Exercise 6.4.2

Proof. Let $\phi :\mathrm{AGL}(1,{𝔽}_p)\to {𝔽}_p^{\text{∗}},{\gamma }_{a,b}\mapsto \phi ({\gamma }_{a,b})=a$.

(a)

This map is well defined, since $$f={\gamma }_{a,b}={\gamma }_{c,d}\in \mathrm{AGL}(1,{𝔽}_p)\Rightarrow \forall <!--\; FUNCTION\; APPLICATION\; -->u\in {𝔽}_p,\mathit{au}+b=\mathit{cu}+d\Rightarrow a=c,b=d.$$

$\phi$ is a group homomorphism: if $f={\gamma }_{a,b},g={\gamma }_{c,d}\in \mathrm{AGL}(1,{𝔽}_p)$, then

$$\phi (g\circ f)=\phi ({\gamma }_{c,d}\circ {\gamma }_{a,b})=\phi ({\gamma }_{\mathit{ac},\mathit{bc}+d})=\mathit{ac}=\phi (g)\phi (f).$$

This homomorphism is surjective, since every $a\in {𝔽}_p^{\text{∗}}$ satisfies $a=\phi ({\gamma }_{a,0})$, with ${\gamma }_{a,0}\in \mathrm{AGL}(1,{𝔽}_p)$.

${\gamma }_{a,b}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺a=1$: the kernel of $\phi$ is $T=\{{\gamma }_{1,b}|b\in {𝔽}_p\}$, so $T$ is a normal subgroup.

As the image of the group homorphism $\phi$ is ${𝔽}_p^{\text{∗}}$, and its kernel $T$, the Isomorphism Theorem shows that

$$\mathrm{AGL}(1,{𝔽}_p)∕T\simeq {𝔽}_p^{\text{∗}}.$$

(b)

The map $\psi :T\to {𝔽}_p,{\gamma }_{1,b}\mapsto b$ is bijective, and satisfies $$\psi ({\gamma }_{1,b}\circ {\gamma }_{1,d})=\psi ({\gamma }_{1,b+d})=b+d=\psi ({\gamma }_{1,b})+\psi ({\gamma }_{1,d}),$$

So $\psi$ is a group homomorphism: $T\simeq {𝔽}_p$.