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Math Help - Symmetric group of P

  1. #1
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    Symmetric group of P

    [Dummit p132, Q34]

    Prove that if p is a prime and P is a subgroup of S_p of order p, then |N_{S_p}(P)|=p(p-1).
    [Argue that every conjugate of P contains exactly p-1 p-cycles and use the formula for the number of p-cycles to compute the index of N_{S_p}(P) in S_p].
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    [Dummit p132, Q34]

    Prove that if p is a prime and P is a subgroup of S_p of order p, then |N_{S_p}(P)|=p(p-1).
    [Argue that every conjugate of P contains exactly p-1 p-cycles and use the formula for the number of p-cycles to compute the index of N_{S_p}(P) in S_p].
    Let P be a subgroup of S_p of order p. It means that P is generated by a p-cycle. Since p is prime and |S_p| = p! it means that P is a Sylow p-subgroup. Consequently, if Q is any other subgroup of order p then Q has to be congugate to P i.e. Q = \tau P\tau^{-1}. Let P=P_1,P_2,...,P_n be all the Sylow p-subgroups. By above each one is conjugate to P_1. However, if \sigma is a p-cycle then \tau \sigma \tau^{-1} is a p-cycle. Furthermore, any p-cycle has order p and must be contained in one of the Sylow p-subgroups. Therefore, P_1,...,P_n contain all the p-cycles among them. Since these are distinct subgroups of order p (prime) it means P_i \cap P_j = \{ (1) \} for i\not = j. Each P_i contains p-1 p-cycles (because we are overlooking the identity). Thus, in total we have (p-1)n p-cycles where n, as above, is the number of conjugate subgroups to P. But we also know that n is the index of the normalizer of P in G, in less fancy language, n = [S_p:N(P)]. Therefore, |N(p)| = |S_p|/n. It remains to compute n. By above we said that there are (p-1)n p-cycles, we also know by counting that there are (p-1)! cycles. Thus, (p-1)n = (p-1)! \implies n = (p-2)!.

    Finally, |N(p)| = \frac{|S_p|}{n} = \frac{p!}{(p-2)!} = p(p-1).
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