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If P is normal in H, and H normal in G, then P is normal in G

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Now, G is a finite group, and P is a Sylow P-Subgroup of G, and H is a subgroup of G with P≤H≤G. So if P is normal in H, and H is normal in G, then P is normal in G.


    2. Relevant equations



    3. The attempt at a solution
    I know I want to show that gpg^-1 is contained in P but that's about as far as I can get. I know gPg^-1 is going to be another Sylow P-Subgroup, but other than that I'm kind of stuck. When I try going about showing that gpg^-1 q, where q is an element in p, I have trouble showing the two elements are in P. I get as far as q=hph^-1 since q is in P and i think my best bet for showing it is contained in P is for closure...but that's all I get. I know I need to use the P is Sylow P-Subgroup since the property normally isn't transitive, but I'm not sure how to.

    Thanks for any help.
     
  2. jcsd
  3. Mar 17, 2012 #2

    Dick

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    All sylow p subgroups are conjugate, right?
     
  4. Mar 17, 2012 #3
    Ya, gPg^-1 is going to be a conjugate to P, but why does that help me show that it is P? Couldn't it be another Sylow P-Subgroup?
     
  5. Mar 18, 2012 #4

    Dick

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    If it's another Sylow P-Subgroup then that subgroup is contained in H. What next?
     
  6. Mar 18, 2012 #5
    Why is it necessarily going to be contained in H? Just because of orders and conjugate? Are you going to get that P is unique and so it's normal?
     
  7. Mar 18, 2012 #6

    Dick

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    Yes, I am. gHg^(-1)=H and P is contained in H.
     
  8. Mar 18, 2012 #7
    So is gPg^(-1)=P because of that?

    I'm sorry, I've always been dreadful at showing that something is a normal subgroup.
     
  9. Mar 18, 2012 #8

    Dick

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    No, that's not the reason. P and gPg^(-1) are two p-sylow subgroups of H. Think about that.
     
  10. Mar 19, 2012 #9
    Since they're normal in H they have to be the same subgroup.
     
  11. Mar 19, 2012 #10

    Dick

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    There's a little more to it than that. P is normal in H. Why is gPg^(-1)?
     
  12. Mar 19, 2012 #11
    Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that
    ). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such it's normal.
     
  13. Mar 19, 2012 #12

    Dick

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    Yeah, that does it. Since P and gPg^(-1) are both sylow p-subgroups of H. Then they are conjugate in H. So gPg^(-1)=hPh^(-1) where h is in H. Since P is normal in H, gPg^(-1)=P.
     
  14. Mar 20, 2012 #13
    Awesome!!! Thank you do much for walking me through it!! :smile:
     
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