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Math Help - p-Sylow subgroup congruence

  1. #1
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    p-Sylow subgroup congruence

    Let p be a prime, G a finite group, and P a p-Sylow subgroup of G. Let M be any subgroup of G which contains N_G(P). Prove that [G:M]\equiv 1 (mod p). (Hint: look carefully at Sylow's Theorems.)
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    Let p be a prime, G a finite group, and P a p-Sylow subgroup of G. Let M be any subgroup of G which contains N_G(P). Prove that [G:M]\equiv 1 (mod p). (Hint: look carefully at Sylow's Theorems.)
    The number of conjugates to P is [G:N(P)].
    By Sylow's third theorem [G:N(P)]\equiv 1(\bmod p).
    But since [G:M] divides [G:N(P)] it means [G:M]\equiv 1(\bmod p) also.
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  3. #3
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    Here is essentially a similar proof to that one:

    It's important to note though that N_M(P) = N_G(P). By Sylow's Theorem Part 3, we have that [G: N_G(P)] \equiv 1 \mod{p}. Then P is a p-Sylow group of M, and N_M(P) is the normalizer of P in M, so [M: N_M(P)] \equiv 1 \mod{p}. Thus [G: M] = \frac {[G: N_G(P)]}{[M: N_M(P)]} \equiv 1 \mod{p}.
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