Symmetric group of P

**aliceinwonderland** · November 18th 2008, 07:59 PM

[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 $S_p$ ].

**ThePerfectHacker** · November 19th 2008, 05:40 AM

Originally Posted by aliceinwonderland

[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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