Exercise 4.20

Proof. Here $p$ is a prime number, and $d\mid p-1$. Let

Then $f$ is a group homomorphism, and $\mathrm{im}(f)$ is the set of $d$th powers, and consequently is a subgroup of $U({𝔽}_p)={𝔽}_p^{\text{∗}}$. $\ker <!--\; FUNCTION\; APPLICATION\; -->(f)$ is the group of the roots of $x^d-1$. As $d\mid p-1$, the polynomial $x^d-1$ has exactly $d$ roots (Prop. 4.1.2), so $|\ker <!--\; FUNCTION\; APPLICATION\; -->(f)|=d$.

As $\mathrm{im}(f)\simeq {𝔽}_p^{\text{∗}}∕\ker <!--\; FUNCTION\; APPLICATION\; -->(f)$,

So there exist exactly $(p-1)∕d$ $d$th powers in ${(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$.

From Prop. 4.2.1, as $d\mid p-1,d\wedge (p-1)=d$, for all $x\in {𝔽}_p^{\text{∗}}$,

So the group of $d$th powers is the group of the roots of $x^{(p-1)∕d}-1$.

$\text{∙}$ If $p=11,d=5$, $\mathrm{im}(f)=\{1,-1\}$.

$\text{∙}$ If $p=17,d=4$, $x\in \mathrm{im}(f)⟺x^4=1$ : $\mathrm{im}(f)=\{1,-1,4,-4\}$.

$\text{∙}$ If $p=19,d=6$, $x\in \mathrm{im}(f)⟺x^3=1$ : $\mathrm{im}(f)=\{1,7,7^2=11\}$,

where $7\equiv 2^6(\mathit{mod}19)$. □