Ternary parallelogram scales are MOS substitution: Difference between revisions
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==== Lemma 2: If ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, then {{nowrap|{''a'', 2''a'', ..., (''n'' - 1)''a''}}} and {{nowrap|[-''m'' + 1 : ''m'']}} are not disjoint ==== | ==== Lemma 2: If ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, then {{nowrap|{''a'', 2''a'', ..., (''n'' - 1)''a''}}} and {{nowrap|[-''m'' + 1 : ''m'']}} are not disjoint ==== | ||
This is a technical lemma. | |||
==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ==== | ==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ==== | ||
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}.}} | ||