Equivalence of Galois groups of two different splitting fields of the same polynomial

Question

All fields are in $\mathbb{C}$

Let $f$ be a polynomial with coefficients in the field $F$.
Let $F_1$ be a Galois extension of $F$ such that its Galois group $G(F_1/F)$ is cyclic and has prime order.

Let $K$ be the splitting field of $f$ over $F$ and $K_1$ be the splitting field of $f$ over $F_1$.

We are given that $G(K/F)$ is simple and non-abelian.

Question: Is $G(K/F) \cong G(K_1/F_1)$?

Comments: We know that $[F_1:F] = p$ for some prime $p$ and this should be the key to the proof but I can't see where to use it. And we know that $K/F$ and $K_1/F_1$ are Galois extensions because they are splitting fields over the lower fields. This question arises from the proof of proposition (9.8) of Algebra by Artin 1991 reproduced below:

Answer 1

The answer appears in the next paragraph after the underlined sentence in my picture. I misunderstood the format of the overall proof. I didn't see where cyclic-ness was ever used though.