Exercise 14.4.17

Proof. (a) Write $e=1_{{𝔽}_{p^2}}\in {\mathrm{Aut}}_{{𝔽}_p}({𝔽}_{p^2})$ the identity of ${𝔽}_{p^2}$, which corresponds to $I_2\in \mathrm{GL}(2,{𝔽}_p)$. We want to show that

First $\mathit{ae}+b{\gamma }_{\alpha }\in C({\gamma }_{\alpha })$, since

As in the proof of Lemma 14.4.3, we take $m\in C({\gamma }_{\alpha })\subset {\mathrm{Aut}}_{{𝔽}_p}({𝔽}_{p^2})$.

Since $\alpha \notin {𝔽}_p$, $(1,\alpha )$ is a basis of ${𝔽}_{p^2}$ over ${𝔽}_p$, thus there are $r,s$ such that

Then, using $m\circ {\gamma }_{\alpha }={\gamma }_{\alpha }\circ m$, we obtain

therefore

Since $m$ is ${𝔽}_p$-linear, and $(1,\alpha )$ is a basis, we obtain that $m=\mathit{ae}+b{\gamma }_{\alpha }$. We have proved that

Moreover, for all $u\in {𝔽}_{p^2}$,

and $\beta \ne 0$, since $\mathit{ae}+b{\gamma }_{\alpha }$ is a linear automorphism. Conversely, if $\beta \in {𝔽}_{p^2}^{\text{∗}}$, ${\gamma }_{\beta }$ is a ${𝔽}_p$ linear automorphism. The decomposition of $\beta$ on the basis $(1,\alpha )$ gives $a,b\in {𝔽}_p$ such that $\beta =a+\mathit{b\alpha }$, and

To conclude,

Now we prove (14.39):

By the proof of Theorem 14.4.6 (p. 455), if $m\in N(C({\gamma }_{\alpha })$, we know that

$\text{∙}$

If $\beta =\alpha$, then $m\circ {\gamma }_{\alpha }={\gamma }_{\alpha }\circ \gamma$, thus $m\in C({\gamma }_{\alpha })$. We have proved above that $m={\gamma }_{\delta },$ where $\delta \in {𝔽}_{p^2}$, and $m(1)={\gamma }_{\delta }(1)=\delta$. Therefore $m={\gamma }_{\delta }={\gamma }_{\delta ,e}\in \mathrm{\Gamma L}(1,{𝔽}_{p^2})$.

$\text{∙}$

If $\beta =\sigma (\alpha )$, then $m\circ {\gamma }_{\alpha }={\gamma }_{\sigma (\alpha )}\circ m$. For all $u\in {𝔽}_{p^2}$, $m(\mathit{\alpha u})=\sigma (\alpha )m(u)$, thus, using $\sigma (1)=1$, and $\delta =m(1)$, $$\begin{array}{ccc}m(1)\hfill & =\delta \hfill & =({\gamma }_{\delta }\circ \sigma )(1),\hfill \\ m(\alpha )\hfill & =\sigma (\alpha )\delta \hfill & =({\gamma }_{\delta }\circ \sigma )(\alpha ).\hfill \end{array}$$

This proves that $m={\gamma }_{\delta }\circ \sigma$. Moreover, for all $u\in {𝔽}_{p^2}$, $m(u)=\mathit{\delta \sigma }(u)={\gamma }_{\delta ,\sigma }(u)$, therefore

$$m={\gamma }_{\delta ,\sigma }\in \mathrm{\Gamma L}(1,{𝔽}_{p^2}).$$

Since $\mathrm{Gal}({𝔽}_{p^2}∕{𝔽}_p)=\{e,\sigma \}$, $m={\gamma }_{\delta }$, or $m={\gamma }_{\delta ,\sigma }={\gamma }_{\delta }\circ \sigma$ for some $\delta \in {𝔽}_{p^2}$.

$\text{∙}$

If $m={\gamma }_{\delta }$, since $n\in C({\gamma }_{\alpha })$, $m\circ n\circ m^{-1}={\gamma }_{\alpha }\circ n\circ {\gamma }_{\alpha }^{-1}=n\in C({\gamma }_{\alpha })$.

$\text{∙}$

If $m={\gamma }_{\delta }\circ \sigma$, then for all $u\in {𝔽}_{p^2}$, $m(u)=\mathit{\delta \sigma }(u)$, thus $$\begin{array}{llll}\hfill (m\circ {\gamma }_{\alpha })(u)& =\mathit{\delta \sigma }(\mathit{\alpha u})=\mathit{\delta \sigma }(\alpha )\sigma (u),& \hfill & \\ \hfill ({\gamma }_{\sigma (\alpha )}\circ m)(u)& =\sigma (\alpha )\mathit{\delta \sigma }(u),& \hfill & \end{array}$$

therefore $m\circ {\gamma }_{\alpha }={\gamma }_{\sigma (\alpha )}\circ m$, and $m\circ {\gamma }_{\alpha }\circ m^{-1}={\gamma }_{\sigma (\alpha )}.$ Then, using the linearity of $m,n$,

$$\begin{array}{llll}\hfill m\circ n\circ m^{-1}& =m\circ (ae+b{\gamma }_{\alpha })\circ m^{-1}& \hfill & \\ \hfill & =ae+bm\circ {\gamma }_{\alpha }\circ m^{-1}& \hfill & \\ \hfill & =ae+b{\gamma }_{\sigma (\alpha )}& \hfill & \\ \hfill & ={\gamma }_{\beta },& \hfill & \end{array}$$

where $\beta =a+\mathit{b\sigma }(\alpha )\in {𝔽}_{p^2}$.

By part (a), $m\circ n\circ m^{-1}={\gamma }_{\beta }\in C({\gamma }_{\alpha })$. We have proved

$$\mathrm{\Gamma L}(1,{𝔽}_{p^2})\subset N(C({\gamma }_{\alpha })).$$