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Math Help - Self Normalizing groups

  1. #1
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    Self Normalizing groups

    Let p denote a prime number. Let G be a finite group. Let P denote a p-sylow subgroup of G. Finally, let H be any subgroup of G that contains N_G(P). Prove that [G:H]≡1 mod p.

    Well here is what I know right away: N _G(P) is in H. This means that gPg ^-1. Since it is in H, this also means that N _H(P)=hPh^-1 So ghP(gh)^-1. This is about where I'm stuck.
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  2. #2
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    |G: N_GP| is equal to the p-sylow subgroups in G, likewise with |H: N_GP|. For some p-Sylow subgroup L \leq N_GH, there is a g such that L=gPg-1 \leqgHg-1=H. We know that's true because there is some theorem that says that if a subgroup contains the normalizer of a p-sylow subgroup it is self normalizing.

    I'm stuck on pulling the G:H ≡1 mod p

    This is refining my first post
    Last edited by cmj1988; November 25th 2009 at 06:49 PM. Reason: correcting...unsuccessfully some formatting
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  3. #3
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    Quote Originally Posted by cmj1988 View Post
    Let p denote a prime number. Let G be a finite group. Let P denote a p-sylow subgroup of G. Finally, let H be any subgroup of G that contains N_G(P). Prove that [G:H]≡1 mod p.

    Well here is what I know right away: N _G(P) is in H. This means that gPg ^-1. Since it is in H, this also means that N _H(P)=hPh^-1 So ghP(gh)^-1. This is about where I'm stuck.
    Lemma 1. Every subgroup H of G which contains the normalizer of a p-Sylow subgroup is its own normalizer.

    Lemma 2. If a group H of order p^n (p prime) acts on a finite set S, then |S| = |S_0|(\text{mod p}), where S_0 is the number of fixed elements by the given action.

    Let S be the set of all left cosets of H in G; let P acts S by conjugation. We see that K \in S_0 \leftrightarrow xKx^{-1}=K for all x \in P, where K is a left coset of H in G. This condition only holds if and only if P < N_G(K). This further implies that N_G(P) < N_G(N_G(K)). We know that N_G(P) is contained in H and N_G(H)=H by lemma 1. Thus S_0=\{H\} and we conclude that [G:H] \equiv 1 (\text{mod p}) by lemma 2.
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  4. #4
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    Quote Originally Posted by aliceinwonderland View Post
    Lemma 1. Every subgroup H of G which contains the normalizer of a p-Sylow subgroup is its own normalizer.

    Lemma 2. If a group H of order p^n (p prime) acts on a finite set S, then |S| = |S_0|(\text{mod p}), where S_0 is the number of fixed elements by the given action.
    I'm just curious, is this a well known lemma.
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  5. #5
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    Quote Originally Posted by cmj1988 View Post
    I'm just curious, is this a well known lemma.
    It seems like you already saw the lemma 1. You can find the lemma 2 in Hungerford's algebra (p93).
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  6. #6
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    okay, one more jerk type question

    why is S_0=1
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  7. #7
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    Quote Originally Posted by cmj1988 View Post
    okay, one more jerk type question

    why is S_0=1
    It is because P cannot be contained in H and aH (a in G, but not in H) at the same time. So, S_0={H} only and |S_0|=1.
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  8. #8
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    I follow your proof except for K, is K the arbitrary p-sylow subgroup contained in gPg^-1 contained in gHg^-1=H
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