Finding element of order in the symmetric group












2














What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?



For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?



EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?






group-theory permutations







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  • $35=5times 7$.
    – Lord Shark the Unknown
    2 hours ago










  • First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
    – toric_actions
    2 hours ago
















2














What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?



For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?



EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?






group-theory permutations







share|cite|improve this question
























  • $35=5times 7$.
    – Lord Shark the Unknown
    2 hours ago










  • First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
    – toric_actions
    2 hours ago














2












2








2


1





What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?



For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?



EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?






group-theory permutations







share|cite|improve this question















What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?



For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?



EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?






group-theory permutations




group-theory permutations






share|cite|improve this question















share|cite|improve this question













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edited 2 hours ago









Shaun

8,722113680




8,722113680










asked 2 hours ago









vesii

855




855












  • $35=5times 7$.
    – Lord Shark the Unknown
    2 hours ago










  • First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
    – toric_actions
    2 hours ago


















  • $35=5times 7$.
    – Lord Shark the Unknown
    2 hours ago










  • First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
    – toric_actions
    2 hours ago
















$35=5times 7$.
– Lord Shark the Unknown
2 hours ago




$35=5times 7$.
– Lord Shark the Unknown
2 hours ago












First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
– toric_actions
2 hours ago




First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
– toric_actions
2 hours ago










2 Answers
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If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$






share|cite|improve this answer





















  • Thanks I get it now. How can I find the maximum order possible?
    – vesii
    2 hours ago










  • $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
    – Ross Millikan
    2 hours ago





















0














Hint: $35 = 5 cdot 7$ and $5+7=12$.






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






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    active

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    active

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    3














    If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$






    share|cite|improve this answer





















    • Thanks I get it now. How can I find the maximum order possible?
      – vesii
      2 hours ago










    • $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
      – Ross Millikan
      2 hours ago


















    3














    If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$






    share|cite|improve this answer





















    • Thanks I get it now. How can I find the maximum order possible?
      – vesii
      2 hours ago










    • $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
      – Ross Millikan
      2 hours ago
















    3












    3








    3






    If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$






    share|cite|improve this answer












    If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Ross Millikan

    292k23196371




    292k23196371












    • Thanks I get it now. How can I find the maximum order possible?
      – vesii
      2 hours ago










    • $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
      – Ross Millikan
      2 hours ago




















    • Thanks I get it now. How can I find the maximum order possible?
      – vesii
      2 hours ago










    • $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
      – Ross Millikan
      2 hours ago


















    Thanks I get it now. How can I find the maximum order possible?
    – vesii
    2 hours ago




    Thanks I get it now. How can I find the maximum order possible?
    – vesii
    2 hours ago












    $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
    – Ross Millikan
    2 hours ago






    $35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
    – Ross Millikan
    2 hours ago













    0














    Hint: $35 = 5 cdot 7$ and $5+7=12$.






    share|cite|improve this answer


























      0














      Hint: $35 = 5 cdot 7$ and $5+7=12$.






      share|cite|improve this answer
























        0












        0








        0






        Hint: $35 = 5 cdot 7$ and $5+7=12$.






        share|cite|improve this answer












        Hint: $35 = 5 cdot 7$ and $5+7=12$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        lhf

        163k10167386




        163k10167386






























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