= Solution
Choose a primitive root $\zeta_n$. Since every root of $\Phi_n$ is a power of $\zeta_n$, its splitting field is $L=K(\zeta_n)$. Each $\sigma\in G$ sends $\zeta_n$ to another primitive root, uniquely of the form
$$
\sigma(\zeta_n)=\zeta_n^{a_\sigma},
\qquad
a_\sigma\in(\mathbb Z/n\mathbb Z)^\times.
$$
The assignment $\sigma\mapsto a_\sigma$ is a homomorphism. It is injective because an automorphism fixing $\zeta_n$ fixes $K(\zeta_n)=L$.
If $\Phi_n$ is irreducible over $K$, the orbit of $\zeta_n$ has all $\varphi(n)$ primitive roots, so the injection has image of size $\varphi(n)$ and is surjective. Conversely, surjectivity makes every primitive root a conjugate of $\zeta_n$. Its minimal polynomial over $K$ then has at least $\varphi(n)$ roots and divides the degree-$\varphi(n)$ polynomial $\Phi_n$, so it equals $\Phi_n$. This proves the <Galois embedding for a cyclotomic polynomial> and the claimed equivalence.
Back to article page
