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Math Help - Sylow p-subgroups

  1. #1
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    Sylow p-subgroups

    Show that a Sylow p-subgroup of the dihedral group is cyclic and normal for every odd prime.
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  2. #2
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    Let p be an odd prime, D_{p} the dihedral group with 2p elements and e its identity element.

    A p-Sylow in D_{p} has an order of p, so it is cyclic: indeed, the order of an element of such a p-Sylow subgroups is either 1 or p (because it divides p) so if the element is different than e, then it's a generator, and since p is different from 1, there are generators.
    Furthermore, D_{p} has 2p elements. The number N_{p} of p-Sylow and 1 are congruent modulo p, and N_{p} divides \frac{|Dp|}{p}, so divides 2. So N_{p}=1. But all p-Sylow subgroups are conjugate, so when there is only one p-Sylow H, it is normal.
    Indeed, \forall x \in D_{p},\ xHx^{-1} has the same order than H. Thus it's a p-Sylow, and xHx^{-1}=H.
    Last edited by clic-clac; November 10th 2008 at 09:10 AM. Reason: voc
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  3. #3
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    clic-clac, the dihedral group is not necessarily in the form D_p. it could be D_n with p \mid n. i'll prove that in fact every sugroup of odd order in D_n is normal and cyclic. to see this, let

    D_n=<a,b: \ a^2=b^n=1, \ aba=b^{-1}> and suppose H \subset D_n is a subgroup with odd order. then H \subseteq <b>, because ab^j, \ 0 \leq j < n, have order 2 and hence they cannot be in

    H. so H is cyclic. besides a cyclic group has only one subgroup of any order, i.e. H is normal. Q.E.D.
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