Exercise 4.3.34 (Normalizer of a subgroup of $S_p$ of order $p$ )

Proof. Let $p$ be a prime and consider a subgroup $P$ of $S_p$ of order $p$. Let $Q=\mathit{𝛾𝑃}{\gamma }^{-1}$ be any conjugate subgroup of $P$. Then $|Q|=|P|=p$.

Since $p$ is prime, $Q$ is cyclic, so $Q=⟨\sigma ⟩$, where $\sigma$ is a permutation of order $p$. Since the order of $\sigma$ is the l.c.m. of the length of the cycles in its cycle decomposition, where $p$ is prime, every non trivial cycle in this decomposition has length $p$, thus the support of these cycles contains all the elements $1,2,\dots ,p$. Since the cycle are disjoint, there is only one cycle in this cycle decomposition, so $\sigma$ is a $p$-cycle. Then ${\sigma }^2,\dots ,{\sigma }^{p-1}$ are also $p$-cycles, therefore $Q$ contains $p-1$ $p$-cycles (and the identity).

So every conjugate $Q$ of $P$ contains exactly $p-1$ $p$-cycles, and

where $\sigma$ is a $p$-cycle.

In particular, $P=⟨𝜃⟩$, where $𝜃$ is a $p$-cycle.

Let $A$ be the set of conjugates of $P=⟨𝜃⟩$, and $C$ be the set of $p$-cycles in $S_p$.

The number of $m$-cycles in $S_n$ is $\frac{n(n-1)(n-m+1)}{m}$, so the number of $p$-cycles in $S_p$ is $(p-1)!$. Therefore

By Proposition 6,

We can compute $|A|$ in another way.

Note that for every $p$-cycle $\sigma \in C$, $\sigma$ and $𝜃$ are conjugate, thus the subgroup $⟨\sigma ⟩$ is conjugate to $P=⟨𝜃⟩$, so $⟨\sigma ⟩\in A$.

Consider the map

Then $\phi$ is surjective: if $Q\in A$, then $Q$ is conjugate to $P$, so $Q=⟨\sigma ⟩$ for some $p$-cycle $\sigma \in C$ by the first part. Since $⟨\sigma ⟩=⟨\sigma ⟩=\cdots =⟨{\sigma }^{p-1}⟩$, the pre-image of some subgroup $⟨\sigma ⟩$ is

of cardinality $p-1$. Therefore

thus, using (1),

Then, by (2),

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