# Thread: Symmetric group of P

1. ## 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 $S_p$].

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