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Math Help - Sylow's theorem and direct product.

  1. #1
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    Sylow's theorem and direct product.

    Let G be a finite p-group of order p^m. Show that G has normal subgroups G0, G1 ....Gm such that {e}=G0 < G1 <.......< Gm=G and O(Gi)=p^i for all i=0, 1, 2.....m.
    (e is identity).
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  2. #2
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    Quote Originally Posted by niranjan View Post
    Let G be a finite p-group of order p^m. Show that G has normal subgroups G0, G1 ....Gm such that {e}=G0 < G1 <.......< Gm=G and O(Gi)=p^i for all i=0, 1, 2.....m.
    (e is identity).
    You can use Sylow's theorem, which is straightforward and your last exercise. But I think it is better to actually prove this result without the use of Sylow's theorem. We will use the same theorem we proved here.

    Obviously G_0 exists. Now by Cauchy's theorem there is a subgroup G_1. We will show how to prove G_2 exists. By the theorem we proved we know that N(G_1)\not = G_1. Since G_1\triangleleft N(G_1) we can form N(G_1)/G_1 and we see that p divides |N(G_1)/G_1|. By Cauchy's theorem there is a subgroup K\subseteq N(G_1)/G_1 with |K|=p. If \pi : N(G_1)\to N(G_1)/G_1 is the natural projection then \pi^{-1}(K) is a subgroup of N(G_1) with |\pi^{-1}(K)| = p^2. Just keep on repeating this argument for p^2,p^3,.. and so form and this gives us a chain G_0\subseteq G_1 \subseteq ... \subseteq G_n with G_j \triangleleft G_{j+1} and (G_{j+1}:G_j)=p.
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