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Thread: p-Sylow subgroup congruence

  1. #1
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    p-Sylow subgroup congruence

    Let $\displaystyle p$ be a prime, $\displaystyle G$ a finite group, and $\displaystyle P$ a $\displaystyle p$-Sylow subgroup of $\displaystyle G$. Let $\displaystyle M$ be any subgroup of $\displaystyle G$ which contains $\displaystyle N_G(P)$. Prove that $\displaystyle [G:M]\equiv 1$ (mod $\displaystyle p$). (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 $\displaystyle p$ be a prime, $\displaystyle G$ a finite group, and $\displaystyle P$ a $\displaystyle p$-Sylow subgroup of $\displaystyle G$. Let $\displaystyle M$ be any subgroup of $\displaystyle G$ which contains $\displaystyle N_G(P)$. Prove that $\displaystyle [G:M]\equiv 1$ (mod $\displaystyle p$). (Hint: look carefully at Sylow's Theorems.)
    The number of conjugates to $\displaystyle P$ is $\displaystyle [G:N(P)]$.
    By Sylow's third theorem $\displaystyle [G:N(P)]\equiv 1(\bmod p)$.
    But since $\displaystyle [G:M]$ divides $\displaystyle [G:N(P)] $ it means $\displaystyle [G:M]\equiv 1(\bmod p)$ 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 $\displaystyle N_M(P) = N_G(P)$. By Sylow's Theorem Part 3, we have that $\displaystyle [G: N_G(P)] \equiv 1 \mod{p}$. Then $\displaystyle P$ is a $\displaystyle p$-Sylow group of $\displaystyle M$, and $\displaystyle N_M(P)$ is the normalizer of $\displaystyle P$ in $\displaystyle M$, so $\displaystyle [M: N_M(P)] \equiv 1 \mod{p}$. Thus $\displaystyle [G: M] = \frac {[G: N_G(P)]}{[M: N_M(P)]} \equiv 1 \mod{p}$.
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