Exercise 14.1.7

Proof. Assume that $\tau \in S_p$ satisfies $\mathit{\tau 𝜃}{\tau }^{-1}\in \mathrm{AGL}(1,{𝔽}_p)$. Then, since $⟨𝜃⟩$ is a group of order $p$, $⟨\mathit{\tau 𝜃}{\tau }^{-1}⟩=\tau ⟨𝜃⟩{\tau }^{-1}$ is a subgroup of $\mathrm{AGL}(1,{𝔽}_p)$ of order $p$ and each element of this subgroup has order $p$ (or $1$).

By part (a) of Lemma 14.1.2, $\mathrm{AGL}(1,{𝔽}_p)$ is the normalizer of $⟨𝜃⟩$ in $S_p$, therefore $⟨𝜃⟩$ is normal in $\mathrm{AGL}(1,{𝔽}_p)$, with $[\mathrm{AGL}(1,{𝔽}_p):⟨𝜃⟩]=p-1$. The order of each element of $\tau ⟨𝜃⟩{\tau }^{-1}$ is relatively prime to $p-1$, then, by Lemma 14.1.3, $\tau ⟨𝜃⟩{\tau }^{-1}\subset ⟨𝜃⟩$, therefore $\tau ⟨𝜃⟩{\tau }^{-1}=⟨𝜃⟩$, since both groups have the same order $p$.

Thus $\tau$ normalizes $⟨𝜃⟩$, hence $\tau \in \mathrm{AGL}(1,{𝔽}_p)$. □