Finding number of subgroups of order 3 in an abelian group of order 24












1












$begingroup$


Problem



Find the number of subgroup of order 3 in an abelian group of order 24?



Attempt
Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
$n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
Is this correct?










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$endgroup$

















    1












    $begingroup$


    Problem



    Find the number of subgroup of order 3 in an abelian group of order 24?



    Attempt
    Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
    $n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
    Is this correct?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Problem



      Find the number of subgroup of order 3 in an abelian group of order 24?



      Attempt
      Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
      $n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
      Is this correct?










      share|cite|improve this question









      $endgroup$




      Problem



      Find the number of subgroup of order 3 in an abelian group of order 24?



      Attempt
      Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
      $n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
      Is this correct?







      group-theory sylow-theory






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      asked Jan 16 at 14:48









      blue boyblue boy

      1,243613




      1,243613






















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          $begingroup$

          Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.



          Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.






          share|cite|improve this answer









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            $begingroup$

            Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.



            Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.



              Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.






              share|cite|improve this answer









              $endgroup$
















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                1





                $begingroup$

                Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.



                Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.






                share|cite|improve this answer









                $endgroup$



                Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.



                Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 20:19









                Sam HughesSam Hughes

                741114




                741114






























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