Exercise 4.4.11 (If $|P| = p$, then $N_{S_p}(P)/C_{S_p}(P) \simeq \mathrm{Aut}(P)$)

Proof. By Exercise 4.3.34,

Since $|P|=p$ is prime, $P$ is cyclic. By Proposition 16, the automorphism group of $P$ is isomorphic to ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$, so

Since $P$ is cyclic of order $p$, $P=⟨𝜃⟩$, where $𝜃$ is a $p$-cycle by Exercise 4.3.34:

where the support of $𝜃$ is $\{a_1,a_2,\dots ,a_p\}=\{1,2,\dots ,p\}$.

It remains to find the order of $C_{S_p}(P)$. Note first that, since $P$ is abelian,

Conversely, if $\lambda \in C_{S_p}(P)$, then $\mathit{𝜆𝜎}=\mathit{𝜎𝜆}$ for all $\sigma \in P$. In particular, $𝜃=\mathit{𝜆𝜃}{\lambda }^{-1}$, thus

Then $\lambda (a_1)$ is in the support of $𝜃$, so $\lambda (a_1)=a_{k+1}$ for some $k\in [[0,p-1]]$, where $a_{k+1}={𝜃}^k(a_1)$. Therefore

where $\lambda (a_1)={𝜃}^k(a_1)$.

Since $\{a_1,a_2,\dots ,a_p\}=\{1,2,\dots ,p\}$, this shows that $\lambda ={𝜃}^k\in ⟨𝜃⟩=P$, thus $C_{S_p}(P)\le P$, so

We obtain

The equalities (1),(2) and (3) give

By corollary 15, $N_{S_p}(P)∕C_{S_p}(P)$ is isomorphic to a subgroup $H$ of $\mathrm{Aut}(P)$, and by (4), this subgroup $H$ has order $p-1=|\mathrm{Aut}(P)|$, so $H$ is the whole group $\mathrm{Aut}(P)$.

This proves

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