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Math Help - P-Sylow subgroups, Normalizers, Indexes

  1. #1
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    P-Sylow subgroups, Normalizers, Indexes

    Let G be a group.

    Prove that G has no P-Sylow subgroup 'H' for which [G:N(H)]=2.

    For every subgroup H in G, the Normalizer is defined as N(H):={ g\in G: gH=Hg }

    I need to find a way to prove this. If |G| is infinity, then for every prime p and natural m, we get that |G| is never m*p^n , therefore there is no p-Sylow subgroups anyway (Is this right? I can't think of any other way to 'take care' of the infinite case).

    Thanks
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Let G be a group.

    Prove that G has no P-Sylow subgroup 'H' for which [G:N(H)]=2.

    For every subgroup H in G, the Normalizer is defined as N(H):={ g\in G: gH=Hg }

    I need to find a way to prove this. If |G| is infinity, then for every prime p and natural m, we get that |G| is never m*p^n , therefore there is no p-Sylow subgroups anyway (Is this right? I can't think of any other way to 'take care' of the infinite case).

    Thanks
    The finite case is an application of Sylow's theorem. Look up the part of the theorem which tells you how many Sylow p-subgroups there are and work out which bit is of interest here. Then think about why there can't just be 2 Sylow p-subgroups.

    For the infinite case you need to remember that, well, there is no infinite case! Essentially, you were right with what you said; you are working with Sylow subgroups, which are only applicable to finite groups!
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    Quote Originally Posted by Swlabr View Post
    The finite case is an application of Sylow's theorem. Look up the part of the theorem which tells you how many Sylow p-subgroups there are and work out which bit is of interest here. Then think about why there can't just be 2 Sylow p-subgroups.

    For the infinite case you need to remember that, well, there is no infinite case! Essentially, you were right with what you said; you are working with Sylow subgroups, which are only applicable to finite groups!
    Oh, I see now that if H is a P-Sylow group of G, then [G:N(H)]=n_p, therefore if [G:N(H)]=2, we get that 2=1mod(p), which is impossible (considering that p has to be natural, since p is prime).

    That's it ?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Oh, I see now that if H is a P-Sylow group of G, then [G:N(H)]=n_p, therefore if [G:N(H)]=2, we get that 2=1mod(p), which is impossible (considering that p has to be natural, since p is prime).

    That's it ?
    Yup, that's all there is to it!
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    Quote Originally Posted by Swlabr View Post
    Yup, that's all there is to it!
    Thanks

    Just to be 'formal', is a=b mod(p) , what does it mean (about the existance of an integer c for which...)? I want to know it, and also to complete the proof.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Thanks

    Just to be 'formal', is a=b mod(p) , what does it mean (about the existance of an integer c for which...)? I want to know it, and also to complete the proof.
    Well, a \equiv b \text{ mod }p \Rightarrow a=pn+b for some n \in \mathbb{Z}. Thus, 2=pn+1 \Rightarrow pn=1 \Rightarrow p|1 \Rightarrow p= \pm 1 \Rightarrow is not prime, a contradiction.
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