Exercise 7.4

Proof. Here $n\mid q-1$.

Let $\phi =F^{\text{∗}}\to F^{\text{∗}}$ be the application defined by $\phi (x)=x^n$. Then $\phi$ is a homomorphism of groups, and $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ is the set of solutions of $x^n=1$. As $n\mid q-1$, $x^n=1$ has exactly $n$ solutions (Prop 7.1.1, Corollary2, or Ex 7.3 with $\alpha =1$). So $|\ker <!--\; FUNCTION\; APPLICATION\; -->\phi |=n$.

Thus $\mathrm{Im}\phi \simeq F^{\text{∗}}∕\ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ is a subgroup with cardinality $|F^{\text{∗}}|∕|\ker <!--\; FUNCTION\; APPLICATION\; -->\phi |=(q-1)∕n$, and $\mathrm{Im}\phi$ is the set of $\alpha$ such that $x^n=\alpha$ is solvable.

Conclusion: the set of $\alpha \in F^{\text{∗}}$ such that $x^n=\alpha$ is solvable is a subgroup with $(q-1)∕n$ elements. □