How many permutations of order $8$ in $S_{10}$?












3












$begingroup$



Find the number of permutations of order $8$ in $S_{10}$. If $sigma$ is one of them, find the number of subgroups in the subgroup generated by $sigma$.







group-theory discrete-mathematics permutations







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$endgroup$








  • 3




    $begingroup$
    Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
    $endgroup$
    – Mark
    Jan 22 at 19:42








  • 1




    $begingroup$
    Where is your input?
    $endgroup$
    – Did
    Jan 22 at 19:50










  • $begingroup$
    I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
    $endgroup$
    – Chris Custer
    Jan 22 at 19:51








  • 2




    $begingroup$
    @Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
    $endgroup$
    – Mark
    Jan 22 at 19:56










  • $begingroup$
    @Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
    $endgroup$
    – Chris Custer
    Jan 23 at 1:25


















3












$begingroup$



Find the number of permutations of order $8$ in $S_{10}$. If $sigma$ is one of them, find the number of subgroups in the subgroup generated by $sigma$.







group-theory discrete-mathematics permutations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
    $endgroup$
    – Mark
    Jan 22 at 19:42








  • 1




    $begingroup$
    Where is your input?
    $endgroup$
    – Did
    Jan 22 at 19:50










  • $begingroup$
    I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
    $endgroup$
    – Chris Custer
    Jan 22 at 19:51








  • 2




    $begingroup$
    @Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
    $endgroup$
    – Mark
    Jan 22 at 19:56










  • $begingroup$
    @Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
    $endgroup$
    – Chris Custer
    Jan 23 at 1:25
















3












3








3


1



$begingroup$



Find the number of permutations of order $8$ in $S_{10}$. If $sigma$ is one of them, find the number of subgroups in the subgroup generated by $sigma$.







group-theory discrete-mathematics permutations







share|cite|improve this question











$endgroup$





Find the number of permutations of order $8$ in $S_{10}$. If $sigma$ is one of them, find the number of subgroups in the subgroup generated by $sigma$.







group-theory discrete-mathematics permutations




group-theory discrete-mathematics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:49









Did

248k23224463




248k23224463










asked Jan 22 at 19:39









JackJack

344




344








  • 3




    $begingroup$
    Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
    $endgroup$
    – Mark
    Jan 22 at 19:42








  • 1




    $begingroup$
    Where is your input?
    $endgroup$
    – Did
    Jan 22 at 19:50










  • $begingroup$
    I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
    $endgroup$
    – Chris Custer
    Jan 22 at 19:51








  • 2




    $begingroup$
    @Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
    $endgroup$
    – Mark
    Jan 22 at 19:56










  • $begingroup$
    @Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
    $endgroup$
    – Chris Custer
    Jan 23 at 1:25
















  • 3




    $begingroup$
    Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
    $endgroup$
    – Mark
    Jan 22 at 19:42








  • 1




    $begingroup$
    Where is your input?
    $endgroup$
    – Did
    Jan 22 at 19:50










  • $begingroup$
    I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
    $endgroup$
    – Chris Custer
    Jan 22 at 19:51








  • 2




    $begingroup$
    @Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
    $endgroup$
    – Mark
    Jan 22 at 19:56










  • $begingroup$
    @Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
    $endgroup$
    – Chris Custer
    Jan 23 at 1:25










3




3




$begingroup$
Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
$endgroup$
– Mark
Jan 22 at 19:42






$begingroup$
Do you know about cycle decomposition? Do you know how to find the order of a permutation given its cycle decomposition?
$endgroup$
– Mark
Jan 22 at 19:42






1




1




$begingroup$
Where is your input?
$endgroup$
– Did
Jan 22 at 19:50




$begingroup$
Where is your input?
$endgroup$
– Did
Jan 22 at 19:50












$begingroup$
I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
$endgroup$
– Chris Custer
Jan 22 at 19:51






$begingroup$
I can think of ${10choose 8}=45$ for starters. That is, the $8$-cycles. But there are surely others. Oh yeah, you can change their order. So, multiply by $8!$?
$endgroup$
– Chris Custer
Jan 22 at 19:51






2




2




$begingroup$
@Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
$endgroup$
– Mark
Jan 22 at 19:56




$begingroup$
@Chris Custer, no, you multiply by $7!$. Once you choose the image of $7$ elements in the cycle the last element has just one option.
$endgroup$
– Mark
Jan 22 at 19:56












$begingroup$
@Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
$endgroup$
– Chris Custer
Jan 23 at 1:25






$begingroup$
@Mark actually I think the reason is you can cyclically permute each $8$-cycle $8$ ways
$endgroup$
– Chris Custer
Jan 23 at 1:25












2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint:



The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think this needs some editing
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:42










  • $begingroup$
    @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
    $endgroup$
    – Bernard
    Jan 22 at 20:46










  • $begingroup$
    You are welcome !
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:51



















0












$begingroup$

An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.



There are are ${10choose 8}cdot7!$ of type $(8,1,1)$. There are ${10choose 8}cdot 7!$ of type $(8,2)$.



Together we get $900cdot7!$



If $sigma$ is one of them, $langlesigma rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than ${e}$.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint:



    The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think this needs some editing
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:42










    • $begingroup$
      @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
      $endgroup$
      – Bernard
      Jan 22 at 20:46










    • $begingroup$
      You are welcome !
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:51
















    0












    $begingroup$

    Hint:



    The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think this needs some editing
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:42










    • $begingroup$
      @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
      $endgroup$
      – Bernard
      Jan 22 at 20:46










    • $begingroup$
      You are welcome !
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:51














    0












    0








    0





    $begingroup$

    Hint:



    The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.






    share|cite|improve this answer











    $endgroup$



    Hint:



    The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 22 at 20:45

























    answered Jan 22 at 19:53









    BernardBernard

    121k740116




    121k740116












    • $begingroup$
      I think this needs some editing
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:42










    • $begingroup$
      @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
      $endgroup$
      – Bernard
      Jan 22 at 20:46










    • $begingroup$
      You are welcome !
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:51


















    • $begingroup$
      I think this needs some editing
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:42










    • $begingroup$
      @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
      $endgroup$
      – Bernard
      Jan 22 at 20:46










    • $begingroup$
      You are welcome !
      $endgroup$
      – J. W. Tanner
      Jan 22 at 20:51
















    $begingroup$
    I think this needs some editing
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:42




    $begingroup$
    I think this needs some editing
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:42












    $begingroup$
    @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
    $endgroup$
    – Bernard
    Jan 22 at 20:46




    $begingroup$
    @J.W.Tanner: Oh! yes. I've fixed it. Thanks for pointing the mistyping!
    $endgroup$
    – Bernard
    Jan 22 at 20:46












    $begingroup$
    You are welcome !
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:51




    $begingroup$
    You are welcome !
    $endgroup$
    – J. W. Tanner
    Jan 22 at 20:51











    0












    $begingroup$

    An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.



    There are are ${10choose 8}cdot7!$ of type $(8,1,1)$. There are ${10choose 8}cdot 7!$ of type $(8,2)$.



    Together we get $900cdot7!$



    If $sigma$ is one of them, $langlesigma rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than ${e}$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.



      There are are ${10choose 8}cdot7!$ of type $(8,1,1)$. There are ${10choose 8}cdot 7!$ of type $(8,2)$.



      Together we get $900cdot7!$



      If $sigma$ is one of them, $langlesigma rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than ${e}$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.



        There are are ${10choose 8}cdot7!$ of type $(8,1,1)$. There are ${10choose 8}cdot 7!$ of type $(8,2)$.



        Together we get $900cdot7!$



        If $sigma$ is one of them, $langlesigma rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than ${e}$.






        share|cite|improve this answer











        $endgroup$



        An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.



        There are are ${10choose 8}cdot7!$ of type $(8,1,1)$. There are ${10choose 8}cdot 7!$ of type $(8,2)$.



        Together we get $900cdot7!$



        If $sigma$ is one of them, $langlesigma rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than ${e}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 24 at 17:04

























        answered Jan 23 at 1:18









        Chris CusterChris Custer

        13.9k3827




        13.9k3827






























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