Ternary parallelogram scales are MOS substitution: Difference between revisions
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Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk''<sub>'''v'''</sub> {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k''<sub>'''v'''</sub>) = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}} | Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk''<sub>'''v'''</sub> {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k''<sub>'''v'''</sub>) = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}} | ||
Claim: ''k''<sub>'''w'''</sub> has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}} | Claim: ψ(''k''<sub>'''w'''</sub>) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}} | ||
Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. | Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. | ||
* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k''<sub>'''w'''</sub> = ''m'' which is of order 2. | * If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ψ(''k''<sub>'''w'''</sub>) = ''m'' which is of order 2. | ||
* If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}} | * If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}} | ||
** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''. | ** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''. | ||