Exercise 6.4.1

Proof.

Let $a,b\in {𝔽}_p$, and ${\gamma }_{a,b}:{𝔽}_p\to {𝔽}_p,u\mapsto {\gamma }_{a,b}(u)=\mathit{au}+b$.

(a)

If $a=0$, ${\gamma }_{a,b}$ is the constant function $b$, thus ${\gamma }_{0,b}$ is not a bijection.

Suppose that $a\ne 0$. Then , for all $u,v\in {𝔽}_p$,

$$v=\mathit{au}+b⟺u=a^{-1}v-a^{-1}b.$$

So every $v\in {𝔽}_p$ has a unique antecedent, therefore ${\gamma }_{a,b}$ is bijective.

(b)

If $a\ne 0$, by part (a), ${\gamma }_{a,b}$ is bijective, and the unique antecedent $u$ of any $v\in {𝔽}_p$ is given by $u=a^{-1}v-a^{-1}b={\gamma }_{a^{-1},-a^{-1}b}(v)$. Consequently $${\gamma }_{a,b}^{-1}={\gamma }_{a^{-1},-a^{-1}b}.$$

(c)

We show that $\mathrm{AGL}(1,{𝔽}_p)$ is a subgroup of $(S({𝔽}_p),\circ )$.

$\text{∙}$ By part (a), if $f\in \mathrm{AGL}(1,{𝔽}_p)$, then $f={\gamma }_{a,b}$, where $a\ne 0$, thus $f$ is bijective : $\mathrm{AGL}(1,{𝔽}_p)\subset S({𝔽}_p)$, and $1_{{𝔽}_p}={\gamma }_{1,0}\in \mathrm{AGL}(1,{𝔽}_p)$, so $\mathrm{AGL}(1,{𝔽}_p)\ne \varnothing$.

$\text{∙}$ If $f,g\in \mathrm{AGL}(1,{𝔽}_p)$, then $f={\gamma }_{a,b},g={\gamma }_{c,d},a,b,c,d\in {𝔽}_p,a\ne 0,c\ne 0$.

For all $u\in {𝔽}_p$,

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

Therefore $g\circ f={\gamma }_{c,d}\circ {\gamma }_{a,b}={\gamma }_{\mathit{ac},\mathit{bc}+d}$ and $\mathit{ac}\ne 0$, so $g\circ f\in \mathrm{AGL}(1,{𝔽}_p)$.

$\text{∙}$ If $f\in \mathrm{AGL}(1,{𝔽}_p)$, $f={\gamma }_{a,b},a\ne 0$, then $f^{-1}={\gamma }_{a^{-1},-a^{-1}b}\in \mathrm{AGL}(1,{𝔽}_p)$.

$\mathrm{AGL}(1,{𝔽}_p)$ is a group under composition.