85b","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261023,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0024","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b52108381a40d55984e5c396aa640f9add8c89d9bec968e19bbcc223ee61e06a","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261024,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0025","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04d3950ad50e0a7ae6711146d885142faed56ab9b72ab572913cae0217ec8414","family_id":"qf-mod26-abelian-commutativity-audit","seed":202610261025,"semantic_template":"qf-mod26-abelian-commutativity-audit/{\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]]}"},"fields":{"title":"Audit Every Pair of a Qualified Finite Group","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. Give (n,identity,all noncommuting [a,b,a*b,b*a] records with a<b in label order,abelian?).","answer":{"kind":"expression","value":"(6, 0, [], Yes)","explanation":"Compare both operation orders for every distinct unordered label pair. The diagonal commutes automatically, but does not establish the off-diagonal obligations. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, [], Yes)."},"skills":["Audit Every Pair of a Qualified Finite Group"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0026","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"988539870a3d9f6b892c81b286afe7ed09c39a4ba19f666939d28021a0aa7dfe","family_id":"qf-mod26-abelian-element-order","seed":202610261026,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":1,\"n\":6,\"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]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley 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]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(1, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0027","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a03798e138f5a7146bdf2d4ee4ac314bd11711ae1c03146fb7614e5e2e059535","family_id":"qf-mod26-abelian-element-order","seed":202610261027,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(0, 0, 1, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 1, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0028","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e38e725005686d6c5e2acca5a3f8dbefd4f9dc63860a50facaf74934e5963d00","family_id":"qf-mod26-abelian-element-order","seed":202610261028,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(5, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0029","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a16b7f087eb786606f19784ce5498a8dd7711adb54679aa9429abc755cce6de","family_id":"qf-mod26-abelian-element-order","seed":202610261029,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(4, 0, 2, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 2, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0030","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3294bacfa7ee49015d07326f6998b6df0b21388adde0df249aad704d9fb10090","family_id":"qf-mod26-abelian-element-order","seed":202610261030,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(3, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0031","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"353108e888ff6a6c6ac377a4bb23e0e0aa8276d53aa991d961f9eb158d352ec4","family_id":"qf-mod26-abelian-element-order","seed":202610261031,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(2, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0032","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c86767eef14f8fcd64190b0776e2e355ff29c3c4018090c0d696488e8b6208b","family_id":"qf-mod26-abelian-element-order","seed":202610261032,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(1, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0033","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1c4d4306f63f16824c3696b9200e1efed912e681dceb06bca370f97d09ed5d1","family_id":"qf-mod26-abelian-element-order","seed":202610261033,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(0, 0, 1, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 1, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0034","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f202b06b881787eccc0a0a2484411600a3ffec3b22f279be3b61f71acb29999","family_id":"qf-mod26-abelian-element-order","seed":202610261034,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(5, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0035","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd3d40567f7e8c2a2f8dba12fc71e194d2e4cbf71968b4e4ffc926cc5a6de290","family_id":"qf-mod26-abelian-element-order","seed":202610261035,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(4, 0, 2, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 2, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0036","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5f578d405be92b8e2831d7905d33a8b181da3fb99fc1bd9220ad57d893a9888","family_id":"qf-mod26-abelian-element-order","seed":202610261036,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(3, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0037","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a836eafe42d906bba39dac73c0c87cbaceafe9e9946af40459647ea70c309a5","family_id":"qf-mod26-abelian-element-order","seed":202610261037,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(2, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0038","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed3d2ba6932f12551a259ad70e6bee16f879c2319a1170913b8ac5c1f3d1d4bc","family_id":"qf-mod26-abelian-element-order","seed":202610261038,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(1, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0039","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42621c79008c65adaa06d00efb6a926e6acff84c81e9c88c635aed96cf5165f3","family_id":"qf-mod26-abelian-element-order","seed":202610261039,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(0, 0, 1, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 1, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0040","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"901273f975a5a2ebda1c028b199289300746814ba4dcc1914c50a63181d57578","family_id":"qf-mod26-abelian-element-order","seed":202610261040,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(5, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0041","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0eab6fdc4822d6cac480baa51b50b073cd6546c4ab47a28bb145588aa69eeddf","family_id":"qf-mod26-abelian-element-order","seed":202610261041,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(4, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0042","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c0ce0aa7ce3e9fdd9cc599014015e684003b8bc388c3d4e03613a7b563fe350","family_id":"qf-mod26-abelian-element-order","seed":202610261042,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(3, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0043","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d4c8f32bfa5b2840676dfadf7354a2a0ce6a5bdce060f291f6da516ef33cbc3","family_id":"qf-mod26-abelian-element-order","seed":202610261043,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(2, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0044","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3940d9368bdabb976a05568fea9ed59218d3bc3fb27bead1f2499b5f5fe1ac7","family_id":"qf-mod26-abelian-element-order","seed":202610261044,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(1, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0045","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5862a202e45b068959a7aa7fb3a2378a9f03343ef3731f437c2371aca7dbe70e","family_id":"qf-mod26-abelian-element-order","seed":202610261045,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(0, 0, 1, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 1, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0046","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"216ccf9e9dc9d00ad02de6595ee34adf809d732955a845836783008faf3bc377","family_id":"qf-mod26-abelian-element-order","seed":202610261046,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(5, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0047","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6bd12b1d21919cd14f5dcdbcba336abcaf905e37129bdce7f94c07c02ae7766","family_id":"qf-mod26-abelian-element-order","seed":202610261047,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(4, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0048","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f46a7f52d3d60c66b195dfa81dec5d6243c656fadc722eb7d03a0cc93cc924b6","family_id":"qf-mod26-abelian-element-order","seed":202610261048,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(3, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0049","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7cc85f93bd336cd6f6e9adea82311a3b87ed4b730a7d97506130fbbaa8659bc3","family_id":"qf-mod26-abelian-element-order","seed":202610261049,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(2, 0, 6, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 6, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0050","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5919855cc95753c62e0704a5940f3ff4eb361b4d66daca5fe964cc648f053ff1","family_id":"qf-mod26-abelian-element-order","seed":202610261050,"semantic_template":"qf-mod26-abelian-element-order/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]]}"},"fields":{"title":"Element Order Starts from the Actual Identity","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,least positive order,returns to identity?).","answer":{"kind":"expression","value":"(1, 0, 3, Yes)","explanation":"Iterate the given operation from its qualified identity until the first positive return; do not substitute an assumed zero label. This calculation does not itself assert the group is abelian. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 3, Yes)."},"skills":["Element Order Starts from the Actual Identity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0051","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc579fdf55bf41986e04c5117443754bce30f9220f4d28f4d43eda5196aa3778","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261051,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":1,\"multiplier\":3,\"n\":6,\"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]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley 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]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1, nonnegative multiplier=3. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(1, 3, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 3, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0052","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"552180547d656e91d99813ef2bcec4d091952070e818acb2e99e5baaa55a114f","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261052,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":0,\"multiplier\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0, nonnegative multiplier=4. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(0, 4, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 4, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0053","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a40d33801762c45dbcea3e8b5dc676de5aee97fb77b89f796d644eb57e50dcf3","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261053,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":5,\"multiplier\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5, nonnegative multiplier=5. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(5, 5, 0, 1)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 5, 0, 1)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0054","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e9f85124dd2bb997592a94493678a6bc60870701f6061db1f9491379e97e5c7","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261054,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":4,\"multiplier\":6,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4, nonnegative multiplier=6. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(4, 6, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 6, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0055","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef2de858779ff8b87175130f89e3a96c3f2ae8a7cf4c7f1399f9fc3deeb00e5f","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261055,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":3,\"multiplier\":7,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3, nonnegative multiplier=7. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(3, 7, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 7, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0056","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"077b5d91df903faa5f8f78cfb283319aa3b978d9950c722229378572c1ec19af","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261056,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":2,\"multiplier\":8,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2, nonnegative multiplier=8. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(2, 8, 0, 4)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 8, 0, 4)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0057","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2409c7a0be08308e62b48f81f5503af0d6ec9e940769ddf6639fd3bfd01692e","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261057,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":1,\"multiplier\":9,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1, nonnegative multiplier=9. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(1, 9, 0, 2)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 9, 0, 2)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0058","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06f826a1737325a137128ac4562525a1fa401f16f2af5f466b1afa353b3e1a41","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261058,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":0,\"multiplier\":10,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0, nonnegative multiplier=10. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(0, 10, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 10, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0059","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98d432bef943131273d4db5dc1dc9b8a6678fc44b0c633fec221506f182c44c6","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261059,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":5,\"multiplier\":11,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5, nonnegative multiplier=11. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(5, 11, 0, 1)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 11, 0, 1)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0060","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e143d33ffe87c327770df1455690983fed89f5d8576d9436283f3616d40c5565","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261060,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":4,\"multiplier\":12,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4, nonnegative multiplier=12. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(4, 12, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 12, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0061","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75a0f4ad26a8b92edb07dc82e78f8a3c31fa313872492a715e6d28c834e4a02e","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261061,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":3,\"multiplier\":13,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3, nonnegative multiplier=13. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(3, 13, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 13, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0062","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb9bc14f1d45df9c6c457faa08a9c21d187e3e3ce4fb121d1baf9d38809f948a","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261062,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":2,\"multiplier\":14,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2, nonnegative multiplier=14. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(2, 14, 0, 4)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 14, 0, 4)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0063","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1e625be4a95726ee253fe2b1505650540b73901cdc71010ae3bf8ac75e155c5","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261063,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":1,\"multiplier\":15,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1, nonnegative multiplier=15. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(1, 15, 0, 2)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 15, 0, 2)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0064","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24b8b65bb49b6b0de0a86710bc7154da56317fdae27f69b75f0e1e29e69b4576","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261064,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":0,\"multiplier\":16,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0, nonnegative multiplier=16. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(0, 16, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 16, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0065","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6cf3268e1a8580b2ad463e26916463de15d4d7b6f50b73e291d27a4a869d2b8","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261065,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":5,\"multiplier\":17,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5, nonnegative multiplier=17. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(5, 17, 0, 1)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 17, 0, 1)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0066","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5330516f8a45b181e9362f28fc33611ca3055619856a53b0697910677136bfaa","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261066,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":4,\"multiplier\":18,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4, nonnegative multiplier=18. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(4, 18, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 18, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0067","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c183d54f6a6facbc91825e80718f5742b9bc5183ee067c7fc560c2c64c769780","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261067,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":3,\"multiplier\":19,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3, nonnegative multiplier=19. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(3, 19, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 19, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0068","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3330955b8358f76c05d8ee1dad90be815e8d5466aaf2b0fe104f3d0d000625fe","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261068,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":2,\"multiplier\":20,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2, nonnegative multiplier=20. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(2, 20, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 20, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0069","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acf6bc1024625b6ba28fb58b7337a66b4261aadc2ea3714ace7771a087ae7dcc","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261069,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":1,\"multiplier\":21,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1, nonnegative multiplier=21. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(1, 21, 0, 2)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 21, 0, 2)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0070","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cbc1a95d92b33f485e96598d6fb11aa6bbc3463894eaf3aec85fd185236bd48","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261070,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":0,\"multiplier\":22,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0, nonnegative multiplier=22. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(0, 22, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 22, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0071","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8cecdc59a11e795c59125f798907a1de8abcd41d2bdd2df1ed1246a987ddf96","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261071,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":5,\"multiplier\":23,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5, nonnegative multiplier=23. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(5, 23, 0, 2)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 23, 0, 2)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0072","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f3cca326c1016b05f68373e983c83fd8ae42ef1d4364f97be8bc8f142bc820d","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261072,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":4,\"multiplier\":24,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4, nonnegative multiplier=24. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(4, 24, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 24, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0073","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a64f0e9364f54382466caa048e23809e912a4b33be3716e89959e2b51338e7ff","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261073,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":3,\"multiplier\":25,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3, nonnegative multiplier=25. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(3, 25, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 25, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0074","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60da241c97c91a21b5e995f09220c0468a96998b65398abf7b1f65598c41dea0","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261074,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":2,\"multiplier\":26,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2, nonnegative multiplier=26. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(2, 26, 0, 3)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 26, 0, 3)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0075","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dbab38d1da415e9524471babd4ad1c20762e3e03cdea599954138c243b8faaa","family_id":"qf-mod26-abelian-integer-multiple","seed":202610261075,"semantic_template":"qf-mod26-abelian-integer-multiple/{\"element\":1,\"multiplier\":27,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]]}"},"fields":{"title":"Repeated Group Operation with a Zero-Multiple Boundary","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1, nonnegative multiplier=27. Give (element,multiplier,identity,result).","answer":{"kind":"expression","value":"(1, 27, 0, 0)","explanation":"Apply the operation the declared number of times from the identity. Zero repetitions give the actual identity, not necessarily label zero. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 27, 0, 0)."},"skills":["Repeated Group Operation with a Zero-Multiple Boundary"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0076","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"229b3a9c3b14b5722b62f8e74c47d74b5d948af8da3bd4f4d96942ba4bc10286","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261076,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":1,\"n\":6,\"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]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley 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]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(1, 0, [0, 1, 2, 3, 4, 5], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [0, 1, 2, 3, 4, 5], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0077","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44e83e5d6dbe0253915b54383e759de27e8ffa6b65309e17bf6f6bc302416701","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261077,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(0, 0, [0], 0, 1)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, [0], 0, 1)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0078","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62cc0e2e42534c5cd0d89c88bb7b87a7d0f788e726bc203cf86ca75d195f776b","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261078,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(5, 0, [0, 5, 3, 4, 2, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, [0, 5, 3, 4, 2, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0079","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c4e0f7e01cdb4ea107157ec92a9b016a8c4bcf70725214a8ce63f02e8686b72","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261079,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(4, 0, [0, 4], 0, 2)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, [0, 4], 0, 2)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0080","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91fba4e39da5380c70d933b0095295f7a956018b26330c654330a61cd4c19f83","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261080,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(3, 0, [0, 3, 2], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, [0, 3, 2], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0081","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92b8e441229bed01c9fd1dc29f6453633b0aac15a8e16399977e4aaf11ff085b","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261081,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(2, 0, [0, 2, 4], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, [0, 2, 4], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0082","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d7f560425de87656c56174314dad622617ef291580332cba4236c2170a9b666","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261082,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(1, 0, [0, 1, 3, 2, 4, 5], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [0, 1, 3, 2, 4, 5], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0083","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5876436b760b7d72e398c4c5fee4a8094f179d872800edcb66f35701023e8d2","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261083,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(0, 0, [0], 0, 1)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, [0], 0, 1)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0084","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f9ab8867ddc8b7b1a4f3341b30655cbcdee40e0f61e0a291e130ecc4f4f5bd3","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261084,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(5, 0, [0, 5, 2, 4, 3, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, [0, 5, 2, 4, 3, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0085","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f44384cd2d900ba88208fc827548c91566d31605bb793f0cb159b37eef4111f","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261085,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(4, 0, [0, 4], 0, 2)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, [0, 4], 0, 2)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0086","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7ba5f73472b5a98822675361cd4385b2355873c67f2ee4561fa2b941cf3e5e7","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261086,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(3, 0, [0, 3, 2], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, [0, 3, 2], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0087","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"375acc59011a9d147eef40a9a5ae54e54cca99e6bcff701b724b681c82cb34db","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261087,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(2, 0, [0, 2, 4, 5, 3, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, [0, 2, 4, 5, 3, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0088","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e8f253d16b1d63dd88ef252f6bd996583ec773b1bf6c0178b7cf9881d732551","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261088,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(1, 0, [0, 1, 4, 2, 3, 5], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [0, 1, 4, 2, 3, 5], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0089","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"adc6ae02c994cb79a86ebc64d80ef75ef743244e84c848d73618c97339b6b106","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261089,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(0, 0, [0], 0, 1)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, [0], 0, 1)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0090","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78a9602ea559b83b4a515f464f606027ef534aff708203a17cea022f7a92f228","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261090,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(5, 0, [0, 5, 2, 3, 4, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, [0, 5, 2, 3, 4, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0091","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68954619569574aacec6f85801e224b2633d1bf7da064d8412cf794fa6793f4c","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261091,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(4, 0, [0, 4, 5], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, [0, 4, 5], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0092","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed94960a3246362f8f98b861767f5522e06bb06a9c29ec8ca43fa290b460baf5","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261092,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(3, 0, [0, 3, 2, 5, 4, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, [0, 3, 2, 5, 4, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0093","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11ea3abff2dc6ec6c6cb99008a04d9c0010417ca5267b3ae5e3085e82a48edf2","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261093,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(2, 0, [0, 2, 3, 5, 4, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, [0, 2, 3, 5, 4, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0094","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa63908a36f3f1ab05b41c521adefa3f3b69a74c6fd414f1c17b75b55ef9fb28","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261094,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(1, 0, [0, 1, 5, 2, 3, 4], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [0, 1, 5, 2, 3, 4], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0095","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b67d99abad727119c017100f88b4106f1c514a448b3638f9fd3ad5322eb4039d","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261095,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":0,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=0. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(0, 0, [0], 0, 1)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, [0], 0, 1)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0096","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b79e084c6d00e8280a2992209e9eb5bde75fa51c82981fad9a1f70a67c54c497","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261096,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":5,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=5. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(5, 0, [0, 5, 2], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, [0, 5, 2], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0097","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afd0eb0a11dbb464c3d374a91427d10190b297469cec41b626909cfe1d1d54a5","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261097,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":4,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=4. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(4, 0, [0, 4, 5], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, [0, 4, 5], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0098","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2851511495bf88a625feea5230b8e5b22bd059943ca8966c314d411985123d9d","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261098,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":3,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=3. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(3, 0, [0, 3, 2, 4, 5, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, [0, 3, 2, 4, 5, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0099","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8aa8e3c193260c507c73ca8383ff45089b892da0694b4d4116fea2245e561bb","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261099,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":2,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=2. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(2, 0, [0, 2, 3, 4, 5, 1], 0, 6)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, [0, 2, 3, 4, 5, 1], 0, 6)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261026-r01-finite-abelian-group-0100","batch_id":"batch-20261013-advance-20261026-r01-finite-abelian-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe89610f2d5cb8cae433a3656d0d846210c8031988c1032a970e51c82f0cc2a5","family_id":"qf-mod26-abelian-cyclic-subgroup","seed":202610261100,"semantic_template":"qf-mod26-abelian-cyclic-subgroup/{\"element\":1,\"n\":6,\"table\":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]]}"},"fields":{"title":"Preserve the Generation Order of One Cyclic Subgroup","stem":"Use the complete Cayley table=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] on Fin(6)={0,…,5}, in the given row/column label order. Closure, associativity, a unique two-sided identity and inverses must first qualify on the whole table. The identity need not be label 0. Do not assume commutativity from the family name or a few samples; these results concern only the specified finite group. element=1. Give (element,identity,generated elements in iteration order,next repeated value,order).","answer":{"kind":"expression","value":"(1, 0, [0, 1, 4], 0, 3)","explanation":"Begin at the identity, retain each new power in encounter order and stop at the first repeat. Do not sort this cycle as a label set or infer that it is the entire group. An exact finite or explicitly certified slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [0, 1, 4], 0, 3)."},"skills":["Preserve the Generation Order of One Cyclic Subgroup"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}