In a group of order 400, is the normalizer of one of the 16 Sylow 5-subgroups itself?












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$begingroup$


In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?



I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!






abstract-algebra group-theory sylow-theory







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    1












    $begingroup$


    In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?



    I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!






    abstract-algebra group-theory sylow-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?



      I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!






      abstract-algebra group-theory sylow-theory







      share|cite|improve this question









      $endgroup$




      In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?



      I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!






      abstract-algebra group-theory sylow-theory




      abstract-algebra group-theory sylow-theory






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      asked Jan 6 at 22:06









      pendermathpendermath

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          $begingroup$

          This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.






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            active

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            $begingroup$

            This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.






                share|cite|improve this answer









                $endgroup$



                This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 22:13









                Matt SamuelMatt Samuel

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