 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 formThe 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 $\phi (n)$ primitive roots, so the injection has image of size $\phi (n)$ and is surjective. Conversely, surjectivity makes every primitive root a conjugate of ${\zeta }_n$. Its minimal polynomial over $K$ then has at least $\phi (n)$ roots and divides the degree-$\phi (n)$ polynomial ${\Phi }_n$, so it equals ${\Phi }_n$. This proves the Galois embedding for a cyclotomic polynomial and the claimed equivalence.