Why is a unique Sylow p-subgroup normal?












2












$begingroup$


I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.






abstract-algebra normal-subgroups sylow-theory







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$endgroup$












  • $begingroup$
    I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:14










  • $begingroup$
    $N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
    $endgroup$
    – Randall
    Nov 2 '17 at 16:14






  • 4




    $begingroup$
    You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:15












  • $begingroup$
    I think @Bungo put it better than I did :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:16










  • $begingroup$
    @Foobanana Your remark is quite correct, I just put a bit more detail :-)
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:17
















2












$begingroup$


I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.






abstract-algebra normal-subgroups sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:14










  • $begingroup$
    $N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
    $endgroup$
    – Randall
    Nov 2 '17 at 16:14






  • 4




    $begingroup$
    You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:15












  • $begingroup$
    I think @Bungo put it better than I did :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:16










  • $begingroup$
    @Foobanana Your remark is quite correct, I just put a bit more detail :-)
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:17














2












2








2





$begingroup$


I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.






abstract-algebra normal-subgroups sylow-theory







share|cite|improve this question











$endgroup$




I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.






abstract-algebra normal-subgroups sylow-theory




abstract-algebra normal-subgroups sylow-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 2 '17 at 16:28









Teddy38

2,1912520




2,1912520










asked Nov 2 '17 at 16:12









mikkelmkmikkelmk

695




695












  • $begingroup$
    I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:14










  • $begingroup$
    $N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
    $endgroup$
    – Randall
    Nov 2 '17 at 16:14






  • 4




    $begingroup$
    You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:15












  • $begingroup$
    I think @Bungo put it better than I did :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:16










  • $begingroup$
    @Foobanana Your remark is quite correct, I just put a bit more detail :-)
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:17


















  • $begingroup$
    I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:14










  • $begingroup$
    $N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
    $endgroup$
    – Randall
    Nov 2 '17 at 16:14






  • 4




    $begingroup$
    You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:15












  • $begingroup$
    I think @Bungo put it better than I did :)
    $endgroup$
    – Foobanana
    Nov 2 '17 at 16:16










  • $begingroup$
    @Foobanana Your remark is quite correct, I just put a bit more detail :-)
    $endgroup$
    – Bungo
    Nov 2 '17 at 16:17
















$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14




$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14












$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14




$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14




4




4




$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15






$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15














$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16




$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16












$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17




$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17










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$begingroup$

Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.






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    8












    $begingroup$

    Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.






        share|cite|improve this answer









        $endgroup$



        Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 2 '17 at 16:14









        Tsemo AristideTsemo Aristide

        57.9k11445




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