The normalizer of a $p$ sylow subgroup is itself












0












$begingroup$


By the orbit-stabilizer theorem since the action is transitive then an orbit ${gPg^{-1}: gin G} =n_p$ is equal to the number of $p$ sylow subgroups in a group $|G|=p^{alpha}s$ with $(p^alpha,s)=1$ and we get that $$|G/Stab_G(P)|=|G/{gin G: gPg^{-1}=P}|=frac{p^alpha s}{|N_G(P)|}=n_p text{ with } begin{cases} n_pequiv 1mod{p} \ n_pmid s end{cases}$$



The second condition on $n_p$ implies that $|N_G(P)|=p^beta$ for some $betalealpha$



The first condition seems imply that $|N_G(P)|=p^alphaimpliesbeta=alpha$



I'm not sure where I'm going with this but I'd like to know what characterizes a group that acts transitively/not transitively...






group-theory sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
    $endgroup$
    – Derek Holt
    Jan 7 at 11:01












  • $begingroup$
    The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
    $endgroup$
    – P Vanchinathan
    Jan 7 at 11:04










  • $begingroup$
    Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
    $endgroup$
    – John Cataldo
    Jan 7 at 11:49










  • $begingroup$
    No, that is nonsense. The cyclic group of order $6$ is a counterexample.
    $endgroup$
    – Derek Holt
    Jan 7 at 12:06






  • 1




    $begingroup$
    How do you get that the order of $N_G(P)$ must be a power of $p$?
    $endgroup$
    – the_fox
    Jan 7 at 12:50
















0












$begingroup$


By the orbit-stabilizer theorem since the action is transitive then an orbit ${gPg^{-1}: gin G} =n_p$ is equal to the number of $p$ sylow subgroups in a group $|G|=p^{alpha}s$ with $(p^alpha,s)=1$ and we get that $$|G/Stab_G(P)|=|G/{gin G: gPg^{-1}=P}|=frac{p^alpha s}{|N_G(P)|}=n_p text{ with } begin{cases} n_pequiv 1mod{p} \ n_pmid s end{cases}$$



The second condition on $n_p$ implies that $|N_G(P)|=p^beta$ for some $betalealpha$



The first condition seems imply that $|N_G(P)|=p^alphaimpliesbeta=alpha$



I'm not sure where I'm going with this but I'd like to know what characterizes a group that acts transitively/not transitively...






group-theory sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
    $endgroup$
    – Derek Holt
    Jan 7 at 11:01












  • $begingroup$
    The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
    $endgroup$
    – P Vanchinathan
    Jan 7 at 11:04










  • $begingroup$
    Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
    $endgroup$
    – John Cataldo
    Jan 7 at 11:49










  • $begingroup$
    No, that is nonsense. The cyclic group of order $6$ is a counterexample.
    $endgroup$
    – Derek Holt
    Jan 7 at 12:06






  • 1




    $begingroup$
    How do you get that the order of $N_G(P)$ must be a power of $p$?
    $endgroup$
    – the_fox
    Jan 7 at 12:50














0












0








0





$begingroup$


By the orbit-stabilizer theorem since the action is transitive then an orbit ${gPg^{-1}: gin G} =n_p$ is equal to the number of $p$ sylow subgroups in a group $|G|=p^{alpha}s$ with $(p^alpha,s)=1$ and we get that $$|G/Stab_G(P)|=|G/{gin G: gPg^{-1}=P}|=frac{p^alpha s}{|N_G(P)|}=n_p text{ with } begin{cases} n_pequiv 1mod{p} \ n_pmid s end{cases}$$



The second condition on $n_p$ implies that $|N_G(P)|=p^beta$ for some $betalealpha$



The first condition seems imply that $|N_G(P)|=p^alphaimpliesbeta=alpha$



I'm not sure where I'm going with this but I'd like to know what characterizes a group that acts transitively/not transitively...






group-theory sylow-theory







share|cite|improve this question











$endgroup$




By the orbit-stabilizer theorem since the action is transitive then an orbit ${gPg^{-1}: gin G} =n_p$ is equal to the number of $p$ sylow subgroups in a group $|G|=p^{alpha}s$ with $(p^alpha,s)=1$ and we get that $$|G/Stab_G(P)|=|G/{gin G: gPg^{-1}=P}|=frac{p^alpha s}{|N_G(P)|}=n_p text{ with } begin{cases} n_pequiv 1mod{p} \ n_pmid s end{cases}$$



The second condition on $n_p$ implies that $|N_G(P)|=p^beta$ for some $betalealpha$



The first condition seems imply that $|N_G(P)|=p^alphaimpliesbeta=alpha$



I'm not sure where I'm going with this but I'd like to know what characterizes a group that acts transitively/not transitively...






group-theory sylow-theory




group-theory sylow-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 11:50







John Cataldo

















asked Jan 7 at 10:58









John CataldoJohn Cataldo

1,1831316




1,1831316












  • $begingroup$
    The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
    $endgroup$
    – Derek Holt
    Jan 7 at 11:01












  • $begingroup$
    The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
    $endgroup$
    – P Vanchinathan
    Jan 7 at 11:04










  • $begingroup$
    Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
    $endgroup$
    – John Cataldo
    Jan 7 at 11:49










  • $begingroup$
    No, that is nonsense. The cyclic group of order $6$ is a counterexample.
    $endgroup$
    – Derek Holt
    Jan 7 at 12:06






  • 1




    $begingroup$
    How do you get that the order of $N_G(P)$ must be a power of $p$?
    $endgroup$
    – the_fox
    Jan 7 at 12:50


















  • $begingroup$
    The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
    $endgroup$
    – Derek Holt
    Jan 7 at 11:01












  • $begingroup$
    The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
    $endgroup$
    – P Vanchinathan
    Jan 7 at 11:04










  • $begingroup$
    Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
    $endgroup$
    – John Cataldo
    Jan 7 at 11:49










  • $begingroup$
    No, that is nonsense. The cyclic group of order $6$ is a counterexample.
    $endgroup$
    – Derek Holt
    Jan 7 at 12:06






  • 1




    $begingroup$
    How do you get that the order of $N_G(P)$ must be a power of $p$?
    $endgroup$
    – the_fox
    Jan 7 at 12:50
















$begingroup$
The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
$endgroup$
– Derek Holt
Jan 7 at 11:01






$begingroup$
The statement that any two Sylow $p$-subgroups of $G$ are conjugate in $G$ is logically equivalent to the statement that the conjugation action of $G$ on its set of Sylow $p$-subgroups is transitive.
$endgroup$
– Derek Holt
Jan 7 at 11:01














$begingroup$
The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
$endgroup$
– P Vanchinathan
Jan 7 at 11:04




$begingroup$
The transitivity of action on p-Sylow subgroups is one part of Sylow's theorem. As the action is by conjugation, this means any two p-Sylow subgroups are conjugate.
$endgroup$
– P Vanchinathan
Jan 7 at 11:04












$begingroup$
Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
$endgroup$
– John Cataldo
Jan 7 at 11:49




$begingroup$
Ok but do you agree that the normalizer of a $p$-sylow subgroup is always itself? Because $|N_G(P)|=p^alpha$ and $Psubset N_G(P)$
$endgroup$
– John Cataldo
Jan 7 at 11:49












$begingroup$
No, that is nonsense. The cyclic group of order $6$ is a counterexample.
$endgroup$
– Derek Holt
Jan 7 at 12:06




$begingroup$
No, that is nonsense. The cyclic group of order $6$ is a counterexample.
$endgroup$
– Derek Holt
Jan 7 at 12:06




1




1




$begingroup$
How do you get that the order of $N_G(P)$ must be a power of $p$?
$endgroup$
– the_fox
Jan 7 at 12:50




$begingroup$
How do you get that the order of $N_G(P)$ must be a power of $p$?
$endgroup$
– the_fox
Jan 7 at 12:50










1 Answer
1






active

oldest

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1












$begingroup$

The second condition implies that $lvert N_G (P) rvert = p^alphaleft(dfrac{s}{n_p}right)$.
This does not imply that $lvert N_G (P) rvert$ is a power of the prime $p$.
What it does imply, however (since $gcd (p^alpha, s) = 1$), is that the highest power of $p$ that divides the integer $lvert N_G (P) rvert$ is $p^alpha$.



P.S. The correct relationship between a $p$-Sylow subgroup $P$ of $G$ and its normaliser is that $[G:N_G (P)] = n_p$.
There is a bijection between all conjugates of $P$ (i.e. all the $p$-Sylow subgroups of $G$) and the left cosets of $N_G (P)$.
The bijection is $gPg^{-1} mapsto gN_G (P)$.






share|cite|improve this answer









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    $begingroup$

    The second condition implies that $lvert N_G (P) rvert = p^alphaleft(dfrac{s}{n_p}right)$.
    This does not imply that $lvert N_G (P) rvert$ is a power of the prime $p$.
    What it does imply, however (since $gcd (p^alpha, s) = 1$), is that the highest power of $p$ that divides the integer $lvert N_G (P) rvert$ is $p^alpha$.



    P.S. The correct relationship between a $p$-Sylow subgroup $P$ of $G$ and its normaliser is that $[G:N_G (P)] = n_p$.
    There is a bijection between all conjugates of $P$ (i.e. all the $p$-Sylow subgroups of $G$) and the left cosets of $N_G (P)$.
    The bijection is $gPg^{-1} mapsto gN_G (P)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The second condition implies that $lvert N_G (P) rvert = p^alphaleft(dfrac{s}{n_p}right)$.
      This does not imply that $lvert N_G (P) rvert$ is a power of the prime $p$.
      What it does imply, however (since $gcd (p^alpha, s) = 1$), is that the highest power of $p$ that divides the integer $lvert N_G (P) rvert$ is $p^alpha$.



      P.S. The correct relationship between a $p$-Sylow subgroup $P$ of $G$ and its normaliser is that $[G:N_G (P)] = n_p$.
      There is a bijection between all conjugates of $P$ (i.e. all the $p$-Sylow subgroups of $G$) and the left cosets of $N_G (P)$.
      The bijection is $gPg^{-1} mapsto gN_G (P)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The second condition implies that $lvert N_G (P) rvert = p^alphaleft(dfrac{s}{n_p}right)$.
        This does not imply that $lvert N_G (P) rvert$ is a power of the prime $p$.
        What it does imply, however (since $gcd (p^alpha, s) = 1$), is that the highest power of $p$ that divides the integer $lvert N_G (P) rvert$ is $p^alpha$.



        P.S. The correct relationship between a $p$-Sylow subgroup $P$ of $G$ and its normaliser is that $[G:N_G (P)] = n_p$.
        There is a bijection between all conjugates of $P$ (i.e. all the $p$-Sylow subgroups of $G$) and the left cosets of $N_G (P)$.
        The bijection is $gPg^{-1} mapsto gN_G (P)$.






        share|cite|improve this answer









        $endgroup$



        The second condition implies that $lvert N_G (P) rvert = p^alphaleft(dfrac{s}{n_p}right)$.
        This does not imply that $lvert N_G (P) rvert$ is a power of the prime $p$.
        What it does imply, however (since $gcd (p^alpha, s) = 1$), is that the highest power of $p$ that divides the integer $lvert N_G (P) rvert$ is $p^alpha$.



        P.S. The correct relationship between a $p$-Sylow subgroup $P$ of $G$ and its normaliser is that $[G:N_G (P)] = n_p$.
        There is a bijection between all conjugates of $P$ (i.e. all the $p$-Sylow subgroups of $G$) and the left cosets of $N_G (P)$.
        The bijection is $gPg^{-1} mapsto gN_G (P)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 13:13









        Chaitanya TappuChaitanya Tappu

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