Question Number 192340 by Mastermind last updated on 15/May/23
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{any}\:\mathrm{permuta}− \\ $$$$\mathrm{tion}\:\theta\:\mathrm{is}\:\mathrm{the}\:\mathrm{least}\:\mathrm{common}\:\mathrm{multiple} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its}\:\mathrm{disjoint}\:\mathrm{cycles}. \\ $$$$ \\ $$$$\:\mathrm{hi} \\ $$
Answered by aleks041103 last updated on 15/May/23
$${let}\:\theta={c}_{\mathrm{1}} ...{c}_{{k}} ,\:{where}\:{c}_{\mathrm{1}} ,...,{c}_{{k}} \:{are}\:{disjoint} \\ $$$${cycles}. \\ $$$${Since}\:{they}\:{are}\:{disjoint},\:{they}\:{commute}. \\ $$$$\Rightarrow\theta^{{n}} ={c}_{\mathrm{1}} ^{{n}} ...{c}_{{k}} ^{{n}} \\ $$$${By}\:{def}.\:\mid\theta\mid={n},\:{where}\:{n}\:{is}\:{the}\:{smallest}\:{positive} \\ $$$${integer}\:{such}\:{that}\:\theta^{{n}} ={id}. \\ $$$${Since}\:{c}_{\mathrm{1}} ,...,{c}_{{k}} \:{are}\:{disjoint},\:{c}_{\mathrm{1}} ^{{n}} ,...,{c}_{{k}} ^{{n}} \:{are}\:{also} \\ $$$${disjoint}. \\ $$$$\Rightarrow\theta^{{n}} ={c}_{\mathrm{1}} ^{{n}} ...{c}_{{k}} ^{{n}} ={id}\Leftrightarrow{c}_{{j}} ^{{n}} ={id}\:{for}\:\forall{j}=\mathrm{1},...,{k} \\ $$$${it}\:{is}\:{obvious}\:{that}\: \\ $$$${c}_{{j}} ^{{n}} ={id}\:\Leftrightarrow\:{l}\left({c}_{{j}} \right)\mid{n},\:\forall{j} \\ $$$$\Rightarrow\theta^{{n}} ={id}\:\Leftrightarrow\:{lcm}\left({l}\left({c}_{\mathrm{1}} \right),...,{l}\left({c}_{{k}} \right)\right)\mid{n} \\ $$$$\Rightarrow\mid\theta\mid={lcm}\left({l}\left({c}_{\mathrm{1}} \right),...,{l}\left({c}_{\mathrm{2}} \right)\right) \\ $$
Commented by Mastermind last updated on 18/May/23
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\: \\ $$