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

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