Exercise 14.1.1

Proof.

(a)

If $\tau$ lies in the normalizer of $⟨𝜃⟩=\{e,𝜃,{𝜃}^2,\dots ,{𝜃}^{p-1}\}$, then $$\mathit{\tau 𝜃}{\tau }^{-1}\in \tau ⟨𝜃⟩{\tau }^{-1}=⟨𝜃⟩,$$

hence

$$\mathit{\tau 𝜃}{\tau }^{-1}={𝜃}^l\text{ for some }l=0,1\dots ,p-1.$$

If $l=0$, then $\mathit{\tau 𝜃}{\tau }^{-1}=e$, thus $\mathit{\tau 𝜃}=\tau$, and $𝜃=e$, which is false. Therefore $l\ne 0$.

$$\mathit{\tau 𝜃}{\tau }^{-1}={𝜃}^l,1\le l\le p-1.$$

(b)

By induction suppose that $\tau (i+j)=\tau (i)+\mathit{jl}$, then $\tau (i+j+1)=\tau (i+j)+l=\tau (i)+(j+1)l$. Case $j=1$ is valid by the identity (14.1). Hence, $\tau (i+j)=\tau (i)+\mathit{jl}$ for all positive integers $j$.