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Math Help - Unique Sylow p-Subgroups

  1. #1
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    Unique Sylow p-Subgroups

    If p is an odd prime, prove the following.
    a) If G is a group of order (p-1)p^2, then G has a unique Sylow p-subgroup.
    b) There are at least four groups of order (p-1)p^2 which are pairwise nonisomorphic.


    I know little about Sylow's subgroups.

     |G|=(p-1)p^2

     n_i\equiv 1 \  (\text{mod} \  i)
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    MHF Contributor Drexel28's Avatar
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    Re: Unique Sylow p-Subgroups

    Quote Originally Posted by olgashukina View Post
    If p is an odd prime, prove the following.
    a) If G is a group of order (p-1)p^2, then G has a unique Sylow p-subgroup.
    You know that the number of Sylow p-subgroups of G divides p-1 and is equivalent to 1 modulo p. Now, tell me, how many numbers strictly less than any given n are equivalent to n modulo 1?

    b) There are at least four groups of order (p-1)p^2 which are pairwise nonisomorphic.
    Ok, once you take care of the obvious abelian ones, the important observation is that p-1\mid |\text{Aut}(\mathbb{Z}_{p^2})|=\varphi(p^2)=p(p-1) and so there exists non-trivial ____ products.
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    Re: Unique Sylow p-Subgroups

    it might be helpful to note that \text{Aut}(\mathbb{Z}_{p^2}) \cong U(\mathbb{Z}_{p^2}) = (\mathbb{Z}_{p^2})^{\times}

    and here i'm a bit confused. it seems to me, that this only gives 3 subgroups of order (p-1)p^2 for the case p = 3.

    so i believe that you also must investigate where you have a homomorphism into

    \text{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p) as well.
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    MHF Contributor Drexel28's Avatar
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    Re: Unique Sylow p-Subgroups

    Quote Originally Posted by Deveno View Post
    it might be helpful to note that \text{Aut}(\mathbb{Z}_{p^2}) \cong U(\mathbb{Z}_{p^2}) = (\mathbb{Z}_{p^2})^{\times}

    and here i'm a bit confused. it seems to me, that this only gives 3 subgroups of order (p-1)p^2 for the case p = 3.

    so i believe that you also must investigate where you have a homomorphism into

    \text{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p) as well.
    Right, \text{Aut}(\mathbb{Z}_p^2)\cong \text{GL}_2(\mathbb{F}_p) and |\text{GL}_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)=p(p-1)^2(p+1). I forgot about the case when p=3, etc. Good catch.
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