Question about p-Sylow subgroups being maximal












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I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G



Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?






abstract-algebra group-theory sylow-theory







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  • $begingroup$
    Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
    $endgroup$
    – Chris Custer
    Dec 8 '18 at 4:55
















1












$begingroup$


I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G



Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?






abstract-algebra group-theory sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
    $endgroup$
    – Chris Custer
    Dec 8 '18 at 4:55














1












1








1





$begingroup$


I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G



Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?






abstract-algebra group-theory sylow-theory







share|cite|improve this question











$endgroup$




I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G



Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?






abstract-algebra group-theory sylow-theory




abstract-algebra group-theory sylow-theory






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 1 at 20:19









Shaun

8,818113681




8,818113681










asked Dec 8 '18 at 4:22









Richard VillalobosRichard Villalobos

1567




1567












  • $begingroup$
    Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
    $endgroup$
    – Chris Custer
    Dec 8 '18 at 4:55


















  • $begingroup$
    Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
    $endgroup$
    – Chris Custer
    Dec 8 '18 at 4:55
















$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55




$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55










2 Answers
2






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$begingroup$

An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.



In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.






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$endgroup$





















    0












    $begingroup$

    I would take almost the same exemple than Randall,in S4 (symmetric group)
    You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

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      1












      $begingroup$

      An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.



      In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.



        In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.



          In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.






          share|cite|improve this answer











          $endgroup$



          An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.



          In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 4:28

























          answered Dec 8 '18 at 4:25









          RandallRandall

          9,31111129




          9,31111129























              0












              $begingroup$

              I would take almost the same exemple than Randall,in S4 (symmetric group)
              You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I would take almost the same exemple than Randall,in S4 (symmetric group)
                You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I would take almost the same exemple than Randall,in S4 (symmetric group)
                  You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.






                  share|cite|improve this answer









                  $endgroup$



                  I would take almost the same exemple than Randall,in S4 (symmetric group)
                  You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 21:54









                  Cauchy is my masterCauchy is my master

                  92




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